问:
Algorithm to solve n variable problem with range
During Software designing our team come across the following problem: we have linear equation with n variables with each variable have positive permissible range or allowed range.

In This Problem the constant a_i are the number of items person wants to be bought of type x_i which has minimum value of purchase l_i and maximum value of purchase r_i and now, buyer know about this ranges that why he smartly ask that is this possible within value b. So now we have to choose one solution if possible and give the invoice accordingly.

So in math form,

```<code> a1x1+a2x2+....+anxn = b
 known parameters:
 a1,a2,...,an and b
 unknown parameter x1,x2,....,xn
 but we know that
 x1 belongs to range [l1,r1]
 x2 belongs to range [l2,r2]
 .
 .
 .
 xn belongs to range [ln,rn]
 0<li<ri
 ai>0
 b>0
</code>```

and we need to find values of all xi in above equation(may be float or integer but must be range). I need to solve this problem as soon as possible.
Example :

```<code>  Let n be 5
  And equation is
  10*x_1+15*x_2+10*x_3+11*x_4+17*x_5=2365
   And allowed ranges are
   For
   X_1 [30,60]
   X_2 [20,40]
   X_3 [20,80]
   X_4 [50,120]
   X_5 [10,140]
  So one of the solution will be
  For
  X_1 =40
  X_2 =36
  X_3 =34
  X_4 =60
  X_5 =25
</code>```

Want to find only one solution.
I need solution in bigO(nlogn)
Or nearly linear as possible as.

答:
This is an example of finding a feasible solution to a linear program. By introducing extra variables $s_i$ and $e_i$ so that

$$x_i-e_i=l_i\\x_i+s_i=r_i\\e_i,s_i\ge0$$

all of the constraints are now equations and every variable must be positive.

One way to find a feasible solution is to use the 2-phase simplex method. This involves introducing even more variables so that every equation becomes trivial to solve initially.

\begin{align}&\text{minimize}&y_0+y_1+\dots+y_n&=z\\&\text{subject to}&a_1x_1+\dots+a_nx_n+y_0&=b\\&&x_1-e_1+y_1&=l_1\\&&x_2-e_2+y_2&=l_2\\&&\vdots\quad\\&&x_n-e_n+y_n&=l_n\\&&x_1+s_1&=r_1\\&&x_2+s_2&=r_2\\&&\vdots~\\&&x_n+s_n&=r_n\\&&x_i,y_i,e_i,s_i&\ge0\end{align}

Note that $y_i$ are essentially the errors of the first $n+1$ equations. If you get $z=0$ in the end, then you have a feasible solution. Otherwise no such solution exists. With some more details, you can even find all of the solutions where $z=0$.

There are many tutorials online you can find on how to solve the above problem using the simplex method as well as plenty of software for solving such problems, such as excel.


问:
How many numbers between 1 and 10,000,000 don't have the sequence 12? Inclusion-exclusion problem
I got the following question:

How many numbers between 1 and 10,000,000 don't have the sequence 12? This is an inclusion-exclusion problem. Sadly I didn't fully understand its concept, so I tried solving it logically. I would like to know if my solution is correct, and if possible, how to solve it using the inclusion-exclusion principle.

Thanks!

So I said lets first find the numbers that do include the sequence 12, and divided it into cases:

Between 1 and 99: $1\cdot 1=1$

Between 100 and 999: $1\cdot 1 \cdot 10$ + $9\cdot 1 \cdot 1=19$

Between 1000 and 9999: $1 \cdot 1 \cdot 10 \cdot 10 + 2\cdot 9 \cdot 10 \cdot 1 \cdot 1 = 280$

Between 10000 and 99999:  $1 \cdot 1 \cdot 10 \cdot 10 \cdot 10 + 3\cdot 9 \cdot 10 \cdot 10\cdot 1\cdot 1=3700$

Between 100000 and 999999: $1 \cdot 1 \cdot 10 \cdot 10 \cdot 10\cdot 10 + 4\cdot 9 \cdot 10 \cdot 10\cdot 10\cdot 1\cdot 1=46000$

Between 1000000 and 9999999: $1 \cdot 1 \cdot 10 \cdot 10 \cdot 10\cdot 10\cdot 10 + 5\cdot 9 \cdot 10 \cdot 10\cdot 10\cdot 10 \cdot 1\cdot 1=550000$

So the total of numbers that don't include the number 12 is:
10000000-550000-46000-3700-280-19=9400001

I don't havea final answer so I don't know if that correct. Obviousy this solution lacks elegance, so I would like to know how solve this problem using the inclusion-exclusion principle.

Thank in advance!

答:
Keep track of the amount of $n$-digit numbers ending in $1$ and not ending in $1$, starting with $n=1$

$$
v_1 = \begin{pmatrix} 1 \\ 8 \end{pmatrix}
$$

so $1+8 = 9$ numbers in total.  The vector for $(n+1)$-digit numbers ending in $1$ and not ending in $1$ that do not contain the sequence "$12$" can be obtained by a recursion:

$$
v_{n+1} = \begin{pmatrix} 1 & 1 \\ 8 & 9 \end{pmatrix} v_n.
$$

If $M$ is the matrix in the recursion above then there are
$$
\begin{pmatrix} 1 & 1 \end{pmatrix}(M^{n-1} + M^{n-2} + \dotsc + 1) \begin{pmatrix} 1 \\ 8 \end{pmatrix}
$$
numbers of at most $n$ digits that do not contain the sequence "$12$".  For $n=7$ this becomes

$$
\begin{pmatrix} 1 & 1 \end{pmatrix}\begin{pmatrix}96031 & 106821 \\
854568 & 950599 \end{pmatrix} \begin{pmatrix} 1 \\ 8 \end{pmatrix} = 9409959
$$
and including the number $10^7$ the final answer is $9409960$.  Based on the above one can find a simpler linear recursion for the amount $a_n$ of numbers of at most $n$ digits that do not contain the sequence "$12$", namely

$$a_0 = 0, a_1=9 \textrm{ and } a_{n+2} = 10 a_{n+1} - a_n + 8.$$

This produces the sequence

$$0, 9,
98,
979,
9700,
96029,
950598,
9409959,
93149000,
922080049, \dotsc$$

ContentLicense: CC BY-SA 3.0


问:
Explicitly solving for inverse function
Consider a mapping $f:\mathbb{R}^2\to \mathbb{R}^2$ given by
$$ f(x,y) = \left(-x+\sqrt{x^2+y^2},-x-\sqrt{x^2+y^2}\right). $$
I want to find an explicit inverse mapping, on the neighborhoods for which $f$ is locally bijective.

A computation shows that the derivative of $f$ is given by
$$ Df(x,y) = \frac{1}{\sqrt{x^2+y^2}}\begin{bmatrix} -\sqrt{x^2+y^2}+x & y \\
 -\sqrt{x^2+y^2}+x & -y \end{bmatrix} $$
so the partial derivative vectors $\partial_1f(x,y)$ and $\partial_2 f(x,y)$ are orthogonal. Also, the derivative has determinant
$$ \det Df(x,y) = \frac{2y}{\sqrt{x^2+y^2}} $$
which is nonzero if and only if $y\neq0$. Hence the inverse function theorem tells us that for each $(x,y)$ with $y\neq0$, the map $f$ is a bijection on some neighborhood of $(x,y)$. However, I am asked to explicitly find an inverse, which I cannot seem to do. I've bashed the problem with algebra for a while, and nothing nice comes out. My professor calls the system of functions $f_1,f_2:\mathbb{R}^2\to \mathbb{R}$ a "parabolic coordinate system". The graphs of the level sets of $f_1$ and $f_2$ make this clear. Can you help me explicitly calculate what the inverse mapping must be, on a neighborhood $U$ where $f|_U$ is a bijection?

(Also, the wikepedia entry and every other page on "parabolic coordinates" uses a very different set of functions to define what they are labeling parabolic coordinates.)

The algebra I've tried: Set
$$ -x+\sqrt{x^2+y^2} = u \text{ and } -x-\sqrt{x^2+y^2}=v$$
like in high-school algebra; we want to express $x$ and $y$ in terms of $u,v$.
Adding the first equation to the second yields
$$ -2x = u + v $$
which implies $ x= -(u+v)/2 $.
Now subtracting the second equation from the first gives
$$ 2\sqrt{x^2+y^2} = u-v $$
so we square both sides and obtain
$$ 4(x^2+y^2) = u^2-2uv+v^2. $$
Now using the formula we derived for $x$ gives
$$ 4(\left(-\frac{u+v}{2}\right)^2+y^2) = u^2-2uv+v^2 $$
which simplifies to
$$ 4y^2 + u^2 + 2uv + v^2 = u^2-2uv+v^2 $$
and again to $ 4y^2 = -4uv $
which means $y=\pm\sqrt{-uv}$.
Is this correct? This attempt, I got further than any other.

答:
$$ \begin{cases}u+v=-2x,\\uv=-y^2,\end{cases}$$

so

$$ \begin{cases}x=-\dfrac{u+v}2,\\y=\pm\sqrt{-uv}.\end{cases}$$

Check:

$$ \begin{cases}-x+\sqrt{x^2+y^2}=\dfrac{u+v}2+\left|\dfrac{u-v}2\right|=u,\\-x-\sqrt{x^2+y^2}=\dfrac{u+v}2-\left|\dfrac{u-v}2\right|=v.\end{cases}$$

The last equalities are correct because $u\ge v$. The sign of $y$ is indeterminate. This is no surprise, as $f$ is even in $y$.

ContentLicense: CC BY-SA 4.0

Title: Explicitly solving for inverse function


问:
The composition of a positive definite operator and a self-adjoint operator is positive definite?
Let $A: \mathbb{R}^n \to \mathbb{R}^n$, $n>1$ be a positive definite operator that is invertible, and it is not the identity map. And let $B$ be a self-adjoint operator.

My question is, $A^{-1}\circ B$ is positive definite?

I have the feeling that this can be true, but I don't know how to prove it.

Thank you very much.

答:
Counter example

Take $n=1$, so any linear map is self-adjoint, $A(x) = 2 x$, $B(x) = - x$.

Then $A^{-1} \circ B (x) = - \dfrac{1}{2} x$, which is clearly negative definite.

What you could say is that $B \circ A^{-1} \circ B$ is positive semi-definite (depending on the kernel of $B$).


问:
number of simplices in barycentric subdivision
Let $K$ be a simplicial complex. Is there a way to calculate the number of k-simlices in the barycentric subdivision $K'$ of $K$? Given the number of $l$-simplices in $K$, for any $l$, of course.

(I am suposed to show that the Euler characteristic does not change, directly from the definition. I can not see how to do this without calculating the number of k-simplices...)

答:
I suppose this should be tagged as homework, since this is problem (1) on http://kurser.math.su.se/file.php/853/Top11.pdf.

The following might be of some help. I am fairly certain that it is correct. Let $K_n$ denote the simplicial complex consisting of a simplex of dimension $n$, together with all its faces. Let $K_n'$ denote its barycentric subdivision. Let $s(n)$ be the number of simplexes in $K_n$. Let $s(m,n)$ be the number of $m$-simplexes in $K_n$. Let $s'(m,n)$ be the number of $m$-simplexes in $K_n'$. Let for $n > 0$ $s''(m,n)$ be the number of $m$-simplexes in $K_n'$ not spanned by the barycenter of the $n$-simplex. Then

$$
    s(n)    = 2^{n+1}-1,
$$

and

$$
    s(m,n)  = \binom{n+1}{m+1},
$$

and

$$
    s'(m,n) =
    \begin{cases}
        s(n)                    &\text{if } m=0,    \\
        s''(m,n) + s''(m-1,n)   &\text{otherwise},
    \end{cases}
$$

and

$$
    s''(m,n) =
    \begin{cases}
        s(n)-1                  &\text{if } m=0,    \\
        \sum\limits_{i=m}^{n-1}\left( s(i,n) \left[s'(m,i) - s''(m,i) \right] \right)
                                 &\text{if } 0 < m < n, \\
         0                       &\text{if } m = n.
    \end{cases}
$$

It is possible that the formulas could be considerably simplified. I can provide derivations of the above after homework deadline at May 13.

UPDATE: The expressions for $s(n)$ and $s(m,n)$ are stated at http://en.wikipedia.org/wiki/Simplex.

For $s'(m,n)$ reason as follows. The case $m = 0$ is clear. For $m > 0$, in addition to the $s''(m,n)$ $m$-simplexes not spanned by the barycenter of the $n$-simplex, there are $s''(m-1,n)$ $m$-simplexes spanned by the barycenter and the vertices of an $(m-1)$-simplex not spanned by the barycenter. There are no other $m$-simplexes.

For $s''(m,n)$ reason as follows. The cases $m = 0$ and $m = n$ are clear. For $0 < m < n$ and $m \leq i \leq n-1$, there are $s(i,n)$ $i$-complexes in $K_n$, each contributing to $K_n'$ with $s'(m,i)$ $m$-simplexes not spanned by the barycenter of the $n$-simplex. There are no other contributions of such simplexes. In the summation, discount for $m$-simplexes already contributed by simplexes of lower dimension: of the $s'(m,i)$ $m$-simplexes contributed by an an $i$-simplex, $s''(m,i)$ $m$-simplexes have already been counted (since these are not spanned by the barycenter of the $i$-simplex).

HOWEVER: Visualizing the complexes $K_n'$ for $n=0,1,2,3$, I get

$$
s'(0,0)=1 \\
s'(0,1)=3, \quad s''(0,1)=2, \\
s'(1,1)=2, \quad s''(1,1)=0, \\
s'(0,2)=7, \quad s''(0,2)=6, \\
s'(1,2)=12, \quad s''(1,2)=6, \\
s'(2,2)=6, \quad s''(2,2)=0, \\
s'(0,3)=15, \quad s''(0,3)=14, \\
s'(1,3)=50, \quad s''(1,3)=36, \\
s'(2,3)=?, \quad s''(2,3)=24, \\
s'(3,3)=24, \quad s''(3,3)=0.
$$

Using the formulas above, I get differing results $s'(1,3)=68$ and $s''(1,3)=54$ (and $s'(2,3)=78$). It seems the expression for $s''$ above is counting some contributions more than once (or I am counting simplexes wrong when visualizing $K_3'$).

UPDATE 2: Formula updated, gives correct results at least for $n \leq 3$ now. I think it is correct for all $n$ now.


问:
How to understand the Riemann surfaces described by Ahlfors complex analysis
I just can not understand how the author build the riemann surfaces. The details are following picture:

I searched on Google, read the definition of the Riemann surface and can understand the usual definition for it. But I cannot understand the construction of the Riemann surfaces by this book especially the different sheet and cut, etc.

答:
The way to understand Riemann surfaces is to start with the square root function and go no further until mastering it.  And begin with Riemann's original notion:  create a new coordinate surface such that a multi-valued function becomes a single-valued function of the points on this new surface.  The new surface is the Riemann surface. Now suppose we wanted to do just that without any regards to the notion of a "Riemann surface".

Consider the function $w=z^2=r^2 e^{2i\theta}$.  That is a mapping of the entire $z$-plane to two copies of the $w$ plane is it not?  Consider $r^2 e^{2i\theta}$ for $-\pi \leq \theta <0$.  This maps one-to-one, the lower $z$-plane to all of the $w$-plane.   Likewise, the upper $z$-plane maps to another copy of the $w$-plane.  We can label these $w$-plane copies $k=1$ and $k=2$ or more commonly, $k=0$ and $k=1$.  We can identify every point in each copy of the two $w$-planes by $(w,\sqrt{w})$.  For example, copy $k=0$ would have the point $(1,1)$ and copy $k=1$ would have the point $(1,-1)$.  So that every point $(u,v)$ in this $2$-copy $w$-plane we call a Riemann surface is unique and further, there is a one-to-one mapping of the points $(u,v)$ in the Riemann surface to the $z$-plane:
$$\displaystyle f(w,r^{1/2}e^{i/2(\theta+2k\pi)})= r^{1/2}e^{i/2(\theta+2k\pi)}$$

Now we would like to make this Riemann surface continuous.  We can easily do this by making a radial cut anywhere on each surface but most commonly along the negative real axis (or positive in Ahlfor's construction) and then splice the lower half of sheet $k=0$ to the upper half of sheet $k=1$ along the respective cut edges, and likewise, splice the upper half of sheet $k=0$ to the lower half of $k=1$ along those cut edges.  So that as we move along this new coordinate system (or surface) say $w=2 e^{i\theta}$ from $\theta=\pi-0.1$ to $\pi+0.1$ we smoothly move across sheet $k=0$ to $k=1$ as moving from the red sheet to the blue sheet in the figures below.

The function $w=z^3$ follows naturally right?  It's just three sheets spliced together.  Try to edit the Mathematica code below to create a plot.

```<code>fun1[z_, n_] := Abs[z]^(1/2) Exp[I/2 (Arg[z] + 2 n
Pi)];
sheet1 = ParametricPlot3D[{Re[z], Im[z], Im@fun1[z,
0]}z -> r Exp[I t], {r, 0.1, 2}, {t, -Pi, Pi},
PlotStyle -> Red];
sheet2 = ParametricPlot3D[{Re[z], Im[z], Im@fun1[z,
1]} /.z -> r Exp[I t], {r, 0.1, 2}, {t, -Pi, Pi},
PlotStyle -> Blue];
sheet3 = Show[{sheet1, sheet2}]
</code>```

Body:
The way to understand Riemann surfaces is to start with the square root function and go no further until mastering it.  And begin with Riemann's original notion:  create a new coordinate surface such that a multi-valued function becomes a single-valued function of the points on this new surface.  The new surface is the Riemann surface. Now suppose we wanted to do just that without any regards to the notion of a "Riemann surface".

Consider the function $w=z^2=r^2 e^{2i\theta}$.  That is a mapping of the entire $z$-plane to two copies of the $w$ plane is it not?  Consider $r^2 e^{2i\theta}$ for $-\pi \leq \theta <0$.  This maps one-to-one, the lower $z$-plane to all of the $w$-plane.   Likewise, the upper $z$-plane maps to another copy of the $w$-plane.  We can label these $w$-plane copies $k=1$ and $k=2$ or more commonly, $k=0$ and $k=1$.  We can identify every point in each copy of the two $w$-planes by $(w,\sqrt{w})$.  For example, copy $k=0$ would have the point $(1,1)$ and copy $k=1$ would have the point $(1,-1)$.  So that every point $(u,v)$ in this $2$-copy $w$-plane we call a Riemann surface is unique and further, there is a one-to-one mapping of the points $(u,v)$ in the Riemann surface to the $z$-plane:
$$\displaystyle f(w,r^{1/2}e^{i/2(\theta+2k\pi)})= r^{1/2}e^{i/2(\theta+2k\pi)}$$

Now we would like to make this Riemann surface continuous.  We can easily do this by making a radial cut anywhere on each surface but most commonly along the negative real axis (or positive in Ahlfor's construction) and then splice the lower half of sheet $k=0$ to the upper half of sheet $k=1$ along the respective cut edges, and likewise, splice the upper half of sheet $k=0$ to the lower half of $k=1$ along those cut edges.  So that as we move along this new coordinate system (or surface) say $w=2 e^{i\theta}$ from $\theta=\pi-0.1$ to $\pi+0.1$ we smoothly move across sheet $k=0$ to $k=1$ as moving from the red sheet to the blue sheet in the figures below.

The function $w=z^3$ follows naturally right?  It's just three sheets spliced together.  Try to edit the Mathematica code below to create a plot.

fun1[z_, n_] := Abs[z]^(1/2) Exp[I/2 (Arg[z] + 2 n
Pi)];
sheet1 = ParametricPlot3D[{Re[z], Im[z], Im@fun1[z,
0]}z -> r Exp[I t], {r, 0.1, 2}, {t, -Pi, Pi},
PlotStyle -> Red];
sheet2 = ParametricPlot3D[{Re[z], Im[z], Im@fun1[z,
1]} /.z -> r Exp[I t], {r, 0.1, 2}, {t, -Pi, Pi},
PlotStyle -> Blue];
sheet3 = Show[{sheet1, sheet2}]


问:
How do I come up with a transformation matrix
What is the matrix that rotates by (pi/4) about an axis that is parallel to Y and passing through the point (9, 0, 0)?

Does anyone knows what this question means?

答:
In fact, such a rotation can be decomposed into a rotation into the $xOz$ plane and the identity on $y$-axis.

A 2D Rotation $R$ with angle $+\tfrac{\pi}{4}$  into the $x0z$ plane (in fact it is a rotation on the projection on $x0z$: see figures):

Let $a:=\tfrac{\sqrt{2}}{2}=\cos\tfrac{\pi}{4}=\sin\tfrac{\pi}{4}$.

$$\underbrace{\left(
\begin{array}{rrr}
x'-9 \\
z'-0
\end{array}
\right)}_{\vec{CM'_1}}=\underbrace{\left(
\begin{array}{rrr}
a & -a \\
a & a
\end{array}
\right)}_R\underbrace{\left(
\begin{array}{rrr}
x-9 \\
z-0
\end{array}
\right)}_{\vec{CM_1}}\iff \begin{cases}x'&=&ax-az-9a+9\\z'&=&ax+az-9a\end{cases}\tag{1}$$

where the rotation matrix $R$ comes from the classical form $\left(
\begin{array}{rrr}
\cos(\theta) & -\sin(\theta) \\
\sin(\theta) & \cos(\theta)
\end{array}
\right)$.

"statu quo" in the $y$ direction:

$$y'=y.\tag{2}$$

The final answer is the affine transformation obtained by grouping (1) and (2):

$$\left(
\begin{array}{r}
x'\\
y'\\
z'
\end{array}
\right)=\left(
\begin{array}{rrr}
a & \ \ 0 & -a \\
0 & \ \ 1 & 0  \\
a & \ \ 0 & a
\end{array}
\right)\left(
\begin{array}{r}
x\\
y\\
z
\end{array}
\right)+\left(
\begin{array}{c}
-9a+9\\
0\\
-9a
\end{array}
\right) \ \ \text{with} \ \ a=\tfrac{\sqrt{2}}{2}.$$


问:
Prove that for every $X \in \mathscr{P}(A)$ there is exactly one $Y \in [X]_R$ such that $Y \cap B = \emptyset$
This is an exercise from Velleman's "How To Prove It". I strugled with this problem for a while, so I just want to make sure that it is correct.

Suppose $B \subseteq A$, and define a relation $R$ on $\mathscr{P}(A)$ as follows:
$$R = \{ (X,Y) \in \mathscr{P}(A) \times \mathscr{P}(A) | X \vartriangle Y \subseteq B\}$$
Prove that for every $X \in \mathscr{P}(A)$ there is exactly one $Y \in [X]_R$ such that $Y \cap B = \emptyset$.

Proof: Let $X \in \mathscr{P}(A)$ be arbitrary. Let $Y = X \setminus B$. We must show that $X \vartriangle Y \subseteq B$ and $Y \cap B = \emptyset$.

Let $y \in X \vartriangle Y$ be arbitrary. Then $y \in X \setminus (X \setminus B) \cup (X \setminus B) \setminus X$. But since $(X \setminus B) \setminus X = \emptyset$, $y\in X \setminus (X \setminus B)$. This means that $y \in X$ and $y \in X \rightarrow y \in B$. It immediately follows that $y \in B$. Since $y$ was arbitrary, $X \vartriangle Y \subseteq B$, so $Y \in [X]_R$.
Now suppose that $Y \cap B \neq \emptyset$. Then we can choose a $y$ such that $y \in Y \cap B = (X\setminus B) \cap B$. Then $y \in X \setminus B$ and $y \in B$. But then we have $y \in B$ and $y \notin B$, which is a contradiction. Thus, $Y \cap B = \emptyset$.

To show that $Y$ is unique, let $Y' \in [X]_R$ be arbitrary such that $Y' \cap B = \emptyset$. So $X \vartriangle Y' \subseteq B$. Since $X \vartriangle Y' \subseteq B$ and $X \vartriangle Y \subseteq B$, $(X \vartriangle Y) \vartriangle (X \vartriangle Y') = (Y \vartriangle X) \vartriangle (X \vartriangle Y') = Y \vartriangle (X \vartriangle X) \vartriangle Y' = Y \vartriangle Y' \subseteq B$. Suppose that $Y \vartriangle Y' \neq \emptyset$. Then we can choose an $x$ such that $x \in Y \vartriangle Y'$. Then  either $x \in Y\setminus Y'$ or $x \in Y' \setminus Y$. In either case, we would have $x \in Y \cap B$ or $x\in Y' \cap B$, which are contradctions. Thus, $Y \vartriangle Y' = \emptyset$. It immediately follows that $Y = Y'$. $\square$

答:
Your proof is okay.

Alternative for second part:

Suppose that $Y\in\left[X\right]_{R}$ and $Y\cap B=\varnothing$.

Then $Y-X\subseteq B$ and combining that fact with $Y\cap B=\varnothing$
we conclude that $Y-X=\varnothing$ or equivalently that $Y\subseteq X$.

Again combining with $Y\cap B=\varnothing$ we find that $Y\subseteq X-B$.

If this inclusion is proper then some $x$ must exist with $x\notin Y$,
$x\in X$ and $x\notin B$.

This however contradicts that $X-Y\subseteq B$ so we conclude that
the inclusion is not proper so that: $$Y=X-B$$


问:
Calculate the MSE of T = $\frac{1}{4}\overline{X}$ with respect to $$ where X~$Gamma(4,)$
Suppose that $X_1,...,X_n$ is an iid sample that is distributed as $Gamma(4,)$ and suppose that $T = \frac{1}{4}\overline{X}$ is an estimator of $$. How do you calculate $MSE(T; )$?

I don't know how to start this problem because I know $MSE(T; )$ = $Var(T) - Bias(T; )^2$ but I don't know how to calculate $Var(T)$. I also calculated $Bias(T; )$ as follows:

$Bias(T; )$ = $E(T)$ - $$ = $E(\frac{1}{4}\overline{X})$ - $$ = .... = $\frac{1}{}$ - $$ = $\frac{1-}{}$
But I'm not sure if that calculation is even correct. Can someone confirm if it's correct as well?

答:
First, when you say that the distribution is a $Gamma(4;\theta)$ it is not well determinated because Gamma distribution has different parametrizations.

Second when you write

Suppose that $X_1,...,X_n$ is an iid sample that is distributed as $Gamma(4,\theta)$

I hope you mean that the random sample is "from a population with distribution"... or "where every single observation is distributed as..."

Third,

$MSE=VAR + (BIAS)^2$ and not $VAR - (BIAS)^2$ as you wrote: this looks like the great is the Bias, the best is the Estimator...

Fourth,

if the problem is only in calculating

$$V(T)=V\Bigg[\frac{1}{4}\overline{X}_n\Bigg]$$

it is simply

$$V(T)=V\Bigg[\frac{1}{4}\frac{\Sigma_i X_i}{n}\Bigg]=\frac{1}{(4n)^2}\cdot n V(X_1)$$

But I'm not sure if that calculation is even correct

this depends which is the parametrization you used for the Gamma distribution.  In the link you can find the two more common parametrizations

IMHO, if you use $T$ to estimate the parameter $\theta$ the parametrization that makes more sense is the one in the first column showed in the link, so that $\mathbb{E}[T]=\theta$ and you estimator is unbiased...but this is only my opinion. Only you can know the exact parametrization you have to use.


问:
How to describe arbitrary directed graph if I have a set?
I have an arbitrary set $A$ which is a subset of another set $A \subset B$, the elements of the set $A$ constitue a directed graph. If I want to describe it via notation (algebra) without mentioning the concrete elements, how can I do that? Is that even possible?

答:
There is no notation for the statement "something is a directed graph". This is probably because it's hard to imagine a situation where you have an object, but you're not sure if it's a directed graph or not. (There is no larger class of objects, some of which are directed graphs and some of which are not, such that you want to say "that object in the larger class? it's actually a directed graph".)

You can use notation for the other relationship in your problem. For example, if you have a directed graph $D$, you could write $V(D) \subseteq B$ to indicate that the vertices of $D$ are a subset of $B$, or $E(D) \subseteq B$ to indicate that the edges of $D$ are a subset of $B$.


问:
Example of a rational function such that : $(f(x))^{3} + (g(x))^{3} + (h(x))^{3}=x$
Can any one give me example of: rational functions $f, g$ and $h$ with rational coefficients such that

$$(f(x))^{3} + (g(x))^{3} + (h(x))^{3}=x$$

Also, if anyone knows a procedure for constructing such examples, I would be happy to learn them as well.

Added: Can we find rational functions $f$ and $g$ such that $(f(x))^3+(g(x))^3=x$

答:
To the question added later:

$f(x)^3 + g(x)^3 = x$

It is not possible.

Setting $x$ to be $1^3, 2^3, \dots $ gives us a non-trivial solution to the diophantine equation $x^3 + y^3 = z^3$, since $f$ and $g$ can only have finitely many roots.


问:
Application of the transversality theorem
I am trying to do this question in Bredon's Topology and geometry about using the transversality theorem to show that the intersection of two manifolds is a manifold.

Now it goes as follows:

Let $f(x,y,z)=(2-(x^2+y^2)^{1/2})+z^2$ on $\mathbb{R}^3 - (0,0,z)$. Then one can show that $M=f^{-1}(1)$ is a manifold.

Now let $N = \{ (x,y,z) \in \mathbb{R}^3 | x^2+y^2=4 \}$. I need to show that $M \cap N$ is a manifolds by using the transversality theorem, but I'm not quite sure how to do this.

I thought about finding the tanget planes to each to these manifolds and then showing that there a points in one of them that aren't in the other, but I dont know if this will work. So I was wondering if I could get some hints on this.

Thank you

答: Try writing down a function $g\colon\mathbb R^3-\{z\text{-axis}\}\to \mathbb R^2$ so that, for example, $g^{-1}(1,4) = M\cap N$ and $(1,4)$ is a regular value of $g$.</p>


问:
What does it mean to take derivative of vector in a direction?
Suppose I have a vector field $ \vec{B} (x,y,z)$ then do $ \frac{ \partial B}{ \partial n}$ where n is the direction vector of a line denote the directional derivative of the vector in the direction of $n$?

The reason I ask is that I recently encountered this in a physics textbook but all the gradients and directional derivatives that I've seen till now were defined for scalar fields.

Edit:

The real quantity that I started with was the one from this mse post:

$$ (\nabla B_i) n_i $$

I thought this would be the directional derivative since it looked like one but then I later realized this is actually a vector field.

A picture from the book:

Page-158, I.e irodov basic laws of electromagnetism

答:
Suppose I have a vector field $\vec{B}(x,y,z)$ then do $\frac{\partial\vec B}{\partial n}$ where $n$ is the direction vector of a line denote the directional derivative of the vector in the direction of $n$?

Yes. So is also stated for Force by Dipole in electric field in the same book. Page number 20, I.E. Irodov. Although the author explicitly states that the operation is rather complex and will not be discussed in the book.

The formal definitions can be noted from here and $\downarrow$.

Derivatives of vector-valued functions of vectors:

Let ${\displaystyle \mathbf {f} (\mathbf {v} )}$ be a vector-valued function of the vector ${\displaystyle \mathbf {v} }$ . Then the derivative of ${\displaystyle \mathbf {f} (\mathbf {v} )}$ with respect to ${\displaystyle \mathbf {v} }$  (or at ${\displaystyle \mathbf {v} }$) in the direction ${\displaystyle \mathbf {u} }$ (which maybe $\hat i,\hat j,\hat k$)  is the second-order tensor defined as:

$${\displaystyle {\frac {\partial \mathbf {f} }{\partial \mathbf {v} }}\cdot \mathbf {u} =D\mathbf {f} (\mathbf {v} )[\mathbf {u} ]=\left[{\frac {d}{d\alpha }}~\mathbf {f} (\mathbf {v} +\alpha ~\mathbf {u} )\right]_{\alpha =0}}$$
for all vectors ${\displaystyle \mathbf {u} }$ .

Properties:

If ${\displaystyle \mathbf {f} (\mathbf {v} )=\mathbf {f} _{1}(\mathbf {v} )+\mathbf {f} _{2}(\mathbf {v} )}$ then ${\displaystyle {\frac {\partial \mathbf {f} }{\partial \mathbf {v} }}\cdot \mathbf {u} =\left({\frac {\partial \mathbf {f} _{1}}{\partial \mathbf {v} }}+{\frac {\partial \mathbf {f} _{2}}{\partial \mathbf {v} }}\right)\cdot \mathbf {u} .}$

If $\mathbf{f}(\mathbf{v}) = \mathbf{f}_1(\mathbf{v})\times\mathbf{f}_2(\mathbf{v})$ then ${\displaystyle {\frac {\partial \mathbf {f} }{\partial \mathbf {v} }}\cdot \mathbf {u} =\left({\frac {\partial \mathbf {f} _{1}}{\partial \mathbf {v} }}\cdot \mathbf {u} \right)\times \mathbf {f} _{2}(\mathbf {v} )+\mathbf {f} _{1}(\mathbf {v} )\times \left({\frac {\partial \mathbf {f} _{2}}{\partial \mathbf {v} }}\cdot \mathbf {u} \right).}$

If $\mathbf{f}(\mathbf{v}) = \mathbf{f}_1(\mathbf{f}_2(\mathbf{v}))$ then ${\displaystyle {\frac {\partial \mathbf {f} }{\partial \mathbf {v} }}\cdot \mathbf {u} ={\frac {\partial \mathbf {f} _{1}}{\partial \mathbf {f} _{2}}}\cdot \left({\frac {\partial \mathbf {f} _{2}}{\partial \mathbf {v} }}\cdot \mathbf {u} \right).}$


问:
finding integrating factor and solving ordinary differential equation
Justify that a differential equation of the form: $$[y+xf(x^2+y^2)]dx+[yf(x^2+y^2)-x]dy=0$$ , where $f(x^2+y^2)$ is an arbitrary function of $(x^2+y^2)$, is not an exact differential equation and $1/(x^2+y^2)$ is an integrating factor for it. Hence, solve this differential equation for $$f(x^2+y^2)=(x^2+y^2)^2$$

I am unable to solve after making it an exact equation, having difficulty in the integration of this question.

Body:
$$(y+xf(x^2+y^2))dx+(yf(x^2+y^2)-x)dy=0$$
$$f(x^2+y^2))(xdx+ydy)+ydx-xdy=0$$
$$\dfrac 12f(x^2+y^2))(d(x^2+y^2))+ydx-xdy=0$$
Divide by $x^2+y^2$:
$$\dfrac {f(x^2+y^2)d(x^2+y^2)}{2(x^2+y^2)}+\dfrac {ydx-xdy}{x^2+y^2}=0$$
Note that:
$$d \arctan (y/x)=\dfrac {xdy-ydx}{x^2+y^2}$$
$$\dfrac {f(x^2+y^2)d(x^2+y^2)}{2(x^2+y^2)}-d(\arctan  (y/x)) =0$$
Integrate for the given function $f$.


问:
Show this sum is uniformly bounded in $N$ and in $i$.
For $N>0, \ d>0$, I am considering $N$ points $(y_1,...,y_N)$ in $(\mathbb{R}^d)^N$ such as there exists a constant $c>0$ with :

$$\underset{1 \leq i,j \leq N \atop i \neq j}{\min} |y_i-y_j| \geq c, \quad \forall N>0.$$

I am trying to show that there exists a constant $M>0$ such as :

$$\underset{1 \leq i \leq N}{\sup} \sum_{j=1}^N \frac{1}{(c+|y_i-y_j|)^4} \leq M, \quad \forall N>0$$

I have tried to use my hypothesis which gives me :

$$\frac{1}{(c+|y_i-y_j|)^4} \leq \frac{1}{16 c^4}$$
but this leads me nowhere since $(\sum_{j=1}^N \frac{1}{16 C^4})_{N \in \mathbb{N}}$ is divergent.

Update 1

I've proposed a solution to this question, I'm not sure weither it's true or false, if anyone wants to proof read it I would be very thankful.

答:
I think I found a second answer which is easier. I'm assuming $d=3$ for this answer.

Let fix $i$ between $1$ and $N$. We found easily that there exists a constant $C$ independant of $i$ and $N$ such as :

$$\frac{1}{(c + |y_i-y_j|)^4} \leq C \int_{B(y_j,\frac{c}{2})} \frac{1}{(c + |y_i-y|)^4} \mathrm{d}y, \quad \forall j$$

Therefore when we sum :

$$\sum_{j=1}^N \frac{1}{(c + |y_i-y_j|)^4} \leq \sum_{j=1}^N \int_{B(y_j,\frac{c} {2})}\frac{1}{(c + |y_i-y|)^4} \mathrm{d}y$$
or since all the balls $B(y_j,\frac{c} {2})$ are disjoinct we found that :

$$\sum_{j=1}^N \frac{1}{(c + |y_i-y_j|)^4} \leq \int_{\bigcup_{j=1}^N B(y_j,\frac{c} {2})} \frac{1}{(c + |y_i-y|)^4} \mathrm{d}y \leq \int_{\mathbb{R}^3} \frac{1}{(c + |y_i-y|)^4} \mathrm{d}y  = \int_{\mathbb{R}^3} \frac{1}{(c + |y|)^4} \mathrm{d}y  $$
and this integral is well defined since we are on $\mathbb{R}^3$. We then pose $M:=\int_{\mathbb{R}^3} \frac{1}{(c + |y|)^4} \mathrm{d}y$ and the result should follow.


问:
How to show these groups are isomorphic through finding an isomorphism
Show, by finding a suitable isomorphism, that the two groups $(\Bbb Z,+)$ and $(\Bbb Z,∗)$, where $a∗b=a+b−1$, are isomorphic.

答:
The first step should be to take a closer look on the group $(\mathbb{Z},\ast)$. In particular what is the neutral element?

A group homomorphism has to send the neutral element to the neutral element. That is a good starting point.

Turns out the neutral element of $(\mathbb{Z},\ast)$ is $1$.

As $a\ast 1=a+1-1=a$ and $1\ast a=1+a-1=a$.

So $\varphi(0)=1$.
The easiest function that obeys this rule, is $x\mapsto x+1$.

So now we have to show that $\varphi(x)=x+1$ is an isomorphism.
That is a bijection (easiest shown by giving an inverse, but showing injective and surjective should be easy enough too), and that it is a homomorphism.

So we have to show that $\varphi(a+b)=\varphi(a)\ast\varphi(b)$.

I leave the rest to you.


问:
Decomposition of the variation of a signed measure as $|\mu|(A) = \int_A |\frac{d\mu_{1a}}{d\mu_2}-1|d\mu_2 + \mu_{1s}(A)$, where $\mu=\mu_1-\mu_2$
Let $\mu_1$ and $\mu_2$ be two finite measures on $(\Omega, \mathcal{F})$. Let $\mu_1 = \mu_{1a}+\mu_{1s}$ be the Lebesgue decomposition of $\mu_1$ w.r.t. $\mu_2$, that is, $\mu_{1a} \ll \mu_2$ and $\mu_{1s}\perp \mu_2$. Let $\mu = \mu_1 - \mu_2$. I'd like to show that for all $A\in \mathcal{F}$,
$$ |\mu|(A) = \int_A |h-1|d\mu_2 + \mu_{1s}(A) $$
where $h = \frac{d\mu_{1a}}{d\mu_2}$ is the Radon-Nikodym derivative of $\mu_{1a}$ w.r.t. $\mu_2$.

I know that we have the following decomposition: $$|\mu|=\mu_+ + \mu_- $$
Here, $\mu_+(A) = \mu(A \cap \Omega_+)$ and $\mu_-(A)=\mu_-(A \cap \Omega_-)$, where $\Omega = \Omega_+ \cup \Omega_-$ is the Hahn decomposition of $\Omega$ w.r.t $\mu$. Also, there exists a finite measure $\lambda$ such that $\mu_1 = \mu_+ + \lambda$ and $\mu_2 = \mu_-+\lambda$ with $\lambda = 0$ iff $\mu_1 \perp \mu_2$. By the definition of Radon-Nikodym derivative, we have $\mu_{1a}(A) = \int_A h d\mu_2$ for all $A\in \mathcal{F}$.

I am unable to use these facts to prove the desired result. Any hint as to how I should proceed would be highly appreciated.

Edit: This problem is exercise 4.13 from the book "Measure theory and Probability Theory" by Krishna B. Athreya and Soumendra N. Lahiri.

答:
While you know the definitions, it is always good to realize that upon experimentation, we can actually stumble upon what the Hahn-decomposition, and the Jordan decomposition of $|\mu|$ are.

How? Let us start with $\mu = \mu_1 - \mu_2$. We substitute $\mu_1 = \mu_{1a} + \mu_{1s}$ to get $\mu = \mu_{1a} + \mu_{1s}-\mu_2$.. Let us rearrange to get $\mu = (\mu_{1a} - \mu_2) + \mu_{1s}$. Let $h = \frac{d \mu_{1a}}{d \mu_2}$.Finally, evaluated at any set $A$, the equality reads :
$$
\mu(A) = \mu_{1a}(A) - \mu_2(A) + \mu_{1s}(A) = \int_{A} h d \mu_2 - \int_A 1 d \mu_2 + \mu_{1s}(A) \\ = \int_{A} (h-1)d \mu_2 + \mu_{1s}(A) \\ = \color{green}{\int_{A \cap \{h \leq 1\}} (h-1) d \mu_2} + \color{blue}{\int_{A\cap\{h > 1\}} (h-1) d \mu_2  + \mu_{1s}(A)}
$$

Recognize that the green part is negative for all $A$ and the blue part is positive for all $A$ (note that $\mu_{1s}$ is a non-negative measure, why?). This immediately hints at the following Jordan decomposition : $-\mu^-(A)$ is the green part and $\mu^+(A)$ is the blue part.

To show this, since the negative-positive sign part is clear, it's enough to show that the green and blue part are mutually singular as measures (functions of $A$, if you like).

That's not too difficult : let $\mu_{2}$ concentrate on $B$ and $\mu_{1s}$ concentrate on $B^c$. The green part then concentrates on $B \cap \{h \leq 1\}$, while the blue concentrates on the rest. To see this, if $C \subset B \cap \{h \leq 1\}$ then the blue part is zero. On the other hand, if either $C \subset B^c$ or if $C \subset \{h > 1\}$ the green part is zero.

Hence, the Hahn decomposition is also established, and the green part is $-\mu^-$ with the blue part being $\mu^+$. The result follows from $\mu = \mu^- + \mu^+$.


答:
It's not true in general for any $\ n\ $. Unless $\ B\ $ is positive definite it must have at least one non-positive eigenvalue $\ \lambda\ $.  If $\ u\ $ is a corresponding eigenvector, then
\begin{align}
u^\top A^{-1}Bu&=\lambda u^\top A^{-1}u\\
&=\lambda(A^{-1}u)^\top A(A^{-1}u)\\
&\le0\ .
\end{align}

Title: The composition of a positive definite operator and a self-adjoint operator is positive definite?

Body:
Let $A: \mathbb{R}^n \to \mathbb{R}^n$, $n>1$ be a positive definite operator that is invertible, and it is not the identity map. And let $B$ be a self-adjoint operator.

My question is, $A^{-1}\circ B$ is positive definite?

I have the feeling that this can be true, but I don't know how to prove it.

Thank you very much.


问:
compositions of n into even parts
I have found here {https://math.stackexchange.com/questions/2167885/compositions-of-n-into-odd-parts} that the number of integer compositions of n into k odd parts would be ${\frac{n+k-1}{2} \choose k-1}$.

I would like to find the number of integer compositions of n into k even parts. My guess is that it would be the same, but I do not see how to prove it.

答:
Each positive, even number is at least equal to two, so you can subtract one to get an odd, but still positive, number.

Therefore, the number of ways to decompose $n$ into $k$ even numbers is the same as the number of compositions of $n-k$ into $k$ odd numbers; and for that you already have a formula.

Title: compositions of n into even parts


问:
computational strategy for solving convex-concave minmax problem
Assume f(x,y) is convex in $x$ and concave in $y$.
Then \begin{equation}\min_x \max_y f(x,y)\end{equation} is globally solvable, because f is convex in x (max of convex is convex.)
But can we find a globally optimal $x$ with coordinate descent? (aka alternating: i.e Fix x, find optimal y. Fix $y$ find optimal $x$ till convergence.) Why or why not?

答: The update cycles, but when averaged over time the average does converge to the optimum. That can be seen in the example and holds quite generally.</p>


问:
Questions about an exercise about the irreducible adjoint representation
The following exercise I want to ask is

Show that if every adjoint representation $G$ is irreducible, there
is no nontrivial subspace $V \subset T_{e}G$ is invariant under
$Ad(g)$, for all $g \in G$ , then any normal subsgroup $H$ in $G$ has
a dimension $0$ or $n = dim G$.

My soultion is here : assume $(m=)dimH \neq 0$. Then claim : $dimH=n$. My main thought is considering codimension of $H$.  Let $K \subset G$ be a codimesion of $G$. If $e_{1},.....,e_{m}$ is a basis of $H$, then $e_{m+1},...,e_{n}$ is a basis of $K$.  Now, put $Ad(g) : G \to End(H \oplus K) $ and $\psi:= Ad(g)|_{K}$. Since $G$ is irreducible, the invariant subspace $V \in T_{e}G$ such that $\psi(k)(v) \in V$ for all $k \in K$ does not exist except for trivial subspace 0. But, That's all I can do it.
I expect to logically induce $m=n$ from the above result, but I think that I am a wrong way to sort it out.

Another detour that I can imagine is the definition of the normal subgroup, left and right coset is coincident, yet I have not found strategy.

答:
First, I'll prove the hint in my comment:

The adjoint action of $G$ on its Lie algebra $\mathfrak{g}$ is defined as follows.  For each $g\in G$, let $C_g:G\rightarrow G$ be defined by $C_g(h) = ghg^{-1}$, that is, by conjugation.  The map $C_g$ is a Lie isomorphism with inverse $(C_g)^{-1} = C_{g^{-1}}$.
Then for the identity $e\in G$, $C_g(e) = geg^{-1} = e$, so $d_e(C_g):T_e G\rightarrow T_g G$.

The adjoint action is given by the map $G\rightarrow Gl(\mathfrak{g})$ with $g\mapsto d_e(C_g)$.

We claim that if $H\subseteq G$ is a normal subgroup, then this action preserves $T_e H\subseteq T_e G$.  To see this, let $v\in T_e H$ and let $\gamma:(-\epsilon,\epsilon)\rightarrow H$ be a smooth curve with $\gamma(0) = e$ and $\gamma'(0) = v$.

Because $\gamma(t)\in H$ for each $t$ in the domain, and because $H$ is normal in $G$, $C_g(\gamma(t)) = g\gamma(t) g^{-1}\in H$.  In other words, the curve $\alpha(t)=g \gamma(t) g^{-1}$ is a curve in $H$.  Moreover, $\alpha(0) = g\gamma(0)g^{-1} = geg^{-1} = e$, and $\alpha'(0)$ is, by definition, $d_e(C_g)(v)$.  Because $\alpha(t)\in H$ for all $t$ in the domain, $d_e(C_g)(v) = \alpha'(0)\in T_e H$.  This shows that $T_e H$ is preserved by the adjoint action of $G$.

Now, to actually answer the question in your post, if $H\subseteq G$ is normal, then $T_e H$ is an invariant subspace of $T_e G$.  Since the adjoint action is irreducible by assumption, it follows that $T_e H$ either has dimension $0$ or dimension $n$.

Now, if $\exp:\mathfrak{g}\rightarrow G$ is the group exponential map, then recall that $\exp$ is a diffeomorphism onto its image when restricted to a small enough open set $U\subseteq T_e G$.  It follows that $\exp|_{U\cap T_e H}:(U\cap T_e H)\rightarrow H$ is a diffeomorphism onto its image.  It's image is an open subset of $H^0$, the identity component of $H$, so it follows that $H^0$ has dimension $0$ or $n$.

Lastly, note that in a disconnected Lie group, all the components are diffeomorphic.  In fact, if $h$ is in some component of $H$, then left multiplication by $h$ is a diffeomorphsim which maps the identity component onto the component containing $h$.  In particular, all the components of $H$ have the same dimension, so that dimension is $0$ or $n$.


问:
Does $x_k$ identically distributed with finite mean imply $x_k/k$ converges to 0 $a.s.$?
I am not sure whether the following statement is true. Let $x_k$ be identically distributed, and only assumed to have finite mean. Can we conclude that $x_k/k$ converges a.s.? Clearly it converges in probability. And if we compute
$$
 P(|x_k|\geq k\epsilon)\leq \frac{E|x_k|}{k\epsilon}$$

However, since we only assume finite mean, it is not sufficient to conclude the almost everywhere convergence. I can come up with counterexamples when $x_i$ are not identically distributed. But not for the identically distributed case. Does anyone have any idea about this?

Thanks in advance.

答:
I think not.

Let $X_i \sim Poisson(\lambda)$.

Then $P(X_k \le k) =\sum_{i=1}^k \frac{\lambda^i}{i!}e^{-\lambda}$.

For every $n$, we will have that $P(X_n \le n, X_{n+1}\le n+1, ...) = P(X_n \le n) P(X_{n+1} \le n+1) \cdots < P(X_n \le n) < 1$.

Hence the limit of $X_i / i$ is not probabilistically guaranteed to stay below $1$ for big enough $n$, so $P(\lim_{n\to \infty} \frac{X_n}{n} = 0) < 1$


问:
Show that the domain is closed
I have to show that the domain defined by the half unit circle: $\left \{ \left ( x,y \right )\in \mathbb{R}^2: x^2+y^2\leq 1, x\geq 0 \right \}$

for the function $f:D\rightarrow \mathbb{R}$ is closed (the function is $f(x,y)=2xy^2-2x^2$), so that I can argue the function has a min/max value. I have shown it is bounded and continuous.

答:
Hint : You can write$$\left \{ \left ( x,y \right )\in \mathbb{R}^2: x^2+y^2\leq 1, x\geq 0 \right \} = g^{-1}([0,1]\times [0, +\infty))$$

where $g : \mathbb{R}^2 \rightarrow  \mathbb{R}^2$ is the continuous function defined for all $(x,y) \in \mathbb{R}^2$ by
$$g(x,y)=(x^2+y^2,x)$$

Title: Show that the domain is closed


问:
Convergence of a sequence of RVs
Question:
A sequence of RVs $X_1,X_2 ... $ converges by distribution to a RV if $\forall x \in \Bbb R, i \in \Bbb N: P(X_i=x) \to P(X=x)$.
Prove that these series uniformly converge (so you need to prove: $\forall x\in \Bbb R \forall \varepsilon >0 \exists N \forall  n>N: |P(X_n=x)-P(X=x)| \le \varepsilon)$

What I did:
Since this is the first time I met this combination of calculus and probability, I tried proving it the same as I would a regular limit (Looking for an N that satisfies this). But I don't really know anything about these sequences apart from the fact:
$\sum_{\forall x \in \Bbb R}P(X_n=x)=1$ and $\sum_{n=1}^{\infty}\sum_{\forall x \in \Bbb R}P(X_n=x)=\infty$
So I don't know how I can find such an N

Thanks.

答:
This is a long, long answer.

One of two things has happened, either whoever wrote the question has some serious holes in his understanding of probability theory, or more likely they wrote the question last thing on a Friday and weren't really paying attention.

Personally I'd just have skipped the question and not worried about it.

If you want to talk to the lecturer about it be very polite and say you don't understand the question and you think there might be a typo. Hopefully he didn't write the questions himself, but if he did you don't want to accuse him of not understanding the subject he's teaching. (That might be the case but he'll get defensive if you say it directly.)

The statement of the question is incorrect in at least three ways.

First the statement
$$\forall x\in\mathbb R, i\in \mathbb N:P(X_i = x)\to P(X = x)$$
is meaningless.  I'm assuming that this should read
$$\forall x\in\mathbb R:P(X_i = x)\to P(X = x)\text{ as }i\to\infty.$$

a random variable with domain $\mathbb R$ is usually continuous.
A random variable is continuous if its probabililty distribution is given by its density function $f_X$ and you get statements like
$$P(a\leq X \leq b) = \int_a^b f_X(x). dx$$
But $\int_a^a f_X(x) dx = 0$ for every $a$ so we must have $P(X=a) = 0$ for every continuous random variable.

So if $X_n$ is any sequence of continuous random variables and $Y$ is any other continuous random variable then  $P(X_i = x) \to P(Y=x)$ for every $x$ because a sequence of zeros converges to zero.  You can't have a definition of convergence that says every sequence converges to every limit.

One definition of convergence in distribution is as follows.

A sequence of random variables $X_n$ converges in distribution to a continuous random variable $X$ if and only if
$$\forall x\in\mathbb R:P(X_i \leq x)\to P(X \leq x)\text{ as }i\to\infty.$$

(I've had to restrict the definition to continuous limits here to keep my corrections looking even vaguely like the question. This is not a very good definition of converges in distribution.)

If you remember your real analysis classes this statement is shorthand for
$$\forall x\in\mathbb R, \varepsilon >0 \exists N\in\mathbb N\text{ such that }\forall n>N: \left|P(X_i \leq X) - P(X_i \leq X)\right| <\varepsilon.$$

So if we replace $P(X_i = X)$ and  $P(X_i = X)$ with $P(X_i \leq X)$ and $P(X_i \leq X)$ in both the question and the answer then the question and the answer are the same.

The definition of a uniform convergence of functions is as follows
We say a sequence of functions $f_i:\mathbb R \to\mathbb R$ converges uniformly to a function $f$ if
$$\forall\varepsilon>0\exists N\in\mathbb N\text{ such that } \forall x\in\mathbb R: \left|f_i(x) - f(x)\right|. $$

So for $f_i(x) =P(X_i\leq x)$ and $f(x) = P(X\leq X)$ the statement you should be trying to prove is

$$\forall\varepsilon>0\exists N\in\mathbb N\text{ such that } \forall x\in\mathbb R: \left|P(X_i\leq x) - P(X\leq x)\right|. $$

So the difference between the (corrected) statement in the question and the one you are trying to prove is that the $\forall x$ has been moved to the right of the "such that".

You can try to prove this yourself if you want, but I think it's much harder than the question was supposed to be. You don't need to know anything about the random variables themselves apart from the fact that $X$ is a continuous random variable. That means that for every $x\in\mathbb R$ we have $P(X=x)=0$.
Equivalently you can say that for every sequence $x_i$  of real numbers with $x_i\to x$ as $x\to\infty$ we have $P(X < x_i) \to P(X<x)$.

You need to use this fact in the proof, consider the following example of what can go wrong if I don't assume that $X$ is continuous.

Let $X_i$ be the random variable such that $X_i = 0$ with probability $0.5$ and $X_i = 1-\frac 1i$ with probability $0.5$.

So if $x<1$ then $P(X_i\leq x) = 0.5$ for sufficiently large $i$ and for $x\geq 1$; $P(X_i\leq x) = 1$ for every $i$.  So the definition above is satisfied by the random variable $X$ which is $0$ with probabiltiy $0.5$ and $1$ with probability $0.5$.

But for $\varepsilon = \frac 13$ I can set $x_i = 1-\frac1{i+1}$
and
$$\left|P(X_i < x_i) - P(X < x_i)\right| = \frac 12 >\frac 13.$$

So for this value of $\varepsilon$ I can't choose an $N$ where $\left|P(X_n < x) - P(X < x)\right| <\varepsilon$ for every $x$ whenever $n>N$. Therefore the sequence doesn't converge uniformly.

If you were to try to prove the statement you should argue that if $P(X_i < x)$ converges for every $x$, but not uniformly then you must be able to choose a sequence $x_i$ and an $\varepsilon$ for which $\left|P(X_i < x_i) - P(X < x_i)\right| >\varepsilon$. From here you can choose a convergent subsequence of $x_i$ and use a diagonal argument to show that there must be some value of $x$ such that $P(X=x)\geq\varepsilon$. (Be careful though I'm allowing $x=\infty$.)

I don't recommend doing that, it's not something anyone would expect an undergraduate to be able to do on their own.

I don't think that's actually going to help you much in this situation, but I hope that it helps with something.

Title: Convergence of a sequence of RVs


问:
Stochastic integration by parts for random point processes
I'm trying to understand this proof of the following specifing integration by parts.

Introduction

Let $\Omega=Point_{\mathbb{R}}$ the set of point distributions in $\mathbb{R}^3$ (i.e an element $w \in \Omega$ is a locally finite subset of $\mathbb{R}^3$). We equip this space with a canonic tribe $\mathcal{A}$ and a probability $\mathbb{P}$.

Let $\phi : \Omega \rightarrow \mathbb{R}^3$ a measurable function. For $w \in \Omega$, we define the realization of $\phi$ the mapping $R_\omega[\phi] : \mathbb{R}^3 \rightarrow \mathbb{R}^3$ :

$$R_\omega[\phi](y):=\phi(y+w)=\phi(\tau_yw), \quad y \in\mathbb{R}^3$$
where $\tau_y :\Omega \rightarrow \Omega$ is the shift application (which is measurable in this case).

Let $\varphi : \Omega \rightarrow\mathbb{R^3}$ a smooth function and $w \in \Omega$, we define the stochastic gradient by $$\bar{\nabla}(\varphi)(\omega):=\nabla (t \mapsto \varphi(t+w))(0).$$
where $\nabla$ is the regular spatial gradient. Using this definition, we define the partial derivative $\bar{\partial}_i$.

the integration by parts

Let $u,v$ smooths functions from $\Omega$ to $\mathbb{R}^3$. According to the text I'm reading, we have the following integration by parts : $$\mathbb{E} \left[ \bar{\partial}_iu v\right]=-\mathbb{E} \left[ u \bar{\partial}_i v\right]$$
The proof starts with the following equality, that I'm strugling to understand :
\begin{aligned}
\mathbb{E}   \left[\bar{\partial}_iu v\right]= \mathbb{E} \left[\int_{K_1} \partial_i R_\omega[u](y) \ R_\omega[v](y) \ \mathrm{d}y\right] \quad \quad (\star)
\end{aligned}
with $K_1=\left[-\frac{1}{2},\frac{1}{2}\right]^3$. Then, we can use the regular integration by parts formula for the $K_1$ integral, which gives us :

$$- \mathbb{E} \left[\int_{K_1}  R_\omega[u](y)  \ \partial_i R_\omega[v](y) \ \mathrm{d}y\right]+ \underbrace{\mathbb{E} \left[\int_{ \partial K_1} n_i R_\omega[u](y) \ R_\omega[v](y) \ \mathrm{d}y \right]}_{:=A}$$
and then it is said that $A=0$. The rest of the proof follows easily using $(\star)$ again.

My questions

Where does the first equality $(\star)$ comes from ?

Why do we have $A=0$ ? In the usual case it is because the test functions have a their support include in a compact but we do not have such hypothesis here.

Any helps or hint are welcomed !

答:
Actually it is necessary to add one hypothesis so that the result holds.

We need to assume that the application shift $\tau_y : \Omega \rightarrow \Omega$ is $\mathbb{P}$ preserving for any vector $y \in \mathbb{R}^3$. Using this stationarity hypothesis, we see that :

$$\mathbb{E}\left[ \bar{\partial}_i v u \right]= \mathbb{E}\left[ \bar{\partial}_i (v \circ \tau_y) \ (u \circ \tau_y) \right], \quad \forall y \in \mathbb{R}^3. $$

Integrating this equality over $y$ on the domain $K_1$ which has a Lebesgue measure  equals to 1, we get by Fubini's theorem the equality $(\star)$.

A similar argument on the boundary integral leaves us with $A=\mathbb{E}[uv] \int_{\partial K_1} n_i$, and it is well known that the latter integral worths $0$, hence $(\star \star)$.


问:
Is it possible to evaluate $\lim_{x\to\infty}\frac{\ln(x+e^x+e^{2x})}{x}$ without applying L'Hopital rule?
$$\lim_{x\to\infty}\frac{\ln(x+e^x+e^{2x})}{x}$$
In the book I am reading, evaluated the limit with L'Hopital rule like this:
$$\lim_{x\to\infty}\frac{\ln(x+e^x+e^{2x})}{x}=\frac{\infty}{\infty}\rightarrow \lim_{x\to\infty}\frac{1+e^x+2e^{2x}}{x+e^x+e^{2x}} $$

Then it used equivalence and write $\lim_{x\to\infty}\cfrac{2e^{2x}}{e^2x}=2$.

My approach to evaluate $\lim_{x\to\infty}\frac{\ln(x+e^x+e^{2x})}{x}$ is different:

in the numerator of the fraction and inside the $\ln()$ the function $e^{2x}$ goes faster to infinity than $x$ , $e^x$. therefore we can ignore these two:
$$\lim_{x\to\infty}\frac{\ln(e^{2x})}{x}=\lim_{x\to\infty}\frac{2x}{x}=2$$
Is my approach right? and was it really necessary to use L'Hopital Rule?

答:
We have that for any $\varepsilon>0$ eventually

$$2=\frac{\ln(e^{2x})}{x}\le \frac{\ln(x+e^x+e^{2x})}{x}\le\frac{\ln(e^{(2+\varepsilon)x})}{x}=2+\varepsilon$$

therefore since $\varepsilon$ is arbitrarily small

$$\frac{\ln(x+e^x+e^{2x})}{x} \to 2$$


问:
Calculating number of customers in a queue
I am trying to calculate the average wait time for a customer in a queue where there is a single server. I have the distribution for the arrival probability of the customer at time (t) as

$F(t) = 1 - e^{-t/200}$

And it's given that the customers arrive one at a time.

I also have the beta distributions for the processing time of the customer. I can calculate the average and max processing time of the customers using the beta distributions. Now, I need to calculate the average and max waiting time of customer (time the customer has to wait before they reach the server).

```<code>(avg/max)Wait = (number of customers in the queue) * ((avg/max) * processing time for a customer)
</code>```

I can calculate the second part of the equation using the beta distribution given to me. But how do I calculate the average number of customers in the queue?

答:
Use Little's Law.

$$L = \lambda W$$

where,

$L$ is the long-term average number of customers in the stationary system,

$λ$ is the long-term average effective arrival rate,

$W$ is the average time that a customer spends in the system


问:
Quadrilateral with given angles

We are looking for angles x and y.

I have found the values of the following angles:
BEA = 74,
BDA = 64,
ACD = 68,
ECD = 112,

plus the relationship $x+y = 68$.

All other angles equations, from triangles or the sum of angles in the quadrilateral (360) end up in the same equation!

I have found through Geogebra that $x=18$ and $y=50$ but I can't figure out a second relationship to determine them geometrically!

Does anyone have any ideas?

Thank you!

答:
Sinx / Sin(x + 64) = Sin38 / Sin84 *Sin22 / Sin48

Multiply both denom and nom by 4Sin82

Sinx /Sin(x + 64) = 4Sin22Sin38Sin82 / (Cos6Sin48Sin82)

(4Sin22Sin38Sin82 = Sin66 REPLACEMENT)

Sinx / Sin(x + 64) = Cos24 / (4Sin48Cos6Sin82)

Sinx / Sin(x + 64) = 1 / (8Sin24Cos6Sin82)

(2Sin24*Cos6 = Sin54 REPLACRMENT)

Sinx / Sin(x + 64) = 1 / (4Sin54*Sin82)

(1/4Sin54 = Sin18 REPLACEMENT)

Sinx / Sin(x + 64) = Sin18 / Sin82

X = 18

Y = 50

Title: Quadrilateral with given angles


问:
Let $Z$ be a positive random variable. Find $ \lim_{n \to \infty} \frac{E[Z^{n+1}]}{ (n+1)E[Z^n]}$
Suppose that $Z$ is a positive random variable.  I am interested in whether the following limits characterizes some properties of the distribution of $Z$:
\begin{align}
 \lim_{n \to \infty} \frac{E[Z^{n+1}]}{ (n+1)E[Z^n]}
\end{align}

For example, we know that $ \lim_{n \to \infty} E^{1/n}[Z^n]$ characterizes the essential supremum.

What I have done.   I have a feeling that it characterized some kind of tail behavior.
For example, if $Z$ is bounded by $A$ then it can be shown that the limit is equal zero:
\begin{align*}
 \lim_{n \to \infty} \frac{E[Z^{n+1}]}{ (n+1)E[Z^n]}  \le  \lim_{n \to \infty} \frac{ A E[Z^{n}]}{ (n+1)E[Z^n]} =\lim_{n \to \infty} \frac{ A }{ (n+1)} =0.
\end{align*}

More generally, I found the following behavior. Suppose that $Z \sim e^{-z^k}$, then
\begin{align}
\lim_{n \to \infty} \frac{E[Z^{n+1}]}{ (n+1)E[Z^n]}= \left \{ \begin{array}{ll}
0& k>1\\
1& k=1\\
\infty & 0<k<1
\end{array} \right.
\end{align}

答:
Let $m(t) = E(e^{tZ}) = \sum_k a_k t^k$ be the moment generating function of $Z$. Your limit is $\lim_k \frac{a_{k+1}}{a_k}$, i.e. you are applying the ratio test for convergence.

Essentially $m(t)$ is expected to exist on an interval $(-L,L)$ where $L^{-1}$ is your limit.


问:
Probability of $X$ out of $N$ dice landing on $M$
The problem is as follows:

We have $N$ dice and we throw them on a table. What is the probability that $M$ will fall $X$ times?

Specific example:
We have $10$ dice and we throw them on a table. What is the probability that $3$ will occur $7$ times (so there are seven dice out of $10$ which have $3$ on top)?

I am really bad at probability. I've checked a few sites about how I could solve this but it seems too hard figure it out.

答:
Suppose first that the ten dice are all different colors.  If you want to get 3 on the blue and red dice only, then you want a non-3 on the other eight colors.  The probability of this happening is $(\frac 16)^2(\frac 56)^8$.

But the dice don't have all different colors.  So blue+red is just one way to pick two colors out of ten.  There are ${10\choose 2}$ ways to do that.  Combining, the total probability is $(\frac 16)^2(\frac 56)^8{10 \choose 2}$.


问:
if $x^2=y^3$, then $x=a^3$ and $y=b^2$
We are given that $x^2 = y^3$ and that $x,y$ are positive integers.

We need to show that there are $a,b$ positive integers such that

$x=a^3$, and $y=b^2$.

I tried to do modular arithmetic in order to solve this problem but I got nowhere. I can't see how one would proceed for this question.

Any help is greatly appreciated

答:
If your question is correct, then:

$$
\begin{align}
x^2 &= y^3 \\
\left(a^3\right)^2 &= \left(b^3\right)^2 \\
a^6 &= b^6 \\
a &= \pm b
\end{align}
$$

Because $a, b \in \mathbb{N}$, then $a = b$.

So in conclusion: $x^2 = y^3$ if and only if $a = b$ with $a, b \in \mathbb{N}$.

Title: if $x^2=y^3$, then $x=a^3$ and $y=b^2$


问:
Use residue formula to calculate integration
Show that for $n\geq 1$, $$\int_{\infty}^{\infty}\frac{dx}{{(1+x^2)}^{n+1}}=\frac{1 \cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)}\cdot\pi$$

My approach: I constructed a semi-circle on the upper half plane, where R>1. I think the order of pole at $z=i$ is $n+1$, because $\frac{1}{(z-i)^{n+1}}=(z+i)^{n+1}(\frac{1}{1+z^2})^{n+1}$ is holomorphic and never goes to zero in some neighborhood of $z=i$.
Then I used the formula,
$res_i f=\lim_{z \to i} \frac{1}{n!}(\frac{d}{dz})^n(z+i)^{-(n+1)}$. Then I plgged into $\int_{-\infty}^{\infty}f(x)dx=2\pi ires_i f$. Here obviously, on the upper circle contour, the integration vanishes. And when I calculate it, the result I get is far from the answer above.

Where could I possibly do wrong?

答:
What was your result? Following your post, I would expect that you arrived at
$$
2 \pi i \,\textrm{Res}(f,i) =\frac{(n+1)(n+2)\cdots (n+n)}{n! 2^{2n}} \pi,
$$

which is the correct result, and is equivalent to the proposed formula. Maybe you got the right result and did not notice!


问:
How to integrate $\sqrt\frac{\arctan(x)}{(1+x^2)}$
The problem is: ($x≥0$ is a real number)

$$\int\sqrt\frac{\arctan(x)}{(1+x^2)}dx$$

I tried u-substitution technique where I set $u=\arctan(x)$ and therefore $dx = (1+x^2)du$. The problem is that it doesn't cancel out the $x$ since there $1+x^2$ is inside a square root.
I tried the partial integration method where I set

$
F(x) = \sqrt{\arctan(x)}
$

$
G'(x) = \sqrt\frac{1}{1+x^2}
$

And then calculated:

$
F'(x) = \frac{1}{2(1+x^2)(\sqrt{\arctan(x)})}
$

$
G(x) = \ln(\sqrt{1+x^2}+x) + C
$

And you can see where this goes, it becomes even more difficult to solve. Reverse substitution, I was unable to make that work either. How do I do this? Is there a simple trick that I'm missing here? If it gives more insight into solving this, the problem is to find volume when this is rotated about the $X$-axis at an interval $[a,b]$ where $a>0$ and $b>0$

答:
When rotating about $x$-axis, if $b>a>0$, the integral to solve is
$$V=\pi  \int_a^b \frac{\arctan x}{x^2+1} \, dx=\frac{1}{2} \pi  \left(\arctan^2 b-\arctan^2 a\right)$$


问:
Is the integer polynomial ring $\mathbb Z[X]$ an Euclidean domain?
Though seemed to be a common question, I couldn’t find any information on that.

I can’t imagine the Euclidean function would be like, if there exists one, so my guess is not. But how to prove it?

答: No. Since $\langle 2, x \rangle$ is not an ideal generated by one element, $\mathbb Z[x]$ is not even a PID, let alone an Euclidean ring.


问:
How many numbers between 1 and 10,000,000 don't have the sequence 12? Inclusion-exclusion problem
I got the following question:

How many numbers between 1 and 10,000,000 don't have the sequence 12? This is an inclusion-exclusion problem. Sadly I didn't fully understand its concept, so I tried solving it logically. I would like to know if my solution is correct, and if possible, how to solve it using the inclusion-exclusion principle.

Thanks!

So I said lets first find the numbers that do include the sequence 12, and divided it into cases:

Between 1 and 99: $1\cdot 1=1$

Between 100 and 999: $1\cdot 1 \cdot 10$ + $9\cdot 1 \cdot 1=19$

Between 1000 and 9999: $1 \cdot 1 \cdot 10 \cdot 10 + 2\cdot 9 \cdot 10 \cdot 1 \cdot 1 = 280$

Between 10000 and 99999:  $1 \cdot 1 \cdot 10 \cdot 10 \cdot 10 + 3\cdot 9 \cdot 10 \cdot 10\cdot 1\cdot 1=3700$

Between 100000 and 999999: $1 \cdot 1 \cdot 10 \cdot 10 \cdot 10\cdot 10 + 4\cdot 9 \cdot 10 \cdot 10\cdot 10\cdot 1\cdot 1=46000$

Between 1000000 and 9999999: $1 \cdot 1 \cdot 10 \cdot 10 \cdot 10\cdot 10\cdot 10 + 5\cdot 9 \cdot 10 \cdot 10\cdot 10\cdot 10 \cdot 1\cdot 1=550000$

So the total of numbers that don't include the number 12 is:
10000000-550000-46000-3700-280-19=9400001

I don't havea final answer so I don't know if that correct. Obviousy this solution lacks elegance, so I would like to know how solve this problem using the inclusion-exclusion principle.

Thank in advance!

答:
All right you are at the first step of the inclusion-exclusion principle. You have just counted those numbers in which 12 appears at least once but in the process have overcounted. Your solution counts $1212$ twice, $121212$ thrice and so on.

By inclusion-exclusion principle,
No. of numbers w/o $12$ = Total no. of numbers - No. of numbers with $1$2 at least once + No. of numbers with $12$ twice - No. of numbers with $12$ thrice + No. of numbers with $12$ four times and so on.

Total no: $10,000,000$

$12$ at least once: $6*10^5$

$12$ at least twice: $(4+3+2+1)*10^3$

$12$ at least thrice: $((2+1)+(1))*10^1$

$12$ four times and so on $=0$

so ans= $10000000-600000+10000-40=9409960$


问:
Mathematics- mechanics
Show that an object accelerating with acceleration a from velocity $u$ to velocity $v$, where $0<u<v$,over a time $t$ is traveling at a velocity of $(u+v)/2$ at time $t/2$; that is, that at the time halfway through the motion the velocity of the object is the mean of the initial and final velocities

答:
Assuming constant acceleration, we can write it from the definition $$a=\frac{v-u}t$$
Calculating $w$, the velocity at $t/2$ $$w=u+a\frac t2=u+\frac{v-u}t\frac t2=\frac{u+v}2$$

Title: Mathematics- mechanics


问:
Trouble with a differential equation
Could somebody please help me solve the differential equation.I entered the equation in various online differential equation calculators but they said the input in invalid

$qv_oe^\frac{-bt}{m}B-\frac{mv_o^2e^\frac{-2bt}{m}}{r} =m\frac{d^2r}{dt^2}$

as $t→\infty,r→0$
I am posting the question from which this equation arose as follows

I am assuming small values of the constant b

$m\frac{dv}{dt}=-bv$
solving this we get that

$v(t)=v_0e^\frac{-bt}{m}$
now

$ma_{radial}=qv(t)B-\frac{mv^2(t)}{r}$

now substituting v(t) I arrived at the above equation.
please help!

答:
With the usual notation we get
$$m\mathbf {\dot v}=q \mathbf v \times \mathbf B-b \mathbf v$$
Now assume that $\mathbf B$ is in the z direction (that is $\mathbf B = B \mathbf k$), and that the particle is launched in the x – y plane. Express $\mathbf v$ and $\mathbf {\dot v}$ in terms of $\mathbf i$ and $\mathbf j$ and you will have two equations each mixing $v_x, \dot{v}_x, v_y, \dot{v}_y$. It's easy to eliminate (say) $v_x\ \text{and}\ \dot{v}_x$, leaving a second order ODE in $v_y$ with constant coefficients.

Title: Trouble with a differential equation


问:
Connections between number theory and abstract algebra.
I haven't taken abstract algebra yet, but I am curious about connections between number theory and abstract algebra.  Do the proofs of things like Fermat's little theorem, the law of quadratic reciprocity, etc. rely upon techniques found in abstract algebra?

Thanks.

答:
Historically, there are many results in number theory that were proved without modern algebra, but are now viewed algebraically.

For example, the Chinese Remainder Theorem (CRT) states that any system of congruences $$\begin{array}{rcl}x&\equiv& a_1 \pmod{b_1}\\x&\equiv& a_2 \pmod{b_2}\\ &\vdots& \\ x&\equiv& a_n \pmod{b_n}\end{array}$$ for which $b_1,b_2,\ldots ,b_n$ are pairwise coprime has a solution $x$, and that all such solutions are congruent $\mod {\prod_{i=1}^n b_i}$.  The CRT was first proved by Sun Tzu$^\star$ around the year $300$.

It turns out that the Chinese remainder theorem is equivalent to the group theoretic statement  that, whenever $b_1,b_2,\ldots ,b_n$ are pairwise coprime and $b=b_1b_2\cdots b_n$, $$\mathbb{Z}_{b}\cong \mathbb{Z}_{b_1}\times \mathbb{Z}_{b_2}\times \cdots \times \mathbb{Z}_{b_n}.$$  In the completely unbiased opinion of this finite group theorist, the above is a much more intuitive form of the CRT.  In fact, seeing the theorem in this form inspired a generalization: if $I_1,I_2,\ldots, I_n$ are pairwise coprime$^{\star\star}$ ideals of a commutative ring $R$, and $I$ is the product of these ideals, then $$R/I\cong R/{I_1}\times R/{I_2}\times \cdots \times R/{I_n}.$$  With this generalization we can use the CRT in other commutative rings, like the Gaussian integers $\mathbb{Z}[i]$, polynomial rings $R[x]$, formal power series rings $R[[x]]$, and so on.  All of these rings are objects of interest in algebraic number theory, a discipline which (obviously) emphasizes the blurry line between what is abstract algebra and what is number theory.

$\hspace{2pt}^\star\hspace{2pt}$ To my great disappointment, this was not the Art of War guy.

$^{\star\star}$ Here "pairwise coprime" means that $I_j+I_k=(1)$ for each $j\neq k$.


问:
Continuity of a two variable function at$(0,0)$
I have to prove or disprove the continuity of the function at $(0,0)$ which is given by:

$$f(x,y)= \begin{cases}\frac{\sin^{-1}(x+2y)}{ \tan^{-1}(2x+4y)}&\text{ if }
 0<\sqrt{x^2+y^2} \\  \frac 12 &\text{ if } (x,y)=(0,0)\end{cases}.$$

I consider $x+2y=t$ where $t$ approaches to $0$. This reduces the two variable function to that of a single variable in $t$. The resulting function is:

$g(t)= \frac{\sin^{-1}(t)}{\tan^{-1}(2t)}$ for non-zero $t$ and $\frac12$ for $t=0$.

This function is continuous at $t=0$. Does it imply the continuity of the original two-variable function at$(0,0)$. What are the other ways of checking continuity at the said point?

答:
Yes your way is fine, since we have that by $t=x+2y \to 0$

$$\frac{\sin^{-1}(t)}{\tan^{-1}(2t)}=\frac12\frac{\sin^{-1}(t)}{t}\frac{2t}{\tan^{-1}(2t)} \to \frac12$$

we can conclude that the give function is continuous at $(0,0)$.


答:
While you do not need abstract algebra to formulate many interesting number-theoretic problems, you will find it very useful in solving those problems.
Many theorems in elementary number theory are very simple in their formulations, yet their known proofs require advanced algebraic machinery, for instance Fermats Last Theorem. Personally I find algebraic proofs of many elementary number theory statements much more natural than their "elementary" proofs, see for example Fermats Small Theorem or Chinese Remainder Thorem and their proofs. Once you get some training in basic group theory and basic ring theory those proofs are very natural.</p>

Title: Connections between number theory and abstract algebra.

Body:
I haven't taken abstract algebra yet, but I am curious about connections between number theory and abstract algebra.  Do the proofs of things like Fermat's little theorem, the law of quadratic reciprocity, etc. rely upon techniques found in abstract algebra?

Thanks.


问:
Rotating an L shape of 3 points so that it always faces a certain way.
I have 3 points in 2d space that roughly make an L shape. I want the L shape to face a certain way. I'm having trouble coming up with the logic to figure out which angle I should rotate my shape.

Here are the 3 main orientations I encounter and their slight variations;

Orientations

I want my L to always be like 1-A (top left).

I know which point is the corner point. What I do is I get the angle to the two other points.

In the case of 1-B these might be (-2, -94) and in 1-C (3, -88). So, for 1-B I'd need to rotate the whole shape by -2 degrees. For 1-C I'd need to rotate by 3 degrees.

In the case of 2-B, my angles are something like (-87, -177) and for 2-C they are (-91, 178). For 2-B, I'd need to rotate the shape by -87 degrees. For 2-C I'd need to rotate by -91 degrees.

I've tried calculating the smaller angle by magnitude and rotating the shape by that.

```<code>var smaller_angle;

if(math.abs(angle1) < math.abs(angle2))
   smaller_angle = angle1;
else
   smaller_angle = angle2;

shape.rotate(smaller_angle);

</code>```

This works for the majority of the time but it won't work, for example, the shape is similar to 3-A with the angles (0, 88). The above logic rotates by 0 and it fails.

What should my logic be so I can find the angle I'm supposed to rotate this shape, given that I know the corner point and I know the angle between corner and other points?

答:
I see that your angles are between $-180^\circ$ and $+180^\circ$. The difference between your two angles is about $90^\circ$. If you want to rotate such that one side is along the positive $x$ axis, and the other one towards the negative $y$ axis, all you need to do is rotate by the opposite of the larger angle.
So $$(-2, -94)\to 2\to(0,-92)\\(3, -88)\to -3\to(0, -91)$$Then $$(0,88)\to -88\to (-88,0)$$

Note that this is valid if the negative $x$ axis is not in the region between the legs.

How to check for that? look at the angle difference between the larger and smaller If it is smaller than $180^\circ$ your inner region does not overlap the $-x$ axis, so you are fine. If it is greater that $180^\circ$ you are overlapping.

What do you do if you overlap the $-x$ axis? In that case subtract $360^\circ$ from the larger value. The previous smaller value is now larger, and continue as before

Note since the question uses a reference frame upside down (positive x to the right, positive y down), the following is a possible python implementation

```<code>def rotate(a1,a2):
    sm, la = a1, a2
    if sm>la:
        sm, la = la, sm
    if la-sm>180:
        sm, la = la, sm+360
    return -sm
</code>```

Body:
I see that your angles are between $-180^\circ$ and $+180^\circ$. The difference between your two angles is about $90^\circ$. If you want to rotate such that one side is along the positive $x$ axis, and the other one towards the negative $y$ axis, all you need to do is rotate by the opposite of the larger angle.
So $$(-2, -94)\to 2\to(0,-92)\\(3, -88)\to -3\to(0, -91)$$Then $$(0,88)\to -88\to (-88,0)$$

Note that this is valid if the negative $x$ axis is not in the region between the legs.

How to check for that? look at the angle difference between the larger and smaller If it is smaller than $180^\circ$ your inner region does not overlap the $-x$ axis, so you are fine. If it is greater that $180^\circ$ you are overlapping.

What do you do if you overlap the $-x$ axis? In that case subtract $360^\circ$ from the larger value. The previous smaller value is now larger, and continue as before

Note since the question uses a reference frame upside down (positive x to the right, positive y down), the following is a possible python implementation

def rotate(a1,a2):
    sm, la = a1, a2
    if sm>la:
        sm, la = la, sm
    if la-sm>180:
        sm, la = la, sm+360
    return -sm


问:
Regularity of solution of $u_t -\Delta u=f$ where $f$ is a continuous function
Let $f \in C(\bar \Omega \times [0,T])$ and consider the following problem:

$\begin{align}
\partial_tu -\Delta u&=f \quad \text{in} \; \Omega \times (0,T)\\
u(\cdot,0)&=u_0 \quad \text{in} \; \Omega
\end{align}$

Assume that $u_0 \in L^2(\Omega)$. My question is: Can we deduce $u\in C^{1+a, \frac{1+a}{2}}(\Omega \times (0,T))$ for some $0<a<1$? Or can we at least obtain a Hoelder estimate only in terms of the space variable?

My thought was that since $f \in C(\bar \Omega \times [0,T])$ implies $f \in L^2(\Omega \times (0,T))$, we could apply the known $L^2-$regularity and obtain $u\in L^2(0,T;H^1_0(\Omega))\cap H^1(0,T;H^{-1}(\Omega)) \cap W^{2,1}_p(\Omega \times (\sigma,T))$ for all $1 \leq p <\infty\;$ and for any $\sigma>0$. Then, by a t-Sobolev embedding we could obtain the desired regularity.

Is the above correct or did I miss anything? I would appreciate any help or hints!

Thanks in advance!

答:
The fact that $u \in W^{2,1}_p(\Omega \times (\sigma,T))$ does not follow from $L^2$-theory using energy estimates; you need to use the fact that $f$ lies in $L^p.$ Your idea is correct to show this and deduce the result using Sobolev embedding, but you would need to use a suitable $L^p$ estimates for the form
$$ \lVert u \rVert_{W^{2,1}_p(\Omega \times (\sigma,\rho))} \leq C \left( \lVert f \rVert_{L^p(\Omega \times (0,T))} + \lVert u \rVert_{L^2(\Omega \times (0,T))} \right),$$
where $0 < \sigma < \rho < T.$ Note this uses a global estimate, which requires some regularity of $\partial\Omega,$ but you can work around this since the question only asks for interior regularity. Then you would use the embedding
$$ W^{2,1}_p(\Omega \times (\sigma,\rho)) \hookrightarrow C^{1+\alpha,\frac{1+\alpha}2}(\overline \Omega \times [\sigma,\rho]) $$
for $p > n,$ where $\alpha = 1-\frac{p}n.$


问:
Let $f \colon A \rightarrow B$ be a function and $S$ a subset of $A$.
Let $f \colon A \rightarrow B$ be a function and $S$ a subset of $A$.

Are the following containments always valid?.

$(1)    f^{-1} (f(S))⊆ S$

$(2)     S⊆f^{-1} (f(S))$

Attempt

$S ⊆ A$

$f^{-1}(S) = \{a ∈ B : f(a) ∈ S\}$

$ a ∈ f^{-1}(S) ⟺ f(a) ∈ S$

$ f^{-1}(S) ⊆ B$

I just need a little help finishing the proof. Am i wrong in the attempt? Thanks.

答:
There is a problem with your proof. $f^{-1}(S)$ doesn't exist. This is because $S$ is a subset of $A$. It only makes sense to take a preimage under $f$ of a subset of $B$. On the other hand, $f(S)$ makes perfect sense. Indeed, $f(S)=\{f(a)\in B|a\in A\}$. Note that $f(S)$ is a subset of $B$. Then, it now makes sense to write $f^{-1}(f(S))=\{a\in A|f(a)\in f(S)\}$. From this it should now be clear that only one of the two options you presented in your question is always true. Can you tell which one it is?

On the other hand, the other option will be true if we add a further assumption on $f$. Can you tell which assumption we need to add? (hint: recall the notion of an injective/surjective function).


问:
Prove a sequence consists a subsequence of consecutive terms whose product is a perfect square

Let $S$ be a sequence of $N$ numbers containing $n$ distinct positive integers. Prove that
if $N>2^n-1$ then $S$ has a consecutive subsequence whose product is a perfect square. Use Pigeonhole principle.

I have been thinking that for a set that has $n$ distinct numbers will have $2^n$ number of subsets. And by the pigeonhole principle, there will be 2 sequence share the same combination. But I got stuck to find a perfect square.

答:
Suppose your $n$ distinct numbers are $\{a_1,\cdots, a_n\}$ and that your sequence is $\{s_1,\cdots s_N\}$.  Of course each $s_i$ equals some $a_j$.

For each consecutive subsequence $T$ we define the $n$ tuples of parities, $\Delta (T)=(\delta_1, \cdots, \delta_n)$ where $\delta_i$ is $0$ if $a_i$ appears an even number of times in the subsequence and it is $1$ otherwise.    Note that there are $2^n$ possible values for $\Delta(T)$.

Consider first the $N$ consecutive subsequences $S_i=\{s_1,\cdots, s_i\}$.  Since there are at least $2^n$ of these we see that either every possible value for $\Delta (T)$ is hit or one value is hit twice.

If every value is hit then we have some $i$ for which $\Delta (S_i)=(0,\cdots, 0)$ and clearly that $S_i$ works.

If some value is hit twice then we have a pair $i<j$ such that $\Delta(S_i)=\Delta(S_j)$ and then the consecutive sequence $\{s_{i+1}, \cdots, s_j\}$ works.  And we are done.


答:
So we have our $N$ number of dice. Given that the dice has $A$ number of sides.

For any single die, the probability that $M$ falls is $\frac{1}{A}$. However, we need $M$ to fall $X$ times. Suppose we want the first $X$ dice to have $M$, then the probability of that happening is $(\frac{1}{A})^X * (\frac{A-1}{A})^{N-X}$. Order does not matter though, so the dice with $M$ can be in any of the $N$ positions, giving us a ${N\choose X}$ possibilities.

We end up with ${N\choose X}*(\frac{1}{A})^X * (\frac{A-1}{A})^{N-X}$

Title: Probability of $X$ out of $N$ dice landing on $M$

Body:
The problem is as follows:

We have $N$ dice and we throw them on a table. What is the probability that $M$ will fall $X$ times?

Specific example:
We have $10$ dice and we throw them on a table. What is the probability that $3$ will occur $7$ times (so there are seven dice out of $10$ which have $3$ on top)?

I am really bad at probability. I've checked a few sites about how I could solve this but it seems too hard figure it out.


问:
What is the complement of the acceptance problem of a Turing machine?
I know that the acceptance problem of a Turing machine is the problem to decide if for any turing machine $M$, given a string $w$ the Turing machine accepts $w$ or not. If the not acceptance of a string is present on the definition of the acceptance problem of a Turing machine, then what is the complement of this problem?

答:
The set of strings for which a TM halts is semi-decidable.

The complement of this set is not semi-decidable.

Because if a set and its complement are both semi-decidable, the set will be decidable.


答: This ring has Krull dimension $2$ but Euclidean domains necessarily have Krull dimension no greater than $1$.

Title: Is the integer polynomial ring $\mathbb Z[X]$ an Euclidean domain?

Body:
Though seemed to be a common question, I couldn’t find any information on that.

I can’t imagine the Euclidean function would be like, if there exists one, so my guess is not. But how to prove it?


问:
Weighted integral of a convolution
Afer some experiments, it seems to me that:

the integral (in the entire domain) of a convolution $(f*s)(x)$, of a filter function $f(x)$ with a a signal function $s(x)$, multiplied by a weight function $w(x)$,

$$ \int (f*s)(x) \cdot w(x) \, dx  $$

, when all functions are integrable,

is equal to the integral of the signal function $s(x)$ weighted by the cross-correlation $(f\star w)(x)$ of the filter function $f(x)$ with the weight function $w(x)$,

$$ \int (f\star w)(x) \cdot s(x) \, dx $$

Is that right?

If it is, what is the proof?

If it is not, is there something similar that let one merge the filter and the weight functions to create a new weight function?

答:
If I am not mistaken, this is just changing the order of integration (which has to be justified by integrability conditions):
\begin{align}
\int (f*s)(x)w(x) dx & = \int\int f(y-x)s(y)dy\, w(x)dx\\
& = \int\int f(y-x)w(x)dx\, s(y)dy\\
& = \int (f\star w)(y)s(y)dy.
\end{align}

Title: Weighted integral of a convolution


问:
General form of centralizer
If we say $B = J_{a_1}(\lambda_1) \oplus J_{a_2}(\lambda_2) ... \oplus  J_{a_n}(\lambda_n)$ I am trying to find the general form of the centralizer of $B$. First when our lambdas are distinct but then also when they aren't and what the dimension is on each of the cases.

So I know B should be an $\sum a_i \times \sum a_i $ matrix with entries $\lambda_i$ on the diagonals but then I am not sure how to find the centralizer for each case?

I am not really sure how to start and if someone could point me in a good direction that would be appreciated.

答:
The result in terms of matrices is well-known among researchers. I have also seen it documented in an undergraduate textbook, but I cannot recall the book's title.

Suppose $B$ is the direct sum of $r$ Jordan forms $J_1,J_2,\ldots, J_r$, each with a different eigenvalue, and suppose each Jordan form $J_k$ has $b_k$ Jordan blocks  for the eigenvalue $\lambda_k$ of multiplicities $m_{k1}\ge m_{k2}\ge\cdots\ge m_{kb_k}$ respectively. That is, suppose
$$
B=\bigoplus_{k=1}^rJ_k=\bigoplus_{k=1}^r\underbrace{\left(J(\lambda_k;m_{k1})\oplus\cdots\oplus J(\lambda_k;m_{kb_k})\right)}_{J_k}\tag{1}
$$
where $J(\lambda;m)$ denotes an upper triangular Jordan block with an eigenvalue $\lambda$ of multiplicity $m$.

For example, in the $9\times9$ Jordan form $B$ below, we have $J_1=J(1;3)\oplus J(1;2)$ and $J_2=J(0;2)\oplus J(0;1)\oplus J(0;1)$.
$$
B=\left[\begin{array}{ccc|cc|cc|c|c}
1&1&0\\
0&1&1\\
0&0&1\\
\hline
&&&1&1\\
&&&0&1\\
\hline
&&&&&0&1\\
&&&&&0&0\\
\hline
&&&&&&&0\\
\hline
&&&&&&&&0
\end{array}\right].
$$
The equation $BA=AB$ thus reduces blockwise to equations of the form
$$
J(\lambda;m)X=XJ(\mu;n)
$$
for an $m\times n$ matrix $X$. Assume that $m\ge n$ (the case $n\le m$ can be treated similarly). By subtracting both sides by $\mu I$, we get $J(\lambda-\mu;m)X=XJ(0;n)$. When $\lambda\ne\mu$, we have $J(\lambda-\mu;m)^nX=XJ(0;n)^n=0$ and hence $X$ must be zero. When $\lambda=\mu$, we have $J(0;m)X=XJ(0;n)$. In particular, we obtain two equalities:
\begin{align}
x_{i+1,j+1}=e_i^TJ(0;m)Xe_{j+1}&=e_i^TXJ(0;n)e_{j+1}=x_{ij},\\
J(0;m)Xe_1&=XJ(0;n)e_1=0.\\
\end{align}
The first one means that $X$ is Toeplitz while the second one means that the up shift of the first column of $X$ is zero. It folows that all entries below the main diagonal of $X$ are zero, i.e. $X$ is an upper triangular Toeplitz matrix. Conversely, one may verify that every upper triangular Toeplitz matrix $X$ satisfies the equation $J(0;m)X=XJ(0;n)$.

Therefore, the solutions to $BA=AB$ has the general form $\bigoplus_{k=1}^rM_k$, where each $M_k$ has the same size as $J_k$, and, when it is partitioned in the same way as $J_k$ (so that the diagonal sub-blocks of $M_k$ has sizes $m_{k1},\ldots,m_{kb_k}$ respectively), its $(i,j)$-th sub-block is an upper triangular Toeplitz matrix in the form of
$$
M_{kij}=\begin{cases}
\pmatrix{T_{kij}\\ 0}&\text{when } m_{ki}>m_{kj},\\
T_{kij}&\text{when } m_{ki}=m_{kj},\\
\pmatrix{0&T_{kij}}&\text{when } m_{ki}<m_{kj},\\
\end{cases}\tag{2}
$$
where $T_{kij}$ denotes an upper triangular Toeplitz square matrix of size $\min(m_{ki},m_{kj})$.

For instance, the centralisers of our previous $9\times9$ example have the following general form:
$$
A=\left[\begin{array}{ccc|cc|cc|c|c}
a&b&c&d&e\\
0&a&b&0&d\\
0&0&a&0&0\\
\hline
0&f&g&h&l\\
0&0&f&0&h\\
\hline
&&&&&p&q&s&t\\
&&&&&0&p&0&0\\
\hline
&&&&&0&u&v&w\\
\hline
&&&&&0&x&y&z\\
\end{array}\right].
$$
With $(1)$ and $(2)$, the dimension of the centraliser subspace is given by
$$
\sum_{k=1}^r\sum_{i,j=1}^{b_k} \min (m_{ki}, m_{kj})
=\sum_{k=1}^r\sum_{i=1}^{b_k}(2i-1)m_{ki},
$$
which is equal to $(1\times3+3\times2)+(1\times2+3\times1+5\times1)=19$ in the above example.

Title: General form of centralizer


问:
Show $\frac{1}{x}$ is differentiable on $\Bbb R\setminus\{0\}$
From $0<|x-x_0|<\delta$,
how can I go to $|\frac{\frac{1}{x} -\frac{1}{x_0}}{x-x_0} - (-\frac{1}{x_0^2})| < \epsilon$

答:
Take $x_0\in\Bbb R\setminus\{0\}$.

$\displaystyle\lim_{x\to x_0}\dfrac{\frac{1}{x}-\frac{1}{x_0}}{x-x_0}=\lim_{x\to x_0}\dfrac{x_0-x}{xx_0(x-x_0)}=\lim_{x\to x_0}-\dfrac{1}{xx_0}=-\dfrac{1}{x_0^2}$

If you still want to use $\varepsilon$-$\delta$ definition of limit, then we'll use that $\displaystyle\lim_{x\to x_0}\dfrac{\frac{1}{x}-\frac{1}{x_0}}{x-x_0}=-\dfrac{1}{x_0}\lim_{x\to x_0}\dfrac{1}{x}$ as shown before, so we only need to prove $\displaystyle\lim_{x\to x_0}\dfrac{1}{x}=\dfrac{1}{x_0}$.

Take some $\varepsilon>0$. We need $\left|\dfrac{1}{x}-\dfrac{1}{x_0}\right|<\varepsilon$. Now, $\left|\dfrac{1}{x}-\dfrac{1}{x_0}\right|=\left|\dfrac{x_0-x}{xx_0}\right|=\dfrac{1}{|x_0|}\left|\dfrac{x_0-x}{x}\right|=\dfrac{1}{|x_0|}\dfrac{|x-x_0|}{|x|}$.

Since $x_0\neq0$ we can consider $\delta_1=|x_0|>0$, so if $x$ satisfies $|x-x_0|<\delta_1$ then $x_0-\delta_1<x<x_0+\delta_1$, which implies $x\neq0$. Do you see why? (Consider the cases $x_0>0$ and $x_0<0$).

Therefore $\dfrac{1}{|x|}<M=\dfrac{1}{N}$, where $N=\min\{|x_0-\delta_1|,|x_0+\delta_1|\}$ for all $x$ satisfying $|x-x_0|<\delta_1$. To see this use the previous paragraph and consider the cases $x_0<0$ and $x_0>0$ as before (remember $\dfrac{1}{x}$ is strictly increasing/decreasing for negative/positive values).

Then you can take $\delta=\min\{\delta_1,\dfrac{\varepsilon|x_0|}{M}\}$. Since $\delta\le\delta_1$, all $x$ satisfying $|x-x_0|<\delta\le\delta_1$ imply $\dfrac{1}{|x|}<M$.

Then, for $x$ such that $|x-x_0|<\delta$, we have $\left|\dfrac{1}{x}-\dfrac{1}{x_0}\right|=\dfrac{1}{|x_0|}\dfrac{1}{|x|}|x-x_0|<M\dfrac{1}{|x_0|}|x-x_0|<M\dfrac{1}{|x_0|}\dfrac{\varepsilon|x_0|}{M}=\varepsilon$.


问:
Definition of deterministic matrix
What is the definition of a deterministic matrix? It seems stupid but I cannot find it anywhere.
Thanks!</p>

答: Deterministic matrix is a matrix that doesn't include any random elements. More precisely, a matrix $A \in \mathbf{R}^{m \times n}$ is a deterministic matrix, iff for all $i=1, \dots, m$ and $j=1, \dots n$ the values $A_{ij}$ are not random, i.e. they take a certain value with probability $1$.

Title: Definition of deterministic matrix


Title: Show $\frac{1}{x}$ is differentiable on $\Bbb R\setminus\{0\}$

Body:
From $0<|x-x_0|<\delta$,
how can I go to $|\frac{\frac{1}{x} -\frac{1}{x_0}}{x-x_0} - (-\frac{1}{x_0^2})| < \epsilon$

Body:
$$|\frac{\frac{1}{x} -\frac{1}{x_0}}{x-x_0} - (-\frac{1}{x^2_0})| < \epsilon\\
|\frac{\frac{x_0-x}{xx_0} }{x-x_0} - (-\frac{1}{x^2_0})| < \epsilon\\
|\frac{\frac{-{(x-x_0)}}{xx_0} }{x-x_0} - (-\frac{1}{x^2_0})| < \epsilon\\
|\frac{\frac{-1}{xx_0} }{1} - (-\frac{1}{x^2_0})| < \epsilon\\
|\frac{1}{x^2_0}-\frac1{xx_0}| < \epsilon\\
|\frac1{x_0}||\frac{1}{x_0}-\frac1{x}| < \epsilon\\
|\frac{x-x_0}{x^2_0x}| < \epsilon\\$$ can you take over ?


问:
Can a Poisson distribution be understood as a discrete Gaussian?
I'm coming to this from a Machine Learning perspective. In ML, we have to compute what we call the loss when our model makes a prediction. This loss gives us a number, allowing us to tell our far or close from the "right answer" (assumed to be provided in some form). Then it can be use to tweak the weights of our model to hopefully make better predictions next time similar inputs are provided.

Therefore one could see each inputs with its associated answer as a probabilistic distribution of events, hence a number of loss functions are probabilistic.

Since these tend to be discrete in nature, a Poisson distribution can be used for regression predictions.

I'm not familiar with Poisson's distribution, but I have extensively used Gaussian. My current grasp is to see Poisson's as essentially a discrete Gaussian. I understand there's a bit of a skew for extreme values (e.g. a true Gaussian has a pretty long tail, which real values may not allow, such as a price of -10$ for instance).

Is that idea far off from reality? If this is far off, any better mental model to get an intuitive grasp of the Poisson distribution?

答: You are not very far off - in fact, for sufficiently large values of $\lambda$ it is common practice to approximate the Poisson with a normal distribution.


问:
average no of vertices per triangle in a graph of infinite triangles

An inﬁnite two-dimensional pattern is indicated above.

The smallest closed ﬁgure made by the lines is called a unit triangle. Within every unit triangle, there is a mouse. At every vertex there is a "sweet". What is the average number of "sweets" per mouse?

答:
The answer is obviously 1.

Since the pattern is infinite, there's a countably infinite number of mice and a countably infinite number of sweets. Thus there's a bijection between the sets of mice and the sets of sweets, hence there's exactly one sweet per mouse and one mouse per sweet.

Wait a moment... I made a mistake. I forgot to take into account the insatiable hunger these mice have for sweets - each single mice eats all the sweets which lie on a single line. Luckily, since there are countably infinitly many parallel lines, each mice gets its own full line to eat, thus there are actually are countably infinitely many sweets per mice.

I do hope that after eating all those sweets, one of the mice will have the decency to eat this exercise  too...


问:
System with $x + ay = 3 , 2x - y = b$
Quesstion about this system:
$x + ay = 3 , 2x - y = b$

Where $a, b \in\mathbb R$. The system admits as solution $(1, 2)$.
I'm struggling with the logic of getting a final result. Is there an efficient way of calculating this

答:
If $a=-\frac{1}{2}$, then the first equation becomes $x-\frac{1}{2}y=3$ or $2x-y=6$. So if $b\neq6$, we have no solutions and if $b=6$ we have infinitely many solutions.

Suppose that $a\neq -\frac{1}{2}$. Then from the first we have $x=3-ay$ and substituting into the second gives $$2x-y=6-2ay-y=6-y(2a+1)=b$$ or $$y=\frac{6-b}{2a+1}$$

and then we have $$x=3-ay=3-\frac{a(6-b)}{2a+1}=\frac{ab+3}{2a+1}$$ So for $a\neq -\frac{1}{2}$ the set of all solutions is given by $$(x,y)=\big(\frac{ab+3}{2a+1},\frac{6-b}{2a+1}\big)$$

Setting $a=1$ and $b=0$ we obtain one solution $(x,y)=(1,2)$.

If $a=0$ we have $(x,y)=(3,6-b),$ etc.

Title: System with $x + ay = 3 , 2x - y = b$


问:
Why are those inequalities always true?
Here are two inequalities that i don't know how to explain.

$\text{For }n\in\mathbb N^* \text{ and } (a_1,...,a_n)\in(\mathbb R^{+})^n\text{ , }
\sqrt{\sum_{k=1}^{n}k^2a_k}\le\sum_{k=1}^{n}\sqrt{\sum_{s=k}^{n}a_s}$

$\text{For }n\in\mathbb N^* \text{ , } (a_1,...,a_n)\in\mathbb R^n\text{ and }(b_1,...,b_n)\in\mathbb R^n\text{ such that }a_1\le...\le a_n \text{ and } b_1\le...\le b_n , \left(\frac{1}{n}\sum_{k=1}^{n}a_k\right)\left(\frac{1}{n}\sum_{k=1}^{n}b_k\right)\le\left(\frac{1}{n}\sum_{k=1}^{n}a_kb_k\right)$

Why are those inequalities always true?

答:
The second inequality results from multiple applications of the rearrangement inequality, which gives us
$$
a_1b_k+a_2b_{k+1}+\cdots+a_{n-k+1}b_n+a_{n-k+2}b_1+\cdots+a_nb_{k-1}\leq a_1b_1+a_2b_2+\cdots+a_nb_n
$$
for all $2\leq k\leq n$.

Title: Why are those inequalities always true?


问:
if $x^2=y^3$, then $x=a^3$ and $y=b^2$
We are given that $x^2 = y^3$ and that $x,y$ are positive integers.

We need to show that there are $a,b$ positive integers such that

$x=a^3$, and $y=b^2$.

I tried to do modular arithmetic in order to solve this problem but I got nowhere. I can't see how one would proceed for this question.

Any help is greatly appreciated

答: As $x^2=y^3$, $y^3$ is a perfect square. Hence all its prime factors have an even multiplicity. At the same time, these multiplicities are multiple of $3$, hence they are multiple of $6$. Both $x^2$ and $y^3$ are perfect sixth powers, from which the claim follows.

Title: if $x^2=y^3$, then $x=a^3$ and $y=b^2$


问:
CNF/ Create a cnf variable from some forumals of CNF
I have the next CNF: $(A \lor C) \land (B \lor C)$

and also: $(D \lor E) \land (F \lor G)$

Now, I want to be sure that only one of the CNF is true.

Meaning, I want to declare two new variable, T and S, when T = $(A \lor C) \land (B \lor C)$, and S = $(D \lor E) \land (F \lor G)$, and than be sure that only one of them is satisfied (using at most one and at least one methods).

My problem is that I don't know how decalre S and T as new CNF variable.

Any help?

答:
Correct me if I am wrong, but I am assuming you want something like S XOR T: $S \oplus T$.

This is equivalent to $T \lor S \land \lnot (T \land S)$:

To get this in CNF:

$$T \lor S \land \lnot (T \land S) \equiv (T \lor S) \land (\lnot T \lor \lnot S)\tag{1}$$

Now, if we want to express this using your assignments of $T = (A \lor C) \land (B \lor C)$ and $S=(D \lor E) \land (F \lor G)$, we get, by substitution into the right-hand side of $(1)$:

$$ \lbrace[(A \lor C) \land (B \lor C)] \lor [(D \lor E) \land (F \lor G)]\rbrace \land \lbrace \lnot[(A \lor C) \land (B \lor C)] \lor \lnot[(D \lor E) \land (F \lor G)]\rbrace\tag{2}$$

Let me know if you need help with changing $(2)$ into CNF, and/or if I've misunderstood your question altogether!


问:
Is the integer polynomial ring $\mathbb Z[X]$ an Euclidean domain?
Though seemed to be a common question, I couldn’t find any information on that.

I can’t imagine the Euclidean function would be like, if there exists one, so my guess is not. But how to prove it?

答:
The other answers are already great. Let me nevertheless state the following fact, which is a nice exercise:

$R[x]$ is euclidean if and only if $R$ is a field.

One direction is usually proven in standard algebra courses/books. Namely, if $k$ is a field, then $k[x]$ is euclidean with the degree as euclidean function.

On the other hand (to give you a hint): We can show that if $R[x]$ is a pid (euclidean domains are pid's), then $R$ has to be a field. For this consider the ideal $(a,x)$ for some non-zero $a \in R$. Now we can find some $f \in R[x]$ with $(a,x) = (f)$. Using the notion of degree and evaluation morphisms, try to show that $a$ has an inverse in $R$.


问:
Proof on 3 equations crossing at one point $(a+b+c=0)$ WITHOUT determinant of the matrix
So, as in title, I have to prove that if there are 3 linear equations:

$ax + by + c = 0$

$bx + cy + a = 0$

$cx + ay + b = 0$

and those cross at one point, then $a + b + c = 0$. I know that's true because it is easy to do the proof using matrix determinants. However, in that case I was told that it is possible to do the proof using only linear algebra in matrix. I have no idea how to do it and I'm curious.

Thank you in advance for your effort.

edit:
I think I know the answer. You can correct me if I'm wrong. If we have a matrix of those equations it looks like this: r_1 = [a b c], r_2 =[b c a], r_3 = [c a b]. If we add r_1 and r_2 to r_3 we get a follownig equation: (a+b+c)x + (b+c+a)y + (c+a+b) = 0. The only way this equaton can be true is if a+b+c=0.

答:
Let $A$ be the matrix of the coefficients and $A|B$ the augmented matrix.

Overdeterminate system like this have solution only if $\text{rank}(A)=\text{rank}(A|B)$.

As $\text{rank}(A)=2$ must be $\text{rank}(A|B)=2$ which means that
$$\det(A|B)=\left|
\begin{array}{ccc}
 a & b & -c \\
 b & c & -a \\
 c & a & -b \\
\end{array}
\right|=0$$
which is $$a^3-3 a b c+b^3+c^3=0\to (a+b+c) \left(a^2-a b-a c+b^2-b c+c^2\right)=0$$
satisfied when $a+b+c=0$.
$$.......................................$$
Alternatively
$$
\begin{cases}
 ax + by + c = 0\\
bx + cy + a = 0\\
cx + ay + b = 0\\
\end{cases}
$$
Solve the first two equations
$$x=\frac{c^2-a b}{b^2-a c};\;y=\frac{a^2-b c}{b^2-a c}$$
plug in the third
$$\frac{a^3-3 a b c+b^3+c^3}{b^2-a c}=0\to a^3-3 a b c+b^3+c^3=0\\ (a+b+c) \left(a^2-a b-a c+b^2-b c+c^2\right)=0\to a+b+c=0$$


问:
Probability of $X$ out of $N$ dice landing on $M$
The problem is as follows:

We have $N$ dice and we throw them on a table. What is the probability that $M$ will fall $X$ times?

Specific example:
We have $10$ dice and we throw them on a table. What is the probability that $3$ will occur $7$ times (so there are seven dice out of $10$ which have $3$ on top)?

I am really bad at probability. I've checked a few sites about how I could solve this but it seems too hard figure it out.

答:
This is problem represents a binomial distribution.

We have $N$ trials of $M$, and each of possible $M$ has a $\frac{1}{S}$ chance of occurring, where $S$ is the number of sides on each die. $X$ will represent how many times you want to roll $M$.

So, we can represent this probability as follows:

$$\binom{N}{X}\left(\frac{1}{S}\right)^{X}\left(\frac{S - 1}{S}\right)^{N - X}$$

Expanded, this looks like this:

$$\frac{N!}{X!(N - X)!}\left(\frac{1}{S}\right)^{X}\left(\frac{S - 1}{S}\right)^{N - X}$$

Note that $x!$ is the factorial function, where $n! = \prod_{i = 1}^{n}i$

So, for your specific example, we have:

$$\frac{10!}{7!(10 - 7)!}\left(\frac{1}{6}\right)^{7}\left(\frac{6 - 1}{6}\right)^{10 - 7}$$
$$\frac{10!}{7!3!}\left(\frac{1}{6}\right)^{7}\left(\frac{5}{6}\right)^{3}$$

Which is about $0.0002480725753188030279937001473352639333434943860183$


问:
Finding the volume of $\{(x,y,z) \in \mathbb{R}^3 \vert (x+y+z)^2 + (x-y)^2 \leq 1, 0 \leq z \leq 3 \}$ by integration
I have been trying to solve this integral for some time now without success:

$$ T = \{(x,y,z) \in \mathbb{R}^3 \vert (x+y+z)^2 + (x-y)^2 \leq 1, 0 \leq z \leq 3 \}$$
$$ I = \int_{T} ~dx~dy~dz $$

I have tried simplifying the domain, by writing:
$$ (x+y+z)^2 + (x-y)^2 \leq 1 \to $$
$$ 2x^2 + 2y^2 + z^2 + 2xz + 2yz \leq 1 \to $$
$$ x^2 + y^2 + \frac{z^2}{2} + xz + yz \leq \frac{1}{2} $$
Now, I tried doing this change of variable: $ x = \rho \cos\theta$, $y = \rho \sin\theta$ and $z = \sqrt 2 w$
With $\rho \in (0,\infty)$, $\theta \in [0,2\pi]$. They are basically cylindrical coordinates.
I do some calculation and I have the following:
$$ \rho^2 + w^2 + (\rho \cos\theta + \rho \sin \theta) \cdot \sqrt{2}w \leq \frac{1}{2}$$
But I don't see how this is helping. I have also tried spherical coordinates substitution without success.

Can I have some hints?

答: Define the transformation $S$ by $$S(x,y,z)=(x+y+z,x-y,z)=(u,v,w)$$ Then the image of your set $T$ under $S$ is the truncated cylinder $$S(T)=\{u^2+v^2 \leq1\}\cap\{0\leq w\leq3\}\cap \mathbb{R}^3$$ It's easy to see how $$\frac{\partial(x,y,z)}{\partial(u,v,w)}=-\frac{1}{2}$$ Therefore $$\int_Tdxdydz=\int_{S(T)}\Bigg|\frac{\partial(x,y,z)}{\partial(u,v,w)}\Bigg|dudvdw=\frac{\text{Volume of } S(T)}{2}=\frac{3\pi}{2}$$