问:
For congruence classes, what does $b = g^{–1}ag$ mean?
Under Wikipedia's definition of congruence classes, (which I got to via studying space groups, point groups and crystallography) it says:

In mathematics, especially group theory, two elements $a$ and $b$ of a group are conjugate if there is an element g in the group such that $b = g^{–1}ag$. This is an equivalence relation whose equivalence classes are called conjugacy classes.

That looks like matrix similarity.... How does that translate to group symmetry?

答:
Let me explain conjugacy in the setting of a planar crystallographic group (which is a little simpler than space groups, and the concepts are pretty much identical).

The key idea of conjugacy in this setting is that if two rigid motions in the plane are conjugate then they are symmetries of "the same type", where by "same type" I am referring (somewhat vaguely) to the classification of planar rigid motions according to their geometric properties: rotation, translation, reflection, or glide reflection; and more numerical properties such as rotation angle and translation distance. Let me put some meat on it by going through the classification of planar rigid motions.

For the proof, let $a,b$ be planar rigid motions that are conjugate to each other. It follows that there exists a planar rigid motion $g$ such that $b = g^{-1} a g$.

If $a$ is a rotation around some point then $b$ is also a rotation around some point. Furthermore, the rotation angles of $a$ and $b$ are equal.

Here's the detailed proof.

First, let $p$ be the point around which $a$ rotates. It follows that $a(p)=p$, i.e. the point $p$ is fixed by $a$. From that, it follows that the point $q = g^{-1}(p)$ is fixed by $b$, i.e. $b(q)=q$, because
$$b(q) = g^{-1} a g(q) = g^{-1} a (p) = g^{-1}(p) = q
$$
That example pretty much captures the whole strategy: one keeps using the conjugacy equation over and over again, to prove that the "classification" properties of $a$ and of $b$ are the same.

Let's move on to the rotation angles of $a$ and $b$. Let $\theta$ be the rotation angle of $a$, which means that if $\overrightarrow{pr}$ is any ray based at $p$, then $\theta$ is equal to the angle between the ray $\overrightarrow{pr}$ and the image ray $a(\overrightarrow{pr})$. Since $g^{-1}$ is a rigid motion, it follows that $\theta$ is equal to the angle between the rays $g^{-1}(\overrightarrow{pr})$ and $g^{-1}(a(\overrightarrow{pr}))$. Now let's re-express those last two rays.
\begin{align*}
g^{-1}(\overrightarrow{pr}) &= \overrightarrow{g^{-1}(p) g^{-1}(r)} \\
  &= \overrightarrow{q g^{-1}(r)} \\
  &=  \overrightarrow{qs}
\end{align*}
where we set $s = g^{-1}(r)$.

Also,
\begin{align*}
g^{-1}(a(\overrightarrow{pr})) &= (g^{-1} a)(\overrightarrow{pr}) \\
   &= (b g^{-1})(\overrightarrow{pr}) \\
   &= b \left(\overrightarrow{g^{-1}(p) g^{-1}(r)}\right) \\
   &= b(\overrightarrow{q s})
\end{align*}
Thus we have shown that $\theta$ is also the angle between the the ray $\overrightarrow{qs}$ and its image ray $b(\overrightarrow{qs})$ under $b$, and therefore $b$ also has rotation angle $\theta$.

That's probably enough detailed proof so you can get the idea, but here's the rest of the rigid motion classification, with proof outlines:

If $a$ is a reflection across a line then $b$ is also a reflection across a line.

Proof outline: Let $L$ be the line that $a$ reflects across; then $g^{-1}(L)$ is the line that $b$ reflects across.

If $a$ is a translation, with translation distance $d$, then $b$ is also a translation with translation distance $d$.

Proof outline: For any point $p$, $d$ equals the distance from $x$ to $a(p)$. Letting $q = g^{-1}(p)$, it follows that $d$ equals the distance from $q=b(p)$ to $b(q)=b(g^{-1}(p)=g^{-1}(a(p))=g^{-1}(q)$.

If $a$ is a glide reflector along a line $L$, translating that line along itself by distance $d$, then $b$ is a glide reflector along the line $g^{-1}(L)$, translating the line along itself by distance $d$.


问:
Does $\ell^3$ norm preserving linear transforms exist?
Does there exist a linear transform from $\mathbb{R}^2\rightarrow\mathbb{R}^2$ that preserves the $\ell^3$ norm?

It is immediately obvious that combinations of permutations and change of sign of individual coordinates satisfy it.

I am wondering if there are more linear transforms that preserves the norm.

Thanks in advance.

答:
Given a linear transformation $\,(x,y)\to (X,Y) := (a x+b y,c x+d y)\,$
then the condition for preserving $\,|x|^3+|y|^3\,$ splits
into cases depending on the signs of $\,x\,$ and $\,y.\,$
and also $\,X\,$ and $\,Y.$
For the case $\,x>0,\,y>0,\,X>0,\,Y>0$ we must have
$$ a^3+c^3=b^3+d^3=1, \quad a\,b^2+c\,d^2=a^2 b+c^2 d=0.$$
The two real solutions of this system of equations is
$\,(X,Y) = (x,y)\,$ and $\,(X,Y) = (y,x).$

The other cases are similar and allows to negate $\,X\,$
or $\,Y.$

ContentLicense: CC BY-SA 4.0


问:
If $x,y,z>0$, prove that: $\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\ge \sqrt{2}\sqrt{2-\frac{7xyz}{(x+y)(y+z)(x+z)}}$
If $x,y,z>0$, prove that: $\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\ge \sqrt{2}\sqrt{2-\frac{7xyz}{(x+y)(y+z)(x+z)}}$

The solution goes as follows:

$a=\frac{x}{y+z}$, $b=\frac{y}{z+x}$, $c=\frac{z}{x+y}$. We can see that:

$ab+bc+ac+2abc=1$

It's enough if: $(a+b+c)^2\ge 4-14abc$

$(a+b+c)^2\ge 4(ab+bc+ab+2abc)-14abc$

From Schur it is enough if: $6abc\ge\frac{9abc}{a+b+c}$ which is true from Nesbit.

Could you please provide a more intuitive and easier approach?

答:
if $x,y,z > 0 $, say $a = \frac{x}{y+z} ,  b=\frac{y}{x+z} , c=\frac{z}{x+y} $
then we'll proof that $$(a+b+c)^2 \ge 4 - 14\cdot a \cdot b \cdot c $$
$(a+b+c)^2 = a^2+2.a.b+b^2+2.a.c+2.b.c+c^2 = a^2+b^2+c^2+2(a.b+b.c+a.c) $

as you said $a.b+a.c+b.c+2.a.b.c = 1$

$a^2+b^2+c^2 +2(1-2.a.b.c) \ge 4 - 14.a.b.c $

$ a^2+b^2+c^2 +2 - 4.a.b.c \ge 4 - 14.a.b.c $

$ a^2+b^2+c^2 +10.a.b.c   \ge 2 $

since $x,y,z > 0 $  means that $x$ or $y$ or $z$ are positive numbers, their lowest value is $ 0 + 10^{-n} $ and since our inequality speaks also minimum value, we can get the result by assuming the minimum value of $x,y,z$ there

for the simplest case scenario, this happens when $ x= y= z$ at the lowest level

$ x^2/(z+y)^2+y^2/(z+x)^2+z^2/(y+x)^2 + (10.x.y.z)/((y+x)(z+x)(z+y)) \ge 2 $

so if i say $x=y=z$, the result is $2$ this means that even if $x , y , z$ was $ < 0 $ our inequality would still be greater than 2


问:
Create Gaussian distribution from random number generator output
I have a random number generator with output between -1 and +1 having a uniform distribution.

Is there a simple mathematical formula that would convert this into an estimated Gaussian distribution (around 0)?

答:
Another approach is to use Box–Muller transform. To be precise if $U_1$ and $U_2$ are independent samples from $U[-1,1]$, then

$$
Z_1 = \sqrt{-2\ln{|U_1|}} \cos(2\pi |U_2|), Z_2 = \sqrt{-2\ln{|U_1|}} \sin(2\pi |U_2|)
$$
are independent standard normal random variables. It is slightly easier than using the integral transform in the context of simulating normal distribution (as integral transform uses the normal CDF which is more cumbersome). But remember that integral transform is applicable for any distribution!!


问:
Simple distance question
If a car travels m kilometres in h hours, how much time will it take to travel M kilometres?

(A) M/m hours

(B) mh/M hours

(C) M/h hours

(D) Mh/m hours

(E) Mm/h hours

I honestly do not understand what the question is asking.

I think the formula is m.h = H.M so (m.h)/M is the time it takes.

But the answer is D.

答:
Let's swap the letters for round numbers e.g. if it first travels 10km (for $m$) in say 3 hours, how long would it take to travel 20km (for $M$).

$\frac{20}{10} = 2$ i.e. it has to travel twice as far, so it would take twice as long $2h$, at the same speed, i.e. 6 hours.

Now we put the letters in

$\frac{M}{m}\times h$

it's the same as $D$

Title: Simple distance question


问:
Why are r.v $X$, $Y$ independent given that their joint pdf factors as follows: $f(x, y) = k(x) g(y)$?
I am currently trying to wrap my head around why given the r.v. $X$, $Y$ and their joint pdf $f(x, y) = k(x) g(y)$ we can say that the r.v. are independent.

I can see that the marginal distribution function of X is
$f_X(x) = k(x) \int^{\infty}_{-\infty} g(y) dy$. However, I am not sure if and why we can say that $\int^{\infty}_{-\infty} g(y) dy$ is 1.

答: as f(x,y) is a joint pdf, its integral must add up to one, therefore.... the product of the integral of k and the integral of g must be 1, but from the given assumptions you cannot say that either integral has any specific value, as long as their product is 1. For statistical independence of X and Y, you only need to prove that f_X is independent of the value of y and vice versa, which is implied by f_X(x) = k(x) * any constant.</p>


答:
The value $\int^{\infty}_{-\infty} g(y) \mathrm dy$ is not $1$ in general, as $g$ is not the marginal density in general. But if $\int^{\infty}_{-\infty} g(y) \mathrm dy\neq 1$ simply choose
$$ \tilde{k}(x):=k(x) \int^{\infty}_{-\infty} g(y) \mathrm dy$$
and
$$ \tilde{g}(x):=\frac{g(x)}{\int^{\infty}_{-\infty} g(y) \mathrm dy} .$$
As $\int^{\infty}_{-\infty} \tilde{k}(x) \mathrm dx=\int^{\infty}_{-\infty}k(x) \int^{\infty}_{-\infty} g(y) \mathrm dy\mathrm dx=\int^{\infty}_{-\infty} f(x,y) \mathrm d(x,y)=1$, these are your marginal densities. Note: the equation
$$f(x, y) = \tilde{k}(x) \tilde{g}(y)$$ is still true. And as you correctly showed $f_X(x)=k(x) \int^{\infty}_{-\infty} g(y) \mathrm dy=\tilde{k}(x)$. This brings you to
$$f(x, y) =f_X(x)f_Y(y).$$

Title: Why are r.v $X$, $Y$ independent given that their joint pdf factors as follows: $f(x, y) = k(x) g(y)$?

Body:
I am currently trying to wrap my head around why given the r.v. $X$, $Y$ and their joint pdf $f(x, y) = k(x) g(y)$ we can say that the r.v. are independent.

I can see that the marginal distribution function of X is
$f_X(x) = k(x) \int^{\infty}_{-\infty} g(y) dy$. However, I am not sure if and why we can say that $\int^{\infty}_{-\infty} g(y) dy$ is 1.


问:
Degree vs Valence of a vertex in a graph
To my understanding, "degree" and "valence" mean the same thing: the number of edges incident to a vertex (including multiplicity for loops).

Is there a difference in the contexts in which they are used?

答: Valence also has a meaning in complex dynamics -- the number of K-rays that land on each point of a periodic orbit (found in Milnor's Periodic Orbits, External Rays, and the Mandelbrot Set: An Expository Account).</p>


问:
Algebraic proof of an inequality being given different constraints on 2 variables
Let $x,y \in \mathbb R$ and $$(1+x+x^2)(1+y+y^2)=2x^2y^2-1$$ and assume that $2<x<y$. I want to algebraically show that $xy<16$ but  I cannot get anywhere. I have tried substitution and nothing seems to help. Graphically I know that it is try because I can simply graph and check visually but I want an actual proof if any one can offer any hints or proof thank you.

答:
Here is a proof which, I am conscious of that, is graphical but has the merit to transform the issue  into more natural variables for the problem at hand, i.e., by taking:

$$\begin{cases}S=x+y\\ P=xy\end{cases} \ \ \iff \ \ \begin{cases}x&=&\frac12(S-\sqrt{S^2-4P})\\y&=&\frac12(S+\sqrt{S^2-4P})\end{cases}\tag{1}$$

(think to relationship $t^2-St+P=0$).

Indeed, with this change of variables, the problem is easily shown to be equivalent (due to symmetry between variables $x$ and $y$) to:

$$\text{knowing that} \ \ -P^2+PS+S^2-P+S+2=0 \tag{2}$$

$$\text{show that} \ \  P<16$$

taking into account constraint $x>2$ (see (1)) and the  implied constraint $S^2-4P \ge 0$.

Now, have a look at fig. 1 below with coordinates axes  $(S,P)$:

Fig. 1: Only the region defined by $(S>4,P>4)$ makes sense. We have extended the "scene" in order to better understand the hyperbolic curve given by equ. (2).

Point $(S,P)$ must

belong to hyperbola (in blue) accounting for constraint (2).

be below parabola (P) (in black) with equation $4P=S^2$ and

be above the red line accounting for condition $x>2.$

The limit point is at the intersection of the blue and red curves (small square) with ordinate $S=16$ as desired.

I am convinced that one can deduce from this graphical representation a fully analytic proof.


问:
Proving Conditional Independence in a Bayesian Network
$$\require{enclose}\begin{array}{c}\enclose{circle}{X_2}&&&&\enclose{circle}{X_1}\\&\searrow&&&&\searrow\\&&\enclose{circle}{X_3}&&&&\enclose{circle}
{X_4}\\&&&\searrow&&\swarrow&&\searrow\\&&&&\enclose{circle}{X_5}&&&&\enclose{circle}{X_6}\end{array}$$

bayesian network

we know

$P(X_1,X_2,X_3,X_4,X_5,X_6)=P(X_1)P(X_2)P(X_3|X_2)P(X_4|X_1)P(X_5|X_3,X_4)P(X_6|X_4)$

I'm trying to prove $X_5$ and $X_6$ are conditionally independent

so I want to prove,

$P(X_5,X_6|X_4)=P(X_5|X_4)P(X_6|X_4)$

I've been trying to use my equation of joint distribution and this more general equation:

$P(X_1,X_2,X_3,X_4,X_5,X_6)=P(X_1)P(X_2|X_2)P(X_3|X_1,X_2)P(X_4|X_1,X_2,X_3)P(X_5|X_1,X_2,X_3,X_4)P(X_6|X_1,X_2,X_3,X_4,X_5)$

$X_3$ seems to complicate the calculations.

Also, I've been trying to disprove the independence of E and F, in general.

In other words, disprove

$P(X_5,X_6)=P(X_5)P(X_6)$

I don't know if I need to use Markov's condition or maybe some re-arrangement would help?

$P(X_5|X_3,X_4)P(X_6|X_4)=\frac{P(X_1,X_2,X_3,X_4,X_5,X_6)}{P(X_1)P(X_2)P(X_3|X_2)P(X_4|X_1)}$ ???

$P(X_5|X_1,X_2,X_3,X_4)P(X_6|X_1,X_2,X_3,X_4,X_5)=\frac{P(X_1,X_2,X_3,X_4,X_5,X_6)}{P(X_1)P(X_2|X_2)P(X_3|X_1,X_2)P(X_4|X_1,X_2,X_3)}$ ???

I tried equating the two expressions for $P(X_1,X_2,X_3,X_4,X_5,X_6)$, but that did not seem to help

$P(X_1)P(X_2)P(X_3|X_2)P(X_4|X_1)P(X_5|X_3,X_4)P(X_6|X_4)=P(X_1)P(X_2|X_2)P(X_3|X_1,X_2)P(X_4|X_1,X_2,X_3)P(X_5|X_1,X_2,X_3,X_4)P(X_6|X_1,X_2,X_3,X_4,X_5)$

useful???

答:
Long Story Short

Investigate the subdiagram for those three nodes. $$\require{enclose}\enclose{box}{X_5}\raise{3ex}{\swarrow\raise{3ex}{\enclose{box}{X_4}}\searrow}\enclose{box}{X_6}$$ This encodes the factorisation: $\mathsf P(X_4,X_5,X_6)=\mathsf P(X_4)\mathsf P(X_5\mid X_4)\mathsf P(X_6\mid X_4)$

So applying the definition of conditional probability immediately produces:

$$\mathsf P(X_5,X_6\mid X_4)=\mathsf P(X_5\mid X_4)\mathsf P(X_6\mid X_4)$$

Long Story Long

To prove the above is valid, we investigate the entire diagram:

$$\begin{matrix}\enclose{box}{X_2}&&&&\enclose{box}{X_1}\\&\searrow&&&&\searrow\\&&\enclose{box}{X_3}&&&&\enclose{box}{X_4}\\&&&\searrow&&\swarrow&&\searrow\\&&&&\enclose{box}{X_5}&&&&\enclose{box}{X_6}\end{matrix}$$

Thus applying the Law of Total Probability, factorise from the diagram, and distribute the summation, we get:

$$\small\begin{align}\mathsf P(X_4,X_5,X_6)&=\sum_{x\in\{X_1,X_1^\complement\}}\sum_{y\in\{X_2
,X_2^\complement\}}\sum_{z\in\{X_3,X_3^\complement\}}\mathsf P(x,y,z,X_4,X_5,X_6)\\&=\sum_{x\in\{X_1,X_1^\complement\}}\sum_{y\in\{X_2
,X_2^\complement\}}\sum_{z\in\{X_3,X_3^\complement\}}\mathsf P(x)\mathsf P(y)\mathsf P(z\mid y)\mathsf P(X_4\mid x)\mathsf P(X_5\mid z,X_4)\mathsf P(X_6\mid X_4)\\&=\mathsf P(X_6\mid X_4)\sum_{z\in\{X_3,X_3^\complement\}}\mathsf P(X_5\mid z,X_4)\sum_{y\in\{X_2
,X_2^\complement\}}\mathsf P(z\mid y)\sum_{x\in\{X_1,X_1^\complement\}}\mathsf P(X_4\mid x)\mathsf P(x)\end{align}$$

Then you just apply the Law of Total Probability again to produce: $$\mathsf P(X_4,X_5,X_6)=\mathsf P(X_4)\mathsf P(X_5\mid X_4)\mathsf P(X_6\mid X_4)$$


问:
What does phi component of a cylindrical coordinate signifies?
So I have a point $(r, \phi, z)$ which I can express in the cartesian coordinate system as $(r cos \phi, r sin \phi, z)$. If I would convert the components of the vector (to cylindrical coordinates)
$
\begin{bmatrix}
r cos \phi \\
r sin \phi \\
z
\end{bmatrix}
$
by multiplying with transformation matrix
$
\begin{bmatrix}
cos \phi & sin \phi & 0 \\
-sin \phi & cos \phi & 0 \\
0 & 0 & 1
\end{bmatrix}
$ we get
$
\begin{bmatrix}
r \\
0 \\
z
\end{bmatrix}
$.
My question is why the $\phi$ component of result zero ? What does it ($\phi$ component) signify here in case of a point on a surface of cylinder ? I thought about this, and I think since the cylindrical coordinate system was made up by a radius and the rotation of point, direction of $\phi$ doesn't have any sense here hence it is zero here. Is this correct ? If so, is there any mathematical definition for this ?

Edit: I posted this question because I was trying to prove divergence theorem on a closed cylindrical surface ($\pi$/4 < $\phi$ < $\pi$/2; 0 < z < 1/2; $r=2$), I took a differential area on the side rectangles $drdz$, but could not figure direction of area in terms of vector component. Should it be $\hat{\phi}$ ($-\hat{\phi}$ for $\pi/4$) or 0 as we found for transformation matrix ? My intuition says it would be $\hat{\phi}$ but I can't explain the 0.

答: In 3D space we may draw a line through the origin, and consider every 2D plane which contains that line. Every point $P$ in the space which lies away from the line belongs to exactly one such plane. To describe $P$ is cylindrical coordinates, we first give its 2D rectangular coordinates $r$ and $z$ in the plane which contains it. Then we specify the plane using an angle $\theta$.


问:
Why does $ x\mathcal{R}y \iff |x| >|y| \text{ or } (|x|=|y| \text{ and } x>y) $ not allow a utility function representation?
Suppose we have the preference relation $\mathcal{R}$ defined on the interval $[-1,1]$ by

$$ x\mathcal{R}y \iff |x| >|y| \text{ or } (|x|=|y| \text{ and } x>y) $$

How do you show that this relation does not allow a utility function representation?

Where a utility function representation should satisfy

$$x\mathcal{R}y \iff u(x)\geq u(y). $$

答: A preference relation that allows a utility function has to be reflexive as $u(x) \ge u(x)$ for all $x \in [-1,1]$. However $1 \mathcal R 1$ doesn't hold.


问:
Proofs in coordinate geometry
Given a parallelogram RSTU in a coordinate plane, where:

R = (b,c)

S = (a + b, c)

T = (a, 0)

U = (0,0)

the statement
$ \sqrt{(b - 0)^2 + (c - 0)^2} = \sqrt{(b +a - a)^2 + (c-0)^2}$
would be used to prove...what?

I thought it would be used to prove that opposite sides were congruent, since when not simplified they both represent the $x\_coord + y\_coord$ of one point minus the $x\_coord + y\_coord$ of the point underneath it.

Is this correct?

答:
The formula they are using is euclidean distance.It is used to find out distance between two points.$A(x_1,y_1)$ and $B(x_2,y_2)$ are points then distance between these two points are

$AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

and you are  correct in your last statement.

So in this question since this is a  parallelogram there are RS=UT and UR=TS and they apply euclidean formula on UR=TS.
this will helpful
http://planetmath.org/euclideandistance

ContentLicense: CC BY-SA 3.0

Title: Proofs in coordinate geometry


问:
Prove that if $n^2$ is divided by 3, then also $n$ can also be divided by 3.

$n\in \Bbb N$

Prove that if $n^2$ is divided by 3, then also n can also be divided by 3.

I started solving this by induction, but I'm not sure that I'm going in the right direction, any suggestions?

答:
Proof by induction: assume $n>3$ and that it is true for $k<n$.

If we have $3|n^2$ then $3$ divides each term of $n^2 - 6n +9 = (n-3)^2$. So we see that  $3| (n-3)^2$ and by the induction hypothesis $3 | (n-3)$ which implies $3 | n$

The statement is vacuously true for $n=1,2$ and obviously true for $n=3$.


问:
Time, Speed and Distance : Trains Partial Distance
Two trains $A$ and $B$ start from station $X$ and $Y$ towards each other. $B$ leaves station $Y$ half an hour after train $A$ leaves station $X$. Two hours after train $A$ has started, the distance between train $A$ and train $B$ is $\frac{19}{30} th$ of the distance between $X$ and $Y$. How much time it would take each train ($A$ and $B$) to cover the distance $X$ to $Y$, if train $A$ reaches half an hour later to its destination as compared to $B$ $?$

My solution approach :-

Let the distance between $X$ and $Y$ be $x$.

Let the speed of train $A$ be $a$ kmph and of train $B$ be $b$ kmph.

As per question $2a + 1.5b = \frac{11x}{30}$ --Eq.(i) (Distance travelled by them i.e. Total distance $-$ Distance left between them $= x-\frac{19x}{30}$

Now we know that train $A$ reaches half an hour later to its destination as compared to $B$, so:-

$x/b + 0.5 = x/a$ --Eq.(ii)

I am stuck here as you can see that I have got three variables and just two equations I can form from the question. What am I missing here? Please help!

答:
As noted in a comment, your second equation is incorrect.  B started half an hour earlier than A, and arrived half an hour sooner, so took one hour less to cover the distance. We have two equations:

$$
\begin{align}\frac{11}{30}x &= 2a+1.5b\\
\frac xa &= 1+\frac xb
\end{align}$$
The point you have missed is that we are not asked to find $a,b,$ and $x$ but $\frac xa$ and $\frac xb$.  If we write $y=\frac xa,\ z=\frac xb$ then the equations become
$$\begin{align}
\frac{11}{30}&=\frac2y+\frac{1.5}{z}\\
y&=1+z
\end{align}$$
Substituting the second in the first, clearing denominators and simplifying gives $$11z^2-94z-45=0$$ whose only positive root is $z=9$.


问:
Proving that $T_t := S_t -\left| x \right| -\frac {n-1}{2} \int _0 ^t \frac {1}{S_u}~du$ is a brownian motion
Consider $B=(B_t)_{t\geq 0}$ $\mathcal F_t$ - brownian motion in $\mathbb R ^n, \  (n\geq 2)$ starting at zero, in a probability space $(\Omega, \mathcal F, (\mathcal F_t)_{t\geq 0}, \mathbb P)$. Then, consider $X_t ^x = x + B_t $ and  $S_t = \left|X^x_t\right|$.
I was trying to show that $(T_t)_{t\geq 0}$ defined by
$$T_t := S_t -\left| x \right| -\frac {n-1}{2} \int _0 ^t \frac {1}{S_u}~du$$

is a real $\mathcal F_t$ - brownian motion.

It's not difficult to see that $W_t = S_t -\left| x \right|$ is a real $\mathcal F_t$ - brownian motion.

Inded, Ito's lemma implies that

$$ W_t = \int _0 ^t \frac{X^x_t}{\left| X^x_t \right| }~ dB_s $$
wich is well defined since $\mathbb P (\exists t>0 :  S_t =\left| X^x_t \right| =0) =0$. Then by Pauls-Lévy Theorem we have that $(W_t)$ is a real $\mathcal F_t$ - brownian motion.

So, $$T_t = W_t  -\frac {n-1}{2} \int _0 ^t \frac {1}{S_u}~du.$$

Have anybody a smarter idea that compute its mean and covariance in order to apply again Paul-Lévy theorem ? If not, any smart advices to simplify the calculation?

Thank's in advance.

Edit: Maybe this other result could helpful to crack this proof.

I could show that $\mathbb E \left\{ \int _0 ^t \frac {1}{S_u}~du\right\} =0$  since $M_t := \int _0 ^t \frac {1}{S_u}~du$ is a bounded local martingale.
Then for the covariation we have
$$ \mathbb E \left \{T_s T_t  \right\}=t \wedge s -c (\mathbb E \left \{W_s M_t  \right\} +\mathbb E \left \{M_s W_t  \right\}) + \mathbb E \left \{M_s M_t  \right\}$$

New Edit:
Could someone please check the question at the end of my solution try at the answer ?

答:
I'd like to show that $M^\lambda = (M_t^\lambda)_{t\geq0}$
$$ M_t^\lambda := \exp\left(i\lambda T_t-\frac{\lambda^2}{2}t \right) $$
 is a complex martingale so $T$ is a brownian motion.

Indeed, by Ito's lemma we have that

$$dM_t ^\lambda = \frac{\lambda^2}{2}M_t ^\lambda dt + i\lambda M_t ^\lambda dT_t - \frac{\lambda^2}{2} d\langle T \rangle_t$$

but also we have that

$$ \langle T \rangle_t = \langle W \rangle_t + \langle \int _0 ^ . \frac{1}{S_u}~du \rangle_t = t$$

so $$dM_t ^\lambda = i\lambda M_t ^\lambda dT_t$$
and  $$\mathbb E \left\{  \int _0 ^t  \lambda^2 (M_t^ \lambda)^2 d\langle T \rangle_t\right\}\leq \exp(\lambda^2 t)< +\infty$$
However, we must remember $T$ is a local martingale. Then even if $\phi_t := i\lambda M_t ^\lambda \in \mathbb H ^2(T)$ can we conclude that $M^\lambda$ is a martingale ?


问:
Multilinear Forms as a Vector Space
if we want to prove that the collection of all multilinear forms is a vector space over $F$, I am having some trouble wrapping my head around some fundamental concepts.

By definition, for some $f:V^k\to F$ which is multilinear, we have $c\cdot f(v_1,...,v_i,...,v_k)=f(v_1,...,c\cdot v_i,...,v_k)$ and $f(v_1,...,v_i,...,v_k)+f(v_1,...,v_i',...,v_k)=f(v_1,...,v_i+v_i',...,v_k)$ - from the link above.

From the first property doesn't this imply that $c\cdot f(v_1,...,v_i,...,v_k) = f(c\cdot v_1,...,v_i,...,v_k)=...=f(v_1,...,c\cdot v_i,...,v_k)=...=f(v_1,...,v_i,...,c\cdot v_k)$. There is no uniqueness to which specific $v_i$ the scalar is applied to?

To prove the collection of all such $f$, $A=\{f\}$ is a vector space, we need to prove that it is closed for addition and scalar multiplication, so for $f,f'\in A$, $(f+f')(v_1,...,v_k)$ is defined as $f(v_1,...,v_k)+f'(v_1,...,v_k)$. How exactly am I supposed to show that the sum of these two is a multilinear form?

My attempt to show that this is a vector space is for some $c\in F$, $c\cdot(f+f')(v_1,...,v_k)=(f+f')(v_1,...,c\cdot v_i,...,v_k)=f(v_1,...,cv_i,...,v_k)+f'(v_1,...,cv_i,...,v_k)=cf(v_1,...,v_k)+cf'(v_1,...,v_k)$.

This would be what's needed to show that it is a vector space but isn't this wrong because it assumes the definition of the sum of two multilinear functions is multilinear?

Any help is appreciated, thank you.

Best,
Adam

答:
The answer to your first question is: yes, the $c$ can be pulled into any of the arguments, since multilinear means that the map is linear in every argument.

For the seconds question, you indeed have to check that for multilinear $f,g\colon V^n\to F$ and some $c\in F$ both $f+g$ and $c\cdot f$ are multilinear again. We can just show that in general $c\cdot f+g$ is multilinear which yields the other two by choosing either $c=1$ or $g=0$. Note that for checking linearity we can use the same trick, when $f(\lambda\cdot v+w)=\lambda\cdot f(v)+f(w)$, we have additivity by choosing $\lambda=0$ and homogeneity by choosing $w=0$.

By definition of $c\cdot f+g$ we have
\begin{align*}
&(c\cdot f+g)(v_1,\dots,\lambda\cdot v_i+v_i',\dots,v_n) \\&\quad= c\cdot f(v_1,\dots,\lambda\cdot v_i+v_i',\dots,v_n) + g(v_1,\dots,\lambda\cdot v_i+v_i',\dots,v_n) \\
\end{align*}
Now both $f$ and $g$ are multilinear, hence this is equal to
\begin{align*}
&= c\cdot\left(\lambda\cdot f(v_1,\dots,v_i,\dots,v_n) + f(v_1,\dots,v_i',\dots,v_n) \right)\\&\quad{}+ \lambda\cdot g(v_1,\dots,v_i,\dots,v_n) + g(v_1,\dots,v_i',\dots,v_n) \\
&= \lambda\cdot \left(c\cdot f(v_1,\dots,v_i,\dots,v_n) + g(v_1,\dots,v_i,\dots,v_n)\right) \\&\quad{}+ c\cdot f(v_1,\dots,v_i',\dots,v_n) + g(v_1,\dots,v_i',\dots,v_n)
\end{align*}
Using the definition of $c\cdot f+g$ again, we obtain
\begin{align*}
&= \lambda\cdot (c\cdot f+g)(v_1,\dots,v_i,\dots,v_n) + (c\cdot f+g)(v_1,\dots,v_i',\dots,v_n).
\end{align*}
Hence, we have shown that $c\cdot f+g$ is linear in the $i$th argument. Since $i$ was arbitrary this shows that $c\cdot f+g$ is indeed multilinear.

Title: Multilinear Forms as a Vector Space


问:
Is this proof by mathematical induction correct?
For any natural numbers $x,y,p$

$x<y$ if and only if $x+p<y+p$. It is asked to prove by induction.

My attempt:
Base case: Let $p=1$. Then $1+x < y+1$.

Inductive step: suppose $x<y$

Then, $S(x)<S(y)$

Adding $p$ to both side of inequality,

$S(x)+p < S(y)+p$

Thanks in advance

答:
ok! so here is another try. I am new to induction so please bear with me! Thanks.

Base case: for p=1

x+1<y+1

this implies x<y.

inductive step: Assume it is true for p.

then x+p<y+p this implies S(x+p)<S(y+p)

this implies (x+p)+1<(y+p)+1

this implies x+(p+1)< y+(p+1)

Is this correct? Thanks for your time!


问:
Determine if set is closed,open, bounded or compact
$S= \{ (x,y)\in \mathbb{R^2}:  x^2  + y^2 = \frac{1}{n^2}, n \in \mathbb{N}$

Help me determine wether the set is closed,open, bounded or compact.

Determining if the set is bounded is trivial, but I am having some trouble determining if the set is closed, open or any of them. I'm pretty sure that the point $(0,0)$ can cause some troubles.

Thanks in advance

答:
The set $S$ is not closed as the origin is a limit point of $S$ that doesn't belong to $S$.

$S$ is not open as you can't find an open ball centered on $(1,0) \in S$ that is included in $S$.


答:
To build on the example given above, the goal is to show that:

$f_{X,Y}(x,y) = k(x)g(y) \implies f_{X,Y}(x,y) =  f_X(x)f_Y(y)$

Start with this equality:

$\int\int f(x,y) dydx= 1$

If $f$ factorizes cleanly into $k(x)$ and $g(y)$, we can write:

$\int \int k(x) g(y) dydx = 1$

$\int k(x)dx \int g(y) dy = 1$

Assuming $\int k(x)dx \neq 0$, we can write:

$\int g(y) dy = \frac{1}{\int k(x)dx }$

In order to get the PDF of $Y$ on the left hand side, multiply both sides by $k(x)$

$k(x) \int g(y) dy = \frac{k(x)}{\int k(x)dx }$

$f_Y(y) = \frac{k(x)}{\int k(x)dx }$

In order to get the PDF of $X$ on the right hand side, multiply by $1$ written sneakily as $\frac{g(y)}{g(y)}$

$f_Y(y) = \frac{k(x)g(y)}{g(y)\int k(x)dx }$

Notice that the numerator is the original joint distribution. So we can write:

$f_Y(y) = \frac{f(x,y)}{f_X(x) }$

And therefore

$f_Y(y)f_X(x) = f(x,y)$

$\square$


问:
Linear Programming Cutting Problem
I'm doing some self studying and I have come across a problem I don't quite understand. The problem only gives me the single constraint:

The first question asks me to show that the given constraint implies the following cuts:

The second questions requires me to come up with a general rule for making cuts like the second cut from the given constraint:

The last question asks me how would I use this constraint to derive such cuts and wants me to convert it to the form in the second question:

I can see the constraint requiring less than four or five "1" values at minimum but I'm not quite sure how I would go about showing it. I haven't been able to find a question of similar nature anywhere else so I turn to people here hoping they might understand what I don't.

答:
Based on comments exchanged with OP.

If it is not the case that $(y_1 + \cdots + y_6) < 5$

then $(y_1 + \cdots + y_6) \geq 6.$

This implies that $9 \times (y_1 + \cdots + y_6) \geq 54.$

However, it is given that

$9 \times (y_1 + \cdots + y_6) \leq
(9)y_1 + (10)y_2 + (11)y_3 + (11)y_4 + (13)y_5 + 16(y_6) = 52 < 54.$

Thus, the original constraint implies that
$9 \times (y_1 + \cdots + y_6) \leq 52.$

Therefore, since the assumption that

it is not the case that $(y_1 + \cdots + y_6) < 5$

led to the conclusion that

$9 \times (y_1 + \cdots + y_6) \geq 54$,

the assumption that it is not the case that $(y_1 + \cdots + y_6) < 5$

has caused a contradiction.

Therefore, the assumption must be false.

The 2nd part is reasoned in the following way.

If $y_2 + \cdots + y_6 \geq 5$, then

$y_2 + \cdots + y_6 = 5 \implies$

$10(y_2 + \cdots + y_6) = 50$.

From the original constraint, this implies that $y_1 = 0.$

Further, from the original constraint, since each variable must be in $\{0,1\}$, it is immediate that $y_5 = 0 = y_6$ (else the original constraint would have to be violated, since the sum would have to be at least $53$).

However, with $0 = y_1, y_5, y_6$, you can't have
$y_2 + y_3 + y_4 = 5$, since the max value of each variable is $1$.

Thus, the assumption that $y_2 + \cdots + y_6 = 5$ has led to a contradiction.

Addendum

Responding to the question/comment of Songaro.

Assume that the following constraints are satisfied:

$$9y_1 + 10y_2 + 11y_3 + 11y_4 + 13y_5 + 16y_6
\leq 52.\tag1$$

$$ y_2 + y_3 + y_4 + y_5 + y_6 = 5. \tag2 $$

$$ y_1, y_2, \cdots, y_6 ~~\text{are each in}~~
\{0,1\}. \tag3$$

From (2) above, you know that
$$ 10y_2 + 10y_3 + 10y_4 + 10y_5 + 10y_6 = 50. \tag4$$

Subtracting (4) from (1) gives
$$9y_1 + y_3 + y_4 + 3y_5 + 6y_6 \leq 2.\tag5$$

Jointly considering (3) and (5) indicates that

If any of $y_1, y_5,$ or $y_6$ are non-zero, (5) above will have to be violated.

Alternatively, assuming that $y_1, y_5, y_6$ are all zero indicates that (2) and (3) above can not jointly be satisfied.

Thus, (2) above has caused a contradiction.

Title: Linear Programming Cutting Problem


答:
You can simplify the working. Say, time taken by $B$ to cover distance $d$ between stations $X$ and $Y$ is $t$ hours. Then time taken by $A$ is $(t+1)$ hours (as $A$ starts $30$ mins earlier and reaches $30$ mins later) and speed of train $A$ is $\displaystyle \frac{d}{t+1}$ and of train $B$ is $\displaystyle \frac{d}{t}$.

So, $\displaystyle \frac{2d}{t+1} + \frac{1.5 d}{t} = \frac{11d}{30}$

Take out $d$ from both sides and solve for $t$ which comes to $9$ hours. That is time taken by train $B$. So time taken by $A$ is $10$ hours.

Note: While the question most likely meant that they have not crossed each other but it should have been more explicit. They can be at a distance of $\frac{19d}{30}$ even after having crossed each other, which is represented by the equation $\displaystyle \frac{2d}{t+1} + \frac{1.5 d}{t} = \frac{49d}{30}$ and it does have a valid solution.

Title: Time, Speed and Distance : Trains Partial Distance

Body:
Two trains $A$ and $B$ start from station $X$ and $Y$ towards each other. $B$ leaves station $Y$ half an hour after train $A$ leaves station $X$. Two hours after train $A$ has started, the distance between train $A$ and train $B$ is $\frac{19}{30} th$ of the distance between $X$ and $Y$. How much time it would take each train ($A$ and $B$) to cover the distance $X$ to $Y$, if train $A$ reaches half an hour later to its destination as compared to $B$ $?$

My solution approach :-

Let the distance between $X$ and $Y$ be $x$.

Let the speed of train $A$ be $a$ kmph and of train $B$ be $b$ kmph.

As per question $2a + 1.5b = \frac{11x}{30}$ --Eq.(i) (Distance travelled by them i.e. Total distance $-$ Distance left between them $= x-\frac{19x}{30}$

Now we know that train $A$ reaches half an hour later to its destination as compared to $B$, so:-

$x/b + 0.5 = x/a$ --Eq.(ii)

I am stuck here as you can see that I have got three variables and just two equations I can form from the question. What am I missing here? Please help!


问:
How to integrate $\int\frac{\sqrt{1-x}}{\sqrt{x}}\ \mathrm dx$
I'm having a bit of trouble solving this integral:  $$\int\frac{\sqrt{1-x}}{\sqrt{x}}dx$$

Here is my attempt at a solution:

I multiplied the numerator and the denominator of $\frac{\sqrt{1-x}}{\sqrt{x}}$ by $\sqrt{x}$, yielding $$\int\frac{\sqrt{x-x^2}}{x}dx.$$
Further simplification resulted in $$\int\frac{\sqrt{\frac{1}{4}-\left(x-\frac{1}{2}\right)^2}}{x}dx.$$
Using trigonometric substitution, I set $$x-\frac{1}{2}=\frac{1}{2}\sin\theta$$ and solving for the differential $dx$ got $$dx=\frac{1}{2}\cos\theta.$$ Substituting this all back into $\int\frac{\sqrt{\frac{1}{4}-(x-\frac{1}{2})^2}}{x}dx$ (and some simplification later) yielded $$\frac{1}{2}\int{\frac{\cos^2\theta}{\sin\theta+1}}d\theta.$$ By substituting $1-\sin^2\theta$ for $\cos^2\theta$ I obtained $$\frac{1}{2}\int{\frac{1}{\sin\theta+1}-\frac{\sin^2\theta}{\sin\theta+1}d\theta}.$$ The issue I'm having is trying to solve this resultant integral. If there is an easier method to solve the problem, that would be graciously accepted.

答:
Integrate $\int \frac{\sqrt{1-x}}{\sqrt x} dx$

let
$$
\begin{equation} \tag 1
x=t^2, \,\,\,\,\, dx=2t\, dt
\end{equation}
$$
now; $\int \frac{\sqrt{1-t^2}}{t} dt$
$$
\begin{equation} \tag{as $dt=\frac{dx}{2t}$}
\int 2 \sqrt{1 - t^2} dt
\end{equation}
$$
$$
\begin{equation} \tag 2
2 \int \sqrt{1 - t^2} dt
\end{equation}
$$
$$
\begin{equation} \tag 3
2 \left[\frac{t}{2}\sqrt{1-t^2}+\frac{1}{2}\sin^{-1} (t) \right]
\end{equation}
$$

put $t=\sqrt{x}$, to get answer.


问:
Summing $\sum_{k=1}^n \binom{2n-2k}{n-k}\frac{H_{2k} - 2H_{k}}{2n-2k-1} \binom{2k}{k}$
The sum in the question has a nice closed-form:
$$
\sum_{k = 1}^{n}\binom{2n - 2k}{n - k}
\frac{H_{2k} - 2H_{k}}{2n-2k-1}\binom{2k}{k} =
\frac{1}{n}\left[4^{n} - 3\binom{2n-1}{n}\right],$$
where $H_k$ is the harmonic number $H_n = \sum_{k=1}^n 1/k.$

My proof is excruciating:

It involved the specialization of a two-variable generating function ( $x$ and $y$ take different functions of $t$ ) so that I could use an identity that seems to have no mere
mortal-constructed proof ( a variant of the WZ method was used, I believe ).

In the process of the proof, I noticed that the expression in big parentheses on the RHS is a sequence of integers, and I searched the OEIS to find $A213119$.

As logarithms are involved, it was a matter of talking the arguments into the logarithm to find that my generating function was indeed equal to implied by the right hand side of the equation.

I seek a simpler, more direct proof.

A contributor wanted to know where the identity originated.  It originated as an attempt to prove the very pretty

$$
\sum_{k = 1}^{n}\binom{n}{k}^{2}H_{k} =
\binom{2n}{n}\left(\,{2H_{n} - H_{2n}}\,\right)
$$

It may be impossible not to use this in any attempted proof, but hopefully, the specialization of a two-variable generating function can be avoided.

答:
With the following I  present a proof to show that it  can be done and
perhaps  inspire  additional  efforts  at  further  simplification.  A
combinatorial proof would be quite nice.

We seek to verify the identity

$$\sum_{k=1}^n {2n-2k\choose n-k}
\frac{H_{2k} - 2H_k}{2n-2k-1} {2k\choose k}
= \frac{1}{n} \left[4^n - 3 {2n-1\choose n}\right].$$

Preliminary. We get for the first piece in $H_{2k}$ call it $A$
that

$$\sum_{k=1}^n {2n-2k\choose n-k}
\frac{1}{2n-2k-1} {2k\choose k}
[z^{2k}] \frac{1}{1-z} \log\frac{1}{1-z}
\\ = \sum_{k=0}^{n-1} {2k\choose k}
\frac{1}{2k-1} {2n-2k\choose n-k}
[z^{2n-2k}] \frac{1}{1-z} \log\frac{1}{1-z}$$

We may raise $k$ to $n$ because the function in $z$ has no constant
term:

$$[z^{2n}] \frac{1}{1-z} \log\frac{1}{1-z}
\sum_{k=0}^{n} {2k\choose k}
\frac{1}{2k-1} {2n-2k\choose n-k} z^{2k}$$

Now the coefficient extractor enforces the  upper limit of the sum and
we get  (in fact  expansions start at  $z^{2k+1}$ which  cancels $k=n$
already)

$$[z^{2n}] \frac{1}{1-z} \log\frac{1}{1-z}
\sum_{k\ge 0} {2k\choose k}
\frac{1}{2k-1} {2n-2k\choose n-k} z^{2k}
\\ = - [z^{2n}] \frac{1}{1-z} \log\frac{1}{1-z}
[w^n] \sqrt{1-4wz^2} \frac{1}{\sqrt{1-4w}}.$$

The same method yields for the second piece in $H_k$ call it $B$

$$- [z^{n}] \frac{1}{1-z} \log\frac{1}{1-z}
[w^n] \sqrt{1-4wz} \frac{1}{\sqrt{1-4w}}.$$

First part. Continuing with piece $B$

$$[w^n] \sqrt{1+\frac{4w(1-z)}{1-4w}}
= - [w^n] \sum_{k\ge 0} {2k\choose k} \frac{1}{2k-1}
(-1)^k \frac{w^k (1-z)^k}{(1-4w)^k}
\\ = - \sum_{k=0}^n {2k\choose k} \frac{1}{2k-1}
(-1)^k [w^{n-k}] \frac{(1-z)^k}{(1-4w)^k}
\\ = - 4^n \sum_{k=0}^n {2k\choose k} \frac{1}{2k-1}
(-1)^k (1-z)^k 4^{-k} {n-1\choose k-1}$$

and extracting the coefficient in $[z^n]$

$$4^n \sum_{k=1}^n {2k\choose k} \frac{1}{2k-1}
(-1)^k 4^{-k} {n-1\choose k-1}
[z^n] (1-z)^{k-1} \log\frac{1}{1-z}
\\ = 4^n \sum_{k=1}^n {2k\choose k} \frac{1}{2k-1}
(-1)^k 4^{-k} {n-1\choose k-1}
\sum_{q=0}^{k-1} (-1)^q {k-1\choose q} \frac{1}{n-q}.$$

Now

$${n-1\choose k-1} {k-1\choose q}
= \frac{(n-1)!}{(n-k)!\times q! \times (k-1-q)!}
= {n-1\choose q} {n-1-q\choose k-1-q}$$

Switching the order of the summation,

$$4^n \sum_{q=0}^{n-1} {n-1\choose q}
\frac{(-1)^q}{n-q} \sum_{k=q+1}^n
{n-1-q\choose k-1-q} {2k\choose k}
\frac{1}{2k-1} (-1)^k 4^{-k}
\\ = \frac{4^n}{n} \sum_{q=0}^{n-1} {n\choose q}
(-1)^q \sum_{k=q+1}^n
{n-1-q\choose k-1-q} {2k\choose k}
\frac{1}{2k-1} (-1)^k 4^{-k}
\\ = - \frac{4^n}{n} \sum_{q=0}^{n-1} {n\choose q}
(-1)^q \sum_{k=q+1}^n
{n-1-q\choose k-1-q} [z^k] \sqrt{1+z}.$$

The inner sum is

$$\sum_{k=0}^{n-1-q}
{n-1-q\choose k} [z^{k+q+1}] \sqrt{1+z}
= \sum_{k=0}^{n-1-q}
{n-1-q\choose k} [z^{n-k}] \sqrt{1+z}
\\ = [z^n] \sqrt{1+z}
\sum_{k=0}^{n-1-q} {n-1-q\choose k} z^k
= [z^n] \sqrt{1+z} (1+z)^{n-1-q}.$$

Substitute into the outer sum to get

$$- \frac{4^n}{n} [z^n] \sqrt{1+z}
\sum_{q=0}^{n-1} {n\choose q} (-1)^q (1+z)^{n-1-q}
= - \frac{4^n}{n} [z^n] \frac{1}{\sqrt{1+z}}
(- (-1)^n + z^n)
\\ = - \frac{4^n}{n}
\left(- 4^{-n} {2n\choose n} + 1\right)
= - \frac{4^n}{n} + {2n\choose n} \frac{1}{n}.$$

Second part. Here we may recycle the first segment
from the easy piece $B$ and obtain for piece $A$

$$4^n \sum_{k=1}^n {2k\choose k} \frac{1}{2k-1}
(-1)^k 4^{-k} {n-1\choose k-1}
[z^{2n}] (1-z^2)^{k-1} (1+z) \log\frac{1}{1-z}.$$

The coefficient extractor in $z$ has two parts, the first of which is

$$\sum_{q=0}^{k-1} (-1)^q {k-1\choose q} \frac{1}{2n-2q}$$

which contributes half the value of the piece $B.$
The second is

$$\sum_{q=0}^{k-1} (-1)^q {k-1\choose q} \frac{1}{2n-1-2q}.$$

This yields

$$-4^n [z^n] \sqrt{1+z} \sum_{q=0}^{n-1} {n-1\choose q}
\frac{(-1)^q}{2n-1-2q} (1+z)^{n-1-q}
\\ = -4^n [z^n] \sum_{q=0}^{n-1} {n-1\choose q}
\frac{(-1)^q}{2n-1-2q} (1+z)^{n-1/2-q}
\\ = -4^n  \sum_{q=0}^{n-1} {n-1\choose q}
\frac{(-1)^q}{2n-1-2q} (n-1/2-q)^{\underline{n}}/n!.$$

We have for the falling factorial

$$  \prod_{p=0}^{n-1} (n-1/2-q-p)
= \frac{1}{2^n}
\prod_{p=0}^{n-1} (2n-1-2q-2p)
\\ = \frac{1}{2^n}
\prod_{p=-(n-1)}^0 (1-2q-2p)
= \frac{(-1)^n}{2^n}
\prod_{p=q-(n-1)}^{q} (2p-1)
\\ = \frac{(-1)^{n+1}}{2^n}
\frac{(2q-1)!}{2^{q-1} (q-1)!}
\prod_{p=q-(n-1)}^{-1} (2p-1).$$

With $2q-2(n-1)-1 = 2q-2n+1$ this finally becomes

$$\frac{(-1)^{q}}{2^n}
\frac{(2q-1)!}{2^{q-1} (q-1)!}
\frac{(2n-1-2q)!}{2^{n-1-q} (n-1-q)!}
\\ = \frac{(-1)^{q}}{2^{2n-1}}
\frac{(2q)!}{q!}
\frac{(2n-1-2q)!}{(n-1-q)!}.$$

This was for $1\le q\le n-1.$ We get for $q=0$

$$\frac{1}{2^n} \prod_{p=-(n-1)}^0 (1-2p)
= \frac{1}{2^n} \frac{(2n-1)!}{2^{n-1} (n-1)!}$$

and we  see that the generic  term in four factorials  represents this
case correctly as well.

Returning to the sum we obtain

$$-\frac{2}{n}
\sum_{q=0}^{n-1} {2q\choose q}
{2n-2-2q\choose n-1-q}
\\ = -\frac{2}{n} [z^{n-1}]
\frac{1}{\sqrt{1-4z}}
\frac{1}{\sqrt{1-4z}}
= -\frac{2}{n} [z^{n-1}] \frac{1}{1-4z}
= -\frac{2}{n} 4^{n-1} = -\frac{1}{2} \frac{4^n}{n}.$$

Conclusion. We now collect the three pieces
with $A$ first then $B:$

$$-\frac{1}{2} \frac{4^n}{n}
- \frac{1}{2} \frac{4^n}{n}
+ \frac{1}{2} \frac{1}{n} {2n\choose n}
\\ + 2 \frac{4^n}{n} - 2 \frac{1}{n} {2n\choose n}
= \frac{4^n}{n} - \frac{3}{2} \frac{1}{n} {2n\choose n}
= \frac{4^n}{n} - 3 \frac{1}{n} {2n-1\choose n-1}.$$

This is indeed

$$\bbox[5px,border:2px solid #00A000]{
\frac{1}{n}
\left[4^n - 3 {2n-1\choose n}\right].}$$


问:
How to prove that $P\mathcal H \perp Q \mathcal H$ if $P,Q$ and $P + Q$ are all projections?

Let $\mathcal H$ be a Hilbert space and $P,Q \in B(\mathcal H)$ are projections. Then $(P+Q)$ is a projection iff $P \mathcal H \perp Q \mathcal H.$

"$\impliedby$" part is easy. How do I prove "$\implies$" part?

Let $(P + Q)$ be an orthogonal projection. Then $$P + Q = (P + Q)^2 = P^2 + PQ + QP + Q^2 = P + Q + PQ + QP.$$

So $PQ + QP = 0.$ Let $x,y \in \mathcal H.$ Then $$\left \langle Px, Qy \right \rangle = \left \langle QPx, y \right \rangle = \left \langle -PQx, y \right \rangle = -\left \langle Qx, Py \right \rangle$$ From here how do I conclude that $\left \langle Px, Qy \right \rangle = 0\ $? Any help in this regard will be warmly appreciated.

Thanks for your time.

答:
If you multiply $PQ+QP=0$ from the left with $(1-P)$, you get $(1-P)QP=0$. If you multiply the same equation with $P$ from the left and the right, you get $2PQP=0$. Thus $QP=(1-P)QP+PQP=0$, which implies
$$
\langle Px,Qy\rangle=\langle QP x,y\rangle=0
$$
for all $x,y\in\mathcal{H}$.


问:
Statement about subspaces. True or false?
If H is a subspace of $R^n$, then there is a n×n matrix A such that H = Col A.

Is this true or false, and why? To be honest I have been looking at this statement for days, but I don't understand it or find anything in my that book explains me something about this.

答: Choose a basis of $H$, use those vectors as columns and add a bunch of zero-columns to get a $n\times n$-matrix.


问:
Find the number of bijective function $g(n):\mathbb{N}\to \mathbb{N}$

Find the number of bijective function  $g(n):\mathbb{N}\to \mathbb{N}$ such that it's satisfies $\sum_{n=1}^{\infty} \frac{g(n)}{n^2}<\infty$

I think  there is no such bijective function exists , suppose $g(n)=n$ then $\frac{g(n)}{n^2}=\frac{n}{n^2}=\frac{1}{n}$ , but $\sum_{n=1}^{\infty}\frac{1}{n} $ is  diverges .
Now again $\sum_{n=1}^{\infty}\frac{1}{n} $ is converges if we omit the term $n$ whose last entries is $9$. But then don't understand how to construct such bijective function .(source:: https://math.stackexchange.com/a/1343048/746904)

This question is came in TIFR GS-2021 .

答:
Abel summation  can be used here.

Let $G(n)$ denote the sum $g(1) + \dots + g(n)$. We then have, for any fixed $M$, $$\sum_{n = 1}^M \frac{g(n)}{n^2} > \sum_{i = 1}^{M - 1} G(i)\left(\frac 1{i^2} - \frac 1 {(i + 1)^2}\right).$$

Using the fact that $g$ is a bijection from $\Bbb N$ to $\Bbb N$, we know that $G(n) \geq 1 + 2 + \dots + n = \frac{n(n + 1)}2$ for any $n$. Putting this into the right hand side of the above equality will give you an estimation that diverges to infinity when $M$ tends to infinity.

As can be seen from the proof, in fact we only used the injectivity of $g$.


答:
For 1, you are right that $c$ can multiply any of the $v_i$, and there is no contradiction. This just shows the action of $c\in F$ could be represented in different ways.

For 2, if $f,f'\in A$, by definition
\begin{align}
(f+f')&(v_1, \dots,v_i+v_i',\dots,v_k) = f(v_1,\dots,v_i+v_i',\dots,v_k) + f'(v_1,\dots,v_i+v_i',\dots,v_k) \\
& = f(v_1,\dots,v_i,\dots,v_k) + f(v_1,\dots,v_i',\dots,v_k) \\
& \hspace{0.5cm} + f'(v_1,\dots,v_i,\dots,v_k) + f'(v_1,\dots,v_i',\dots,v_k) \quad\text{by multilinearity of $f,f'$} \\
& = (f+f')(v_1,\dots,v_i,\dots,v_k) + (f+f')(v_1,\dots,v_i',\dots,v_k).
\end{align}

\begin{align}
c\cdot (f+f')(v_1, \dots,v_i,\dots,v_k) & = c\cdot [f(v_1, \dots,v_i,\dots,v_k)+f'(v_1, \dots,v_i,\dots,v_k)] \\
& = c\cdot f(v_1, \dots,v_i,\dots,v_k) +  c\cdot f'(v_1, \dots,v_i,\dots,v_k)\\
& = f(v_1,\dots,cv_i,\dots,v_k) + f'(v_1,\dots,cv_i,\dots,v_k) \\
& = (f+f')(v_1,\dots,cv_i,\dots,v_k)
\end{align}

These two together prove $f+f'$ is a multilinear form and $A$ is closed under addtion. It is very similar to show $A$ is closed under scalar multiplication. It follows the same process of using the definition, expand by multilinearity, and conclude.

Your attempt after 2 shows the action of $F$ is distributive: $c(f+f')=cf+cf'$. This is part of the definition of $F$-action, and I don't think you have to show it.

Title: Multilinear Forms as a Vector Space

Body:
if we want to prove that the collection of all multilinear forms is a vector space over $F$, I am having some trouble wrapping my head around some fundamental concepts.

By definition, for some $f:V^k\to F$ which is multilinear, we have $c\cdot f(v_1,...,v_i,...,v_k)=f(v_1,...,c\cdot v_i,...,v_k)$ and $f(v_1,...,v_i,...,v_k)+f(v_1,...,v_i',...,v_k)=f(v_1,...,v_i+v_i',...,v_k)$ - from the link above.

From the first property doesn't this imply that $c\cdot f(v_1,...,v_i,...,v_k) = f(c\cdot v_1,...,v_i,...,v_k)=...=f(v_1,...,c\cdot v_i,...,v_k)=...=f(v_1,...,v_i,...,c\cdot v_k)$. There is no uniqueness to which specific $v_i$ the scalar is applied to?

To prove the collection of all such $f$, $A=\{f\}$ is a vector space, we need to prove that it is closed for addition and scalar multiplication, so for $f,f'\in A$, $(f+f')(v_1,...,v_k)$ is defined as $f(v_1,...,v_k)+f'(v_1,...,v_k)$. How exactly am I supposed to show that the sum of these two is a multilinear form?

My attempt to show that this is a vector space is for some $c\in F$, $c\cdot(f+f')(v_1,...,v_k)=(f+f')(v_1,...,c\cdot v_i,...,v_k)=f(v_1,...,cv_i,...,v_k)+f'(v_1,...,cv_i,...,v_k)=cf(v_1,...,v_k)+cf'(v_1,...,v_k)$.

This would be what's needed to show that it is a vector space but isn't this wrong because it assumes the definition of the sum of two multilinear functions is multilinear?

Any help is appreciated, thank you.

Best,
Adam


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

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

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

答:
In general you should avoid L'Hospital's Rule unless the differentiation of numerator and denominator is damn simple. If you need product/quotient/chain rule of derivatives then probably you are better off avoiding it.

Same is the case of Taylor series. If you need to multiply/divide/compose some Taylor series then avoid it.

A better approach is to use limit laws combined with standard limits to either evaluate the limit or to simplify it in a form which makes the application of L'Hospital's Rule or Taylor series easy.

Let's observe that $(1+\arctan x) ^{1/x}\to e$ (prove it if you are not convinced). Thus the numerator is of the form $a-b$ where both $a, b$ tend to $e$. Next we note that $$a-b=b\cdot \frac{\exp(\log a - \log b) - 1}{\log a - \log b}\cdot (\log a - \log b) $$ Since the first two factors tend to $e$ and $1$ we can replace numerator by $$e(\log a - \log b) $$ Thus we need to evaluate the limit of $$e\cdot\dfrac{1-\dfrac{\log(1+\arctan x)} {x}} {x} =e\cdot\frac{x-\log(1+\arctan x)} {x^2}$$ Next we can substitute $x=\tan t$ and the above expression is transformed into $$e\cdot \frac{\tan t-\log(1+t)}{t^2}\cdot\frac{t^2}{\tan^2t}$$ Since the last factor tends to $1$ the desired limit is equal to the limit of $$e\left(\frac{\tan t - t} {t^2}+\frac{t-\log(1+t)}{t^2}\right)$$ Using L'Hospital's Rule or Taylor series one deduces that the first fraction tends to $0$ and the second fraction tends to $-1/2$. Thus the desired limit is $-e/2$.


问:
Show that $n^2\log\left(1+\frac{1}{n}\right)$ does not converge to $1$

How to show that $n^2\log\left(1+\dfrac{1}{n}\right)\to 1$ is false?

I have to show that $\left(1+\dfrac{1}{n}\right)^{n^2}$ doesn't tend to $e.$

答:
$n^2\log\left(1+\dfrac{1}{n}\right)$ tend to " $\infty \cdot 0$" as $n \rightarrow \infty$.

$\frac{\log\left(1+\dfrac{1}{n}\right)}{1/n^2}$ tend to "$\frac{0}{0}$" as $n \rightarrow \infty$.

By L'hopital's rule, the limit is equivalent to the limit of,

$\frac{\left( \frac{-1/n^2}{1+\frac{1}{n}} \right)}{\frac{-2}{n^3}} = \left( \frac{n}{2} \right) \left( \frac{1}{1+\frac{1}{n}} \right)$, which is "$\infty \cdot 1$" $= \infty$.


问:
Inverse of $x^x$ using Lambert W function
I am currently looking into the Lambert W function. From my understanding it is defined as:
$$f(x)=xe^x$$
$$W(x)=f^{-1}(x)$$
So in the application of this I am trying to define the inverse of the function $y=x^x$. This is my working:
$$y=x^x$$
$$\ln(y)=x\ln(x)=e^{\ln(x)}\ln(x)$$
and this fits the form $f(u)=ue^u$

Now I can say that:
$$\ln(x)=W\left[\ln(y)\right]$$
$$\therefore x=e^{W\left[\ln(y)\right]}$$

Is this working correct and give the right answer?

答:
Your derivation is correct.

Consider that your $f^{-1}$ is the inverse relation but not an inverse function.

The functions $\mathbb{R}\to\mathbb{R},x\mapsto xe^x$ and $\mathbb{C}\to\mathbb{C},x\mapsto xe^x$ are not injective. In $\mathbb{R}$, $xe^x$ has the minimum $(-1,-\frac{1}{e})$. $W$ is therefore indeed the inverse relation, but not a function and therefore not an inverse function. See e.g. the top figure in Wikipedia: Lambert W function.

We have:

$$y=x^x$$

$$y=e^{x\ln(x)}$$

We cannot simply apply the $\ln$-function to the equation because $\ln(e^z)=z$ is valid only for $-\pi<\text{Im}(z)\le\pi$ (see e.g. [Abramowitz/Stegun 1970] p. 69 4.2.3).

Because the complex function $\exp$ is periodic, it is not injective and its inverse relation is not a function. The inverse relation of $\mathbb{C}\to\mathbb{C},z\mapsto e^z$ is $z\mapsto\text{Ln}=\ln(z)+2k\pi i$ ($\forall k\in\mathbb{Z}$) $\ \ $ (see e.g. [Abramowitz/Stegun 1970] p. 67 4.1.5).

$\text{Ln}(e^z)=z+2k\pi i$ ($\forall k\in\mathbb{Z}$) $\ \ $ (see e.g. [Abramowitz/Stegun 1970] p. 69 4.2.2).

$$y=e^{x\ln(x)}\tag1$$

$$\text{Ln}(y)=\text{Ln}(e^{x\ln(x)})$$

Because both arguments are equal (acc. to equation (1)), we have to choose the same branch of Ln on both sides of the equation:

$$\ln(y)+2k\pi i=x\ln(x)+2k\pi i\ (\forall k\in\mathbb{Z})$$

$$\ln(y)=x\ln(x)$$

$$...$$

$\ $

[Abramowitz/Stegun 1970] Abramowitz, M.; Stegun, I.: Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables. National Bureau of Standard 1970


问:
Proof that the sum of exterior angles of a complex polygon or a non-planar polygon is greater than 2 Pi
In the context of writing a ray-tracing program, I must be able to determine if a polygon in three-dimensional space is a planar simple convex polygon. My conjecture is that if I take the sum of the exterior angles (absolute values, always in the range $[0, \pi]$) then this sum is equal to $2\pi$ if and only if it is a planar simple convex polygon.

I firmly believe this to be true, but I seek somewhat rigorous proof. I already managed to show that the sum equals $2\pi$ for convex polygons and that it exceeds $2\pi$ for concave ones. However, I do not have any ideas on how to continue to prove that complex and non-planar polygons have a sum of exterior angles greater than $2\pi$.

For complex planar polygons, I managed to show (w.l.o.g.) that the exterior angles which "turn to the right" minus those that "turn to the left" must be a $2\pi n$ for some natural $n$. From there it's trivial to show that if $n>1$ then the total must be greater than $2\pi$, but for $n=0$ and $n=1$ I do not know how to proceed.

For non-planar polygons, I thought about what would happen if we project it to 2-dimensions then somehow show that the projection has an exterior angle sum strictly between $2\pi$ and that of the non-planar version, but I once again cannot fathom how to show that. (Edit: this idea does not work since there are counterexamples of non-planar polygons whose projections onto two-dimensional space have a larger exterior angle sum.)

答:
I think your conjecture is true, but that you do not need it to reach your conclusion programmatically in essentially the same time with a loop through consecutive pairs of edges.

In that loop, first check for planarity by checking that all pairs have the same normal (up to a constant factor). You do that by calculating the cross products. Then find the exterior angles. The polygon will be simple and convex if and only if there are no sign changes and they sum to $\pm 2\pi$.

You may have to worry about numerical instability if consecutive edges are nearly collinear.


问:
How to get $2700$ grams of corn from a bag of $6.8$ kilograms?
The problem is as follows:

A merchant has a kiosk and has one bag with $6.8$ kilograms of corn and a two pan scale and three weights. One measures $5$ kilograms, the other $2$ kilograms, and the other $1$ kilogram. In how many weighing trials at minimum could he serve an order of $2700$ grams of corn?

The choices given are:

$4$ trials

$3$ trials

$1$ trial

$2$ trials

I'm not sure exactly how to get to the right combination for this problem. It would be very nice if the answer could include the steps in a logical and very detailed way so I can understand.

To my best of ability I thought that I could arrange the weights as follows:

$5$, $2$, $1$, $6.8$

To get $2.7$ from these numbers is somewhat convoluted to me. Thus I need some help. What can I do?

The thing is either method which I've tried from subtracting this amount from $6.8$ does give me more than what I need, thus I am going in circles.

答:
A merchant has a kiosk and has one bag with 6.8 kilograms of corn and a two pan scale and three weights. One measures 5 kilograms, the other 2 kilograms, and the other 1 kilogram. In how many weighing trials at minimum could he serve an order of 2700 grams of corn?

One answer has already shown how it can be done in two weighing trials, and another has proven it cannot be done in one. So, my answer will lay out what's possible in each trial, for future reference:

Starting from nothing, we can measure six exact kilogram quantities (or eight if our bag of corn was larger):

$1000$, $2000$, $5000$: place the corresponding weight on one pan, then fill the other pan with corn. Select the pan with corn.

$3000$: place the $1000$ and $2000$ weights together on one pan, then fill the other pan with corn. Select the pan with corn.

$4000$: place the $1000$ weight on one pan and the $5000$ weight on the other pan. Fill the $1000$ side with corn until balance. Select the pan with corn.

$6000$: place the $1000$ and $5000$ weights together on one pan, then fill the other pan with corn. Select the pan with corn.

$7000$ and $8000$ could also be computed if the bag of corn was larger.

Starting with a quantity $q$, we can further make the following quantities:

$q/2$: place half of the quantity on each pan of the balance, then select one side and discard the other (or return to the bag).

$(q+1000)/2$: place the 1000g weight on one side, then arrange part of the starting quantity on each side until the scale is balanced. Select the portion from the side with the weight.

Similarly with any of the other kilogram amounts from section $1$.

Starting with a quantity $q$ on one pan and sufficient corn left in the bag, we can further compute the following:

$q-1000$: Put the $1000$ weight on the second pan, fill that pan until balance, discard $q$ from the first pan and select what's in the second pan.

similarly for any other kilogram amount.

If we have one additional container (or the table or floor1) available to store corn in, then we can add any of the above results to an existing result, by moving the result to the additional container, then computing the desired result, finally adding the second result to the additional container.

1The floor clearly won't work if the substance being measured is extremely fluid like water or flour, but corn should be possible, if a little tedious. Hopefully the floor is clean.


问:
Proof that $1+2+3+4+\cdots+n = \frac{n\times(n+1)}2$
Why is $1+2+3+4+\ldots+n = \dfrac{n\times(n+1)}2$   $\space$ ?

答:
Many good answers; I add a proof of the formula by induction.

Defining $T(n) = \sum_{i=1}^n i$, the claim is that $T(n) = n(n+1)/2$

For the base case of $n=1$, the sum gives us that $T(n) = 1$ and $n(n+1)/2 = 2/2 = 1$ $\quad \color{#0B2}\checkmark$

Assuming the inductive hypothesis that $T(k)=k(k+1)/2$, we have:

$$\begin{align}
T(k{+}1) &= \sum_{i=1}^{k+1} i \\
&= \sum_{i=1}^{k} i  \;+ k{+}1\\
&= k(k+1)/2 \;+ k{+}1 \\
&= k(k+1)/2 \;+ 2(k+1)/2  \\
&= (k+2)(k+1)/2  \\
&= (k{+}1)((k{+}1)+1)/2 \quad \color{#0B2}\checkmark \\
\end{align}$$
as required to completion the induction.


问:
Which of the following two estimators is better?
Let $X=(X_{1},...,X_{n})$ and $Y=(Y_{1},...,Y_{n})$ be independent simple samples from the distributions of $\mathcal{N}(m_{x},\sigma ^2)$ and $\mathcal{N}(m_{y},\sigma ^2)$, respectively. Which of the following two estimators: $T_{1}(X,Y) = \overline{X} \ \overline{Y}$, $T_{2} (X,Y)= \frac{1}{n} \sum \limits_{i=1}^{n} X_{i}Y_{i}$ is a better estimator for the parameter $t=m_{x}m_{y}$?

$(\overline{X} = \frac{1}{n}\sum \limits_{i=1}^{n}X_{i})$

.
.
.

I look forward to guidance. I know that we definitely need to calculate some expected values but I don't know what. How to start?

答:
Both estimators are unbiased and consistent. So, in order to compare which estimator is better, we need a metric. I propose the variance of estimator.

According to CLT and because $X,Y$ are independant (hence, $cov(X,Y) = 0$ ), we have
$$\sqrt{n}\left(
\begin{pmatrix}
\bar{X}  \\ \bar{Y} \end{pmatrix}
-\begin{pmatrix}
 m_x \\  m_y \end{pmatrix}
\right) \xrightarrow{n \to +\infty} \mathcal{N}\left(0,\begin{pmatrix}
\sigma^2 & 0 \\ 0 & \sigma^2 \end{pmatrix}
\right) \tag{1}$$
From $(1)$ we can apply the Multivariate delta method, with $h(x,y) = xy$
$$\sqrt{n}(T_1(X,Y)- m_xm_y) \xrightarrow{n \to +\infty} \mathcal{N}(0,m_y^2\sigma^2+m_x^2\sigma^2) $$

And we know that
$$\sqrt{n}(T_2(X,Y)- m_xm_y) \xrightarrow{n \to +\infty} \mathcal{N}(0,V(XY)) $$

We notice that $(m_x^2+m_y^2)\sigma^2 < V(XY) = E(X^2Y^2)=(\sigma^2+m_x^2)(\sigma^2+m_y^2) - m_x^2m_y^2$, we can conclude that $T_1$ is better than $T_2$


问:
Why is the maximum functional $f \mapsto \max_{t \in [0,1]} f(t)$ continuous on $C[0,1]$?
Define the maximum functional on the normed space $(C[0,1], \|\cdot\|_\infty)$ as follows:
$$
m: C[0,1] \to \mathbb R, \quad f \mapsto m(f) := \max_{t \in [0,1]} f(t).
$$
Why then is $m$ continuous?

答:
The two proofs above are nice. Here's a direct proof of the continuity of $f \mapsto m(f)=\max_{t \in [0,1]}f(t)$ via the $\varepsilon$-$\delta$ definition of continuity.

Let $\varepsilon > 0$. Let $f,g \in C[0,1]$ such that $\|f - g\|_\infty \leq \delta := \varepsilon /4$.

Let $t_0, t_1 \in [0,1]$ be the maximum points of $f$ and $g$ respectively. That is $m(f) = f(t_0)$, $m(g) = g(t_1)$.

Case $m(f) \leq m(g)$:

Then
$$
\begin{aligned}
&g(t_0) - \varepsilon/4 \leq m(f) \leq m(g) \leq f(t_1) + \varepsilon/4 \\
\Rightarrow \quad &|g(t_0) - f(t_1)| \leq \varepsilon /2.
\end{aligned}
$$
$$
\begin{aligned}
|m(f) - m(g)| &= |f(t_0) - g(t_1)| \\
&= |f(t_0) - g(t_1)| \\
&= |f(t_0) - g(t_0) + g(t_0) - f(t_1) + f(t_1) - g(t_1)| \\
&\leq \underbrace{|f(t_0) - g(t_0)|}_{\leq \varepsilon/4} + \underbrace{|g(t_0) - f(t_1)|}_{\leq \varepsilon/2} + \underbrace{|f(t_1) - g(t_1)|}_{\leq \varepsilon/4} \leq \varepsilon. \quad \checkmark
\end{aligned}
$$

Case $m(g) \leq m(f)$:

By symmetry, the same argument holds as in the first case. (Just swap $f$ with $g$ and $t_0$ with $t_1$.) $\quad \checkmark$

In all, we've shown the $\varepsilon$-$\delta$ definition of continuity -- i.e. $\forall \varepsilon > 0 \quad \exists \delta > 0 : \|f-g\|_\infty < \delta \Rightarrow |m(f) - m(g)| \leq \varepsilon$.


问:
What is the fastest way to find the characteristic polynomial of a matrix?
Finding the characteristic polynomial of a matrix of order $n$ is a tedious and boring task for $n > 2$.

I know that:

the coefficient of $\lambda^n$ is $(-1)^n$,

the coefficient of $\lambda^{n-1}$ is $(-1)^{n-1}(a_{11} + a_{22} + \dots + a_{nn})$,

the constant term is $\det{A}$.

When finding the coefficient of the linear term $\lambda$ of the characteristic polynomial of a $3\times 3$ matrix, one has to calculate the determinant of the matrix $A - \lambda I_n$ anyway. (But you don't have to sum all the terms, only the linear terms.)

Does anybody know a faster way?

答:
By hand, I think Ted Shifrin's method is fastest.

By computer, evaluate $\det (\lambda \cdot \mathrm{Id}- A)$ for $n$ values of $\lambda$, and find the characteristic polynomial by Lagrange interpolation.


问:
$B \equiv 1$ or $-1$ (mod n) where $B$ is the product of the elements of $U_n$.
How to prove the product of the elements of $U_n$ , $B$is congruent to $1$ or $-1$.

$U_n$ is the set of all positive numbers less than and relatively prime to $n$.

We have to prove $B \equiv 1$ (mod n) where  $B$ is the product of the elements of $U_n$.

My Attempt:   We can see the statement is true for $n = 1 , 2$.

The number of elements of $U_n$  will be  even for all natural number $n> 3$ as $\phi(n)$  is an even number for all natural number $n \geq 3$.
So we will get odd number of elements of order $2$ for each group of order $n(\geq 3)$.

$B = b_1 b_2 ... b_k$  where $b_i \in U_n$  and $b_i$ is of order $2$ and $k$  is an odd number for all$n \geq 3$.

Now it remains to prove this product is congruent to $1$ or $-1$ mod n.

Can anyone please help me ?

答:
Lemma Let $A$ be a multiplicatively notated finite elementary abelian group of order $2^k$, $k \in \mathbb{N}_{\gt 1}$. Then $\underset{a \in A}\prod a=1$.

Proof Regard $A$ as a $k$-dimensional vectorspace over the field of two elements $\mathbb{F}_2=\{0, 1\}$. Write 0 for the zero vector and 1 for the vector $(1,1, \cdots, 1)$ ($k$-coordinates). With each $v \in A$ we associate the vector 1 $− v$, when
summing all vectors, i.e. all elements of $A$. Since there are $2^{k−1}$ such pairs, the sum
equals $2^{k−1} \cdot $ 1 $\equiv$ 0 (mod $2$).

Corollary Let $A$ a multiplicatively notated finite abelian group and let $x=\underset{a \in A}\prod a$. Then $x=1$, unless the subgroup of all involutions, $I=\{a \in A: a^2=1\} \cong C_2$ in which case $x$ is the unique element of order $2$.

Proof $\underset{a \in A}\prod a=\underset{a \in I}\prod a \cdot \underset{a \notin I}\prod a$. If $a \notin I$, then $a \neq a^{-1}$, so $\underset{a \notin I}\prod a=1$, Now apply Lemma to $I$ and the proof is complete.

Note The corollary can be restated as follows: let $A$ be a finite abelian group and $P \in Syl_2(A)$, then the product $x$ of all elements of $A$ is $1$ if $P$ is trivial or non-cyclic, and if $P$ is non-trivial cyclic, $x$ equals the unique element of order $2$ of $A$.


问:
Create Gaussian distribution from random number generator output
I have a random number generator with output between -1 and +1 having a uniform distribution.

Is there a simple mathematical formula that would convert this into an estimated Gaussian distribution (around 0)?

答:
Standard normal sample in R. In R, the function rnorm (without extra parameters) gives numbers easily distinguished from a random sample from standard normal.

```<code>set.seed(2021)
z = rnorm(100)
</code>```

The sample of 100 observations generated in this way
passes the Shapiro-Wilk normality test, with P-value above 5%.

```<code>shapiro.test(z)$p.val
[1] 0.2116184
</code>```

A normal probability plot (QQ-plot) is very nearly linear (ignoring a few wobbles in the tails, as usual).

```<code>qqnorm(z); qqline(z, col="green")
</code>```

Standard normal sample from uniform data. In R the function runif without extra parameters
gives uniformly generated pseudo-random numbers in $(0,1)$
Also, in R, the function qnorm provides an excellent
rational approximatin to the inverse of the standard normal CDF. So the following R code also gives numbers that
are difficult to distinguish from a random sample from
standard normal.

```<code>set.seed(312)
u = runif(100)
z = qnorm(u)
shapiro.test(z)$p.val
[1] 0.4793456
qqnorm(z); qqline(z, col="green")
</code>```

Using your pseudo-random numbers. You say your pseudorandom numbers are uniformly
distributed in $(-1, 1).$ Such a sample of numbers
v can be generated in R as below. Then by taking
the absolute value as suggested by @tommik (+1)
and using qnorm as above, you can get a sample
that is not easy to distinguish from a random standard
normal sample.

```<code>set.seed(1212)
v = runif(100, -1, 1)
summary(v)
    Min.  1st Qu.   Median     Mean  3rd Qu.     Max.
-0.97538 -0.55026 -0.09479 -0.07188  0.32183  0.98725
u = abs(v)
summary(u)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
0.01182 0.23112 0.44815 0.48566 0.77076 0.98725
z = qnorm(u)
shapiro.test(z)$p.val
[1] 0.3778305
qqnorm(z); qqline(z, col="green")
</code>```

Notes: In the early days of digital computers
the following method of converting pseudo-random numbers on $(0,1)$ that uses only simple arithmetic was used. Sum 12 variates $U_i \stackrel{indep}{\sim}\mathsf{Unif}(0,1),$ then subtract 6. Convergence by the CLT of a sum
of uniform distributions to normal is roughly
satisfactory. The results are nearly normal
but confined to $(-6,6).$

```<code>set.seed(12)
z = replicate(100, sum(runif(12))- 6)
shapiro.test(z)$p.val
[1] 0.2368584
</code>```

For some years the Box-Muller method, suggested by @Suman (+1), was a standard method of generating standard normal deviates. However, double-precision arithmetic of the transcendental functions involved restrict the accuracy of the
resulting 'standard normal' deviates beyond about $\pm 7.$. This inaccuracy is difficult to detect by most methods.

The inverse CDF of standard normal cannot be
expressed directly but the rational approximation implemented in qnorm (due to Michael Wichura)
is accurate within double-precision arithmetic.

Body:
Standard normal sample in R. In R, the function rnorm (without extra parameters) gives numbers easily distinguished from a random sample from standard normal.

set.seed(2021)
z = rnorm(100)

The sample of 100 observations generated in this way
passes the Shapiro-Wilk normality test, with P-value above 5%.

shapiro.test(z)$p.val
[1] 0.2116184

A normal probability plot (QQ-plot) is very nearly linear (ignoring a few wobbles in the tails, as usual).

qqnorm(z); qqline(z, col="green")

Standard normal sample from uniform data. In R the function runif without extra parameters
gives uniformly generated pseudo-random numbers in $(0,1)$
Also, in R, the function qnorm provides an excellent
rational approximatin to the inverse of the standard normal CDF. So the following R code also gives numbers that
are difficult to distinguish from a random sample from
standard normal.

set.seed(312)
u = runif(100)
z = qnorm(u)
shapiro.test(z)$p.val
[1] 0.4793456
qqnorm(z); qqline(z, col="green")

Using your pseudo-random numbers. You say your pseudorandom numbers are uniformly
distributed in $(-1, 1).$ Such a sample of numbers
v can be generated in R as below. Then by taking
the absolute value as suggested by @tommik (+1)
and using qnorm as above, you can get a sample
that is not easy to distinguish from a random standard
normal sample.

set.seed(1212)
v = runif(100, -1, 1)
summary(v)
    Min.  1st Qu.   Median     Mean  3rd Qu.     Max.
-0.97538 -0.55026 -0.09479 -0.07188  0.32183  0.98725
u = abs(v)
summary(u)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
0.01182 0.23112 0.44815 0.48566 0.77076 0.98725
z = qnorm(u)
shapiro.test(z)$p.val
[1] 0.3778305
qqnorm(z); qqline(z, col="green")

Notes: In the early days of digital computers
the following method of converting pseudo-random numbers on $(0,1)$ that uses only simple arithmetic was used. Sum 12 variates $U_i \stackrel{indep}{\sim}\mathsf{Unif}(0,1),$ then subtract 6. Convergence by the CLT of a sum
of uniform distributions to normal is roughly
satisfactory. The results are nearly normal
but confined to $(-6,6).$

set.seed(12)
z = replicate(100, sum(runif(12))- 6)
shapiro.test(z)$p.val
[1] 0.2368584

For some years the Box-Muller method, suggested by @Suman (+1), was a standard method of generating standard normal deviates. However, double-precision arithmetic of the transcendental functions involved restrict the accuracy of the
resulting 'standard normal' deviates beyond about $\pm 7.$. This inaccuracy is difficult to detect by most methods.

The inverse CDF of standard normal cannot be
expressed directly but the rational approximation implemented in qnorm (due to Michael Wichura)
is accurate within double-precision arithmetic.


问:
Finding of an example of a metric space.
Find such a metric space (other than the discrete metric space) that boundary of every subset is empty.

If I take usual metric on $\mathbb{N}$

As, for every subset $\{x_1,x_2,\cdot\cdot\cdot,x_n\}$,
We get $B(x_i,\frac{1}{2}) \cap (\mathbb{N}\setminus \{x_1,x_2,\cdot\cdot\cdot,x_n\})= \varnothing$

And, for an infinite subset $A$ of $\mathbb{N}$ , same logic is also applicable i.e.,  $B(x,\frac{1}{2}) \cap (\mathbb{N}\setminus A)= \varnothing$ ( where, $x$ runs over for every element in $A$)

So, every subset has empty boundary.

Please verify someone.....

答:
Every topological space with the property that every subset has empty boundary must be discrete.
Indeed, any non-closed subset has nonempty boundary, so all subsets must be closed, i.e. the topology is discrete.</p>


问:
On a certain integral related to Euler-Masheroni constant
When trying to derive Mertens' theorem using Perron's formula, I encountered the following integral:

$$
\int_0^1{1-e^{-t}\over t}\mathrm dt-\int_1^\infty{e^{-t}\over t}\mathrm dt
$$

which, according to Titchmarsh's The Theory of Riemann Zeta Function, evaluates to $\gamma$ (a.k.a. Euler-Mascheroni constant).

To begin with, I converted this integral into the following limit:

$$
\lim_{N\to\infty}\left[\int_0^1{1-(1-t/N)^N\over t}\mathrm dt-\int_1^N{(1-t/N)^N\over t}\mathrm dt\right]
$$

so that I can eventually simplify the expression into

$$
\lim_{N\to\infty}\left[\sum_{m=1}^N\frac1m-\log N\right]
$$

which is identical to the definition of $\gamma$.

I wonder what tools can be used to justify the limit step.

答:
METHODOLOGY $1$:

To show that

$$\lim_{N\to\infty}\int_1^N \frac{\left(1-\frac tN\right)^N}{t}\,dt=\int_1^\infty \frac{e^{-t}}{t}\,dt$$

we will show that

$$\lim_{N\to\infty}\int_1^N \frac{e^{-t}}t \left(e^t\left(1-\frac tN\right)^N-1\right)\,dt=0\tag 1$$

Proceeding, we use the inequalities

$$\left(1-\frac tN\right)^N\le e^{-t}\tag 2$$

and

$$\begin{align}
e^t\left(1-\frac tN\right)^N &\ge \left(1+\frac tN\right)^N\left(1-\frac tN\right)^N\\\\
&=\left(1-\frac{t^2}{N^2}\right)^N\\\\
&\ge 1-\frac{t^2}{N}\tag3
\end{align}$$

Using $(2)$ and $(3)$, we have the estimates for the integral in $(1)$

$$-\frac1N \int_1^N te^{-t}\,dt \le \int_1^N \frac{e^{-t}}t \left(e^t\left(1-\frac tN\right)^N-1\right)\,dt\le 0\tag4$$

whence application of the squeeze theorem to $(4)$ yields $(1)$.

To show that

$$\lim_{N\to\infty}\int_0^1 \frac{1-\left(1-\frac tN\right)^N}{t}\,dt=\int_0^1 \frac{1-e^{-t}}{t}\,dt$$

we exploit the uniform convergence of $\left(1-\frac tN\right)^N$ to $e^{-t}$ for $t\in [0,1]$.  Then, we may interchange the limit and integral to obtain the result.

Putting it together, we have

$$\lim_{N\to \infty}\left(\int_1^N \frac{\left(1-\frac tN\right)^N}{t}\,dt-\int_0^1 \frac{1-\left(1-\frac tN\right)^N}{t}\,dt\right)=\int_0^1 \frac{1-e^{-t}}{t}\,dt-\int_1^\infty \frac{e^{-t}}{t}\,dt$$

as was to be shown.

METHODOLOGY $2$:

There is another way to proceed.  We can write the integral of interest as

$$\begin{align}
I&=\int_0^1 \frac{1-e^{-t}}{t}\,dt-\int_1^\infty \frac{e^{-t}}{t}\,dt\\\\
&=\lim_{\varepsilon\to 0^+}\int_\varepsilon^1  \frac{1-e^{-t}}{t}\,dt-\lim_{L\to \infty}\int_1^L \frac{e^{-t}}{t}\,dt\tag1
\end{align}$$

Integrating by parts both of the integrals on the right-hand side of $(1)$ we find

$$\begin{align}
I&=\lim_{\varepsilon\to 0^+}\left(-(1-e^{-\varepsilon})\log(\varepsilon)-\int_\varepsilon^1  e^{-t}\log(t)\,dt\right)\\\\
&-\lim_{L\to \infty}\left(e^{-L}\log(L)-\int_1^L e^{-t}\log(t)\,dt\right)\\\\
&=-\int_0^\infty e^{-t}\log(t)\,dt\\\\
\end{align}$$

Finally, using the development in the NOTE section of THIS ANSWER, we see that $-\int_0^\infty e^{-t}\log(t)\,dt=\gamma$ as was to be shown!


问:
Why does the row method work in finding basis?
Let $S=\left\{
u_1=\begin{pmatrix}1\\2\\2\\1\end{pmatrix}
u_2=\begin{pmatrix}3\\6\\6\\3\end{pmatrix}
u_3=\begin{pmatrix}4\\9\\9\\5\end{pmatrix}
u_4=\begin{pmatrix}-2\\-1\\-1\\1\end{pmatrix}
u_5=\begin{pmatrix}5\\8\\9\\4\end{pmatrix}
u_6=\begin{pmatrix}4\\2\\7\\3\end{pmatrix}
\right\}$ and $V=span(S)$

$A = (u_1,u_2,...u_6)$

$A =\begin{pmatrix}1 & 3 & 4 & -2 & 5 & 4\\
2 & 6 & 9 & -1 & 8 & 2\\
2 & 6 & 9 & -1 & 9 & 7\\
1 & 3 & 5 & 1 & 4 & 3\\
\end{pmatrix}\rightarrow
\begin{pmatrix}1 & 3 & 0 & -14 & 0 & -37\\
0 & 0 & 1 & 3 & 0 & 4\\
0 & 0 & 0 & 0 & 1 & 5\\
0 & 0 & 0 & 0 & 0 & 0\\
\end{pmatrix}$

Hence, basis for V using the column space method is $V=Col(A)$ is $\{u_1,u_3,u_5\}$. Up to this part I can understand because linear relationships between columns are preserved after performing row operations. But I do not understand the following method.

$A^T=\begin{pmatrix}
1 & 2 & 2 & 1\\
3 & 6 & 6 & 3\\
4 & 9 & 9 & 5\\
-2 & -1 & -1 & 1\\
5 & 8 & 9 & 4\\
4 & 2 & 7 & 3\\
\end{pmatrix} \rightarrow
\begin{pmatrix}
1 & 0 & 0 & -1\\
0 & 1 & 0 & 0\\
0 & 0 & 1 & 1\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0\\
\end{pmatrix}$

Then $\left\{
\begin{pmatrix} 1\\0\\0\\-1\end{pmatrix},
\begin{pmatrix} 0\\1\\0\\0\end{pmatrix},
\begin{pmatrix} 0\\0\\1\\1\end{pmatrix}\right\}$ form a basis for V

Why does the row space of the transpose of A form the basis for V?

Also it is noted that this method also make it easier to write any vector in V relative to this basis.

For example, $u_6 =
4\begin{pmatrix} 1\\0\\0\\-1\end{pmatrix}+
2\begin{pmatrix} 0\\1\\0\\0\end{pmatrix}+
7\begin{pmatrix} 0\\0\\1\\1\end{pmatrix}
$

Why does it work?

答:
As you note, the first method works because the row operations do not change the linear relations between the columns, and since the column space of RREF(A) has a basis of columns containing pivot entries, those columns in the original matrix still form a basis for the column space of A.

The second approach works under a different idea: row operations do not change the row space of a matrix.  You can check for each operation.  For example, if $v=v_1 + v_2$ and we replace $v_2$ with $v_3=v_2 + 3v_1$, then $v=-2v_1 + v_3$. So, at the end of doing Gaussian elimination, our rows span the same space, but the non-zero ones are linearly independent.  Now, just notice that taking transposes swaps row space and column space.


问:
Group symmetries of a trigonal bipyramidal molecule
I am trying to apply what I have just learnt in group theory to find the group symmetries of a trigonal bipyramidal molecule. I know that the atoms forming an equilateral triangle is the dihedral group of order 6. However, how do I find the group of symmetries of the whole molcule i.e. including the three atoms that are collinear? Do I split the molecule up and count them separately and add them up?

The attached image is the molecule I wish to find the group of symmetries of.

答: There are two kind of symmetries in your molecule: the symmetries of the equilateral triangle (the dihedral group of order 6 = the symmetric group $S_3$) and the vertical reflexion. Since those commute, the group of symmetries is the direct product $S_3 \times \mathbb{Z}/2\mathbb{Z}$.


问:
An improper integral from PUTNAM preparatory problems
Evaluate the improper integral  $\int^1_0 ln(x)\cdot ln(1-x)dx$.
I got it from this link:https://www.math.ust.hk/~yangwang/Misc/putnam_first100.pdf

Kindly give me a hint to solve it.

Body:
Hint:

$$\ln(1-x) = \sum_{n=1}^{+\infty}\frac{(-1)^n}{n}x^n$$

$$\int_0^1\ln(x)x^ndx =-\frac{1}{(n+1)^2}$$

$$\sum_{n=1}^{+\infty} \frac{1}{n^2} =\frac{\pi^2}{6}$$


问:
Solving linear equations using division of matrix determinants

I did not know that we can use this method to solve linear equations and was wondering what method is called. Thank you so much

答:
The method is called Cramer's rule. Namely, let $A\mathbf{x}=\mathbf{b}$ be a system of linear equations such that $A\in \mathbb{R}^{n\times n}$ with $\det(A)\neq 0$. Then we have a unique solution $\mathbf{x}=(x_1,\dots,x_n)^T$ with
$$x_i=\frac{\det(A_i)}{\det(A)},$$
where $A_i$ is the square matrix obtained by replacing the $i$-th column of $A$ by the column vector $\mathbf{b}$.


问:
Holomorphicity of $\sin(\theta)\cos(\theta)+i\cos(\theta)^2$ imply that $\theta$ is constant
Suppose $\theta(x,y)$ be a smooth function on a connected domain, taking values in $(0,2\pi]$, and define
$
c=\cos(\theta),s=\sin(\theta).
$
Suppose that the function $f(x,y)=sc+ic^2$ is holomorphic.

I have a proof that $\theta$ must be constant. I wonder whether there are other, perhaps easier or more conceptual proofs.

My proof:

The Cauchy–Riemann equations imply that $u=sc,v=c^2$ are harmonic functions. A short computation then gives
$$
0=-\Delta v=2\cos(2\theta)(\theta_x^2+\theta_y^2)+\sin(2\theta)\Delta\theta,
$$
$$
0=\Delta u=-2\sin(2\theta)(\theta_x^2+\theta_y^2)+\cos(2\theta)\Delta\theta,
$$
or equivalently
$$
\begin{pmatrix} 2\cos(2\theta) & \sin(2\theta) \\\ -2\sin(2\theta) & \cos(2\theta) \end{pmatrix}\begin{pmatrix} \theta_x^2+\theta_y^2 \\\ \Delta\theta \end{pmatrix}=0.
$$
The invertibility of the matrix implies that $\theta_x=\theta_y=0$, so $\theta$ is constant.

答: The function $sc +ic^2$ maps $(0,2\pi]$ onto a circle. The range of a nonconstant holomorphic function on a connected domain is open in the plane. It follows that your $f$ is constant, which implies $\theta$ is constant.


问:
Is this true: $(n \log n)^2 \in \Theta(\frac{n^3}{\log n})$
I have to decide whether $(n \log n)^2 \in \Theta(\frac{n^3}{\log n})$ is true or false.

I can't really find a good approach, I'd assume it's wrong, but I can't prove it.

答: if $(n \log n)^2 \in \Theta(\frac{n^3}{\log n})$ there is a constant $C>0$ such that $$n^2\log^2n\ge \frac{Cn^3}{\log n}$$ for $n$ sufficiently large.  This simplifies to $$\log^3n\geq Cn$$ which is false.


问:
Isomorphism between two finite fields
We have $k_1:= \mathbb F_7(\alpha)$ and $k_2 := \mathbb F_7(\beta)$ where $\alpha^2 = 3$ and $\beta^2 = -1$ in $\mathbb F_7$. I have to show that these two are isomorphic.

Let $\phi:k_1 \rightarrow k_2$ be a homomorphism which preserves $1 \in k_1$. Then

$$\phi(\alpha^2)= \phi(3) = 3 = \phi(\alpha)^2$$

where

$$\phi(\alpha) = x + y\beta\;,\;\;x,y \in \mathbb F_7$$

Thus

$$(x+y\beta)^2 = \phi(\alpha)^2  = x^2 + 2xy\beta +y ^2 \beta^2 = x^2 +2xy\beta -y^2=3$$

So $x$ or $y$ must be $0$. But $y$ can't be zero because $3$ has no root in $\mathbb F_7$. So $x = 0$ such that

$$-y^2 = 3 \rightarrow y \in \{-2, 2\}$$

Is the function $\phi$ with $\phi(x) = x\;$ for $\;x \in \mathbb F_7\;$ and $\;\phi(\alpha) = 2\beta\;$ then an isomorphism ?

答: Well, you have a candidate; why don't you check whether $\phi((a+b\alpha)(c+d\alpha)) = \phi(a+b\alpha)\phi(c+d\alpha)$, for all $a,b,c,d \in \mathbb{F}_7$? For sums it's trivial, as you already define it as a linear map over the base field.</p>

Title: Isomorphism between two finite fields


问:
solving equations of density functions
I have an engineering background, but not deep mathematics background. So, I am having difficulty proving an equation regarding density function as shown in the following problem, and I hope to get some help.

- Problem statement:
Given densities $f_1$ and $f_2$, set $g=f_1-f_2$. Then, $c=||g^+||=||g^-||=\frac{1}{2}||g||$ where $||\cdot||$ is L$1$-norm and $c > 0$.

For your reference, density function $f$ is $f \geq 0$, $||f||=1$, and $f \in L^1$ where $L^1$ is L$1$ space corresponding to measure space $(X,\mathcal{A},\mu)$, i.e., all real-valued measurable functions $f:X\rightarrow R$ s.t. $||f|| = \int_X |f(x)| \mu(dx) < \infty$.

- My tries:
I tried the followings, but I am not sure if it is correct.

Since $f_1 \in L^1$ and $f_2 \in L^1$, $g=f_1-f_2 \in L^1$, and, if we assume $||g^+||=||g^-||=c$ with $c > 0$,
\begin{align}
2c=||g^+|| + ||g^-|| &= ||(f_1-f_2)^+|| + ||(f_1-f_2)^-||\\
&= ||(f_1-f_2)^+ + (f_1-f_2)^-||\\
&= ||g||
\end{align}
where $f^+ := \max(0,f)$ and $f^- := \max(0,-f)$, and as a result, $f=f^+-f^-$. Also, one more useful property regarding L$1$ norm is that, if $f_1>0$, $f_2>0$, and $f_1,f_2 \in L^1$, then $||f_1 + f_2|| = ||f_1|| + ||f_2||$.

Thank you for your help in advance.

答:
To prove $\|g^+\|=\|g^-\|,$ we first use the fact that $\|f_1\|=\|f_2\|=1$ and that $f_1,f_2\ge0$ to get
\begin{align}
0=\|f_1\|-\|f_2\|&=\int_X |f_1| \ \mu(dx) - \int_X |f_2| \ \mu(dx) \\
&=\int_X \left(|f_1|-|f_2|\right) \ \mu(dx) \\ &= \int_X \left(f_1-f_2\right) \ \mu(dx).
\end{align}
But
$$\int_X \left(f_1-f_2\right) \ \mu(dx)=\int_X (f_1-f_2)^+ \ \mu(dx)-\int_X (f_1-f_2)^- \ \mu(dx)=0,$$ which implies that $$ \int_X (f_1-f_2)^+ \ \mu(dx)=\int_X (f_1-f_2)^- \ \mu(dx).$$
Since $(f_1-f_2)^+$ and $(f_1-f_2)^-$ are non-negative functions, we have that $\|g^+\|=\|g^-\|$.

The rest of the proof looks correct.

Title: solving equations of density functions


问:
Combinatorics question from an exam in 2015

A committee of $11$ people is to be formed from $8$ males and $5$ females. If $m$ is the number of ways the committee is formed with at least $6$ males, and $n$ is the number of ways the committee is formed with at least $3$ females, then what are $m$ and $n$?

The answer is

 $$m = n = 78$$

It can be solved by classical combination tactics but I wanted to solve it using generating functions and I wrote them like that:

\begin{align*}
m&=(x^6+x^7+x^8)(1+x+\dots+x^5)\\[5pt]
n&=(1+x+\dots+x^8)(x^3+x^4+x^5)\\[5pt]
\end{align*}

However, for $m$ and $n$, the coefficient is $3$. Where did I make mistake? Why didn't generating functions give correct answer?

答: First notice that any committee of 11 must include at least 6 men, since it can only include at most 5 women. Similarly, any committee of 11 must include at least 3 women, since it can only include at most 8 men. So the question is actually asking how many different ways can we choose 11 people from 13 people. In generating function terms, this is the coefficient of $x^{11}$ in the binomial expansion of $(1+x)^{13}$.


问:
Find all normal subgroups of $S_3 \times \mathbb{Z}_6$
I know that if $H_1,H_2$ are normal subgroups of $G_1,G_2$, respectively, then $H_1\times H_2$ is a normal subgroup of $G_1\times G_2$. Since $\mathbb{Z}_6$ is abelian, all its 4 subgroups are normal. The 3 normal subgroups of $S_3$ are $S_3$, $A_3$ and the trivial subgroup. Thus I can get 12 normal subgroups of $S_3 \times \mathbb{Z}_6$, but I do not know whether there are others, or how to find them.

答:
In addition to the 12 normal subgroups mentioned in the initial post, there are two other normal subgroups:
$$
A_1=\{(e,\bar0),\,((12),\bar3),\,((13),\bar3),\,((23),\bar3),\,((123),\bar0),\,((132),\bar0)\};
$$
$$
A_2=\langle((123),\bar0),\,(e,\bar2),((12),\bar3)\rangle.
$$
The subgroup $A_1$ is of order 6, and the subgroup $A_2$ contains 18 elements.


问:
Let $\alpha\in[0,9].$ Is the set of real numbers such that the average of the digits of each of these numbers $=\alpha,$ countable or uncountable?
Let $\alpha \in [0,9]\subset\mathbb{R}.$ Is the following set countable or uncountable:

The set of all (nonnegative) real numbers which, when written in decimal expansion form

$$k_1\ldots k_m\cdot k_{m+1}\ldots$$

where the $k_i$ are digits between $0$ and $9$, satisfy

$$\lim_{n\to\infty}\left( \frac{\sum_{j=1}^n k_j}{n} \right) = \alpha ?$$

答:
There are uncountably many numbers with $\alpha = 0$.

Let $\{a_n\}$ be a strictly increasing sequence of natural numbers. For each sequence, we'll build a number with $\alpha=0$. Since there are uncountably many such sequences (in particular, the set of sequences s.t. $a_{n+1} = a_n + 1$ or $a_{n+1} = a_n + 2$ is uncountable), we can generate uncountably many numbers.

Let $s_n = \sum_{i=1}^n a_i$. Let $x$ be the number such that its $i$'th digit $x_i = 1$ iff $i = s_n$ for some $n$, and $x_i=0$ otherwise.

Then $\alpha(x) = 0$. It suffices to show that $\lim \frac n {s_n} = 0$ (it's the average up to $s_n$'th digit). It follows from the fact that $a_n \ge n$ and therefore $s_n \ge \frac {n^2} 2$.

There are uncountably many numbers for any $\alpha \in (0, 1)$. Handling $\alpha \ge 1$ is trivial (just add $\lfloor\alpha \rfloor$ to all digits).

Consider digits between $s_{n-1}$ and $s_{n}$, There are $a_n$ such digits. Let $c_n \in \mathbb N$ s.t. $\frac {c_n} {a_n}$ is the closest to $\alpha$. Let ${c_n}$ digits between $s_{n-1}$ and $s_n$ be $1$ and the rest of the digits be $0$, then the average of these digits is $\frac {c_n} {a_n}$.

Since $a_n$ strictly increases, $\lim \frac {c_n} {a_n} = \alpha$. Therefore, for any $\epsilon$ there exists $n_0$ s.t. $\frac {c_n} {a_n} \in [\alpha - \epsilon, \alpha + \epsilon]$ for $n \ge n_0$.
Then the average up to $s_n$ is
\begin{align*}
avg_{s_n}
&= \frac {c_1 + \cdots + c_n} {s_n}\\
&= \frac {c_1 + \cdots + c_{n_0}} {s_n} + \frac {c_{n_0 + 1} + \cdots + c_{n}} {s_n} \\
&= \frac {c_1 + \cdots + c_{n_0}} {s_n} - \alpha \frac {s_{n_0}} {s_n} + \frac {\alpha s_{n_0} + c_{n_0 + 1} + \cdots + c_{n}} {s_n} \\
&\to 0 + 0 + \ell \in [\alpha - \epsilon, \alpha + \epsilon],
\end{align*}
since
$$\alpha s_{n_0} + c_{n_0 + 1} + \cdots + c_{n} \ge \alpha s_{n_0} + a_{n_0 + 1} (\alpha - \epsilon) + \cdots + a_{n}(\alpha - \epsilon) \ge \alpha s_n - \epsilon s_n$$
(the same for $\alpha+\epsilon$).
Since this holds for arbitrary $\epsilon$, $\lim avg_n = \alpha$.

Some technicalities

For the limit to exist, we also need to consider all $n' \in [s_{n-1}, s_n]$. It suffices to guarantee that $a_n = o(s_{n-1})$, and I gave an example of such a family of sequences above.

For $\alpha \in (0,1)$, I don't actually guarantee that the generated numbers are distinct. We can guarantee this with some adjustments, but it's simpler to do the following: generate one number, and then change its $s_n$'th digits to be either $0$ or $1$ (all combinations). As shown in the case $\alpha = 0$, it doesn't change the limit of the average.


问:
A different form of $GL(V,B)$.
Let $B:V\times V \to F$ be a bi-linear form.
Suppose that $B$ is given by the matrix $\underline{B}=(B_{ij})$.

How would I show that:
$$GL(V,B):=\{ g \in GL_n(F): B(gu,gv)=B(u,v) \ \ \forall u,v \in V\}$$
satisfies the following:
$$GL(V,B)=\{ g\in GL_n(F): g^t\underline{B}g=\underline{B}\}$$
?

Clueless.

答:
Here's a hint:

We have If the coordinate representations of $u,v$ and $B$ are $\underline{u},\underline{v},\underline{B}$ respectively, $B(u,v)=\underline{u}^T\underline{B}\,\underline{v}$. This means that $B(gu,gv)=(g\underline{u})^T\underline{B} g\underline{v}=\underline{u}^Tg^T\underline{B}\, g\underline{v}$. What does this tell us about matrices that preserve $B$?

Title: A different form of $GL(V,B)$.


问:
How can I show this equation is less than or equal to $\pi$?
I have that $$ \left|\int_\gamma\frac{\mathrm{d}z}{2+z^2}\right| $$
where $\gamma$ is upper half of unit circle

I know that $$\gamma(t) = e^{it}$$
and therefore $$\gamma(t) = e^{it}$$

and as $$ \left|\int_\gamma f(z)\ \mathrm{d}z\right| = \left|\int_a^b f\big(\gamma(t)\big)\gamma'(t)\ \mathrm{d}t\right| $$

I know the originally equation is equal to $$\left|\int_0^\pi\frac{1}{2 + (e^{it})^2}ie^{it}\ \mathrm{d}t\right|$$

how could I show this is $\displaystyle\left|\int_0^\pi\frac{1}{2 + (e^{it})^2}\cdot ie^{it}\ \mathrm{d}t\right|\le \pi$

答:
$$ \bigg | \int_0^\pi \frac{i e^{i\theta} d \theta}{2 + e^{2i\theta}} \bigg | \le   \int_0^{\pi} \frac{d\theta}{|2 + e^{2i \theta}|} \le \pi $$
Where we have written $z = e^{i\theta}$ and the last inequality follows from noting that $|2+e^{2i\theta}| \ge 1$.


问:
Associativity of product of ideals
Let $A$ and $B$ be ideals of a ring $R$. Define the product of ideals as $AB= \Bigg \{ \displaystyle \sum_{k=1}^n a_k b_k \mid a_k \in A, b_k \in B, n \in \mathbb{N} \Bigg \}$. Is there an easy way to show that this product is associative, i.e., $(AB)C=A(BC)$? And in this case can we say that $(AB)C=A(BC)=ABC:=\Bigg \{ \displaystyle \sum_{k=1}^n a_k b_k c_k \mid a_k \in A, b_k \in B, c_k \in C, n \in \mathbb{N} \Bigg \}$?

答: By definition, we have that $$(AB)C = \{x_1 c_1 + \cdots + x_n c_n \,|\, n \geq 1, \, x_1, \dots, x_n \in AB, \text{ and } c_1, \dots, c_n \in C\}.$$ Considering that each of the $x_i$ is of the form $x_i = a_{i1} b_{i1} + \cdots + a_{ik} b_{ik}$ for some integer $k \geq 1,$ it follows that $(AB)C \subseteq ABC$ because each of the elements of $(AB)C$ belongs to $ABC.$ Conversely, any product of the form $abc$ for some elements $a \in A,$ $b \in B,$ and $c \in C$ belongs to $(AB)C$ because it can be written as $x_1 c_1$ for $x_1 = ab$ and $c_1 = c,$ hence for any integer $n \geq 1$ and any elements $a_1, \dots, a_n \in A,$ $b_1, \dots, b_n \in B,$ and $c_1, \dots, c_n \in C,$ the sum $a_1 b_1 c_1 + \cdots + a_n b_n c_n$ belongs to $(AB)C.$ Consequently, we have that $(AB)C = ABC.$ But the analogous argument can be made for $A(BC),$ hence we have that $(AB)C = ABC = A(BC).$

Title: Associativity of product of ideals


问:
Is the following polynomial reducible?
I need to check whether the following polynomial $g(x)=x^2+t^2−1∈\Bbb{C}(t)$ (where $\Bbb{C}(t)$ is the field of all rational expressions over $\Bbb{C}$) is reducible. My attempt: I assume by contradiction that it's reducible, and I write it as:
$$(x+p(t))(x+q(t))=x^2+t^2−1$$

So we get that:
$$p(t)+q(t)=0$$
$$p(t)q(t)=t^2−1$$

Since $\Bbb{C}(t)$ is a field, it's also a UFD. So either $p(t)q(t)=(t−1)(t+1)$ or WLOG, $p(t)=λ(t^2−1)$ for some $0≠λ∈\Bbb{C}$. Either way, we get a contradiction to the first condition over $p(t),q(t)$. Is my approach correct or am I missing something?

答: C is algebraically closed so only polynomes of one degree are irreductible</p>

Title: Is the following polynomial reducible?


问:
How do we prove that given a set S such that |S|=|SxS| then |P(S)|=|P(S)xP(S)| where P(S) is the power set of S?
Given a bijection f from S to SxS, I can't seem to find an appropriate mapping from P(S) to P(S)xP(S). I don't know if I am missing something obvious but I have tried making a connection through P(SxS) as well but to no avail.
Edit: I'm especially interested in the case of S being the natural numbers as I want to use this and prove that |R|=|R^2|.</p>

答:
If $S$ is empty or infinite, then $P(S)$ is respectively a singleton or an infinite set, and so $P(S)$ is in bijection with $P(S) \times P(S)$.

The statement does not hold, however, if $S$ is a singleton. In this case, $\vert P(S) \vert=2$, while $\vert P(S) \times P(S) \vert=4$.


问:
fitting an exponential curve with squared exponent through three points
I apologize if the format is horrible! I am not sure how to do this.

\begin{center}
I am using least squares regression to fit an exponential to my data. Thus, I need to find a good guess for initial parameters. The model function for this data is
\end{center}

\begin{equation}
y\ =\ ae^{bK^{2}} +c
\end{equation}
\begin{center}

where $\displaystyle K\ =b^{-1} \ \sqrt{2( x+3/2)}$ and $\displaystyle b\ =\ 0.15$
\end{center}
Here is the method that i have done, however it does not result in a very good fit.

\begin{gather*}
y_{1} \ =\ ae^{bK^{2}_{1}} +c\\
y_{2} \ =\ ae^{bK^{2}_{2}} +c\\
y_{3} =ae^{bK^{2}_{3}} +c
\end{gather*}
Thus,
\begin{gather*}
y_{1} '\ =\ 2abK_{1} e^{bK^{2}_{1}}\\
y_{2} '\ =\ 2abK_{2} e^{bK^{2}_{2}}
\end{gather*}

\begin{equation*}
\frac{y_{2} '}{y1'} =\frac{K_{2}}{K_{1}} e^{b\left( K^{2}_{2} -K1^{2}\right)}
\end{equation*}
Moreover, note that $\displaystyle y_{2} '\ =\ \frac{dy_{2}}{dK_{2}} =\frac{y_{3} -y_{2}}{K_{3} -K_{2}}$ and similarly, $\displaystyle y_{1} '=\ \frac{y_{2} -y_{1}}{K_{2} -K_{1}}$.

Thus $\displaystyle b$ is equal to

\begin{gather*}
b\ =\ \ln\left(\frac{y_{2} 'K_{1}}{y_{1} 'K_{2}}\right) /\left( K^{2}_{2} -K^{2}_{1}\right)\\
\end{gather*}
Now to find $\displaystyle a$, we will eliminate $\displaystyle c$ by

\begin{gather*}
y_{2} -y_{1} \ =\ a\left( e^{bK^{2}_{2}} -e^{bK^{2}_{1}}\right)\\
y_{3} -y_{2} \ =\ a\left( e^{bK^{2}_{3}} -e^{bK^{2}_{2}}\right)
\end{gather*}
so
\begin{gather*}
a_{1} \ =\ \frac{y_{2} -y_{1}}{e^{bK^{2}_{2}} -e^{bK^{2}_{1}}}\\
a_{2} \ =\ \frac{y_{3} -y_{2}}{e^{bK^{2}_{3}} -e^{bK^{2}_{2}}}
\end{gather*}
and we have $\displaystyle a\ =\frac{a_{1} +a_{2}}{2}$.

Now for $\displaystyle c$ we just plug in all the values into each equation and solve for $\displaystyle c$. i.e

\begin{gather*}
c_{1} \ =\ ae^{bK^{2}_{1}} -y_{1}\\
c_{2} \ =\ ae^{bK^{2}_{2}} -y_{2}\\
c_{3} =ae^{bK^{2}_{3}} -y_{3}
\end{gather*}
so $\displaystyle c\ =\ \frac{c_{1} +c_{2} +c_{3}}{3}$ .

When I plot this fit against the data, the curve only really fits through the first three points. I am not sure what is going on or where the problem is.

DATA:
\begin{gather*}
( 2,2.35448742279577\ \ )\\
( 4,-1.368206934161941)\\
( 6,-3.305390118335089)\\
( 8,-4.482711701457797)\\
( 10,-5.266602875206834)\\
( 12,-5.820768423448095)\\
( 14-6.229391391650249)\\
( 16,-6.540202260195814)\\
( 18-6.782285215077716)\\
( 20-6.974372715890539)
\end{gather*}
THE ATTEMPT PLOTTED IS HERE

答:
You are writing the equation in a tortured way, but I guess that in the end we can write

$$y=e^{ax+b}+c.$$ (My constant aren't yours.)

For three general points, there is no closed-form formula, the equation is transcendental. But for three equidistant abscissas (as the plot seems to show), there is an option:

$$y_0=e^{ax_0+b}+c,\\y_1=e^{a(x_0+h)+b}+c,\\y_2=e^{a(x_0+2h)+b}+c$$

then,

$$\frac{y_2-y_1}{y_1-y_0}=\frac{e^{2ah}-e^{ah}}{e^{ah}-1}=e^{ah}.$$

From this, draw $a$, and the rest follows.


问:
Basic inequality involving modulus
If $0\leq a,b\leq 1$ and $\lvert a-b\rvert\leq r$ for some $r>0$, then $a\leq b+r$ and $a\geq b-r$?

By definition, they are equivalent.

答:
Note that $\forall a,b \in \mathbb{R}$ also $\forall r>0$ we can write this,
$$|X|\leq r \implies -r \leq X \leq r$$so you can rewrite $$\forall a,b \  \in [0,1]: |a-b|\leq r \Leftrightarrow-r\leq a-b\leq r\\ \text{all sides } +b\\
b-r\leq a-b+b\leq b+r\\b-r \leq a\leq b+r$$ or if you have different definition for absolute sign like $\ |x|=\max \{x,-x\}$ you can also do like this
$$\max \{a-b,b-a\}\leq r \\\to
\begin{cases}a-b\leq r & \implies a\leq b+r \\b-a\leq r & \implies b-r\leq a \end{cases}$$ hope it can helps you

Title: Basic inequality involving modulus


问:
Subgroups of $S_n$, contained in $A_n$ or half of elements are odd permutations
good morning. I am unable to resolve this issue, could someone help me, please?

Question: If $H$ is a non-trivial subgroup of $S_{n}$, then or $H \subset A_{n}$ or exacly half of the elements are odd permutations.

答:
We can still reason like this. If $A=A_n\cap H$ and $x\in H\setminus A_n$, then $H=A\cup xA$ and $|A|=|xA|$.
Why does $H=A\cup xA$? If $h\in H$ and $h\notin A$, then $x^{-1}h\in A$. So $h\in xA$.