问:
Compact operator in $L^2$ theory of linear elliptic equation
If
$$
A: L^2(\Omega) \rightarrow H_0^1(\Omega)
$$
is a bounded linear operator, and
$$
E:H_0^1(\Omega) \rightarrow  L^2(\Omega)
$$
is compact imbedding,  then how to show
$$
A=AE: H_0^1(\Omega) \rightarrow  H_0^1(\Omega)
$$
is compact ?

This question is gotten when I read a book about PDE,  I fail to get why $A=AE$ is compact.  And, in fact, I don't know  what is the mean of $A=AE$.

答:
A compact operator is one which sends bounded sets to relatively compact sets. If $X\subset H_0^1(\Omega)$ is bounded and $E$ is a compact operator, then $E(X)$ is compact in $L^2(\Omega)$. Since $A$ is bounded, it sends relatively compact sets to relatively compact sets, meaning that $A(E(X))$ is compact in $H_0^1(\Omega)$. You can also do it through sequences, if you'd like.

In general, if $T:V\rightarrow W$ is compact, $S_1:V_1\rightarrow V$ is bounded, and $S_2:W\rightarrow W_2$ is bounded, then $S_2TS_1:V_1\rightarrow W_2$ is compact, via the same reasoning.


问:
Conditional probability question games
I am having a problem that I am unable to solve. Can you help me with that?

I have a kindergarten and 40% of the kids up to 4 years old play video games. However, for older kids, only 15% of age older than 4 play video games. Kids under 4 account for 10% of the total kids.

a. If I select randomly 15 kids that play video games, what is the chance that at least 11 of them are 4 years old and under?

b. If I select 15 kids randomly, what is the chance that 5 are under age 4 that play video games, 2 do not play video games but smaller than 4, 6 are older than 4 and play video games and the rest do not play games?

c. Compute the expected number of kids that play video games for 15 kids.

答:
Consider the following table that represent your bivariate data

Using this table you can construct the rv required to answer to any question...

Example:

a): Selecting 15 kids at random from those who play Video Games, it results by the table that

$$\frac{4}{17.5}\approx 22.86\%$$ are under 4 Years Old... thus you can use the Binomial to answer

et cetera et cetera

ContentLicense: CC BY-SA 4.0

Title: Conditional probability question games


问:
When is an Integer a Rational Number, and are All Ratios Rational, Even $\frac{\sqrt{7}}{2}$?

$$\Bbb{Q} = \left\{\frac ab \mid \text{$a$ and $b$ are integers and $b \ne 0$} \right\}$$

In other words, a rational number is a number that can be written as one integer over another.

For an integer, the denominator is $1$ in that case. For example, $5$ can be written as $\dfrac 51$.

Is $5$ a rational number? Or is $\dfrac 51$  a rational number? I'm not able to figure out what the definition is actually saying. What are the numbers that cannot be written as one integer over another?

Irrational numbers are the numbers that cannot be written as one integer over another. Roots of numbers that are not perfect squares are examples of irrational numbers.

However, what is this then: $\dfrac {\sqrt 7} {2}$?

答:
$\sqrt{7}$ is not a rational number.

You can prove it by the absurd.

Suppose that you can write it as $\sqrt{7}=\frac{a}{b}$ with $a$ and $b$ natural numbers.

You can suppose without loss of generality that $a$ and $b$ are coprime.
 Decompose $a$ and $b$ in prime factors: $a=a_1^{n_1}\dots a_k^{n_k}$ and $b=b_1^{m_1}\dots b_l^{m_l}$. All the $a_i$ are primes and different of all the $b_j$ (also primes) because $a$ $b$ are coprime.

$\sqrt{7}=\frac{a}{b}$ give you $7=\frac{a^2}{b^2}$ hence $a^2=7\cdot b^2$ hence $a_1^{2n_1}\dots a_k^{2n_k}=7b_1^{2m_1}\dots b_l^{2m_l}$ hence one of the $a_i$ is $7$ (lets say $a_1=7$). We have: $7^{2n_1-1}\dots a_k^{2n_k}=b_1^{2m_1}\dots b_l^{2m_l}$ hence $7$ divide $b_1^{2m_1}\dots b_l^{2m_l}$ hence one of the $b_j$ is $7$ contradiction with $a$ $b$ coprimes.

ContentLicense: CC BY-SA 3.0


问:
How should I evaluate the flux?
I have the following vector field

$F= \frac{11x}{\left(x^2+y^2+z^2\right)^{3/2}} \vec{i} + \frac{11y}{\left(x^2+y^2+z^2\right)^{3/2}} \vec{j} + \frac{11z}{\left(x^2+y^2+z^2\right)^{3/2}} \vec{k}$

I need to calculate the flux, $\iint_{S} F \cdot \vec{n}\ \text{dS}$, through the surface $S: x^2+y^2+z^2 = 16$ (sphere). As $F$ has a singularity at origin, I can't use the Divergent Theorem (right?). So I try to calculate using the surface integral of a vector field, $\iint_{S} F \cdot \vec{n}\ \text{dS} = \iint_{D} F(\mathbf{r}(\mathbf{u},\mathbf{v})) \cdot (\mathbf{r_u} \times \mathbf{r_v})\ \text{dA}$.

I've used the parametric representation $\mathbf{r}(\phi,\theta) = (4 \sin\phi\cos\theta, 4 \sin\phi\sin\theta, 4 \cos\phi)$. When I perform $F(\mathbf{r}(\mathbf{u},\mathbf{v})) \cdot (\mathbf{r_u} \times \mathbf{r_v})$, I'm getting 0. With this result, my flux would be 0, but according to my teacher answer, the flux is $44\pi$. What I'm doing wrong?

答:
In spherical coordinates,

$x = r \cos \theta \sin \phi, y = r \sin \theta \sin \phi, z = r\cos \phi$

$F = \frac{11x}{\left(x^2+y^2+z^2\right)^{3/2}} \vec{i} + \frac{11y}{\left(x^2+y^2+z^2\right)^{3/2}} \vec{j} + \frac{11z}{\left(x^2+y^2+z^2\right)^{3/2}} \vec{k}$

$F = (\frac{11\cos \theta \sin \phi}{r^2}, \frac{11\sin \theta \sin \phi}{r^2}, \frac{11 \cos \phi}{r^2})$

Outward normal vector $\hat{n} = \frac{1}{r}(x, y, z) = (\cos \theta \sin \phi \, , \sin \theta \sin \phi \, , \cos \phi \,)$

In spherical coordinates, $dS = r^2 \sin \phi \ d \theta \, d \phi$

$Flux = \displaystyle \iint_S \vec{F} \cdot \hat{n} dS$

$\vec{F} \cdot \hat{n} dS = (\frac{11\cos \theta \sin \phi}{r^2}, \frac{11\sin \theta \sin \phi}{r^2}, \frac{11 \cos \phi}{r^2}) \cdot (\cos \theta \sin \phi \, , \sin \theta \sin \phi \, , \cos \phi \,) dS$

$ = 11(\cos^2 \theta \sin^2 \phi + \sin^2 \theta \sin^2 \phi + \cos^2 \phi) \sin \phi d\theta d\phi = 11 \sin \phi d\theta d\phi$

So, $Flux = \displaystyle \iint_S \vec{F} \cdot \hat{n} dS = 11 \int_0^\pi \int_0^{2\pi}\sin \phi d\theta d\phi = \fbox {44 $\pi$}$

Title: How should I evaluate the flux?


问:
Computing Ramanujan asymptotic formula from Rademacher's formula for the partition function
I am trying to derive the Hardy-Ramanujan asymptotic formula

$$p(n) \sim \frac{1}{4n\sqrt{3}}e^{\pi\sqrt{\frac{2n}{3}}}$$

from Radmacher's formula for the partition function $p(n)$ given by

$$p(n)=\frac{1}{\pi\sqrt{2}}\sum_{k=1}^{\infty}A_{k}(n)\sqrt{k}\left[\frac{d}{dx}\frac{sinh\left(\frac{\pi}{k}\sqrt{\frac{2}{3}\left(x-\frac{1}{24}\right)}\right)}{\sqrt{x-\frac{1}{24}}}\right]_{x=n}$$
where
$$A_{k}(n)=\sum_{h=0, (h,k)=1}^{k-1}e^{\pi i(s(h,k)-2n\frac{h}{k})}$$ and
$$s(h,k)=\sum_{r=1}^{k-1}\frac{r}{k}\left(\frac{hr}{k}-\left\lfloor\frac{hr}{k}\right\rfloor-\frac{1}{2}\right)$$

G.E. Andrews, and any other literature on this topic, says that we can obtain the H-R asymptotic expression from the first term of the Rademacher series, i.e. for $k=1$. I don't know how to approach this as simply calculating for $k=1$ does not give the desired result. Could we perhaps try and use Lapalce's method, or the method of steepest descent?

答:
$$Q_k(n)=A_{k}(n)\sqrt{k}\left[\frac{d}{dx}\frac{sinh\left(\frac{\pi}{k}\sqrt{\frac{2}{3}\left(x-\frac{1}{24}\right)}\right)}{\sqrt{x-\frac{1}{24}}}\right]_{x=n}$$
$$p(n)=Q_1(n)+\sum_{k=2}^n Q_k(n) + \sum_{k=n+1}^\infty Q_k(n)$$
$$ = Q_1(n)+ O(n Q_2(n))  + \sum_{k=n+1}^\infty O(n^{1/2}k^{-3/2})$$
$$ = Q_1(n)+O(n e^{\frac\pi2\sqrt{\frac{2n}{3}}})$$

($O(n^{1/2}k^{-3/2})$ is not completely obvious, it is because the main term $\frac{\pi}{\sqrt6 k(n-1/24)}$ cancels in the expression for the derivative)


问:
proof: p|b and p|a if $p|a^2+b^2+ab$ and p = 6k+5
Let a,b be two natural numbers and p a prime number and

$p=6k+5 $ (k is an unfixed integer)

Prove that if
$$ p|a^2+ab+b^2$$
Then
$p|b$ and $p|a$

I found this problem in a discrete math book in the number theory section. I tried solving it by multiplying $a^2+ab+b^2$ by $a-b$ which results in
$$p|a^3-b^3$$

But i don't know how to proceed.

答:
As $p|(a^2+ab+b^2)$

if $p|a\iff p|b$

For odd $p,$

$$p|(a^2+ab+b^2)\iff p|(2a+b)^2+3b^2$$

If $(p,ab)=1,$   $$\left(\dfrac{2a+b}b\right)^2\equiv-3\pmod p$$

Now use For what odd prime p is -3 a quadratic residues? Non-residue?  OR

$-3$ is a quadratic residue iff $p \equiv 1 \pmod 3$  OR

Prove that $-3$ is a quadratic residue mod an odd prime $>3$ if and only if $p$ is of the form of $6n+1$


问:
Integrating a sine function that is to an odd power
I've started the chapter in my book where we begin to integrate trig functions, so bear in mind I've only got started and that I do not have a handle on more advanced techniques.

$\eqalign{
  & \int {{{\sin }^3}x} dx  \cr
  &  = \int {\sin x({{\sin }^2}x} )dx  \cr
  &  = \int {\sin x({1 \over 2}}  - {1 \over 2}\cos 2x)dx  \cr
  &  = \int {{1 \over 2}\sin x(1 - \cos 2x)dx}   \cr
  &  = \int {{1 \over 2}\sin x(1 - (2{{\cos }^2}x - 1))dx}   \cr
  &  = \int {{1 \over 2}\sin x(2 - 2{{\cos }^2}x)dx}   \cr
  &  = \int {\sin x - {{\cos }^2}} x\sin xdx  \cr
  & y = {1 \over 3}{\cos ^3}x - \cos x + C \cr} $

I got the right answer but it seems like an awfully long winded way of doing things, have I made things harder than they should be with this method?

答:
You have unnecessary middle steps, just use $\sin^2x = 1-\cos^2x$ after the second line.

To handle other odd powers:

$$\sin^{2k+1}x=(\sin^{2}x)^k(\sin x)=(1-\cos^2x)^k(\sin x)$$

then use $u=\cos x$.


问:
Interesting question about exponential generating function of the number of partitions under a certain statistic.
Find the exponential generating function of the number of partitions of $[n]$ ( according to) under the statistic of the number of blocks that contain an even number of elements (a block $K$ such that $|K|$ is even).

Then find the expectation (average) of blocks with even number of elements.

There are many questions with similar ideas, but are different.

So,  I said

For $n=1$ we have the partition: $\{\{1\}\}$

For $n=2$ we have $\{\{1,2\}\} , \{\{1\},\{2\}\}$

For $n=3$ we have these partitions:
$$\{ \{ 1,2,3 \} \}, \{\{1,2\}, \{ 3 \}\} ,  \{ \{ 1,3 \},\{ 2 \} \} , \{ \{2,3\},\{1\}\}, \{\{ 1 \}, \{ 2 \}, \{3 \} \}$$

If $x$ counts a length of a single number then:

$A_1(x)=x$ ,
$A_2(x)=2x^2/2!$ ,
$A_3(x)=5x^3/3!$

Now let $q$ counts the number of blocks with even number of elements.

Then,

$A_1(q,x)=x$

$A_2(q,x)= \frac{x^2}{2
!}(q+1)$

$A_3(q,x)=\frac{x^3}{3!} (3q+2)$

I was not able to find the recurrence relation for $A_n(q)$ ($A_0(q)=1$) in order to find the exponential generating function.

Can you provide me any kind of support.

答:
If I understand correctly, you want $$A_n(q)=\sum _{\pi \vdash [n]}q^{ev(\pi)},$$
where $ev(\pi):=\text{number of even blocks in }\pi.$

Well, suppose you know $A_n,$ then you want to create all blocks with $n+1$ elements. So you want to choose the elements that are going to be in the block of $n+1.$ This corresponds, without the polynomials to the recursion of Bell numbers $$B_{n+1}=\sum _{i=0}^n\binom{n}{i}B_{n-i}.$$
Now, taking care of the polynomial, notice that there are two cases. Either we place $n+1$ in a block with even elements or in an odd one. If we do place it with odd elements we are forming one even block per partition on $B_{n-i}.$ So you have that
$$A_{n+1}(q)=q\sum _{i\text{ odd}}\binom{n}{i}A_{n-i}(q)+\sum _{i\text{ even }}\binom{n}{i}A_{n-i}(q).$$
Sage obtains the following polynomials for $n$ up to 10:

```<code>x |--> 1
x |--> x + 1
x |--> 3*x + 2
x |--> 3*x^2 + 7*x + 5
x |--> 15*x^2 + 25*x + 12
x |--> 15*x^3 + 60*x^2 + 91*x + 37
x |--> 105*x^3 + 315*x^2 + 329*x + 128
x |--> 105*x^4 + 630*x^3 + 1533*x^2 + 1415*x + 457
x |--> 945*x^4 + 4410*x^3 + 7623*x^2 + 6297*x + 1872
</code>```

Body:
If I understand correctly, you want $$A_n(q)=\sum _{\pi \vdash [n]}q^{ev(\pi)},$$
where $ev(\pi):=\text{number of even blocks in }\pi.$

Well, suppose you know $A_n,$ then you want to create all blocks with $n+1$ elements. So you want to choose the elements that are going to be in the block of $n+1.$ This corresponds, without the polynomials to the recursion of Bell numbers $$B_{n+1}=\sum _{i=0}^n\binom{n}{i}B_{n-i}.$$
Now, taking care of the polynomial, notice that there are two cases. Either we place $n+1$ in a block with even elements or in an odd one. If we do place it with odd elements we are forming one even block per partition on $B_{n-i}.$ So you have that
$$A_{n+1}(q)=q\sum _{i\text{ odd}}\binom{n}{i}A_{n-i}(q)+\sum _{i\text{ even }}\binom{n}{i}A_{n-i}(q).$$
Sage obtains the following polynomials for $n$ up to 10:

x |--> 1
x |--> x + 1
x |--> 3*x + 2
x |--> 3*x^2 + 7*x + 5
x |--> 15*x^2 + 25*x + 12
x |--> 15*x^3 + 60*x^2 + 91*x + 37
x |--> 105*x^3 + 315*x^2 + 329*x + 128
x |--> 105*x^4 + 630*x^3 + 1533*x^2 + 1415*x + 457
x |--> 945*x^4 + 4410*x^3 + 7623*x^2 + 6297*x + 1872


答:
The following is assuming OP's original p|a-b. Otherwise, disregard.

$p|a-b \implies p|(a-b)^{2} = a^{2}-2ab+b^{2}$.

Since $p|a^2+ab+b^2, p|(a^2+ab+b^2)-(a^{2}-2ab+b^{2})=3ab \implies p |ab$ as $p = 6k+5 \implies p \neq 3 \implies p \nmid 3$.

As $p$ is prime $p|ab \implies p|a$ or $p|b$. Using $p|a^2+ab+b^2$, you will find that WLOG if $p|a, p|a(a+b)=a^{2}+ab$ so that $p|b^{2} \implies p|b$. Hence $p|a,b$.

Title: proof: p|b and p|a if $p|a^2+b^2+ab$ and p = 6k+5

Body:
Let a,b be two natural numbers and p a prime number and

$p=6k+5 $ (k is an unfixed integer)

Prove that if
$$ p|a^2+ab+b^2$$
Then
$p|b$ and $p|a$

I found this problem in a discrete math book in the number theory section. I tried solving it by multiplying $a^2+ab+b^2$ by $a-b$ which results in
$$p|a^3-b^3$$

But i don't know how to proceed.


问:
How to find the directrix in order to get intersection between two parabolas and a segment?
So I am trying to implement the Fortune's algorithm in order to have a Voronoi pattern (wonderful post about it here : http://www.ams.org/publicoutreach/feature-column/fcarc-voronoi)

For this, I need to know intersections between two parabolas, which I know how to do (thanks to this article : http://www.zonalandeducation.com/mmts/intersections/intersectionOfTwoParabollas1/intersectionOfTwoParabolas1.htm) but also something that the first article doesn't cover : how to find the value Y of the directrix when I want two parabolas to intersect in a specific X coordinate (while the two parabolas share the same directrix - 'the sweep line').

Let me explain better :
See the image descripting the thing - As I am not allowed to have embed image yet :'(
The thing that the 1st article doesn't cover is that the edges between some sites infinitely continue to the left or to the right. What I want is to limit it into a plane (a square).
So, I have :

A plane with its edges (I know the X and Y coordinates of its edges)

Two focus points, f1 and f2 (for Fortune, those focus points are the sites), of the two parabolas, p1 and p2.

A directrix (for Fortune, 'the sweep line') which is the same for all the focus points, and which I can move up and down as I want.

And I don't have the intersection between p1, p2 and the edges of the plane.

Actually, what I would like to have is the intersections circled in red here : https://imgur.com/a/SY340Dk (I already have those in purple, thanks to the second article), but in order to get them, I need to find the Y-value of the directrix where two parabolas will intersect with the edge of the plane.

So, as shown in the image, how to find the Y value of the directrix, in order to get the intersection between p1, p2 and the left edge ?

Thank you for your time,

Cyril G

答: This circled point $P$ is such that $d:=PF_1=PF_2=PH$ by definition of a parabola (same distance from a focus than to the directrix), you just have to take the directrix at vertical distance $d$ from $P$.


问:
Is there a 'Noetherialization' Process for Rings (Can I take a ring and get a "similar" Noetherian one)
I had asked this question in one of the seminars I attend and they said I should ask it here. Let R be a commutative ring with unity
where $1 \neq 0$
and suppose that R is not Noetherian (i.e. all of the 3 equivalent conditions: max.c, acc, and finitely generated ideals - all fail concurrently).
Can we take this R through some sort of algorithim to obtain a similar ring R(N) which is Noetherian? I get that the usage of "similar" here is really vague but I am not sure how else to phrase it. Also I would like for this new ring R(N) to be commutative with unity, and having $1 \neq 0$ ideally (or at least left and right Noetherian if it is not commutative let's say). If there is really nothing of this sort, then my question becomes, what is the next best thing? Again that is vague but not sure how else to phrase it. If there is such a thing, then can we generalize it to modules over a commutative unital ring with $1 \neq 0$?

(I tried googling this, no luck there).

答:
Let $\newcommand{\CRing}{\mathrm{CRing}}\CRing$ be the category of commutative, unital rings and let $\CRing_N$ be the full subcategory of Noetherian rings. Usually something like a "Noetherialization" would mean an adjoint to the inclusion of categories, so we can ask whether it is a reflective subcategory (which means that the inclusion $\CRing_N\hookrightarrow\CRing$ has a left adjoint). This however fails since this functor does not create limits. For instance, a countable product of the ring $\mathbb Z$ is not Noetherian, so it cannot be reflected to the subcategory of Noetherian rings.

Considering colimits should similarly show that this category is not coreflective either, so inclusion doesn't admit a right adjoint either.


问:
Input in Differential Equations and Difference Equations
According to my understanding:

In differential equations, the input is defined as the entire inhomogeneous part of the equation as it would be written on the right hand side.  For example, the input of $y'' + 2y' - t + te^t + y = 0;\ y(0) = 1;\ y'(0) = 1$ is $t - te^t$.  When solving for the unit impulse response, this input is to be replaced by $\delta$, yielding $y_{_\delta}'' + 2y_{_\delta}' + y_{_\delta} = \delta;\ y_{_\delta}(0) = 0;\ y_{_\delta}'(0) = 0$.

In difference equations, the input is defined as (in general) only part of this inhomogeneous part.  For example, given $y[t + 2] - 7y[t + 1] + ty[t] = \cosh(t + 2) - \cosh(t + 1);\ y[0] = 5;\ y[1] = 10$, it cannot be determined whether:

The input is $f[t] = \cosh(t + 2) - \cosh(t + 1)$ and the difference equation has the form $y[t + 2] - 7y[t + 1] + ty[t] = f[t];\ y[0] = 5;\ y[1] = 10$,

The input is $f[t] = \cosh(t)$ and the difference equation has the form $y[t + 2] - 7y[t + 1] + ty[t] = f[t + 2] - f[t + 1];\ y[0] = 5;\ y[1] = 10$,

The input is $f[t] = \cosh(t + 1)$ and the difference equation has the form $y[t + 2] - 7y[t + 1] + ty[t] = f[t + 1] - f[t];\ y[0] = 5;\ y[1] = 10$,

or any other number of possibilities is correct.  The distinction between these possibilities seems to be non-trivial, since when solving for the unit impulse response, only the input is to be replaced by $\delta$, not the entire inhomogeneous part of the difference equation.  This leads to the seemingly false conclusion that the difference equation and initial conditions jointly fail to uniquely determine a unit impulse response, that additional information defining the input must be provided.

Since differential equations seem to be limiting cases of difference equations with arbitrarily short time steps, why is there this substantial difference between the two in how the inputs are managed, especially when solving for the unit impulse response?

答:
I am a Mathematica user and know the history of the package rather well. They introduced the difference equations in version 7.0 and the differential equations in the first. So there is a big difference in how interested people are in either.

In DifferenceRoot there is last section that is of high interest to the intent of Your question:

```<code>GeneratingFunction[
 DifferenceRoot[
   Function[{y, n}, {-y[n] - y[n + 1] + y[n + 2] == 0, y[0] == 0,
     y[1] == 1}]][n], n, z]
</code>```

$$-\frac{z}{-1 + z + z^2}$$

The built-in is DifferenceRoot. The difference equation in this example is the second list to the function. This is in terms of Mathematica Your similarity between difference equation and differential equation. There is not much.

Differential equation are special cases of limit of a difference quotient over time solely. Difference equation are part of the theory of generating functions GeneratingFunction. There are differential equation for generating functions, GeneratingFunction, and difference equations. Both define the generating function complete.

Most generating functions are rational polynoms or transcendential functions.

There are too exponential generating functions:

Both are mathematics but very different in nature. Do not mix them up.

Mathematica knows some more of this functions categories infinite sums and recurrence equations.

The solution to difference equation are holonomic sequences. The solution to differential equation do not necessarily need to suit more than differentiability of suitable order.

But this is only the Mathematica version. Difference equation appears in very different contexts.

Examples:

Difference_equation as a scheme to set up solvers for differential equations.

The above detailed Mathematica interpretation deals with homogeneous differential equations.

Defnition for scheme for interpolation of functions or data.

The equations in school math with at least one subtraction.

This introduction the relation as in Mathematica with recurrence, What-is-a-difference-equation.
....

sol = RSolve[{y[n + 2] - 7 y[n + 1] + n y[n] == 0, y[0] == 5,
y1 == 10}, y, n]

give a solution in Mathematica. With f there is non.

This is very rapidly increasing.

This shows the homogeneous function diverges much faster than the inhomogenity does and is pausible that this might be an error in question.

I made use of What is a differnce equation? on google.com

Body:
I am a Mathematica user and know the history of the package rather well. They introduced the difference equations in version 7.0 and the differential equations in the first. So there is a big difference in how interested people are in either.

In DifferenceRoot there is last section that is of high interest to the intent of Your question:

GeneratingFunction[
 DifferenceRoot[
   Function[{y, n}, {-y[n] - y[n + 1] + y[n + 2] == 0, y[0] == 0,
     y[1] == 1}]][n], n, z]

$$-\frac{z}{-1 + z + z^2}$$

The built-in is DifferenceRoot. The difference equation in this example is the second list to the function. This is in terms of Mathematica Your similarity between difference equation and differential equation. There is not much.

Differential equation are special cases of limit of a difference quotient over time solely. Difference equation are part of the theory of generating functions GeneratingFunction. There are differential equation for generating functions, GeneratingFunction, and difference equations. Both define the generating function complete.

Most generating functions are rational polynoms or transcendential functions.

There are too exponential generating functions:

Both are mathematics but very different in nature. Do not mix them up.

Mathematica knows some more of this functions categories infinite sums and recurrence equations.

The solution to difference equation are holonomic sequences. The solution to differential equation do not necessarily need to suit more than differentiability of suitable order.

But this is only the Mathematica version. Difference equation appears in very different contexts.

Examples:

Difference_equation as a scheme to set up solvers for differential equations.

The above detailed Mathematica interpretation deals with homogeneous differential equations.

Defnition for scheme for interpolation of functions or data.

The equations in school math with at least one subtraction.

This introduction the relation as in Mathematica with recurrence, What-is-a-difference-equation.
....

sol = RSolve[{y[n + 2] - 7 y[n + 1] + n y[n] == 0, y[0] == 5,
y1 == 10}, y, n]

give a solution in Mathematica. With f there is non.

This is very rapidly increasing.

This shows the homogeneous function diverges much faster than the inhomogenity does and is pausible that this might be an error in question.

I made use of What is a differnce equation? on google.com


问:
Sum of kth roots ($\sum\sqrt[k]{m}$)
I'm trying to find an asymptotic to $$S(n) = \sum_{k=1}^n\sqrt[k]{m}$$ From computational tests, it seems to grow nearly as slowly as $n$. However even $$\sum_{k=1}^\infty\sqrt[k]{m}-1$$ diverges (for $m\neq1$) by the comparison test.

I'm thinking it might be something like $n\log{\log n}$, but I don't know how to show it.

Update: So it turns out to be closer to $n\sqrt[n]{m}$. Does anybody know if there is a nice formula?

答:
We can get some intuition for this by taking the integral
$$\int_1^n m^{1/x} = n + \frac{m}{\log m} + (\log m) (\log n-\log\log m-\gamma+1) + O\left(\frac{m}{\log^2 m} + \frac{\log^2 m}{n}+\frac{1}{m^2}\right)$$.

However, we know that the actual result has to grow at least as fast as $m$.

The best, reasonably simple, formula I've found so far is: $$m + n + \log n \log m - 1$$.

If $m$ and $n$ grows at the same speed, it doesn't quite grow fast enough (loses about 0.2% at $m=n=10^5$), but if one of $m$ or $n$ grows fastest, it is nearly perfect.

Plotted with $m=2$, $0<n\leq10$

Title: Sum of kth roots ($\sum\sqrt[k]{m}$)


问:
Absolute and conditional convergence of series containing trigonometric functions
I have a slight problem with this kind of exercise. For example, if we have:

$$\sum_{n=1}^\infty \frac{\cos{\frac{n\pi}{4}}}{n^a}, a>0$$

Now, for example, say I want to check the absolute convergence of this example. I know that there are negative terms depending on $n$, and the maximum value $\cos{\frac{n\pi}{4}}$ can attain is $1$.

$$\vert\frac{\cos{\frac{n\pi}{4}}}{n^a}\rvert\le \frac{1}{n^a}$$

Was I correct to make this deduction?

Do I proceed like this any time I have a $sin$ or $cos$ function when checking for absolute convergence?

If I am correct, my series absolutely converge for $a>1$. How do we check if it conditionally converges for $0<a<1$?

We are not allowed to use any integration techniques (Abel's and Dirichlet's test are encouraged).

Thanks!

答:
Yes.

It's always a good thing to try.

You are correct about when it converges absolutely, but keep in mind that so far you haven't shown it doesn't converge absolutely for any $a\le 1,$ only that it does converge absolutely for $a>1.$ (Hint: you can lower bound it too, though some terms are zero so it's a little trickier.) The fact that it converges for any $a>0$  follows from one of the two tests you mentioned.


问:
Trigonometric integral using residue complex analysis.
There is this problem where I need to proof the following statement: $$\int_{0}^{\pi}\cos^{2n}\theta d\theta = \pi\frac{(2n)!}{2^{2n}(n!)^2}$$ First I used the fact that the function is ever so I can divide by $2$ and integrate over $(0,2\pi)$. Then I made the substituition: $$z=e^{i\theta}\,dz=iz\,d\theta\quad\text{then}\quad\cos\theta=\frac{z^2+1}{2z},$$ letting $C$ be the unitary circle:$$\frac{1}{i2^{2n+1}}\oint_{C}\frac{(z^2+1)^{2n}}{z^{2n+1}}dz.$$ Now since there is a pole at $z=0$, I chose to just use the Cauchy-formula for $(2n)$ rank pole, taking the numerator $(z^2+1)^{2n}$ for $f(z)$: $$\oint_{C}\frac{(z^2+1)^{2n}}{z^{2n+1}}dz=\frac{2\pi i}{(2n)!}\frac{d^{2n}}{dz^{2n}}f(0).$$ The $2n$ derivatives leaves me $(2n)!2^{2n}$ and then inserting this to the total: $$\int_{0}^{\pi}\cos^{2n}\theta\,d\theta = \pi.$$ I tried the residue theorem, but it is just what i have been doing above. What is wrong here? The problem or my calculations?

答:
Your problem is with formula:

$$\oint_{C}\underbrace{\frac{(z^2+1)^{2n}}{z^{2n+1}}}_{f(z)}dz=\frac{2\pi i }{(2n)!}\left(\frac{d^{2n}}{dz^{2n}}f\right)_{|z=0}.$$

which should be:

$$\oint_{C}\frac{(z^2+1)^{2n}}{z^{2n+1}}dz=\frac{2\pi i }{(2n)!} \left(\frac{d^{2n}}{dz^{2n}}z^{2n+1}f\right)_{|z=0}.$$

due to formula of residue in a pole $c$ with order $n$:

$$\displaystyle \operatorname {Res} (f,c)={\frac {1}{(n-1)!}}\lim _{z\to c}{\frac {d^{n-1}}{dz^{n-1}}}\left((z-c)^{n}f(z)\right).$$


问:
Gradient of holomorphic function equals conjugate of complex derivative
I'm using TensorFlow for some computations with complex variables (and derivatives of these computations). When I compute the derivative of (simple) holomorphic functions, the results obtained with TensorFlow are the conjugate of what I would expect. A simple example:

Given $z = x + yi$, and $f(z) = zz = x^2 + i2xy - y^2$.

We have
$\frac{df}{dz} = \frac12 \left(\frac{\partial f}{\partial x} - i\frac{\partial f}{\partial y}\right) = 2x + i2y$.

Hence, for $z = \frac{1}{5}i$, $\frac{df}{dz} = \frac{2}{5}i$. However, using TensorFlow, I obtain  $\frac{df}{dz} = -\frac{2}{5}i$. While searching for this, I found the following statement: "The gradient of a holomorphic function is the conjugate of its complex derivative." in a Github Issue, however, I don't understand why. In other words: Why is the gradient of a holomorphic functions equal to the conjugate of the complex derivative, and, where is the mistake in this simple example?

答:
It is late but I am writing for anyone encountering the same problem as yours. The explanation focuses on why tf.gradients in Tensorflow yeilds (-2j/5) but not (2j/5). First, tf.gradients only handles real-valued functions, thus it converts your original function $f(z)=x^2+2jxy−y^2$ into a new function, say, $$g(z) = Re\{f(z)\} = x^2 - y^2 .$$

Then, when you use Tensorflow to evaluate the gradient of $f(z)$ at $z=1j/5$, it turns out that Tensorflow will evaluate the gradient of $g(z)$ but not $f(z)$. So, what tf.gradients returns is $$\nabla_z g(z) = 2 \frac{d g(z)}{dz^*} = 2 \times \underbrace{ \frac{1}{2} \left( \frac{\partial g(z)}{\partial x} + j\frac{\partial g(z)}{\partial y} \right) }_{\textrm{eq.(230) in the book [B1]}} = \frac{\partial g(z)}{\partial x} + j\frac{\partial g(z)}{\partial y} = \underbrace{ \left[ \begin{matrix} \partial g(z)/\partial x \\ \\ \partial g(z)/\partial y \end{matrix} \right] }_{\textrm{in Cartersian
coordinates}} .$$

In short, you will end up with $\nabla_z g(z) = 2x + j(-2y)$. As a result, when evaluating  $\nabla_z g(z)$ at $z=1j/5$, you will have $\nabla_z g(z)\bigl|_{z=1j/5} \bigr. = 2\times 0 + j(-2\times (1/5)) = -2j/5$. The bottom line is that tf.gradients deal with real-valued functions of complex variables, but not complex-valued functions of complex variables.

[B1] The Matrix Cookbook [ http://matrixcookbook.com ]


问:
Find $c$ such that $ 0.05 = P \left( \sum_{i=1}^{10}X_i \geq c \right) $
I have to solve, for $c$, the equation

$$
0.05 = P \left( \sum_{i=1}^{10}X_i \geq c \right)
$$

where $X_i~~ i.i.d  \operatorname{Poisson}(1) \Rightarrow \sum_{i=1}^{10}X_i \sim \operatorname{Poisson}(10) $.

If the random variables were continuous I could calculate $c$ from the complement $0.95 = P(\sum_{i=1}^{10}X_i \leq c)$ but that is not the case.

答:
If the random variables were continuous I could calculate cc from the complement

of course you can also in a discrete case! in any case the probability over all the support is 1.

Easy find $C=16$.

To speed up the calculation (I did it with a poket calculator)

$P(X=14)>0.05$ (14 is not good)

$P(X=15)=0.035$ (mmmh...let's see the next)

$P(X=16)=0.022$ (15 is not good)

The other values over $X=17$ have all a probability close to zero....Thus $C=16$

Verify with a Calculator but I'm sure it is correct!


问:
Determine all sets of non-negative integers x,y and z which satisfy the equation $2^x + 3^y = z^2$

Determine all sets of non-negative integers x,y and z which satisfy the equation $2^x + 3^y = z^2$

This came in the 1992 INMO and curiously enough also seems to have been included on the 1996 BMO Round 2? I have never heard of a question being directly copied from another Olympiad so this was a first for me.

Anyways, first, I looked at the case $y=0$. This quickly gave me one solution, viz $(x,y,z)=(3,0,3)$

Next, I considered $x,y,z>0$

We know $2^x + 3^y \equiv (-1)^x+0 \bmod 3$ and that perfect squares are $\equiv 0,1 \bmod 3$. It is easy to see that the only combination that works is $x$ be even and $z=3m+1$ type $\Rightarrow z$ is odd

Also, we know that odd perfect squares are $\equiv 1 \bmod 4$. Further, $3^y\equiv (-1)^y \bmod 4$ and since $x$ is even it implies that $x≥2$ thus $2^x$ is divisible by $4$. This further implies that $(-1)^y \equiv 1 \bmod 4 \Rightarrow y$ is also even.

Let $x=2k$. Then our original expression becomes $$3^y=(z+2^k)(z-2^k)$$
We have two possibilities: first is that $(z-2^k)=1$ and $(z+2^k)=3^y$ and second is $(z-2^k)=3^{y-a}$ and $(z+2^k)=3^a$. But since we previously established that $z=3k±1$ and as $2^k \equiv (-1)^y \bmod 3$, we can quickly discard the second possibility.

So we finally have, $$(z-2^k)=1$$ $$(z+2^k)=3^y$$

Here I got woefully stuck. Another thing I got was that $k$ is also even (which means $x$ is itself a multiple of $4$). One more thing is that since $y$ is even $3^y$ is divisible by $9$. I don't know how we can use this fact right now but I thought it could be worth mentioning.

Any help to proceed would be appreciated, thanks.

答:
First, there are a few minor problems with your proof:

Next, I considered $x,y,z>0$

Have you found all solutions with $xyz=0$? (No!)

We know $2^x + 3^y \equiv (-1)^x+0 \bmod 3$ and that perfect squares are $\equiv 0,1 \bmod 3$. It is easy to see that the only combination that works is $x$ be even and $z=3m+1$ type $\Rightarrow z$ is odd.

It is true that $x$ must be even, but not that $z\equiv1\pmod{3}$. It is also possible that $z\equiv2\pmod{3}$. Fortunately you later state that $z=3k\pm1$, so perhaps this is just a typo. But the conclusion that $z$ is even seems out of place; instead this follows from the simple fact that $x>0$, as then
$$z^2\equiv 2^x+3^y\equiv1\pmod{2}.$$

The rest of the proof is fine. The linked duplicates provide alternative (and quicker) solutions to your original problem, but here's a quick and easy finish to your approach:

You already note that $y$ is even, so
$$2^{k+1}=(z+2^k)-(z-2^k)=3^y-1=(3^{y/2}+1)(3^{y/2}-1).$$
Then both factors on the right hand side are powers of $2$, and they differ by $2$, so $y=2$.

As noted in the comments, this is a special case of Mihăilescu's theorem, previously known as Catalan's conjecture. It was still a conjecture at the time these questions were posed in the IMO contests, so it's safe to say you were not expected to know or use Mihăilescu's theorem. Participants with an interest in number theory might be aware of the conjecture (it's quite famous), so at least they would 'know'  that this should be the only solution.


答:
Using the leading term of Radmacher's formula, we have
\begin{align*}
p(n) & \sim \frac{1}{{\pi \sqrt 2 }}\left[ {\frac{\mathrm{d}}{{\mathrm{d}x}}\frac{{\sinh \left( {\pi \sqrt {\frac{2}{3}\left( {x - \frac{1}{{24}}} \right)} } \right)}}{{\sqrt {x - \frac{1}{{24}}} }}} \right]_{x = n}
\\ &
 = \frac{{4\sqrt 3 }}{{24n - 1}}\cosh \left( {\pi \sqrt {\frac{2}{3}\left( {n - \frac{1}{{24}}} \right)} } \right) - \frac{1}{\pi}\frac{{24\sqrt 3 }}{{(24n - 1)^{3/2} }}\sinh \left( {\pi \sqrt {\frac{2}{3}\left( {n - \frac{1}{{24}}} \right)} } \right).
\end{align*}
Now
\begin{align*}
\pi \sqrt {\frac{2}{3}\left( {n - \frac{1}{{24}}} \right)}  = \pi \sqrt {\frac{2}{3}n} \sqrt {1 - \frac{1}{{24n}}} & = \pi \sqrt {\frac{2}{3}n} \left( {1 + \mathcal{O}\!\left( {\frac{1}{n}} \right)} \right) \\ &= \pi \sqrt {\frac{2}{3}n}  + \mathcal{O}\!\left( {\frac{1}{{\sqrt n }}} \right).
\end{align*}
Thus,
\begin{align*}
& \cosh \left( {\pi \sqrt {\frac{2}{3}\left( {n - \frac{1}{{24}}} \right)} } \right),\sinh \left( {\pi \sqrt {\frac{2}{3}\left( {n - \frac{1}{{24}}} \right)} } \right) \sim \frac{1}{2}\exp \left( {\pi \sqrt {\frac{2}{3}\left( {n - \frac{1}{{24}}} \right)} } \right) \\ & = \frac{1}{2}\exp \left( {\pi \sqrt {\frac{2}{3}n}  + \mathcal{O}\!\left( {\frac{1}{{\sqrt n }}} \right)} \right) = \frac{1}{2}\exp \left( {\pi \sqrt {\frac{2}{3}n} } \right)\left( {1 +\mathcal{O}\!\left( {\frac{1}{{\sqrt n }}} \right)} \right) \\ & \sim \frac{1}{2}\exp \left( {\pi \sqrt {\frac{2}{3}n} } \right).
\end{align*}
And therefore,
\begin{align*}
p(n) & \sim \frac{{4\sqrt 3 }}{{24n - 1}}\frac{1}{2}\exp \left( {\pi \sqrt {\frac{2}{3}n} } \right) - \frac{1}{\pi }\frac{{24\sqrt 3 }}{{(24n - 1)^{3/2} }}\frac{1}{2}\exp \left( {\pi \sqrt {\frac{2}{3}n} } \right)
\\ &
 \sim \frac{{4\sqrt 3 }}{{24n}}\frac{1}{2}\exp \left( {\pi \sqrt {\frac{2}{3}n} } \right) - \frac{1}{\pi }\frac{{24\sqrt 3 }}{{(24n)^{3/2} }}\frac{1}{2}\exp \left( {\pi \sqrt {\frac{2}{3}n} } \right)
\\ &
 = \frac{1}{{4n\sqrt 3 }}\exp \left( {\pi \sqrt {\frac{2}{3}n} } \right)\left( {1 - \frac{{\sqrt 3 }}{{\sqrt 2 \pi n^{1/2} }}} \right) \sim \frac{1}{{4n\sqrt 3 }}\exp \left( {\pi \sqrt {\frac{2}{3}n} } \right).
\end{align*}

Title: Computing Ramanujan asymptotic formula from Rademacher's formula for the partition function

Body:
I am trying to derive the Hardy-Ramanujan asymptotic formula

$$p(n) \sim \frac{1}{4n\sqrt{3}}e^{\pi\sqrt{\frac{2n}{3}}}$$

from Radmacher's formula for the partition function $p(n)$ given by

$$p(n)=\frac{1}{\pi\sqrt{2}}\sum_{k=1}^{\infty}A_{k}(n)\sqrt{k}\left[\frac{d}{dx}\frac{sinh\left(\frac{\pi}{k}\sqrt{\frac{2}{3}\left(x-\frac{1}{24}\right)}\right)}{\sqrt{x-\frac{1}{24}}}\right]_{x=n}$$
where
$$A_{k}(n)=\sum_{h=0, (h,k)=1}^{k-1}e^{\pi i(s(h,k)-2n\frac{h}{k})}$$ and
$$s(h,k)=\sum_{r=1}^{k-1}\frac{r}{k}\left(\frac{hr}{k}-\left\lfloor\frac{hr}{k}\right\rfloor-\frac{1}{2}\right)$$

G.E. Andrews, and any other literature on this topic, says that we can obtain the H-R asymptotic expression from the first term of the Rademacher series, i.e. for $k=1$. I don't know how to approach this as simply calculating for $k=1$ does not give the desired result. Could we perhaps try and use Lapalce's method, or the method of steepest descent?


问:
An elementary embedding which fixes the ordinals, but is not the identity.
In Woodin's "In the Search of Ultimate-$L$", in section $3.5$ on extenders, he uses an example which I have a hard time understanding. He essentially constructs two transitive models of $\mathsf{ZFC}^-$ with an elementary embedding between them, which is cofinal and is the identity on the ordinals but is not the identity!

I would appreciate any clarification on this. First let me quote the relevant bit:

Let $L[G]$ be a generic extension of $L$ for adding $\omega_2^L$ many Cohen reals and let $L[G][g]$ be a generic extension of $L[G]$ for adding $\omega_2^L$-many Cohen reals.

Now define $$M=L_{\omega_2^L}(\mathbb{R}^{L[G]})[G]$$ and define $$N=L_{\omega_2^L}(\mathbb{R}^{L[G][g]})[G][g],$$ where each is viewed as a transitive set. Thus in each case we are constructing
over the reals from an additional predicate. Note that $P(\omega)$ exists in both
$M$ and $N$ but for example, $P(\omega_1)$ does not exist in either $M$ or $N$.

It follows by the homogeneity of Cohen forcing that both $M$ and $N$ are
models of ZFC\Powerset with the usual formulation of the Axiom of Choice
and that the natural map $$\pi:M\rightarrow N,$$ where $\pi(\mathbb{R}^M)=\mathbb{R}^N$ is an elementary embedding.

Now I think I understand the construction and I think that the natural map is given by exactly setting $\pi(\mathbb{R}^M)=\mathbb{R}^N$ and $\pi|\mathbb{R}^M=\operatorname{id}_{\mathbb{R}^M}$ and for other sets in $M$, since we build it level by level with definable subsets and we know where to send each parameter, we know where to send each $x\in M$ level by level.

Now what I have trouble with, would be to check that elementarity and that these are models of $\mathsf{ZFC}^-$. As hinted, I tried to use homogeneity of the Cohen forcing to show elementarity, but for formulas that have parameters in $M$, I don't see how we can get rid of the parameters. And also between the axioms of $\mathsf{ZFC}^-$, choice is the one which is the most elusive for me, as although I don't have a formal argument for the others, but they seem plausible and usually in these circumstances we don't have choice. So my questions are: How can we show that $1) \mbox{ }\pi$ is elementary and $2) \mbox{ } M$ and $N$ are models of $\mathsf{ZFC}^-$, especially choice?

EDIT: I have doubts on why $\pi$ is even well-defined, as I have defined it above. Going level by level we may encounter several formulas with other parameters which define the same $x\in M$. And checking this is crucial in another sense, since later on when we prove elementarity, these different formulas should indeed correspond in a well-defined manner.

EDIT II: I was informed about this post from MO, which actually solves half of my question! By the argument in that proof $\pi$ is readily seen to be well-defined and elementary. But I am still not sure how we can show that $M$ and $N$ satisfy $\mathsf{ZFC}^-$, especially choice.

答:
Edit: I have added an argument that shows choice in $M$ in terms of existence of choice functions.

Replacement in M:
Here is an argument that shows $ZF^-$ in $M$ (the same works for $N$ of course):

Let $\mathbb R^G$ denote $\mathbb R^{L[G]}$. In a first step, we will see that in $M$ any definable map $f:\mathbb R^G\rightarrow \omega_2$ is bounded. In a second step I will explain how replacement follows. In $M$, all sets are definable from ordinal parameters, reals and proper initial segments of $G$. Since any real and any proper inital segment of $G$ is in an intermediate extension $L[G\upharpoonright \alpha]$ and since $L[G]$ is a further $Add(\omega, \omega_2)$-extension from there, we might as well assume that all parameters in the definition of $f$ are from the ground model and will suppress them in what follows. If $\dot x$ is a $Add(\omega, \omega)$-name, then by ccc, there is $\alpha_{\dot x}<\omega_2$ so that $1\Vdash f^{L_{\check\omega_2}(\mathbb R^\dot G)[G]}(\dot x)<\check\alpha_{\dot x}$. If $\dot y$ is any $Add(\omega, \omega_2)$-nice name for a real then there is an automorphism $\pi$ of $Add(\omega, \omega_2)$ that turns $\dot y$ into an $Add(\omega, \omega)$-name $\hat\pi(\dot y)$. But then
$$1\Vdash f^{L_{\check\omega_2}(\mathbb R^\dot G)[\dot G]}(\hat\pi(\dot y))<\check\alpha_{\hat\pi(\dot y)}$$
implies
$$ 1\Vdash f^{L_{\check\omega_2}(\mathbb R^{\widehat{\pi^{-1}}(\dot G)})[\widehat{\pi^{-1}}(\dot G)]}(\dot y)<\check\alpha_{\hat\pi(\dot y)}$$
Note that $f$ does not change as all the parameters in its definiton are in the ground model.
Furthermore, it is easy to see that
$$1\Vdash L_{\omega_2}(\mathbb R^{\widehat{\pi^{-1}}(\dot G)})[\widehat{\pi^{-1}}(\dot G)] = L_{\omega_2}(\mathbb R^{\dot G})[\dot G]$$
using that $\pi$ is definable in $M$.
Thus $f^M(y)<\alpha_{\hat\pi(\dot y)}$ so that $f^M$ is bounded by
$$\operatorname{sup}\{\alpha_{\dot x}\mid \dot x \text{ is a }Add(\omega, \omega)\text{-name}\}<\omega_2$$
We can use this to show that reflection is true in $M$.

Claim: For any $\in$-formula $\varphi$ there is a $M$-definable club of $\alpha<\omega_2$ with $\varphi$ absulute between $L_\alpha(\mathbb R^G)[G]$ and $M$.

Proof: We will do it by induction on the complexity of $\varphi$. The only difficult case is if $\varphi = \exists x \psi(x)$. Note that in $M$, any set is the surjective image of $\mathbb R^G\times\omega_1$ in $M$ (as all $L_\alpha(\mathbb R^G)[G]$, $\alpha<\omega_2$ are such an image). Even better, $\omega_1$ is the surjective image of $\mathbb R^G$ in $M$ so that all sets are the surjective image of a function with domain $\mathbb R^G$. Let $D$ be an $M$-definable club that witnesses our induction hypothesis for $\psi$. We can close some $L_\beta(\mathbb R^G)[G]$ with $\beta\in D$ under witnesses for $\varphi$ as follows: Find a surjection $f:\mathbb R^G\rightarrow [L_\beta(\mathbb R^G)[G]]^{<\omega}$ in $M$ and map a real $x$ to the least level of the hierachy in which a witness to $\varphi(\vec p)$ (if there is one) exists where $f(x)=\vec p$. This function must be bounded so that there is a least $\gamma\in D$ so that $L_\gamma(\mathbb R^G)[G]$ contains witnesses for $\varphi$ with parameters from $L_\beta(\mathbb R^G)[G]$. This map $\beta\mapsto \gamma$ is definable in $M$ and the closure points of this map form a club with the desired properties.$\Box$

Now replacement (and even better, collection) is an easy consequence of the reflection principle above.

Choice in M:
The axiom of choice in $M$ (in terms of existence of choice functions) follows by somewhat similar arguments:

First note that for any $\alpha<\omega_2$ there is a map $C_\alpha:\mathbb R^G\rightarrow\mathbb R^G$ in $M$ so that $C_\alpha(x)$ is (better: codes canonically a) $Add(\omega, \omega)$-generic over $L[G\upharpoonright\alpha, x]$ for any $x$.
For any $x\in\mathbb R^G$ there is some $\alpha\leq\beta<\alpha+\omega_1$ so that the section of $G$ starting at $\beta$ with length $\omega$ is generic over $L[G\upharpoonright\alpha][x]$. Let $C_\alpha(x)$ be this section with $\beta$ as small as possible. Then $C_\alpha$ is definable over $(L_{\alpha+\omega_1}(\mathbb R^G)[G];\in, G)$ and hence is in $M$.

[Here the use of the predicate $G$ is crucial! No such function $C$ exists in $L_{\omega_2}(\mathbb R^G)$, so in particular the map $x\mapsto \{y\mid y$ is $Add(\omega, \omega)$-generic over $L[x]\}$ does not have a choice function there.]

Now let $f\in M$ be a function we want to find a choice function for. Since replacement holds in $M$ and all sets there are surjective images of $\mathbb R^G$ we may assume that $f:\mathbb R\rightarrow\mathcal P(\mathbb R)^M\setminus\{\emptyset\}$. Let $\alpha<\omega_2$ be large enough so that $f$ is definable in $M$ from paraters in $L[G\upharpoonright\alpha]$ (remeber that these parameters can be chosen to be ordinals, reals and initial segments of $G$ so such an $\alpha$ exists).

Claim: If $x\in \mathbb R^G$ and $g\in M$ is $Add(\omega,\omega)$-generic over $L[G\upharpoonright\alpha, x]$ then $f(x)\cap\mathbb R^{L[G\upharpoonright\alpha, x, g]}\neq\emptyset$.

Proof: For notational reasons we assume $\alpha=0$, so that $L[G\upharpoonright\alpha] = L$. Work in $L[x]$. Note that $L[x]$ is either $L$ or a Cohen extension thereof. Also $g$ is generic over $L[x]$. Thus $G$ factors into $h\times H^\prime$ so that $L[x]=L[h]$ and $H^\prime$ is $Add(\omega, \omega_2)$-generic over $L[x]$ with $L[G]=L[x][H^\prime]$. Similarly, we can take an isomorphis $\mu:Add(\omega, \omega_2)\rightarrow Add(\omega, \omega)\times Add(\omega, \omega_2)$ so that $H^\prime$ factors again  (via $\pi^{-1}$) into $g\times H$. In $L[x]$ we can find a nice name $\dot y$ so that
$$1\Vdash^{L[x]}_{Add(\omega, \omega_2)} \dot y \in f(\check x)^{L_{\check\omega_2}(\mathbb R^{\dot G})[\dot G]}$$
Using the ccc, we can further find an automorphism $\pi$ of $Add(\omega, \omega_2)$ so that $\mu\circ\pi$ shifts $\dot y$ completely into the first factor $Add(\omega, \omega)$. With the same strategy as in the boundedness arguments, using that all parameters defining $f$ are in the ground model, we can see that in fact
$$z=\widehat{\mu\circ\pi}(\dot y)^{g\times H^\prime}\in f(x)$$
But the evaluation of $\widehat{\mu\circ\pi}(\dot y)$ depends only on $g$ so that $z\in L[x][g]$ witnesses the claim to hold.

If $\alpha\neq 0$, replace $L$ by $L[G\upharpoonright\alpha]$ in the argument above, work from there and use that $L[G]$ is still a $Add(\omega,\omega_2)$-extension from there. $\Box$

It is now straight forward to define a choice function for $f$ in $M$:
Just map a real $x$ to the $<_{L[x, C_\alpha(x)]}$-least element of $f(x)$ (where $<_{L[x, C_\alpha(X)]}$ denotes the canonical wellorder on the reals of $L[x, C_\alpha(x)]$).


问:
Random variables question - $n$ different balls {$1, ..., n$} are placed randomly in $n$ different urns

$n$ different balls {$1, ..., n$} are placed randomly in $n$ different urns.

$X_n$ is the random variable which marks how many balls are in the "correct" (same index) urn.

Using $\lim_{n \to \infty}\mathbb{P}(X_n = 0)=\frac{1}{e}$  (I already showed this in a different question), show that $\forall k \in \mathbb{N}, \lim_{n \to \infty}\mathbb{P}(X_n = k)=\frac{1}{k!e}$

My idea was to find some relation such as $\mathbb{P}(X_n = k) = \frac{1}{k!} \mathbb{P}(X_{n+k} = 0)$ using combinatorics, but I can't seem to find such a thing. Hints would be appreciated.

答:
The probability that exactly $k$ balls are in the correct urn is the sum over all length $k$ subsets of the $n$ balls of the probability of those exact $k$ balls being in the correct urn and that no other balls are in the correct urns. This is given by $$\sum \frac{1}{\frac{n!}{(n-k)!}}\mathbb{P}(X_{n-k} = 0) = \sum \frac{(n-k)!}{n!}\mathbb{P}(X_{n-k} = 0)$$

The summand will be the same no matter which of the $\binom{n}{k}$ subsets is being summed, so it simplifies to $$\binom{n}{k} \frac{(n-k)!}{n!}\mathbb{P}(X_{n-k} = 0) = \frac{n!}{k!(n-k)!} \frac{(n-k)!}{n!} \mathbb{P}(X_{n-k} = 0) = \frac{1}{k!}\mathbb{P}(X_{n-k}=0)$$

And then you know that $\lim_{n \to \infty} \mathbb{P}(X_n = 0) = \frac{1}{e}$, so then $$\lim_{n \to \infty} \mathbb{P}(X_n = k) = \frac{1}{k!e}$$


问:
Choosing a good series for the comparison test
I'm given the series:

$$\sum_{n=2}^{\infty} \frac{n^2}{n^4-n-3}$$

I know it converges, however I'm meant to show that by the comparison test. What would be a good choice here? $\frac{1}{k^2}$ and $\frac{1}{k}$ don't work here, obviously.

答: Since $\sum_{n=2}^\infty\frac1{n^2}$ converges and since$$\lim_{n\to\infty}\frac{\dfrac{n^2}{n^4-n-3}}{\dfrac1{n^2}}=1,$$your series converges.


问:
Does the number of elements in Vector space and the dimension of Vector space same?
The number of elements in any basis of a vector space is called the dimension of Vector space.

In the Vector space F^n(F) and Null Space; the dimension of Vector Space and the number of Vectors in the Vector Space are distinct.

Does the number of elements in the Vector Space and the Dimension of Vector Space always distinct???

If not,then in which case number of elements in Vector Space and the Dimension of Vector Space both are equal??

答:
We focus on finite-dimensional cases, where we can prove without much effort the dimension and the number of elements never agree.

First suppose we work over an infinite field $\mathbb{F}$ (e.g.$\mathbb{Q}$,$\mathbb{R}$,$\mathbb{C}$). In this case, any finite-dimensional vector space over $\mathbb{F}$ has infinitely many elements. (As an example, we may consider $\mathbb{R}$ itself as a $1$-dimensional vector space over $\mathbb{R}$, which has infinitely many elements. )

Then suppose $\mathbb{F}$ is a finite field of $q$ elements. By the definition of a field, we have $q>1$. It's clear that an $n$-dimensional vector space over $\mathbb{F}$ has $q^n$ elements. So again, we conclude $q^n > n$.

If we do not restrict ourselves to finite-dimensional vector spaces, things can be subtler.


问:
Generalization of Boolean OR?
I have been looking at the Boolean OR function and Im trying to find its integral analogue.

What I mean is:

Boolean AND (x, y) where x and y are Boolean Values with 0 = False, 1 = True is equivalent to multiplication:

xANDy = x*y

xORy = ?

It appears to be equivalent x + y - xANDy = x + y - x*y

Is this a correct intuition? My defense of AND and multiplication comes from their relationship in probability where p of A and B = pA * pB if A and B are independent. Likewise probability p of A or B = pA + pB - pA*pB

答:
Using De Morgan's laws, if you have an expression for "not" and "and", you can get an expression for "or."

The probability intuition is good - indeed, the probability rule for "or" is just a version of De Morgan's law.

Title: Generalization of Boolean OR?


问:
Find area of triangle ABC with $∠CAB = α $ and $∠ABC = β$
From the point M lying inside the triangle ABC, perpendiculars are drawn to the sides BC, AC, AB, the lengths of which are k, l and m, respectively. Find the area of a triangle ABC if ∠CAB = α and ∠ABC = β. α = π / 4, β = 2π / 3, k = 5, l = 3, m = 4. If the answer is not an integer, round it to the nearest integer.

I built circles on the segments MA, MB and MC as on diameters ... they will pass through the bases of the perpendiculars lowered to the sides ...

it turns out the task is to find the area of an inscribed quadrilateral along two sides and the angle between them, if the other two angles adjacent to these sides are straight lines ...

maybe there is some ready-made formula for the solution, but I don't know it ... but there is enough data

答:
Let [] denote areas. Consider the quadrilateral AYMZ

$$[AYMZ] = \frac12 m^2 \cot a_1 + \frac12 l^2 \cot a_2$$
$$l\sin a_1 = m \sin a_1, \>\>\>\>\> a_1+a_2 = \alpha$$

Eliminate $a_1$ and $a_2$ to get
$$[AYMZ]=\frac12(m^2+l^2)\cot \alpha + ml\csc \alpha$$

Similarly,
\begin{align}
[BZMX] &=\frac12(k^2+m^2)\cot \beta+ km\csc \beta\\
[CXMY] &=\frac12(l^2+k^2)\cot \gamma+ lk\csc \gamma\\
\end{align}
where $\gamma = \pi - (\alpha+\beta)$. Thus, the area of the triangle ABC is
\begin{align}
[ABC] &= [AYMZ] + [BZMX] + [CXMY] \\
& = \frac12[ (m^2+l^2)\cot \alpha+(k^2+m^2)\cot \beta+ (l^2+k^2)\cot \gamma]\\
&\hspace{1cm}+ ml\csc \alpha+ km\csc \beta + lk\csc \gamma
\end{align}


问:
Computing Ramanujan asymptotic formula from Rademacher's formula for the partition function
I am trying to derive the Hardy-Ramanujan asymptotic formula

$$p(n) \sim \frac{1}{4n\sqrt{3}}e^{\pi\sqrt{\frac{2n}{3}}}$$

from Radmacher's formula for the partition function $p(n)$ given by

$$p(n)=\frac{1}{\pi\sqrt{2}}\sum_{k=1}^{\infty}A_{k}(n)\sqrt{k}\left[\frac{d}{dx}\frac{sinh\left(\frac{\pi}{k}\sqrt{\frac{2}{3}\left(x-\frac{1}{24}\right)}\right)}{\sqrt{x-\frac{1}{24}}}\right]_{x=n}$$
where
$$A_{k}(n)=\sum_{h=0, (h,k)=1}^{k-1}e^{\pi i(s(h,k)-2n\frac{h}{k})}$$ and
$$s(h,k)=\sum_{r=1}^{k-1}\frac{r}{k}\left(\frac{hr}{k}-\left\lfloor\frac{hr}{k}\right\rfloor-\frac{1}{2}\right)$$

G.E. Andrews, and any other literature on this topic, says that we can obtain the H-R asymptotic expression from the first term of the Rademacher series, i.e. for $k=1$. I don't know how to approach this as simply calculating for $k=1$ does not give the desired result. Could we perhaps try and use Lapalce's method, or the method of steepest descent?

答:
Rough sketch: Take the first term, and you'll get:
$$\begin{align}p\left(n\right)&\sim\frac{1}{\pi\sqrt{2}}A_{1}\left(n\right)\cdot\frac{d}{dx}\left(\frac{\sinh\left(\pi\sqrt{\frac{2}{3}\left(x-\frac{1}{24}\right)}\right)}{\sqrt{x-\frac{1}{24}}}\right)\\&\sim\frac{1}{\pi\sqrt{2}}\frac{d}{dx}\left(\frac{\sinh\left(\pi\sqrt{\frac{2x}{3}}\right)}{\sqrt{x}}\right)\\&\sim\frac{1}{2\pi\sqrt{2}}\frac{d}{dx}\left(\frac{\exp\left(\pi\sqrt{\frac{2x}{3}}\right)}{\sqrt{x}}\right)\\&=\frac{1}{2\pi\sqrt{2}}\left(\frac{\pi\exp\left(\pi\sqrt{\frac{2x}{3}}\right)}{\sqrt{6}x}-\frac{\exp\left(\pi\sqrt{\frac{2x}{3}}\right)}{2x\sqrt{x}}\right)\\&=\frac{\exp\left(\pi\sqrt{\frac{2x}{3}}\right)}{4x\sqrt{3}}\left(1-\frac{\sqrt{6}}{2\pi\sqrt{x}}\right)\\&\sim\frac{\exp\left(\pi\sqrt{\frac{2x}{3}}\right)}{4x\sqrt{3}}\end{align}$$

In the first line, we take the first term (as it is the leading term). We can see $A_1(n)=1$ and we write $x-1/24$ as $x$, since we're looking for as $x \rightarrow \infty$.

In the third-line, we take the leading term of the $\sinh(x)$ function.
$$\sinh\left(x\right)=\frac{e^{x}-e^{-x}}{2}\sim\frac{e^{x}}{2}$$


问:
How to interpret e^i in growth term?
We can interpret e as e^1 i.e. starting with 1 we keep compounding continously at 100% rate  i.e. limit [1 + (1/t)]^t where t -> infinity..for e^2 we are compounding continuosly at 200% rate

We can apply euler's form and get cos(1) + isin(1) = 0.54 + 0.84 i but how do we interpret it, apart from jumping from one-dimension number line (e^1, e^2 etc.) to two-dimension complex plane (e^i)?

答:
Here’s a method I used for trying to get this idea across in the last course I taught. I wasn’t too successful with the students, but maybe I’ll be more successful now.

You know the explanation of $e$ as $\lim_{n\to\infty}(1+\frac1n)^n$. Perhaps you have also seen this:
$$
e^x=\lim_{n\to\infty}\left(1+\frac xn\right)^n
$$
If you haven’t, you can verify it by taking logarithms and doing a little algebraic magic. $(\star)$

Thus, if you want to, you can substitute $i$ for $x$ in the formula and look at $e^{\,i}$ as describing what you get by compounding for a year at an annualized rate of $i$, whatever that could mean.

But look rather at the slightly more general $e^{\,it}$ for $t$ real. When  you consider the complex number $1+\frac{it}n$, it’s just a little off the unit circle, and up almost exactly $t/n$ radians. You’ve probably seen something like this when you saw the proof of the relation $\lim_{h\to0}\frac{\sin h}h=1$. But make sure you draw the picture of $1+\frac{it}n$ in the Wessel-Argand-Gauss plane. Then think of what happens when you raise this to $n$-th power. The power is (almost) $t$ radians from the positive $x$-axis now, because of what happens when you multiply two complex numbers. And how far from the origin? Just a little more than a distance of $1$. And the two errors go to zero as you take $n\to\infty$.

You’ve just seen an argument of why $e^{\,it}=\cos t+i\sin t$. Of course you’d need to keep good control of the error terms in the finite power, and make sure
that in the limit they really go to zero. But it’s a sound argument.

$(\star)$ If you want to see the “algebraic magic”, I’ll add it in an edit.

Title: How to interpret e^i in growth term?


问:
Number of coins of integer weights

$A$ tells $B$ : "I could be visiting you on any day in the next two months and you must give me gold coins of as much total weight in grams as the number of days that would elapse from today."  If gold coins are available in integer gram weights, what is the least number of coins with which $B$ can meet $A$'s demand on any day?

$(1)$ $31$

$(2)$ $7$

$(3)$ $6$

$(4)$ $13$

How to approach this problem? Can anybody shed some light on it? Thanks for your time.

Source $:$ CSIR NTA NET DECEMBER $2019.$

答:
Hint :

Any positive integer can be written as sum of powers of $2$. Every decimal number has a (unique) binary (base-$2$) representation.

Title: Number of coins of integer weights


问:
Good books to Learn General topology
I am quite interested in learning topology from the basics, I would like to find a book that is quite detailed and has a large number of examples.

What are the best books to easily understand topology?

Thanks.

答: Munkres is good, or this book: http://www.ehu.eus/~mtwmastm/TG0910.pdf</p>

Title: Good books to Learn General topology


Title: Choosing a good series for the comparison test

Body:
I'm given the series:

$$\sum_{n=2}^{\infty} \frac{n^2}{n^4-n-3}$$

I know it converges, however I'm meant to show that by the comparison test. What would be a good choice here? $\frac{1}{k^2}$ and $\frac{1}{k}$ don't work here, obviously.

Body:
For $ n\ge 3$, we have

$$n\le \frac{n^4}{3}$$
and

$$3\le \frac{n^4}{3}$$

thus

$$n^4-n-3\ge \frac{n^4}{3}$$

and

$$0\le \frac{n^2}{n^4-n-3}\le \frac{3}{n^2}$$


问:
How to find the variance and mean of a min function
Problem

Let $Y_1, Y_2, . . . , Y_n$ be i.i.d. with p.d.f.

$f(y; θ) = \frac1θ e^\frac{-y}θ , y>0 $

Consider two estimators for the parameter θ,

$\hat θ_1 = n \min_{i=1,2,..,n} Y_i $

$\hat θ_2 = \frac{Y_1 + Y_2 + . . . + Y_n}n$

Show that $\hat θ_1$  is indeed an unbiased estimator, and find its efficiency relative to $\hat θ_2$

My progress

The expected value of $f$ is $θ$ because it is an exponetial distribution  with the rate parameter of $\frac1θ$

Now I need to find the the variance and mean of $\hat θ_1$ but I have no idea how to do that because we have a min function if someone could explain that to me.

For $\hat θ_2$ I can write it a $Y_1$ which is just $f$ because we are adding $n$ $Y$ components and dividing by $n$. So the mean is $θ$ and variance is ${θ^2}$

答:
Now I need to find the the variance and mean of $\hat{\theta}_1$ but I have no idea how to do that because we have a min function if someone could explain that to me.

The CDF of the min of $n$ iid rv is

$$F_{min}=1-(1-F)^n$$

Thus in your case, setting $U=min(\mathbf{y})$

$$F_U(u)=1-e^{-\frac{yn}{\theta}}$$

and consequently, derivating,

$$f_U(u)=\frac{n}{\theta}e^{-\frac{yn}{\theta}}$$

Now you have all the tools to proceed by yourself

$\hat{\theta}_2$ is the sample mean...thus its mean is $\theta$ the same of the population's one but its variance is $\frac{\theta^2}{n}$

In fact:

$$\mathbb{V}\Bigg[\frac{1}{n}\Sigma_i Y_i\Bigg]=\frac{1}{n^2}n\theta^2$$

(The expectation is a linear operator but the variance is not. It's a quadratic operator)


问:
Can someone please explain $e$ in layman's term?
I never really understood what $e$ means and I'm always terrified when I see it in equations. What is it? Can somebody dumb it down for me? I know it's a constant. Is it as simple as that?</p>

答:
The simplest way to understand it is, consider the following equation:

$$f(x)=\left(1+\frac{1}{x}\right)^x$$

As $x$ gets larger and larger, notice what number it gets closer to:

$$f(1)=2$$
$$f(2)=2.25$$
$$f(3)\approx 2.37073$$
$$...$$
$$f(1000)\approx 2.7169$$
$$...$$

It is important in mathematics because it is arguably the single real number most commonly found in nature (from heat flow to battery discharge to population growth/decay models), which is why its inverse is called the natural logarithm.


问:
Smooth function with specific features
Im looking for a function that's continuous everywhere and whose second derivative is continuous everywhere with the following properties:

Near the origin, the function is approximately constant: $$f\approx c$$ with $c \in \mathbb{R}$

Far from the origin (outside some scale), the function is approximately linear: $$f(x) \approx m*x+b$$ with $m \neq 0$

So the function would look like

.

I just defined this function piecewise in mathematica as an example, but the second derivative is not continuous, because the curvy bits are quadratic and the flat bits are linear, so the second derivative is a series of steps.

I will eventually define a vector field from this scalar field which describes a flow in a flat manifold. Moreover, there is an energy functional that is defined in terms of the second derivatives of this scalar field, hence the reason for continuity.

I tried using bump functions and integrating backwards to get the scalar field I want, but the integrals became messy quite fast, as you might expect. Similarly, I tried using hyperbolic tangents, but I wasn't able to smoothly attach the pieces together.

I am also trying to minimize the magnitude of the second derivative (because it relates to the energy functional). So I am hoping to find a function that approaches the origin in a linear way, then curves in such a way that the second derivative has constant sign, then approaches zero again, then has constant opposite sign. Here is an example of such a second derivative using bump functions.

I tried integrating this and using the integration constants to satisfy my required properties, but the integrals are quite messy!

I appreciate any help!

答:
Is approximation of standard $f(x) = x \cdot (1 - e^{-x^2})$ good enough?

If not - then probably $x \cdot (1 - w(x)) + w(x)$ where $w(x)$ is bump function is the best we can get.

Title: Smooth function with specific features


问:
Prove subspace existence
I am trying to solve the following problem:

Given the following equations system

$
\begin{equation}
\begin{cases}
20a + b + c + d + e = 0 \\
a + 20b + c + d + e = 0 \\
a + b + 20c + d + e = 0
\end{cases}
\end{equation},
$

How to prove that its solutions set are a $\mathbb{R}^5$ subspace? How to describe it?

I am trying to solve the system, but once it demands a $\mathbb{R}^5$ subspace (I have never worked with it), I really don't know how to deal with it.

答:
The solution is an element of $\mathbb{R}^5$ since you are working in 5 unknowns and considering only real solution(s), I suppose.

To solve this, convert this into a matrix-vector equation: $$\begin{bmatrix}20&1&1&1&1\\ 1&20&1&1&1\\1&1&20&1&1\end{bmatrix}\begin{bmatrix}a\\b\\c\\d\\e\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$$ or in augmented matrix $$\left[\begin{array}{ccccc|c}20&1&1&1&1&0\\ 1&20&1&1&1&0\\1&1&20&1&1&0\end{array}\right].$$ Perform elementary row operations to get $$\left[\begin{array}{ccccc|c}1&0&0&\frac{1}{22}&\frac{1}{22}&0\\ 0&1&0&\frac{1}{22}&\frac{1}{22}&0\\0&0&1&\frac{1}{22}&\frac{1}{22}&0\end{array}\right],$$ which is equivalent to the system of equation $$\begin{cases}a+\frac{1}{22}d+\frac{1}{22}e=0\\b+\frac{1}{22}d+\frac{1}{22}e=0\\c+\frac{1}{22}d+\frac{1}{22}e=0\end{cases}.$$ Let $d,e$ be free variable. Then $a,b,c$ can be expressed in terms of $d,e$ by $$\begin{cases}a=-\frac{1}{22}d-\frac{1}{22}e\\b=-\frac{1}{22}d-\frac{1}{22}e\\c=-\frac{1}{22}d-\frac{1}{22}e\end{cases}.$$ Thus the solutions are the set $$\left\{\left.\begin{bmatrix}-\frac{1}{22}d-\frac{1}{22}e\\-\frac{1}{22}d-\frac{1}{22}e\\-\frac{1}{22}d-\frac{1}{22}e\\d\\e \end{bmatrix} \in \mathbb{R}^5\right\vert d,e\in\mathbb{R}\right\}$$ This is how you solve it: perform Gaussian-Jordan Elimination, put free variable, write out the solution(s).

Now you can directly prove that the solution set is a subspace. However, the set of solution(s) to a homogeneous system of equations (every line equal to 0) is a well-known subspace called a null space. So you don't have to to tedious work above.

Precisely, for any $m\times n$ matrix $\mathbf{A}$, the set $$\mathcal{N}(\mathbf{A})=\{\mathbf{x}\in\mathbb{R}^n\mid \mathbf{Ax=0}\}$$ is a subspace of $\mathbb{R}^n$.

To prove this, we are left to show that it is closed under addition and scalar multiplication. Let $\mathbf{x,y}\in\mathcal{N}(\mathbf{A})$ and $k\in\mathbb{R}$. We have $$\mathbf{A}(\mathbf{x}+\mathbf{y})=\mathbf{A}\mathbf{x}+\mathbf{A}\mathbf{y}=\mathbf{0}+\mathbf{0}=\mathbf{0},$$ and $$\mathbf{A}(k\mathbf{x})=k\mathbf{A}\mathbf{x}=k\mathbf{0}=\mathbf{0}.$$ Thus, we have that $\mathcal{N}(\mathbf{A})$ is a subspace of $\mathbb{R}^n$.

Title: Prove subspace existence


问:
$ \limsup_{k\to \infty} (a_k)^{1/k} \leq \limsup_{k\to \infty} \frac{a_{k+1}}{a_k} $
I am working on the following problem, let $(a_n)_{n \in \mathbb N}$ be a sequence of positive real numbers, we wish to show that:

$$ \limsup_{k\to \infty} (a_k)^{1/k} \leq  \limsup_{k\to \infty} \frac{a_{k+1}}{a_k}  $$

a similar problem as in (but which does not answer my own questions/confusions):

Inequality involving $\limsup$ and $\liminf$: $ \liminf(a_{n+1}/a_n) \le \liminf((a_n)^{(1/n)}) \le \limsup((a_n)^{(1/n)}) \le \limsup(a_{n+1}/a_n)$

We have:
$$\limsup_{n\to \infty} \frac{a_{n+1}}{a_n}=R$$
I understand that from the limit definition we have for any $\epsilon>0$ there exists an $N$ such that $n>N$ means:
$$\frac{a_{n+1}}{a_n} \leq R+ \epsilon.$$
We can then arrive at the following inequality using the one above for $k\geq N$:
$${a_k} \leq a_N(R+ \epsilon)^{k-N}.$$
and this can be rewritten to:
$$ {a_k} \leq a_N(R+ \epsilon)^k (R+ \epsilon)^{-N} .$$
$$ (a_k)^{\frac{1}{k}} \leq a_N ^{\frac{1}{k}}(R+ \epsilon) (R+ \epsilon)^{\frac{-N}{k}} .$$

If I take the limit and supremum for $k\to \infty$ I find ($1 \cdot (R+\epsilon)\cdot 1$):
$$ \limsup_{k \to \infty } (a_k)^{\frac{1}{k}} \leq R+\epsilon $$
The question tells me to derive that following inequality (why $2\epsilon$?):
$$\limsup_{k\to \infty}(a_k)^{\frac{1}{k}} \leq R + 2\epsilon$$
Which can be used to conclude that (how did they just drop the $\epsilon$?):
$$\limsup_{k\to \infty}(a_k)^{\frac{1}{k}} \leq R. $$

Which is equivalent to the desired equality between the limsups.

答:
You have a typo, there is a $1/N$ exponent that should be $-N$.

The $2\epsilon$, you don't need it the way you wrote the answer.

As for "how did they just drop the $\epsilon$", if $a\leq R+\epsilon$ for all $\epsilon>0$, then $a\leq R$. This is simply the fact that $a-R$ is less than every positive number, so it has to be less than or equal zero.


问:
proof: p|b and p|a if $p|a^2+b^2+ab$ and p = 6k+5
Let a,b be two natural numbers and p a prime number and

$p=6k+5 $ (k is an unfixed integer)

Prove that if
$$ p|a^2+ab+b^2$$
Then
$p|b$ and $p|a$

I found this problem in a discrete math book in the number theory section. I tried solving it by multiplying $a^2+ab+b^2$ by $a-b$ which results in
$$p|a^3-b^3$$

But i don't know how to proceed.

答:
You've already found that $p$ divides $a^3-b^3$, i.e. that $a^3\equiv b^3\pmod{p}$. In particular, if $p\mid a$ then clearly also $p\mid b$ and vice versa. So suppose $p\nmid a,b$. Given that $p\equiv5\pmod{6}$ the order of the group $(\Bbb{Z}/p\Bbb{Z})^{\times}$ is not divisible by $3$. This means it contains no elements of order $3$, and so the map
$$(\Bbb{Z}/p\Bbb{Z})^{\times}\ \longrightarrow\ (\Bbb{Z}/p\Bbb{Z})^{\times}:\ n\ \longmapsto\ n^3,$$
is injective. This means that $a^3\equiv b^3\pmod{p}$ implies $a\equiv b\pmod{p}$, that is to say $p\mid(a-b)$. Then $p$ also divides
$$(a^2+b^2+ab)-(a-b)^2=3ab,$$
and so $p$ divides either $a$ or $b$, or both, contradicting the assumption that $p\nmid a,b$.


答:
1. Forget about the $2\epsilon$ thing. Take upper limits (limsup) and you get $ \limsup_{k \to \infty } (a_k)^{\frac{1}{k}} \leq R+\epsilon$. That's perfect.

2. We can drop the $\epsilon$ because we chose $\epsilon$ arbitrarily. This is a standard technique. For example, if one has difficulty understanding $5 \leq 5$, we may convince him by showing $5 < 5 + 1, 5 < 5 + 0.2, 5 < 5 + 0.0002 \dots$, and finally  $5 < 5 + \epsilon$ for arbitrary $\epsilon$.

Body:
I am working on the following problem, let $(a_n)_{n \in \mathbb N}$ be a sequence of positive real numbers, we wish to show that:

$$ \limsup_{k\to \infty} (a_k)^{1/k} \leq  \limsup_{k\to \infty} \frac{a_{k+1}}{a_k}  $$

a similar problem as in (but which does not answer my own questions/confusions):

Inequality involving $\limsup$ and $\liminf$: $ \liminf(a_{n+1}/a_n) \le \liminf((a_n)^{(1/n)}) \le \limsup((a_n)^{(1/n)}) \le \limsup(a_{n+1}/a_n)$

We have:
$$\limsup_{n\to \infty} \frac{a_{n+1}}{a_n}=R$$
I understand that from the limit definition we have for any $\epsilon>0$ there exists an $N$ such that $n>N$ means:
$$\frac{a_{n+1}}{a_n} \leq R+ \epsilon.$$
We can then arrive at the following inequality using the one above for $k\geq N$:
$${a_k} \leq a_N(R+ \epsilon)^{k-N}.$$
and this can be rewritten to:
$$ {a_k} \leq a_N(R+ \epsilon)^k (R+ \epsilon)^{-N} .$$
$$ (a_k)^{\frac{1}{k}} \leq a_N ^{\frac{1}{k}}(R+ \epsilon) (R+ \epsilon)^{\frac{-N}{k}} .$$

If I take the limit and supremum for $k\to \infty$ I find ($1 \cdot (R+\epsilon)\cdot 1$):
$$ \limsup_{k \to \infty } (a_k)^{\frac{1}{k}} \leq R+\epsilon $$
The question tells me to derive that following inequality (why $2\epsilon$?):
$$\limsup_{k\to \infty}(a_k)^{\frac{1}{k}} \leq R + 2\epsilon$$
Which can be used to conclude that (how did they just drop the $\epsilon$?):
$$\limsup_{k\to \infty}(a_k)^{\frac{1}{k}} \leq R. $$

Which is equivalent to the desired equality between the limsups.


问:
Application of Galois theory
i have a question regarding roots of equation,
find all $a$,such that the cubic polynomial $x^3-bx+a=0$ has three integer roots,
how can you solve these by using galois theory,what does the reducible polynomials,splitting fields,field extensions have to do with these,
explain each of them,in detail,because it serves as a introduction to galois theory,
ok for eg take $b$ to be 3 and list all $a$ such that equation has three integer roots</p>

答:
Suppose the polynomial has three integer roots $r_1, r_2, r_3$. Then $(x - r_1)(x - r_2)(x - r_3) = x^3 - bx + a$, hence

$$r_1 + r_2 + r_3 = 0$$
$$r_1 r_2 + r_2 r_3 + r_3 r_1 = -b$$
$$r_1 r_2 r_3 = -a.$$

Squaring the first equation gives $r_1^2 + r_2^2 + r_3^2 = 2b$, which immediately tells you that for fixed $b$ there are only finitely many possibilities for the roots, and from here it's casework for any fixed $b$.

For example, for $b = 3$ we get $r_1^2 + r_2^2 + r_3^2 = 6$, which has solutions $(\pm 1, \pm 1, \pm 2)$ up to cyclic permutation, and of these solutions only $(-1, -1, 2)$ and $(1, 1, -2)$ add up to zero. Hence the possible values in this case are $a = \pm 2$.

Title: Application of Galois theory


问:
Dagger category generated by $\mathsf{Set}$ viewed as a subcategory of $\mathsf{Rel}$.
Whenever a category $\mathcal{C}$ is being viewed a subcategory of a dagger category $\mathcal{D}$, define that the dagger category generated by $\mathcal{C}$ is the least subcategory of $\mathcal{D}$ that is closed under the dagger operation and includes $\mathcal{C}$.

Question. Viewing $\mathcal{C}=\mathsf{Set}$ as a subcategory of $\mathcal{D}=\mathsf{Rel}$, what is the dagger category generated by $\mathcal{C}$? I'm especially interested to see whether all of $\mathcal{D}$ is generated in this case.

答:
I think that indeed all of $\mathsf{Rel}$ is generated by the following argument:

Consider a relation $R\subseteq X\times Y$, then we have natural maps $i\colon R\to X\times Y$, $\pi_1\colon X\times Y\to X$ and $\pi_2\colon X\times Y\to Y$.
Define $R^\prime = (\pi_2\circ i)\circ(\pi_1\circ i)^\dagger$.

Now $x R^\prime y$ iff $\exists z\in R\colon x(\pi_2\circ i) z\mbox{ and }z(\pi_1\circ i)^\dagger y$, which is true iff $\exists z\in R\colon x = \pi_2(i(z))\mbox{ and }y=\pi_1(i(z))$, which in turn is equivalent to $(x,y)\in R$ or $xRy$.
Hence $R=R^\prime$.

Since $\pi_2\circ i$ is in $\mathsf{Set}$ and $(\pi_1\circ i)^\dagger$ is in $\mathsf{Set}^\dagger$, this proves that $R$ is in the dagger category generated by $\mathsf{Set}$.


问:
Wrong exponential using numpy?
The exponential $e^{-5^2/8}$ is supposed to give $0.04393693362$ but when using the following numpy code:
print(np.exp((-5*5)/2*4.0)), I get "1.9287498479639178e-22". Can you figure out what is wrong? I have stared at this for an hour now :/

答: You need to place brackets around 2*4.0. It's calculating $e^{-50}$ instead of $e^{-3.125}$.

Title: Wrong exponential using numpy?


问:
Characters of the absolute Galois group and their restrictions
I am reading this paper Cremona et. al. In section 3 (pages 5-6), they talk about characters more specifically about Dirichlet characters and characters of the absolute Galois group, $G_{\mathbb{Q}}$. I am a little lost probably because my background is not strong enough.

I have a couple of questions that I got stuck on were:

How to think of characters over $\mathbb{Q}$ as characters of $G_{\mathbb{Q}}$?

How does a character of $G_{\mathbb{Q}}$ restricts to a number field $K$ ?

Could someone recommend a resource to learn about these things?  Should I be reading about Galois representations?

Thank you in advance!

答:
For a finite image (equivalently continuous in the relevant topologies) character $$\rho : Gal(\Bbb{Q}(\zeta_\infty)/\Bbb{Q})\to \Bbb{C}^\times$$ Then the subfield fixed by $\ker(\rho)$ is a finite extension of $\Bbb{Q}$ (of degree $|Image(\rho)|$). Thus for some
$k$, $\ker(\rho)$ contains $Gal(\Bbb{Q}(\zeta_\infty)/\Bbb{Q}(\zeta_k))$ and hence $\rho$ is a character $$Gal(\Bbb{Q}(\zeta_k)/\Bbb{Q})\cong \Bbb{Z}/k \Bbb{Z}^\times\to \Bbb{C}^\times$$  A character $$f:\Bbb{Z}/k \Bbb{Z}^\times\to \Bbb{C}^\times$$ corresponds to a Dirichlet character: $\chi(n)= f(n)$ if $\gcd(n,k)=1$, otherwise $\chi(n)=0$.

Conversely any Dirichlet character gives a character $\Bbb{Z}/k \Bbb{Z}^\times\to \Bbb{C}^\times$ for some $k$ thus a character $Gal(\Bbb{Q}(\zeta_k)/\Bbb{Q})\to \Bbb{C}^\times$ and hence a continuous character $Gal(\overline{\Bbb{Q}}/\Bbb{Q})\to \Bbb{C}^\times$.

The correspondence is not one-to-one, it is if we take the minimum possible $k$ and restrict to the primitive Dirichlet characters.

Any character $Gal(\overline{\Bbb{Q}}/\Bbb{Q})\to \Bbb{C}^\times$ factorizes through the abelianization of $Gal(\overline{\Bbb{Q}}/\Bbb{Q})$ which is $Gal(\Bbb{Q}(\zeta_\infty)/\Bbb{Q})$ (Kronecker Weber theorem).

You probably meant restricting a character $Gal(\overline{\Bbb{Q}}/\Bbb{Q})\to \Bbb{C}^\times$ to a character  $Gal(\overline{\Bbb{Q}}/K)\to \Bbb{C}^\times$.

The continuous characters $Gal(\overline{\Bbb{Q}}/K)\to \Bbb{C}^\times$ correspond similarly to the finite image Hecke characters of $K$, but this is a much more complicated story, the theorems from class field theory.


问:
Distribution of combination of random variables
I am struggling to find a solution on finding the following distribution:
$P(X+Y \leq z) p + P(Y \leq z) (1-p)$
where $X, Y$ are both gaussian random variables with zero mean and variances $s_x$ and $s_Y$, respectively. $Y$ is statistically independent from $X$. Also, $p$ is a probability. Is this sum Gaussian also ? And if so, how can I find mean and variance?
I've reached to the conclusion that X+Y is gaussian with zero mean and varriance $s_x + s_y$, therefore the whole sum is equal to a Gaussian with zero mean and varriance $p^2 (s_x + s_y)+(1-p)^2 s_y $.
Is my thought correct?

答:
What you have created is a weighted mixture of two Gaussian distributions with the same means of $0$ but different variances of $s_1+s_2$ and $s_2$.  You should not expect this to have a Gaussian distribution; it does not.

If they had had different means of $m_1+m_2$ and $m_2$ then your distribution would have mean of $m = p(m_1+m_2)+(1-p)m_2 = pm_1+m_2$ and variance of $s=p(s_1+s_2)+p(m_1+m_2)^2 + (1-p)s_2+(1-p)m_2^2 -m^2 = ps_1+s_2+p(1-p)m_1^2$.

But since $m_1+m_2=m_2=0$, your distribution has mean $m=0$ and variance $s=ps_1+s_2$.


问:
Inverting a function in the parametrisation of Riemann's minimal surface
While writing code to render Riemann's minimal surface – which is now available here as part of my Malibu STL surface generator – the following problem came up. Consider the following function:
$$y(\theta)=\operatorname{Re}\left(\sqrt{\frac{1-e^{2i\theta}}{(\sigma-1)e^{2i\theta}+(\sigma+1)e^{i\theta}}}\right)\qquad\theta\in[0,\pi],\sigma>0$$
This is (after constant factors have been taken out) the $y$-coordinate of a point on the surface corresponding to some complex number $w$, which itself is the image under a conformal map of a point on the unit circle in the upper half-plane, hence the appearance of $e^{i\theta}$. $\sigma$ is a given constant; for all values of $\sigma$, $y$ is strictly increasing up to a point and strictly decreasing after that.

In order for neighbouring patches to match up exactly, right down to the vertices in the output STL file, I needed to invert $y$, i.e. given $y(\theta)$ find the two possibilities for $\theta$. Failing that, I needed to find for a given $\theta_1$ another angle $\theta_2$ such that $y(\theta_1)=y(\theta_2)$. I could not do either analytically and had to resort to numerical methods.

If there was no taking of the real component I would be able to solve this rather easily, but $y$ is the result of using the Weierstrass–Enneper parametrisation and there's no way around it here. The $\operatorname{Re}$ causes SymPy to output a very complicated expression for $y$ that does not use $\operatorname{Re}$ but uses $\sin\theta$ and $\cos\theta$, and I made no headway this way.

Is it possible to solve the "preimage" or "second-preimage" problems for $y$ described above analytically?

Edit: My full parametrisation of a fundamental patch of the Riemann surface is as follows. Let
$$m=\frac1{1+\sigma},m'=1-m=\sigma m,c=\frac{1+\sigma}{\sqrt\sigma}$$
Then
$$x(w)=c\operatorname{Im}(2(m'F(\varphi=\sin^{-1}\sqrt{1-w},m)-E(\varphi,m))+p)$$
$$y(w)=c\operatorname{Re}\left(p=\sqrt{\frac{(\sigma+w)(1-w)}{w(1+\sigma)}}\right)$$
$$z(w)=2\operatorname{Im}(F(\varphi,m))$$
where $w$ lies in the semicircle with diameter $1$ and $\sigma$ in the upper half-plane punctured at $(0,0)$. The conformal map I am using is the Möbius transformation
$$w=\frac{2\sigma v}{(\sigma-1)v+(\sigma+1)}$$
where $v$ lies in the upper half-unit disc. Substituting this map in the $y(w)$ equation and then restricting $v$ to lie on the unit circle ($v=e^{i\theta}$) gives the equation at the top of this question.

The translation vector between unit cells is given by $(x(-\sigma), 0, z(-\sigma))=\left(\frac2{\sqrt\sigma}((1+\sigma)E(m')-K(m')),0,2K(m')\right)$.

答:
I tried a different tack. Let $x=\cos\theta$ so that $x\in[-1,1]$ and $e^{i\theta}=x+i\sqrt{1-x^2}$, then after explicitly determining the real and imaginary parts of the expression under the square root in the formula for $y$ I got the following equivalent representation:
$$2y^2(x)=(1-\sigma)L+\sqrt{2L}\qquad L=\frac{1-x^2}{(x+1)\sigma^2-x+1}$$
Since I only care about finding matching angles that give the same result under $y$, I don't need the square and doubling on the LHS, so from now on define $y(x)=(1-\sigma)L+\sqrt{2L}$. This $y$ is easily inverted, being the composition of two quadratic expressions. Suppose $k=y(x)$ is known, then
$$L=\frac{k(1-\sigma)+1\pm\sqrt{2k(1-\sigma)+1}}{(1-\sigma)^2}$$
$$x=\frac12\left(L(1-\sigma^2)\pm\sqrt{(L(1+\sigma^2)-2)^2-(2L\sigma)^2}\right)$$
Finally $\cos^{-1}x=\theta$.

The previous version of my code in Malibu had assumed that "for all values of $\sigma$, $y$ is strictly increasing up to a point and strictly decreasing after that". This is actually only true for $\sigma\le3$. What is true for all $\sigma$ is that $y(\cos\theta)$ has a local extremum at
$$\theta_0=\cos^{-1}\frac{1-\sigma}{1+\sigma}\qquad\left[y(\cos\theta_0)=\left(\frac2{1+\sigma}\right)^2\right]$$
and this is a maximum for $\sigma\le3$ but a minimum for $\sigma>3$. In the latter case there are two local maxima, one on each side of $\theta_0$:
$$\theta_\pm=\cos^{-1}\frac{1+\sigma\pm\sqrt{3(\sigma-3)(3\sigma-1)}}{4(1-\sigma)}\qquad\left[y(\cos\theta_\pm)=\frac1{2(\sigma-1)}\right]$$
Thus there can be up to four inverses for a given $k$, which corresponds to the patch boundary caving in at $\sigma>3$ ($\sigma=10$ is pictured):

The following table shows how the sign choices produce the different inverses:
$$\begin{array}{c|cccc}
\text{sign choice for }(L,x)&(-,+)&(+,+)&(+,-)&(-,-)\\
\theta\text{ range if }\sigma\le3&[0,\theta_0]& & &[\theta_0,\pi]\\
\theta\text{ range if }\sigma>3&[0,\theta_-]&[\theta_-,\theta_0]&[\theta_0,\theta_+]&[\theta_+,\pi]\\
\end{array}$$


问:
Difficulties with Modulus - Chinese Remainder Theorem, GCD and the Extended Euclidean Algorithm
I am finding it quite difficult to deconstruct the steps for seemingly simple questions and decide which mathematical process to use. For example

If $x \equiv 2$ (mod $11$) and $ x \equiv 1$ (mod $17$), what is the value of $x$ (mod $187$)?

There are many online resources that patiently walk through questions such as this, but I find that some of the numbers suggested in the steps are either random guesses to get closer to an objective (without saying so) or there is no explanation as to how they were calculated or why they were chosen during a step. Then trying to follows the process onwards after that is unproductive because one is trying to puzzle out unknown step calculations instead of the objective. I'd appreciate any advice please. As in all of mathematics, it's more rewarding to solve it oneself, but sometimes productive learning hours turn to despair.

答:
Saying that x is congruent to 2 modulo 11 means that x= 2+ 11m for some integer m.  Saying that x is congruent to 1 modulo 17 means that x= 1+ 17n for some integer n.

So 2+ 11m= 1+ 17n which is the same as 17n- 11m= 1.

Now 11 divides into 17 once with remaider 6: 17- 11= 6.

6 divides into 11  once with remainder 5:  11- 6= 5.

And, of course, 6- 5= 1.

6- 5= 6- (11- 6)= 2(6)- 11= 1.

2(17- 11)- 11= 2(17)- 3(11)= 1.

So one solution to this is n= 2, m= 3. x= 1+ 17(2)= 35 or x= 2+ 11(3)= 35.   Of course any number of the forM x= 35+ (11)(17)k=  35+ 187k is also a solution. So x= 35 (mod 187).


问:
In how many ways can 3 boys and 4 girls sit alternately in a row if a particular boy and a girl are to sit together?
I figured out the solution for alternate arrangement of boys and girls without restrictions. If we fix the positions of the four girls, the three boys can only occupy the spaces between them. Now, we can permute the girls and the boys in their fixed positions, so the number of total permutations is $4!\times 3!=144$.

With the added restriction that a girl and a boy are to sit together, I realized that there were two cases: the girl could be at an end of the row or she could be in the middle. If she is at an end, there is only one position the boy can take (just next to her), and $3!\times 2!$ ways to permute the other 3 girls and 2 boys. If she is in the middle of the row, there are two positions the boy could take (to her right or left) so the total number of permutations now becomes $2\times 3!\times 2!$.

The total number of permutations is $3!\times 2! + 2\times 2!\times 3! + 2\times 2!\times 3! + 2!\times 3! = 72$.

While solving this problem I got the idea to fix the boy-girl pair and permute the other persons about them, but could not develop it further. Could someone provide an alternative solution, using this or some other idea?

答: First, seat the other five. You can do this in $3!\times 2!=12$ ways. Now there are six places to seat the special pair; and for each position, there is only one way that they can sit to maintain the girl-boy alternation. So the answer is indeed $6\times 12=72$.


问:
Why do we need 2 equations to solve 2 variables, 3 equations to solve 3 variables, etc...?
The question put another way: why does solving a system of two equations with 2 variables give an exact answer?

I understand that using graphing, I will get either 2 parallel lines, which means 'no solutions' or intersecting lines, which means '1 solution'. But I'm wondering why this happens in terms of algebra, not graphing. I.E. I am asking for a proof/illustration.

答:
Assume that we have $n$ linear equation in $n$ variables. We can write these as:

\begin{array}{ccc}
a_{1,1}x_1 + \cdots + a_{1,n}x_n &=& c_1 \\
a_{2,1}x_1 + \cdots + a_{2,n}x_n &=& c_2 \\
 &\vdots &  \\
a_{n,1}x_1 + \cdots + a_{n,n}x_n &=& c_n
\end{array}

All of these equations can be grouped together into matrix notation:

$$\left[\begin{array}{ccc} a_{1,1} & \cdots & a_{1,n}  \\
\vdots & \ddots & \vdots \\
a_{n,1} & \cdots & a_{n,n}\end{array}\right]\left[\begin{array}{c}x_1 \\ \vdots \\ x_n \end{array}\right] = \left[\begin{array}{c}c_1 \\ \vdots \\ c_n \end{array}\right]$$

The $n \times n$ matrix, $A:=[a_{i,j}]$ is our object of interest. If $\det(A) \neq 0$ then $A$ is invertible, and we can multiply through on the left: $A{\bf x} = {\bf c} \iff A^{-1}A{\bf x} = A^{-1}{\bf c} \iff {\bf x} = A^{-1}{\bf c}$. If $\det(A) \neq 0$ then there is a unique solution and that is given by expanding $A^{-1}{\bf c}$.

If $\det(A) = 0$ then there are several possibilities for solutions. It all relates to the rank and the nullity of $A$. These two numbers are related by the so-called Rank-Nullity Theorem.

When $\det(A) \neq 0$, the rank of $A$ is $n$ and the nullity is $0$. That means the dimension of the set of ${\bf x}$ which go to ${\bf c}$ is zero, i.e. a point, and so there is a unique solution. If $\det(A) = 0$ then the rank may be anything from as high as $n-1$ (where the solution set will have dimension one, i.e. be a line), to as little as $0$ (where the solution set will have dimension $n$, i.e. be the whole space).

(To illustrate the last case, we could have $n$ equations in $n$ variables, and they are all $0x_1 + 0x_2 + \cdots + 0x_n = 0$. Any old ${\bf x}$ will solve this.)


答:
See that the sitting pattern must be $GBGBGBG$. We construct a combination as follow:

The individual boy chooses among $3$ choice.

The individual girl chooses among the $2$ adjacent seats next to the one in the first step.

The other $2$ boys permute $2!$ ways.

The other $3$ girls permute $3!$ ways.

By product rule, the total number is $3\times 2\times 2!\times 3! =72$.

Title: In how many ways can 3 boys and 4 girls sit alternately in a row if a particular boy and a girl are to sit together?

Body:
I figured out the solution for alternate arrangement of boys and girls without restrictions. If we fix the positions of the four girls, the three boys can only occupy the spaces between them. Now, we can permute the girls and the boys in their fixed positions, so the number of total permutations is $4!\times 3!=144$.

With the added restriction that a girl and a boy are to sit together, I realized that there were two cases: the girl could be at an end of the row or she could be in the middle. If she is at an end, there is only one position the boy can take (just next to her), and $3!\times 2!$ ways to permute the other 3 girls and 2 boys. If she is in the middle of the row, there are two positions the boy could take (to her right or left) so the total number of permutations now becomes $2\times 3!\times 2!$.

The total number of permutations is $3!\times 2! + 2\times 2!\times 3! + 2\times 2!\times 3! + 2!\times 3! = 72$.

While solving this problem I got the idea to fix the boy-girl pair and permute the other persons about them, but could not develop it further. Could someone provide an alternative solution, using this or some other idea?


问:
How to prove that $a^3 + b^3 \geq a^2b + ab^2$?
Um I am solving problems in Arthur Engels book "Problem Solving Strategies". I was doing a problem from inequalities chapter, and I stumbled across a problem which I managed to condense and simplify into this:--

For $a + b> 0$, $$a^3 + b^3 \geq a^2b + ab^2$$

I have no idea how to begin but this is what I did.

By A.M-G.M,
$$\frac{a^3 + a^3 + b^3}{3} \geq \sqrt[3]{a^3a^3b^3}$$
$$\implies a^3 + a^3 + b^3 \geq 3a^2b$$
...(i)
$$$$
$$\frac{b^3 + b^3 + a^3}{3} \geq \sqrt[3]{b^3b^3a^3}$$
$$\implies b^3 + b^3 + a^3 \geq 3ab^2$$
...(ii)

Now adding (i) and (ii) we have,
$$3a^3 + 3b^3 \geq 3a^2b + 3ab^2$$

Dividing everything by 3 we have,
$$a^3 + b^3 \geq a^2b + ab^2$$
Which is exactly what I wanted, but I have no idea whether this is correct. Please check it for me and please also tell if there are other methods to prove this.

(Also could you please invite I am very new to stackexchange and would like increase my reputation. Please.)

Body:
Hint:

$$a^3+b^3-(a^2b+ab^2)=(a+b)(a^2-ab+b^2)-ab(a+b)=(a+b)(a-b)^2\ge0$$ if $a+b\ge0$


问:
Complement of conditional probability (trivial probability question)
This is a trivial question but I just wanted to check whether:

If $P(A|B) = 0.90$ then $P(A^C|B) = 0.10$

答:
Indeed, we have the complement rule for conditional probability.

The prove proceeds as follow:

$$P(A^c|B)=\frac{P(A^c\cap B)}{P(B)}=\frac{P(B)-P(A\cap B)}{P(B)}=1-\frac{P(A\cap B)}{P(B)}=1-P(A|B).$$


问:
Integrating a sine function that is to an odd power
I've started the chapter in my book where we begin to integrate trig functions, so bear in mind I've only got started and that I do not have a handle on more advanced techniques.

$\eqalign{
  & \int {{{\sin }^3}x} dx  \cr
  &  = \int {\sin x({{\sin }^2}x} )dx  \cr
  &  = \int {\sin x({1 \over 2}}  - {1 \over 2}\cos 2x)dx  \cr
  &  = \int {{1 \over 2}\sin x(1 - \cos 2x)dx}   \cr
  &  = \int {{1 \over 2}\sin x(1 - (2{{\cos }^2}x - 1))dx}   \cr
  &  = \int {{1 \over 2}\sin x(2 - 2{{\cos }^2}x)dx}   \cr
  &  = \int {\sin x - {{\cos }^2}} x\sin xdx  \cr
  & y = {1 \over 3}{\cos ^3}x - \cos x + C \cr} $

I got the right answer but it seems like an awfully long winded way of doing things, have I made things harder than they should be with this method?

答:
Isn't it faster to use subsitution like $\cos(x)=t$?
$$\eqalign{
  & \int {{{\sin }^3}x} dx  \cr
  &  = \int {\sin x({{\sin }^2}x} )dx  \cr
  &  =-\int \sin^2(x)d\cos(x)  \cr
  &  = -\int (1-\cos^2(x)) d\cos(x)   \cr
  &  = -\int (1-t^2) dt \cr }
 $$</p>


答:
You can't apply AM-GM, as we need that $a,b\ge 0$ and not $a+b>0$.

Let's use the $a+b>0$ condition. Dividing by $a+b>0$ gives that it's enough to prove
$a^2-ab+b^2 \ge ab$
$\iff (a-b)^2 \ge 0$

The reason we divide is that we remember that $a^3+b^3=(a+b)(a^2-ab+b^2)$ since $a^n+b^n=(a+b)(a^{n-1}-a^{n-2}b+...+b^{n-1})$ for odd positive $n$.

Title: How to prove that $a^3 + b^3 \geq a^2b + ab^2$?

Body:
Um I am solving problems in Arthur Engels book "Problem Solving Strategies". I was doing a problem from inequalities chapter, and I stumbled across a problem which I managed to condense and simplify into this:--

For $a + b> 0$, $$a^3 + b^3 \geq a^2b + ab^2$$

I have no idea how to begin but this is what I did.

By A.M-G.M,
$$\frac{a^3 + a^3 + b^3}{3} \geq \sqrt[3]{a^3a^3b^3}$$
$$\implies a^3 + a^3 + b^3 \geq 3a^2b$$
...(i)
$$$$
$$\frac{b^3 + b^3 + a^3}{3} \geq \sqrt[3]{b^3b^3a^3}$$
$$\implies b^3 + b^3 + a^3 \geq 3ab^2$$
...(ii)

Now adding (i) and (ii) we have,
$$3a^3 + 3b^3 \geq 3a^2b + 3ab^2$$

Dividing everything by 3 we have,
$$a^3 + b^3 \geq a^2b + ab^2$$
Which is exactly what I wanted, but I have no idea whether this is correct. Please check it for me and please also tell if there are other methods to prove this.

(Also could you please invite I am very new to stackexchange and would like increase my reputation. Please.)


答:
Here is a way of using AM-GM in this case:
$$a^3+b^3 = (a+b)(a^2+b^2-ab) \geqslant (a+b)(2ab-ab) = (a+b)ab = a^2b+ab^2$$


答: Your proof is correct for $a,b>0$. This restriction comes from the condition of AM-GM inequality. You need to include more since the problem asks for the cases when $a+b>0$.


问:
$A_n \cap A_m$ has exactly n + m elements
Find an infinite family $A$ of infinite subsets of $\mathbb{N}$ such that for each $A_n, A_m$ holds that $A_n \cap A_m$ has exactly n + m elements

My idea was to split $\mathbb{N}$ in such a way that it is fulfilled, but so far I have not managed to formalize this idea.

答:
I think you cannot find any such example.

That is because of the following reasons.

1-$A_n \cap A_n$ has $n+n=2n$ elements. Hence every $A_n$ must be a finite set

2-$A_n \cap A_{2n}$ has $n+2n=3n$ elements

3-$3n=|A_n \cap A_{2n} |<|A_n|=2n$ since $A_n \cap A_{2n} \subset A_n  \Rightarrow \Leftarrow $


问:
Union of two measurable sets is measurable
I was reading the proof of the following Lemma:

If $E_1, E_2$ are two Lebesgue measurable sets then their union and intersection are also Lebesgue measurable.

The proof is as follows:

Denote by $m^*(E)$ the outer measure of a set $E$.

We are only going to prove the case of union of two sets, the case of intersection is proven similarly.

We know that a set $E$ is called measurable iff for every $A\subset \mathbb{R}$ it holds $$m^*(A)=m^*(A\cap E)+m^*(A\cap E^c)$$ where $E^c$ is the complement of $E$. In our case this relation holds for the sets $E_1$ and $E_2$.

Furthermore for every $A\subset \mathbb{R}$ $$A = (A\cap E) \cup (A\cap E^c)$$ and using the subadditivity of outer measure we have: $$m^*(A)\leq m^*(A\cap E)+m^*(A\cap E^c)$$

So to prove that a set is measurable we only need to prove the reverse inequality (because if they are both true then equality must hold): $$m^*(A)\geq m^*(A\cap E)+m^*(A\cap E^c)$$

Thus, in our case we want to show: $$m^*(A)\geq m^*(A\cap (E_1\cup E_2))+m^*(A\cap (E_1\cup E_2)^c)$$

We know that $E_1$ is measurable so $$m^*(A)= m^*(A\cap E_1)+m^*(A\cap E_1^c)$$

Now, this is where my trouble begins. The book states: As $E_2$ is also measurable it holds $$m^*(A\cap E_1^c)= m^*(A\cap E_1^c\cap E_2)+m^*(A\cap E_1^c\cap E_2^c)$$

I do not seem to understand where this last relation is coming from. The definition states that the first relation must hold for every subset of $\mathbb{R}$ on the left hand side, but the set $A\cap E_1^c$ is always a subset of $E_1^c$ and cannot thus run through all subsets of $\mathbb{R}$. What am I missing? The proof finishes by manipulating the last two relations and the subadditivity of outer measure.

I have searched on SE, but couldn't find any similar questions or methods of proof. It would mean the world if somebody could give me a little insight. Thanks in advance!

答:
You have already stated that $E_2$ is measurable iff for every $B\subset\mathbb{R}$

$m^*(B)=m^*(B\cap E_2)+m^*(B\cap E_2^c)$.

Set $B:=A\cap E_1^C$.


问:
If $|f(z)| \leq |z|^2+\frac{1}{\sqrt{|z|}}$, show f is quadratic polynomial.
Suppose the function f is analytic in the punctured plane $z!=0$ (it means we excluded the zero) and satisfies the above condition, $|f(z)| \leq |z|^2+\frac{1}{\sqrt{|z|}}$, then show f is quadratic polynomial.

I think that if we multiply $z$ on the both side then we get $g(z)=zf(z)$ goes to 0 when z is going to 0. therefore, g is analytic function. Then what we should do to prove this?

答:
As you already noted $zf(z)$ is analytic and zero at $z=0$, hence $f$ is (after removing the singularity) entire and we have $|f(z)|<C|z|^2$ for $|z|>R$ with suitable $C$ and $R$.
Then $g(z)=\frac{f(z)-f(0)}{z}$ is entire and has the property $|g(z)|<C'|z|$ for $|z|>R$ with suitable $C'$.
Finally, $h(z)=\frac{g(z)-g(0)}{z}$ is entire and $|h(z)|<C''$ for $|z|>R$ with suitable $C''$. But then $h$ is bounded on all of $\mathbb C$, hence constant, $h(z)=h(0)$
We conclude
$$ f(z)=(h(0)z+g(0))z+f(0).$$

Remark: If we only had  the weaker condition $|f(z)|\le |z|^2+\frac1{|z|}$, we could only conclude that $zf(z)$ is a cubic polynomial, so $f$ would be a rational funciton but might have a pole at $0$.


问:
$|\partial (A \cup B)| + |\partial (A \cap B) | \le |\partial A| + | \partial B|$
I'm trying to prove this lemma:

$\begin{align}
&\text{Let }A\text{ in }B\text{ be subsets of }V(\Gamma)\text{ for some graph }\Gamma.\\
&\text{If we define }\partial(A)\text{ to be the set of all edges of graph }\Gamma\\
&\text{with one end in }A\text{ and one not in }A\text{, then}\\
&|\partial(A\cup B)|+|\partial(A\cap B)|\le|\partial A|+|\partial B|.
\end{align}$

It's supposed to be an easy proof, but I don't see it. All I manage to get is that $|\partial(A\cup B)|+|\partial(A\cap B)|\le 2\cdot(|\partial A|+|\partial B|)$.

I also have a hint, but I don't know how to use it.

Hint: the difference between the two sides is twice the number of edges joining $\;A\setminus B\;$ to $\;B\setminus A\;$.

This lemma will help me prove the lemma from this question:
Proving that distinct edge atoms of a graph are vertex-disjoint.
And with that lemma I can prove that every vertex-transitive graph has edge connectivity equal to its valency.

答:
For disjoint subsets $C,D \subseteq V$, let $e(C,D)$ denote the number of edges with one endpoint in $C$ and the other in $D$. Also, for clarity, let $X$ denote $V - (A \cup B)$.

Each edge in $\partial A$ has one endpoint in either $A - B$ or $A \cap B$, and the other endpoint in either $B - A$ or $X$. This gives the following identity: $$|\partial A| = e(A - B, B - A) + e(A - B, X) +  e(A\cap B, B - A) + e(A \cap B, X) .$$

Analogously, we get
$$|\partial B| = e(B - A, A - B) + e(B - A, X) +  e(A\cap B, A - B) +  e(A \cap B, X).$$

Note also that

$$|\partial (A \cup B)| = e(A - B, X) + e(B - A, X) + e (A \cap B, X)$$

and

$$|\partial (A \cap B)| = e (A\cap B, A - B) + e(A \cap B, B - A) + e(A \cap B, X).$$

Now if you can avoid getting lost in this jumble of letters, by adding the first two identities you can verify that

$$|\partial A| + |\partial B| = 2 e(A - B, B - A) + |\partial (A \cup B)| + |\partial (A \cap B)|,$$

which implies the inequality in the title.