问:
If $E$ is measurable then $\delta E$ is measurable
Let $\delta =(\delta_1,\delta_2,\cdots,\delta_d)$ where $\delta_i>0$, and $E$ is a subset of $\mathbb{R}^d$. Define $\delta E$ by $\delta E = \{(\delta_1x_1,\delta_2x_2,\cdots, \delta_dx_d)|(x_1,x_2,\cdots,x_d)\in E\}$
Prove that if $E$ is meaurable, then $\delta E$ is measurable and  $m(\delta_E)= \delta_1\cdots\delta_d m(E)$.

It is the same question as this link:
If E is measurable, then $\delta E$ is measurable.

However, the answer on that link works with a linear map, $T$, which has not come in my class. I believe I am supposed to do this by showing $m^{\mathrm{*}}(O - \delta E) < \epsilon$ where $O$ is an open set. Can anyone help me get started on this one?
Edit: We also have not covered Borel sets or Sigma Algebras

答:
To ease discussion, I consider the case $d=2$ only. The general case can be treated in exactly the same way. Define a function
$f:\mathbb{R}^{2}\rightarrow\mathbb{R}^{2}$ by $f(x,y)=(\frac{1}{\delta_{1}}x,\frac{1}{\delta_{2}}y).$
Firstly, we show that $f$ is measurable. Let $\pi_{1},\pi_{2}:\mathbb{R}^{2}\rightarrow\mathbb{R}$
be the canonical projections onto the first coordinate and the second
coordinate respectively. By universal property of product $\sigma$-algebra,
to show that $f$ is $\mathcal{B}(\mathbb{R})\otimes\mathcal{B}(\mathbb{R})/\mathcal{B}(\mathbb{R})\otimes\mathcal{B}(\mathbb{R})$,
it suffices that $\pi_{1}\circ f:\mathbb{R}^{2}\rightarrow\mathbb{R}$
and $\pi_{2}\circ f:\mathbb{R}^{2}\rightarrow\mathbb{R}$ are $\mathcal{B}(\mathbb{R})\otimes\mathcal{B}(\mathbb{R})/\mathcal{B}(\mathbb{R})$-measurable.
Observe that $\pi_{1}\circ f(x,y)=\frac{1}{\delta_{1}}x$, so for any
interval $(\alpha,\beta)$, we have $(\pi_{1}\circ f)^{-1}\left((\alpha,\beta)\right)=(\delta_{1}\alpha,\delta_{2}\beta)\times\mathbb{R}\in\mathcal{B}(\mathbb{R})\otimes\mathcal{B}(\mathbb{R})$.
Since $\{(\alpha,\beta)\mid\alpha<\beta\}$ is a generator of the
$\sigma$-algebra $\mathcal{B}(\mathbb{R})$, it follows that $\pi_{1}\circ f$
is $\mathcal{B}(\mathbb{R})\otimes\mathcal{B}(\mathbb{R})/\mathcal{B}(\mathbb{R})$-measurable.
Similarly, $\pi_{2}\circ f$ is $\mathcal{B}(\mathbb{R})\otimes\mathcal{B}(\mathbb{R})/\mathcal{B}(\mathbb{R})$-measurable.
Therefore, $f$ is $\mathcal{B}(\mathbb{R})\otimes\mathcal{B}(\mathbb{R})/\mathcal{B}(\mathbb{R})\otimes\mathcal{B}(\mathbb{R})$.

Let $E\subseteq\mathbb{R}^{2}$ be a Borel set. Note that $f^{-1}(E)=\{(\delta_{1}x,\delta_{2}y)\mid(x,y)\in E\}$.
For, let $(x',y')\in\mbox{RHS}$, then $(x',y')=(\delta_{1}x,\delta_{2}y)$
for some $(x,y)\in E$. Now $f(x',y')=(\frac{1}{\delta_{1}}x',\frac{1}{\delta_{2}}y')=(x,y)\in E$,
so $(x',y')\in f^{-1}(E)=\mbox{LHS}$. On the other hand, let $(x',y')\in\mbox{LHS}$,
then $f(x',y')\in E$, i.e., $(\frac{1}{\delta_{1}}x',\frac{1}{\delta_{2}}y')\in E$.
Let $x=\frac{1}{\delta_{1}}x'$ and $y=\frac{1}{\delta_{2}}y'$, then
$(x,y)\in E$ and $(x',y')=(\delta_{1}x,\delta_{2}y)$. This shows
that $(x',y')\in\mbox{RHS}$. It follows that $\{(\delta_{1}x,\delta_{2}y)\mid(x,y)\in E\}$
is a Borel set.

We go to show that $m_{2}(f^{-1}(E))=\delta_{1}\delta_{2}m_{2}(E)$
for any Borel set $E\subseteq\mathbb{R}^{2}$, where $m_{2}$ is the
usual Lebesgue measure on $\mathbb{R}^{2}$. Define measures $\mu_{1},\mu_{2}:\mathcal{B}(\mathbb{R}^{2})\rightarrow[0,\infty]$
by $\mu_{1}(E)=m_{2}(f^{-1}(E))$ and $\mu_{2}(E)=\delta_{1}\delta_{2}m_{2}(E)$.
Let $n\in\mathbb{N}$ be arbitrary. We go to show that $\mu_{1}(E)=\mu_{2}(E)$
for any $E\in\mathcal{B}((-n,n]\times(-n,n])$ by invoking Dynkin
$\pi$-$\lambda$ theorem. Let $\mathcal{P}=\{(-n,a]\times(-n,b]\mid a,b\in(-n,n]\}$
and $\mathcal{D}=\{E\in\mathcal{B}((-n,n]\times(-n,n])\mid\mu_{1}(E)=\mu_{2}(E)\}$.
We verify that $\mathcal{P}$ is a $\pi$-class and $\mathcal{D}$
is a $\lambda$-class. Let $A_{1}=(-n,a_{1}]\times(-n,b_{1}]$ and
$A_{2}=(-n,a_{2}]\times(-n,b_{2}]$ for some $a_{1},a_{2},b_{1},b_{2}\in(-n,n],$
then $A_{1}\cap A_{2}=(-n,a_{1}\wedge a_{2}]\times(-n,b_{1}\wedge b_{2}]\in\mathcal{P}$,
so $\mathcal{P}$ is a $\pi$-class (Here, $x\wedge y=\min(x,y)$).
Clearly $\emptyset\in\mathcal{D}$. It can be verified directly that
$(-n,n]\times(-n,n]\in\mathcal{D}$. Let $A\in\mathcal{D}$. Since
$\mu_{1}$ and $\mu_{2}$ are finite when restricted on $\mathcal{B}((-n,n]\times(-n,n])$,
we have that
\begin{eqnarray*}
\mu_{1}(A^{c}) & = & \mu_{1}((-n,n]\times(-n,n])-\mu_{1}(A)\\
 & = & \mu_{2}((-n,n]\times(-n,n])-\mu_{2}(A)\\
 & = & \mu_{2}(A^{c}).
\end{eqnarray*}
This shows that $A\in\mathcal{D}\Rightarrow A^{c}\in\mathcal{D}$.
Finally, let $A_{1},A_{2},\ldots\in\mathcal{D}$ be pairwisely disjoint.
We have that
\begin{eqnarray*}
\mu_{1}(\cup_{n}A_{n}) & = & \sum_{n=1}^{\infty}\mu_{1}(A_{n})\\
 & = & \sum_{n=1}^{\infty}\mu_{2}(A_{n})\\
 & = & \mu_{2}(\cup_{n}A_{n})
\end{eqnarray*}
and hence $\cup_{n}A_{n}\in\mathcal{D}$. This shows that $\mathcal{D}$
is a $\lambda$-class. It can be verified directly that $\mathcal{P}\subseteq\mathcal{D}$.
Now, by Dynkin's $\pi$-$\lambda$ theorem, we conclude that $\sigma(\mathcal{P})\subseteq\mathcal{D}$.
However, it is well-known that $\sigma(\mathcal{P})=\mathcal{B}((-n,n]\times(-n,n])$.
Hence, we actually have $\sigma(\mathcal{P})=\mathcal{D}.$ That is,
for any Borel subset $E\subseteq(-n,n]\times(-n,n]$, we have that
$\mu_{1}(E)=\mu_{2}(E)$. Finally, if $E\subseteq\mathbb{R}^{2}$
is a Borel set, by continuity of measure, we have that
\begin{eqnarray*}
\mu_{1}(E) & = & \lim_{n\rightarrow\infty}\mu_{1}(E\cap(-n,n]\times(-n,n])\\
 & = & \lim_{n\rightarrow\infty}\mu_{2}(E\cap(-n,n]\times(-n,n])\\
 & = & \mu_{2}(E).
\end{eqnarray*}


问:
Is there a formula for $n^{th}$ power of $\cos(ax+b)$?
Is there a closed form formula or trigonometric relation for $\cos^{n}(ax+b)$?

答:
Let's forget $a$ and $b$, as it's not the problem here.

$$\cos^n x=\left(\frac{e^{ix}+e^{-ix}}{2}\right)^n=\frac1{2^n}\sum_{k=0}^n\binom{n}{k}e^{kix}e^{-(n-k)ix}\\=\frac1{2^n}\sum_{k=0}^n\binom{n}{k}e^{(2k-n)ix}\\=\frac1{2^n}\sum_{k=0}^n\binom{n}{k}(\cos(2k-n)x+i\sin(2k-n)x)\\
=\frac1{2^n}\sum_{k=0}^n\binom{n}{k}\cos(2k-n)x$$

ContentLicense: CC BY-SA 4.0


问:
Studying the monotony of an integral
I'm trying to understand the following exercise:

Be $f: \mathbb{R} \rightarrow \mathbb{R}$ a continuous function, such that $f(x) \ge 0,\forall x \ge0$

Study the monotonicity of the following function: $g(x) := \int_{0}^{x^2} f(t)dt$

I'm still new to integrals and I'm lost on how should I start solving this.
Any tips are welcome!

Thanks

答:
By the FTC, if $\;F(x)\;$ is a primitive of $\;f\;$ (it exists since $\;f\;$ is continuous), then

$$\int_0^{x^2}f(t)\,dt=F(x^2)-F(0)\implies\left(\int_0^{x^2}f(t)\,dt\right)'=\left(F(x^2)-F(0)\right)'\stackrel{\text{chain rule}}=2xF'(x^2)$$

Well, can you now say something about the sign of $\;g'(x)\;$ and thus about monotony of $\;g\;$ ?

Title: Studying the monotony of an integral


问:
What is $-i$ exactly?
We all know that $i$ doesn't have any sign: it is neither positive nor negative. Then how can people use $-i$ for anything?

Also, we define $i$ a number such that $i^2 = -1$. But  it can also be seen that $(-i)^2 = -1$. Then in what way is $-i$ different from $i$? If it isn't, then why do we even have something like a  $-i$?

答:
Most of the answers already take the whole structure of the complex numbers as given. Therefore let me go to a more elementary level, where we don't yet assume anything about the complex numbers except that they contain the real numbers, and at least one additional number $i$ which solves the equation $x^2+1=0$.

Now consider the equation $x+i=0$. I think you would agree that a set of numbers where we could not solve this equation would not be terribly useful. By definition, we denote the solution of $x+a=0$ with $-a$, so the solution here is $-i$. However that's up to now only a notation; for example, the same way we get $-0$, which we know is the same as $0$. Therefore the question is: What is $-i$?

First, we note that $-i$ cannot be a real number, because otherwise $i$ would be its negative (also a real number), and we already know $i$ is not a real number. So there remain two possibilities: Either $-i=i$, or $-i$ is a different non-real number than $i$.

We also want that the distributive law continues to hold.

Now assume $-i=i$, that is, $i+i=0$. Then we have $-2 = (-1) + (-1) = i^2 + i^2 = i(i+i) = i\cdot 0 = 0$. Now that's obviously wrong, therefore we cannot have $-i=i$. Thus $-i$ is a distinct number from $i$.

Edit: On A.P.'s request, I also add the proof that $-i=(-1)i$.

To prove that $-i=(-1)i$ means to prove that $i+(-1)i=0$, because that's how $-i$ is defined.

We obviously have, using the fact that $x=1x$ for all $x$:
$$i +(-1)i = 1i + (-1)i =(1+(-1))i = 0i = 0$$

Note that you can prove essentially the same way that $i(-1) = -i$ (in case you take into account that multiplication might not turn out to be commutative, which actually would not be that unusual, although for complex numbers multiplication actually is commutative)- Therefore you also get
$$(-i)^2+1 =  i(-1)(-1)i+1 = i^2+1 = 0$$

Now for the proof of $(-1)i=-i$ as well as above when proving $-i\ne i$, I've used $0i=0$, which seems obvious, but to be sure, let's prove this as well. We have
$$0i = (0+0)i = 0i+0i$$
Now we add $-(0i)$ to both sides of the equation (remember, the demand that for each $x$ we have a $-x$ which fulfils $x+(-x)=0$ was how we arrived at $-i$ at the first place), to get for the left hand side $0i + (-(0i)) = 0$, and for the right hand side $0i + 0i + (-(0i)) = 0i + 0 = 0i$ (note that in this, I've made another assumption, namely that addition is still associative even if expressions with $i$ are involved; but that's also a natural assumption which you'd not give up unless necessary).

Note that ultimately, all I needed was

the rules for real numbers continue to hold unchanged

the fact that $i$ solves the equation $x^2+1=0$

the associativity of addition

$x+0=0$ for all $x$

the existence of an additive inverse $-x$ for any number $x$

the associativity of multiplication

$1x=x$ for all $x$

the distributive law

The first two are given by the problem, while the rest are very fundamental assumptions which you wouldn't give up lightly.

ContentLicense: CC BY-SA 3.0

Title: What is $-i$ exactly?


问:
Vectors in 3D and area of a triangle
I have three points with coordinates: $A (5,-1,0),B(2,4,10)$, and $C(6,-1,4)$.

I have the following vectors $\overrightarrow {CA} = (-1, 0, -4)$ and $\overrightarrow{CB} = (-4, 5, 6)$.

To find the area of the triangle I used the dot product between these vectors to get the angle and then applied the formula $A=0.5ab\sin{C}$ to find the area of the triangle which gave me $15.07(2dp)$.

However in the given solutions the answer is given as $(3*\sqrt(102))/2$

I think they have used the trig identity $\cos^2(\theta) + \sin^2(\theta) = 1$ to find the value of $\sin(\theta)$ rather than $\arccos(\theta)$ to find the angle ACB. However I don't understand why there would be such a discrepancy between the two answers; one using $\arccos$ and the other using the trig identity.

答:
If you calculate the angle, you introduce numerical errors. The most elegant way to calculate the area is to use the cross product. You will not need any trigonometric functions:
$$A=\frac12|\vec{CA}\times\vec{CB}|$$
So all you need will be some multiplications, additions(subtractions), and a square root. Let me know if you can follow this path.
By the way, the answer should be $\frac32\sqrt{101}$, if I did the math right, which is closer to your calculations

Title: Vectors in 3D and area of a triangle


问:
Conditional Probability using an urn.
An urn contains $5$ white balls and $6$ black balls. $3$ balls are drawn sequentially without replacement. What are the probabilities...

a. $P(\text{2nd is black} | \text{1st is white})$?

For this one I did the

$$ P(\text{1st is white})=\frac{5}{11} , $$

and

$$ P(\text{second is black}) = \frac{3}{11}*\frac{3}{11} $$

(that is the probability of WB and BB added together). I multiplied the probability of

$$ \frac{P(\text{2nd black} \cap \text{1st white})}{P(\text{1st is white})} , $$

for this I got

$$ \frac{\frac{5}{11}*\frac{3}{11}}{\frac{5}{11}} = \frac{3}{11} . $$

However since this is without replacement I did not think that the answer could be independent, some feedback would help. Alright so I have this figured out that it should be $\frac{6}{10}$.

b. If I am looking now for $P(\text{3rd is black} | \text{1st is white and 2nd is black})$, I ended up with

$$ \frac{\frac{5}{9}*\frac{3}{11}}{\frac{3}{11}} , $$

this is again an independent answer is this correct?

答:
I'm not sure what you did, but

P(2 is B|1 is W) = P(pick B among 4W6B) = 6/10

Title: Conditional Probability using an urn.


问:
Existence of Simple functions converging uniformly
Let $f$ be a function on $\mathbb{R}^d$. Assume that $f$ is measurable and for every $x \in \mathbb{R}^d$, $f(x) \in [0,M]$; some $M \in \mathbb{R}^+$.

Show there is a sequence of simple functions $f_n$ also taking on values in $[0,M]$ such that

$$f_n \xrightarrow{\text{uniform}} f.$$

My attempt. Take some partition of the interval, and define characteristics on these intervals then you have a sequence of simple functions then apply Egorovs theorem to give us uniform convergence on a closed subset that barely fits in $\mathbb{R}^d$ (by barely fits I mean the measure of the complement is small). Is this a correct approach or? Possibly a nested sequence of subsets of $[0,M]$? I am thinking like $\bigcup_{n=1}^M[0,n]$ but $M$ may not be integer.

答: Define $f_n(x)= \sum\limits_{k=1}^{[2^{n}M]+1} \frac {k-1} {2^{n}} \chi_{\frac {k-1} {2^{n}} \leq f(x) <\frac {k} {2^{n}}}$ and verify that $|f_n(x)-f(x)| <\frac 1 {2^{n}}$ for all $x$. This has the additonal property that $f_n$'s increase to $f$ uniformly.


答:
As mentioned in your comment, if you are only looking to find the period of $cos^n(ax+b)$, it wouldn't be necessary to write a closed form formula or expression for it.
When $n$ is even, We see that $(\cos(\pi - (ax+b)))^n = (\cos(\pi + (ax+b)))^n$ and this turns the period of the function into $\pi/a$, whereas when $n$ is odd, we don't have this luxury anymore. Instead we have $\cos^n(ax+b)$ going its full $2\pi/a$ cycle.

Edit: As @Rob Arthan pointed out in his comment, given $f(x)$ with period $T$ and some function $g(x)$; $h(x) = g(f(x))$ has period $T/k$ for some integer $k$. It is also possible to observe that $T$ is a period of $h$, but often depending upon the function $g$ we can do better.

In our case $f = \cos(ax+b)$ and $g(x) = x^n$. When $n$ is odd, $g(x)$ is injective, and so every input to $g$ has a unique associated output, which means there is no room in $[0,2\pi/a]$ (the original period of $f$) for repetition of the same output values. This is why, when $n$ is odd, the period of $h$ is the same as the period of $f$. However, when $g$ is not injective, we have multiple inputs being mapped to the same output. In the case of an even $n$, we have exactly two inputs in a given period being mapped to the same output, which essentially means we are repeating the graph of $f$ twice in $[0,2\pi/a]$ upon composing, which means the period is halved simply giving us a period of $\pi/a$.

Title: Is there a formula for $n^{th}$ power of $\cos(ax+b)$?

Body: Is there a closed form formula or trigonometric relation for $\cos^{n}(ax+b)$?


问:
Convex set also has convex interior (Corollary)
While studying convex analysis, i came across two statements:

First statement:

Let $S$ be a convex set in $R^{n}$ with a nonempty interior. Let $x_{1} \in \operatorname{cl} S$ and $x_{2} \in$ int
S. Then $\lambda x_{1}+(1-\lambda) x_{2} \in$ int $S$ for each $\lambda \in(0,1)$. I understood the proof regarding this one.

Corollary:

Let $S$ be a convex set. Then int $S$ is convex.

My attempt:

For int $S$ to be convex, one has to let $x_{1}$, $x_{2}$ $\in$ int $S$ and show that $\lambda x_{1}+(1-\lambda) x_{2} \in int(S)$. I see that is almost what the first statement says. However, in statement number one, $x_{1}$ belongs to the closure of the set $S$ and i need it in $int (S)$ to conclude.

In an attempt to solve this issue, i was trying to use set inclusions such as $S \subset$ $cl(S)$ or $cl(S) = S \cup \partial S$ to conclude that $x_{1} \in cl(S)$ $\implies$ $x_{1} \in int (S)$ with the help of the first statement.

However, i just can't figure out this corollary.

Can someone help?

Thanks in advance, Lucas

答:
You cannot conclude $x_{1} \in \text{cl}(S)$ $\implies$ $x_{1} \in \text{int} (S)$ because it is wrong. Just observe that the statement

Let $x_{1} \in \operatorname{cl} S$ and $x_{2} \in$ int
S. Then $\lambda x_{1}+(1-\lambda) x_{2} \in \text{int} (S)$ for each $\lambda \in(0,1)$

is more general than

Let $x_{1}$, $x_{2}$ $\in$ int $S$. Then  $\lambda x_{1}+(1-\lambda) x_{2} \in \text{int} (S)$ for each $\lambda \in(0,1)$.

In fact, if $x_{1}, x_{2} \in$ int $S$, then $x_1 \in \text{int} (S) \subset S \subset \text{cl}(S)$ and $x_2 \in  \text{int} (S)$.


问:
What does the annotation R mean here?
My professor is talking about relations and he gave us this example.

$A=\mathbb{Z}$ and  $B=\mathbb{Z}$.

$R\subset A\times B$.

$aRb$ if and only if $a-b$ is even.

Some examples of subsets were given.

5R3, 3R5, and 6R4. This makes sense because a-b for all the values here are even and they are all contained in $\mathbb{Z}$.

The definition for the image is given by  $\{b\in  B:\exists a\in A:(a,b)\in R\}$. So why is the image  of 5R3, 3R5, and 6R4 B and not {3,5,4}. What am I not understanding about the notation of what R means?

答:
The relation $R$ is the set of all pairs $(a,b)\in\mathbb{Z}$ such that $a-b$ is even.  You're right that the image of the relation $\{(5,3),(3,5),(6,4)\}$  would be just $\{3,4,5\}$, but those are just examples of elements of $R$, not all of them.  There are many (infinitely!) more elements, such as $(0,0)$, $(-1,19)$, $(5,37)$, ...

In particular, notice that for any integer $b$ at all, $(b,b)$ will be an element of $R$, since $b-b=0$ is even.  This proves that $b$ is in the image of $R$, so the image of $R$ actually contains every integer.

Title: What does the annotation R mean here?


问:
By Using the Binomial expansion as follow that the expression for the first three term is mathematically correct or not?
Let the expression is defined as $(1-\frac{1}{i})^{2i}$.

For example: Using the Binomial expansion as follow  $$(1-\frac{1}{i})^{2i}=\sum_{n=0}^{n}\frac{(2i)!}{n!(2i-n)!}(-1)^n(\frac{1}{i})^n$$

$$\approx1-\frac{1}{i}+O((-\frac{1}{i})^3)$$.

Please need your help that the above expression for the first three term is mathematically correct or not, where the big $O$ notation means that the first neglected or unknown term is of the order $(-\frac{1}{i})^3$.

Body:
That's fairly straight forward.  When n= 0, n!= 0!= 1 and 2i- n= 2i so (2i- n)!= 2i!.  The coefficient is $\frac{(2i)!}{n!(2i- n)!}= \frac{(2i)!}{(2i)!}= 1$.  Of course, $(-1)^0= 1$ and $\left(\frac{1}{i}\right)^01= 1$ so the first term is 1

When n= 1, n!= 1!= 1 and $\frac{(2i)!}{(2i- n)!}$$= \frac{2i(2i-1)!}{(2i-1)!}= 2i$, $(-1)^1= -1$ and $\left(\frac{1}{i}\right)^1= \frac{1}{i}$ so the second term is $-2$, not $-\frac{1}{i}$.

When n= 2, n!= 2!= 2 and $\frac{(2i)!}{(2i-2)!}= $$\frac{2i(2i-1)(2i-2)!}{(2i-2)!}= $$2i(2i-1)$,  $(-1)^2= 1$, and $\left(\frac{1}{i}\right)= \frac{1}{i^2}$ so the third term is $\frac{2(2i)(2i-1)}{i^2}= \frac{4(2i- 1)}{i}$.


问:
Measure change in angle during rotation of quaternion
I am observing the rotation of a box. While the rotation is continuous, I've captured the following three quaternions $q_0, q_1, q_2$ to understand its rotation:

\begin{equation}
q = [s, x \boldsymbol{i} + y \boldsymbol{j} + z \boldsymbol{k}] \quad s,x,y,z \in \mathbb{R}
\end{equation}

\begin{equation}
q_0 = \lbrack 0.67156, -  0.018402 \boldsymbol{i} -   0.74072 \boldsymbol{j} - 0.0025122 \boldsymbol{k} \rbrack; \\
q_1 = \lbrack 0.22424, -  0.103290 \boldsymbol{i} -   0.96833 \boldsymbol{j} + 0.0371360 \boldsymbol{k} \rbrack; \\
q_2 = \lbrack 0.25635, + 0.228900 \boldsymbol{i} +  0.93650 \boldsymbol{j} - 0.0696340 \boldsymbol{k} \rbrack; \\
\end{equation}

Visually observing the box, I can tell it is stationary at $q_0$. It is rotated by $\approx 80^{\circ}$ when it arrives to $q_1$. The same is true during rotation from $q_1$ to $q_2$. The total visual change in rotation from $q_0$ to $q_2$ is $\approx 160^{\circ}$ on one axis. I am not sure which axis that is. Consider the below illustration as an example:

At $q_0$, the side of the box that is labeled as $3$ is facing on the ground, and $1$ is facing the sky. At $q_2$, $3$ is now facing predominantly at the sky and $1$ is facing the ground.

GOAL: What I want to do is calculate the change in rotation for each axis to see where the rotation is occurring and how close it comes to my visual interpretation ( is it $160^{\circ}$ ?).

I start by defining a quaternion, $q_s = [1, 0 \boldsymbol{i} + 0\boldsymbol{j} + 0\boldsymbol{k}]$, that is an arbitrary rotation from which I can measure change. I then begin normalizing my axis one at a time. For my first iteration, I focus on the x-axis:

\begin{equation}
q^\prime = \frac{q}{ \sqrt{s^2 + x^2} }
\end{equation}

Completing this step, I have the following unit quaternions and their conjugates:

\begin{equation}
q_{0}^\prime = \lbrack 0.99962, -  0.027391 \boldsymbol{i} -   0 \boldsymbol{j} + 0 \boldsymbol{k} \rbrack; \\
q_{1}^\prime = \lbrack 0.90827, -  0.41838i \boldsymbol{i} -   0 \boldsymbol{j} + 0 \boldsymbol{k} \rbrack; \\
q_{2}^\prime = \lbrack 0.74591, + 0.66604 \boldsymbol{i} +  0 \boldsymbol{j} - 0 \boldsymbol{k} \rbrack; \\
\end{equation}

\begin{equation}
q_{0}^* = \lbrack 0.99962, +  0.027391 \boldsymbol{i} +   0 \boldsymbol{j} + 0 \boldsymbol{k} \rbrack; \\
q_{1}^* = \lbrack 0.90827, +  0.418380 \boldsymbol{i} +   0 \boldsymbol{j} + 0 \boldsymbol{k} \rbrack; \\
q_{2}^* = \lbrack 0.74591, - 0.666040 \boldsymbol{i} +  0 \boldsymbol{j} + 0 \boldsymbol{k} \rbrack; \\
\end{equation}

To measure the rotation, I take the product of $q_s$ and $q_{n}^*$, $z$. The angle of change is $2 * acosd(z_s)$. I repeat this process for each axis, and below are my results:

x-axis: $3.1392^{\circ}, 49.4649^{\circ}, 83.5249^{\circ}$

y-axis: $95.6076^{\circ}, 153.9234^{\circ}, 149.3828^{\circ}$

z-axis: $0.42868^{\circ}, 18.8066^{\circ}, 30.3938^{\circ}$

None of the answers seem correct and I suspect I am doing something wrong. It is rather odd that every angle on the x-axis is about $\frac{1}{2}$ of the rotation. What must I correct to be able to achieve my GOAL?

My sources to get this far:

Quaternions for Computer Graphics 2011th Edition by John Vince

https://stackoverflow.com/questions/5782658/extracting-yaw-from-a-quaternion

Compute Angle Between Quaternions (in Matlab)

答: If your goal is to determine what axis your rotation is occurring about, the good news is that there's a much simpler answer than what you're trying to put together; in fact, this is a large part of why quaternions are used! Specifically, the rotation that carries the orientation represented by $q_0$ to the one represented by $q_2$ is just $q_2q_0^{-1}$. And if a quaternion is given by $a+b\mathbf{i}+c\mathbf{j}+d\mathbf{k}$, the axis of rotation is just (the normalized version of) $\langle b, c, d\rangle$, and the amount of rotation is $2 \arccos(a)$, assuming that the original quaternion is normalized.


问:
Summation and binomial coefficient
I'm studying sampling at the moment, but I can't get the passage below:

When $n$ units are sampled from $N$ units without replacement, then each unit of the population
can occur with other units selected out of the remaining $(N - 1)$ units in the population and each unit occurs in $\binom{N-1}{n-1}$ of the $\binom{N}{n}$ possible samples. So

\begin{equation}
    \sum_{i = 1}^{\binom{N}{n}} \left( \sum_{i=1}^n y_i\right) = \binom{N - 1}{n - 1}\sum_{i = 1}^{N} y_i
\end{equation}

Are there any arguments/calculations that can help me understand better the passage above?

答:
The notation in $$\sum_{i = 1}^{\binom{N}{n}} \left( \sum_{i=1}^n y_i\right) = \binom{N - 1}{n - 1}\sum_{i = 1}^{N} y_i$$ is not good, but I think it is possible to find the intended meaning.

You start with a set of $N$ values presented as $y_i$.

The initial $\sum\limits_{i = 1}^{\binom{N}{n}}$ counts over all the different ways of taking a combination of $n$ items from the big set: this is supposed to represent the possibilities in sampling without replacement.

The $\sum\limits_{i=1}^n y_i$ is then the sum of the $n$ items in each selected combination. I can see no justification for using the same index for the combinations  and each combination's individual items.

The $\sum\limits_{i=1}^N y_i$ on the right-hand side is simply the sum of all the values in the big set.

The equality is not difficult to justify intuitively: each $y_i$ from the big set can appear in a total of ${\binom{N-1}{n-1}}$ different combinations, since if it is in a combination then the other elements of the combination fill be the $n-1$ space left, taken from the remaining $N-1$ values of the big set.

Title: Summation and binomial coefficient


问:
Proving the existence of models with large cardinality
I'm reading through Marker's Model Theory: An Introduction, and I'm confused by the proof of Proposition 2.2.2. The result is as follows:

Let $T$ be an $\mathcal{L}$-theory with infinite models. If $\kappa$ is an infinite cardinal with $\kappa \geq |\mathcal{L}|$, then there is a model of $T$ of cardinality $\kappa$.

The proof goes as follows. Let $\mathcal{L}^*$ be the language $\mathcal{L} \cup \{c_\alpha : \alpha \in \kappa \}$, where the $c_\alpha$ are constant symbols. Let $T^*$ be the $\mathcal{L}^*$-theory $T^* = T \cup \{ c_\alpha \neq c_\beta : \alpha, \beta \in \kappa, \alpha \neq \beta \}$. Any model of $T^*$ will have at least $\kappa$ many elements. To prove that $T^*$ is satisfiable, we use the Compactness Theorem by showing that $T^*$ is finitely satisfiable. If $\Delta \subseteq T^*$ is a finite subset of $T^*$, then $\Delta$ is contained in $T \cup \{ c_\alpha \neq c_\beta : \alpha, \beta \in I \}$ for some finite subset $I$ of $\kappa$. Any infinite model of $T$ would model $\Delta$, so we know that $T^*$ is finitely satisfiable, and therefore satisfiable.

I have two questions about this. Firstly, it's not obvious to me that a model of $T^*$ would be of cardinality exactly $\kappa$, only at least $\kappa$. The $c_\alpha \neq c_\beta$ sentences ensure that $\kappa$ maps injectively into a model $\mathcal{M}$ by $\alpha \mapsto c_\alpha^\mathcal{M}$, but I don't see why that precludes the possibility that $M$ is of cardinality strictly greater than $\kappa$. At first, I took this to mean that there existed a model of cardinality at least $\kappa$ and it was just worded in a way that I misunderstood, but Marker uses this result to prove Vaught's test, where he invokes the existence of a model for $T$ of cardinality exactly $\kappa$.

Secondly, I don't see where the assumption that $\kappa \geq |\mathcal{L}|$ is used.

Any help clearing up either of these confusions would be greatly appreciated.

答:
Your two concerns are related. We use $\kappa \geq |\mathcal{L}|$ in order to show that there is a model of size exactly $\kappa$.

The idea is to pick your favorite set of size $\kappa$ and then close it under all the operations of $\mathcal{L}$.

This is entirely analogous to finding the smallest subgroup containing a subset of a group $G$. We take $S \subseteq G$ and close it under the group operations.

Showing that taking this closure doesn't add too many elements (that is, we end with $\kappa$ many elements) uses the fact that we don't need to do this closure operation too many times -- after all, there are at most $\kappa$ many operations in $\mathcal{L}$ to worry about.

The actual process of taking this closure is a bit tedious, but isn't hard. You can find more information by googling around for the "Downward Lowenheim-Skolem Theorem"

I hope this helps ^_^


问:
Clarification between charts being incompatible and smooth structures being "exotic"
Inspired by this question, $\mathbb{R}$ can be given the atlas $\mathcal{A} = \{f(x) = x : x \in \mathbb{R}\}$ or the atlas $\mathcal{B} = \{g(x) = x^3 : x \in \mathbb{R}\}$.

These are incompatible because there exist combinations of $f,g$ and their inverses which produce non smooth functions $\mathbb{R} \rightarrow \mathbb{R}$, namely $f \circ g^{-1} = \sqrt[3]{x}$, so this fails to be a diffeomorphism.

Obviously $f(\mathbb{R}) = g(\mathbb{R}) = \mathbb{R}$ are homeomorphic as topological spaces. Why does this not count as an "exotic line"?, which is precisely when we have homeomorphic objects but not diffeomorphic ones, the classification theorem says there can be no such thing as an exotic $\mathbb{R}$, since the line is $1$ dimensional and weird things can only happen in starting at dimension $4$.

Perhaps I am struggling to grasp the difference since I have not actually seen what an exotic differential structure looks like, I suppose they require all sorts of advanced constructions and can't be expressed with a nice example, so what makes being exotic so much more bad than just being incompatible?

答:
There is nothing exotic here. Yes, the two atlases give us two distinct smooth structures, i.e. two distinct smooth manifolds with underlying topological space $\mathbb R$. Let us call them $\mathbb R_{id}$ and $\mathbb R_g$. But these smooth manifolds are diffeomorphic: In fact, $g : \mathbb R_g \to \mathbb R_{id}$ is a diffeomorphism since
$$\mathbb R \stackrel{g^{-1}}{\to} \mathbb R \stackrel{g}{\to} \mathbb R \stackrel{id}{\to} \mathbb R \phantom{x} , \phantom{x} \mathbb R \stackrel{id^{-1}}{\to} \mathbb R \stackrel{g^{-1}}{\to} \mathbb R \stackrel{g}{\to} \mathbb R$$
are the identity maps which are smooth.

See $C^{\infty } $ diffeomorphism between 2 nonequivalent manifolds .


问:
Why this formulation is not the official exposition of ZF-?
Specification: if $\phi$ is a formula that doesn't use the symbol $x$, with symbols $``y,a,b"$ as its free variables, then : $$\forall a \forall b:  \forall k \exists! x \forall y (y \in x \leftrightarrow y \in k \land \phi) $$

Reflection: if $\phi$ is a formula with symbols $``a,b"$ as its free variables, that doesn't use the symbol $\mathsf  v$ ; and if $\phi^\mathsf v$ is the formula obtained by bounding all quantifiers in $\phi$ with $\mathsf v$, then: $$\forall a \forall b \ \exists \text{ supertrs } \mathsf v : \phi \implies \phi^\mathsf v$$

Where:

$\mathsf v \text{ is supertrs} \equiv_{df} \forall m \forall n (m \in v \land (n \in m \lor n \subseteq m) \to n \in v)$

That is: a supertransitive set is a set closed under $\in$ and $\subseteq$ relations.

This is a modification of Dana Scott's theory about sets and stages, and I believe this proves all axioms of ZF-Reg., and even without Extensionality it can interpret ZFC.

This exposition of ZF- is actually very simple, neat and elegant. It is much shorter than the usual presentation of ZF- by axioms of Extensionality, Empty, Set union, Power, Separation, Replacement, and Infinity. Moreover, it appears to be pretty much natural and reduces all those diversily looking axioms into just two simple principles "we reflect, then specify!", that's all.

Since this is a well established result decades ago, then given the merits of shortage, elegance, and principled reduction; why it was not adopted as the official exposition of ZF-?

答:
Leaving aside communal inertia ("why change a perfectly fine presentation?"), which is of course significant, I don't see the benefit of this proposed reaxiomatization.

It is much shorter than the usual presentation of ZF- by axioms of Extensionality, Empty, Set union, Power, Separation, Replacement, and Infinity.

Sure, but length isn't an inherent indicator of quality.

This exposition of ZF- is actually very simple

Is it? Someone new to the field would have substantially more difficulty using it, I suspect, than using the usual $\mathsf{ZF-}$ axioms. Simplicity of presentation and simplicity of use are two different things, and in most situations I think the latter is more important than the former.

it appears to be pretty much natural and reduces all those diversily looking axioms into just two simple principles "we reflect, then specify!", that's all

Again, I don't really buy this: why should "we reflect, then specify" be an accepted bit of set-theoretic intuition? Motivating reflection is nontrivial. By contrast, the $\mathsf{ZF-}$ axioms are graded: consider the various levels of "justifiability" of e.g. Pairing, Infinity, Replacement, and Powerset. Someone who doesn't find all of them intuitive will still (probably) find some of them intuitive, and so a larger portion of the theory is "immediately acceptable." I think this is very useful for the pragmatic side of things: to a strong set theory skeptic we can argue that the "inoffensive part" of $\mathsf{ZF-}$ is still strong enough to do what they need, while it's not clear how to whip up something similar for your approach.

There's also the more technical aspect: if you're interested in fragments of $\mathsf{ZF-}$ (which I personally am very much), a more modular approach is significantly easier to use.

To sum up, I think that while this is of course a neat result, as an axiomatization of $\mathsf{ZF-}$ it is inferior to the standard one(s) in terms of intuitive appeal and immediate usability - and I don't think that its advantages make up for that.


问:
Eventually a null column of the matrix $(A-\lambda I)^{\alpha}$
I was doing some calculations with some matrices and I came across an interesting pattern, and I would like to know if what is happening is some fact or if it is coincidence with the matrices I am working on.

I will try to formalize my question in general, but I am working with $3\times 3$ and $4\times 4$ matrices, maybe the result is true in some specific dimensions, I don't know....

Suppose we have an $n\times n$ matrix $A$, I think we can first restrict our case to the real case, with real eigenvalues. Assuming the factored polynomial is given by,

$$p_c(t)=(t-\lambda_1)^{k_1}\cdot(t-\lambda_2)^{k_2}\cdots(t-\lambda_n)^{k_n}$$

Let some eigenvalue $\lambda_j$ with $k_j\ge 2$, and proceed with the calculation of $A-\lambda_j I$, where $I$ is the is the identity matrix.

What I noticed was that for some $\alpha\le k_j$, the matrix $(A-\lambda_j I)^{\alpha}$ has at least one null column.

So the question is this, if I have a eigenvalue with a multiplicity greater or equal than $2$, then there will be an exponent, less than the multiplicity, that the matrix $(A-\lambda_j I)^{\alpha}$ will have at least one null column.

I would like to know if there is any result that supports this observation that I made, or if it really was a coincidence given the matrices I am working on.

答:
Let $A$ be an $n \times n$ matrix with eigenvalue $\lambda$ of multiplicity $n,$ then $A-\lambda I$ has eigenvalue zero with multiplicity $n$, prove that $A^{n} =0.$ (Use Cayley-Hamilton)

Let $A$ be a matrix, then the Jordan form is given by  $J= PAP^{-1}$ for some matrix $P$. Now assume $\lambda$ is an eigenvalue of $A$ of multiplicity $r,$ then
\begin{align}
 P (A-\lambda I)^rP^{-1} &= P(A-\lambda I )P^{-1} P (A-\lambda I )P^{-1} \cdots P(A-\lambda I)P^{-1}\\
&=(P(A-\lambda I )P^{-1})^r = (J- \lambda I)^r.
\end{align}
Note that one block of $(J-\lambda)$ has $0$ on the diagonal(this block has eigenvalues 0), and multiplication is done block-wise, so this block ends up being $0$ in $(J-\lambda I)^r$, and as a consequence the rows and columns of this block in $(J-\lambda I)^r$ are zero, and the other blocks are of nonzero determinant.

$ (A-\lambda I)^r = P^{-1}(J-\lambda I)^rP.$ Conclude


问:
Finding joint probability function of Bernoulli and Discrete uniform.
X is a Bernoulli random variable with parameter p

and Y is a discrete uniform random variable that assumes the values 1, 2 and 3.

I am trying to find the distribution of (X,Y) without assuming independence between X and Y. Is it possible? Thanks in advance.

答:
My guess is you're looking for a distribution of $X$ and $Y$ for which the marginals are Bernoulli and discrete, respectively. Suppose $a, b, c, d, e, f$ are all nonnegative and add to one, and think about distributions looking like
\begin{equation}
\begin{matrix}
\bf{X=0} & a & b & c \\
\bf{X=1} & d & e & f \\
& \bf{Y=1} & \bf{Y=2} & \bf{Y=3}
\end{matrix}
\end{equation}


答:
Note that $a=\sqrt{17}$, $b=\sqrt{77}$ and $\cos C= -\frac{20}{ab}$, which yields the area

$$A= \frac12 ab \sqrt{1-\cos^2C}=\frac12 \sqrt{17\cdot77-400}=\frac{3\sqrt{101}}2$$

Title: Vectors in 3D and area of a triangle

Body:
I have three points with coordinates: $A (5,-1,0),B(2,4,10)$, and $C(6,-1,4)$.

I have the following vectors $\overrightarrow {CA} = (-1, 0, -4)$ and $\overrightarrow{CB} = (-4, 5, 6)$.

To find the area of the triangle I used the dot product between these vectors to get the angle and then applied the formula $A=0.5ab\sin{C}$ to find the area of the triangle which gave me $15.07(2dp)$.

However in the given solutions the answer is given as $(3*\sqrt(102))/2$

I think they have used the trig identity $\cos^2(\theta) + \sin^2(\theta) = 1$ to find the value of $\sin(\theta)$ rather than $\arccos(\theta)$ to find the angle ACB. However I don't understand why there would be such a discrepancy between the two answers; one using $\arccos$ and the other using the trig identity.


答:
Your book got a typo: it's $101$ and not $102$. And that's it, that is approximately your answer.

Hope this helps. Ask anything if not clear :)

Title: Vectors in 3D and area of a triangle


问:
How to prove this subspace of $W^{2,2}$ is Hilbert?
Let $I=(l_0,l_1)\subset \mathbb{R}$

$$H^2_{l_0}=\{u\in H^2(I): u(l_0)=u'(l_0)=0\},$$
endowed with the inner product $$\langle u,v\rangle_{H^2_{l_0}}:=\left\langle u'',v''\right\rangle_{L^2(I)}=\int_I u''v''\,dx,$$ for real valued functions.
Here $H^2(I):=W^{2,2}(I)$

I have proven that $\left<\cdot,\cdot\right>_{H^2_{l_0}}$ defines an inner product, it also defines a norm. So far I have seen $H^2_{l_0}$ is a pre-Hilbert space, but I'm having troubles proving the completeness with the metric induced by the inner product.

I really appreciate your help.

答:
Take a Cauchy Sequence $(u_m)_m $ in $H^2_{l_0}(I)$. By poincaré inequality twice, there is $C_1,C_0>0$ such that
$$C_0||u_m-u_m\|_{L^2} \leq C_1||u'_m-u'_m\|_{L^2} \leq  ||u''_m-u''_m\|_{L^2} \quad \forall n,m \in \mathbb{N} $$

Because $L^2(I)$ is complete there is a $w_0,w_1,w_2 \in L^2(I)$ such that
\begin{align}
\|u_m''-w_2\|_{L^2} & \to 0 \\
\|u_m'-w_1\|_{L^2} & \to 0 \\
\|u_m-w_0\|_{L^2} & \to 0 \\
\end{align}
We are going to prove that $w''_0=w_2$.
\begin{align}
\int_0^{l_0} w_0 v'' &= \int_0^{l_0} (w_0-u_m) v''+\int_0^{l_0} u_mv'' \\
 & = \underbrace{\int_0^{l_0} (w_0-u_m)v''}_{\to 0 ~ \text{when} ~ m \to \infty  } + \underbrace{ u_m(x)v'(x)-u'_m(x)v(x) |^{l_0}_{0}}_{0} +\int_0^{l_0} u''_mv \quad (\text{Integration by parts).} \\
 & = \lim_m\int^{l_0}_0 u''_m v = \int^{l_0}_0 w_2 v \quad \forall
v \in C^{\infty}_0(I)\end{align}

Hence, $w_0''=w_w.$

Observe that the Poincare inequality gives
$w_0'=w_1$. Additionally, $\|u_m-w_0\|_{H^2}\to 0$ when $m \to \infty$. Therefore  $w_0 \in H^2(I)$. Now we must check the boundary conditions.

\begin{align}
  (w(x)v'(x)-w'(x)v(x)) |^{l_0}_{0} &= \int_{0}^{l_0} w''(x)v(x)-w(x)v''(x) \\
&= \lim_m \int_{0}^{l_0} u_m''(x)v(x)-u_m(x)v''(x) \\
&= \lim _m ~ (u_m(x)v'(x)-u_m'(x)v(x) )|^{l_0}_{0} \\
&= \lim_m (u_m(0) v'(0)-u_m'(0)v(0)) \quad \forall v \in H^2(I) \tag{*}
\end{align}
Because $H^2(I)$ is embedded into $C^{1}(\bar I)$. You can select $v$ with $v(0)=v(0)=v'(0)=0$ and $v(l_0) \neq 0 $. By (*) results $w(0)v(l_0)=0$, consequently $w(l_0)=0$, analogously $w'(l_0)=0$. Then, $w_0 \in H^2_{l_0}(I)$. In other words, $H^2_{l_0}(I)$ is closed.


问:
Understanding an almost convergence theorem
I am reading a proof of the following result :
$$X_n \, \xrightarrow{\mathrm{a.s.}} \, X. \iff \forall \varepsilon > 0, \lim_{n \rightarrow + \infty }
\mathbb{P} ( \sup_{k \geq n} |X_k - X | \geq \varepsilon ) = 0 $$

to prove $\implies$, we fix a $\varepsilon$ and we define $A_n := \{ \sup_{k \geq n} |X_k - X | \geq \varepsilon \} $ , $C  := \{ \omega, \lim X_n (\omega ) = X( \omega) \} $ and $B_n := C \cap A_n$
The proof states that $B_{n+1} \subset B_n$ and this is true because of the construction of $A_n$ and $\mathbb{P}(C) = 1$ which is due to the almost convergence.

Afterwards, the proof stats that $\cap_{n} B_n = \varnothing $ and I don't understand why. Can someone help me understand this point ?

Thank you for your help

答: If $X_n(\omega) \to X(\omega) $ then $\sup_{k \geq n} |x_n(\omega)-X(\omega)| <\epsilon$ for some $n$ . For that $n$, $\omega \notin A_n$ hence not in $\cap_k A_k$.


问:
Any good upper bounds for $\sum_{i = 0}^{m} 2^i {k+i \choose i}$?
I am trying to solve a recursion (I faced it when I wanted to analyze the running time of an algorithm precisely). I reduced it to compute a "good" upper bound for $\sum_{i = 0}^{m}2^i {k + i \choose i}$. We have these assumptions that $1 < m \leq Ck$, for some $C\in \mathbb{N}$.
I have tried many approximations but the bounds are not what I want. They are very loose. On the other hand, it doesn't look like a very sophisticated summation but I have so much difficulty solving it!
Does anyone have any ideas?

答:
First step is that  we can actually give to the sum a closed expression in terms of the Hypergeometric function

In fact can rewrite the sum as follows (allow me to change a bit your notation)
$$
\eqalign{
  & S(n,m,x) = \sum\limits_{i = 0}^m {\left( \matrix{ n + i \cr i \cr}  \right)x^{\,\,i} }
  = \sum\limits_{i = 0}^m {\left( { - 1} \right)^{\,\,i} \left( \matrix{ - n - 1 \cr i \cr}  \right)x^{\,\,i} }  = \quad \quad (1)  \cr
  &  = \sum\limits_{0\, \le \,i\, \le \,m} {\left( \matrix{ - n - 1 \cr i \cr}  \right)\left( { - x} \right)^{\,\,i} }
  = \left( { - x} \right)^{\,\,m} \sum\limits_{0\, \le \,k\,\left( { \le \,m} \right)}
 {\left( \matrix{ - n - 1 \cr  m - k \cr}  \right)\left( { - x} \right)^{\, - k} }  = \quad \quad (2)  \cr
  &  = x^{\,\,m} \sum\limits_{0\, \le \,k\,\left( { \le \,m} \right)}
 {\left( \matrix{n + m - k \cr m - k \cr}  \right)x^{\, - k} }
  = x^{\,\,m} \sum\limits_{0\, \le \,k\, \le \,m}
 {{{\left( {n + m - k} \right)!} \over {n!\left( {m - k} \right)!}}x^{\, - k} }  = \quad \quad (3)  \cr
  &  = x^{\,\,m} {{\left( {n + m} \right)!} \over {m!n!}}\sum\limits_{0\, \le \,k\, \le \,m}
 {{{\left( {n + m - k} \right)!m!} \over {\left( {n + m} \right)!\left( {m - k} \right)!}}x^{\, - k} }  = \quad \quad (4)  \cr
  &  = x^{\,\,m} \left( \matrix{ n + m \cr m \cr}  \right)\sum\limits_{0\, \le \,k\,\left( { \le \,m} \right)}
 {{{\left( \matrix{m \cr k \cr}  \right)} \over {\left( \matrix{n + m \cr k \cr}  \right)}}x^{\, - k} }  = \quad \quad (5)  \cr
  &  = x^{\,\,m} \left( \matrix{n + m \cr  m \cr}  \right)\sum\limits_{0\, \le \,k\,\left( { \le \,m} \right)}
 {{{m^{\,\underline {\,k\,} } } \over {\left( {n + m} \right)^{\,\underline {\,k\,} } }}x^{\, - k} }  = \quad \quad (6)  \cr
  &  = x^{\,\,m} \left( \matrix{n + m \cr m \cr}  \right)\sum\limits_{0\, \le \,k\,\left( { \le \,m} \right)}
 {{{\left( { - m} \right)^{\,\overline {\,k\,} } }
 \over {\left( { - n - m} \right)^{\,\overline {\,k\,} } }}x^{\, - k} }  = \quad \quad (7)  \cr
  &  = x^{\,\,m} \left( \matrix{  n + m \cr m \cr}  \right)
 {}_2F_{\,1} \left( {\left. {\matrix{ {1,\; - m}  \cr { - \left( {n + m} \right)}  \cr } \;} \right|\;{1 \over x}} \right)
 \quad \quad (8) \cr}
$$

where $a^{\,\underline {\,k\,} } ,\quad a^{\,\overline {\,k\,} } $ represent respectively the Falling and Rising Factorial
and the various steps are:

-1) upper negation;

invert the summation index;

binomial expressed as factorials;

extraction of the leading binomial component;

, 6), 7) rewriting of the binomial fraction

conversion to Hypergeometric.
Note the upper bound of the sum being indicated in brackets when it is implicit in the binomial or falling factorial.

For what concerns the establishing of an upper bound, it seems that you are referring
to the case in which $n << m$.

If it is so, we can use one of the expression above, eg. the (6). The coefficient in the sum can be rewritten in various ways
$$
\eqalign{
  & {{m^{\,\underline {\,k\,} } } \over {\left( {m + n} \right)^{\,\underline {\,k\,} } }}\,
 = {{\left( {m + n - k} \right)^{\,\underline {\,n\,} } } \over {\left( {m + n} \right)^{\,\underline {\,n\,} } }}
 = {{\left( {m + 1 - k} \right)^{\,\overline {\,n\,} } } \over {\left( {m + 1} \right)^{\,\overline {\,n\,} } }} =   \cr
  &  = {{\Gamma \left( {m + 1 + n - k} \right)\Gamma \left( {m + 1} \right)}
 \over {\Gamma \left( {m + 1 - k} \right)\Gamma \left( {m + 1 + n} \right)}} =   \cr
  &  = \prod\limits_{j = 0}^{n - 1} {{{m + 1 + j - k} \over {m + 1 + j}}}
  = \prod\limits_{j = 0}^{n - 1} {\left( {1 - {k \over {m + 1 + j}}} \right)}  \cr}
$$
and from these we can proceed with various approximations, depending on the required acurracy.

For instance
$$
{{m^{\,\underline {\,k\,} } } \over {\left( {m + n} \right)^{\,\underline {\,k\,} } }}
 = \prod\limits_{j = 0}^{n - 1} {\left( {1 - {k \over {m + 1 + j}}} \right)}
  < \mathop  \approx \limits_{n <  < m} \left( {1 - {k \over {m + \left( {n + 1} \right)/2}}} \right)^n
$$
and then proceeding to take the log and simplify (with the accuracy needed), e.g.
$$
 \approx \exp \left( { - {{nk} \over {m + \left( {n + 1} \right)/2}}} \right)
$$
which is suitable to be summed in $k$.

Or you can start from the expression through the Gamma function and apply the Stirling approximation, ...


问:
Showing that for every $e > 0$, there exist quasi-equilateral triangles with error $e$ whose vertices have integer coordinates in the plane
I'm currently working on the problem stated below, but i'm still kind of a beginner with proofs, so I would appreciate some tips (not a solution) on how to approach this proof.

A triangle is quasi-equilateral with error $e > 0$, if, for every angle $A$, $B$, $C$ in the triangle, we have that $\vert{A-60^{\circ}}\vert < e$, $\vert{B-60^{\circ}}\vert < e$ and $\vert{C-60^{\circ}}\vert < e $. Show that for every $e > 0$, there exist quasi-equilateral triangles with error $e$ that have vertices with integer coordinates in the plane.

I know how to prove that there are no equilateral triangles with integer coordinates, however, this looks much more difficult and I can't find a way to begin.

答:
Idea: set the origin $O$ be the circumcenter of the triangle $ABC$ and use the Pythagorean triple theorem.

Hint 1

Set the vertice $C$ on the axis $\vec{Ox'}$ (the coordinate of $C$ is $(-c,0)$ with $c\in \Bbb N^*)$. The origin $O$ is the circumcenter of the triangle $ABC$. The coordinate of points $A$ and $B$ are $(a,b)$ and $(a,-b)$ with $a,b \in \Bbb N^*$

Hint 2

 The length $OA = OB = \sqrt{a^2+b^2}$ must in $\Bbb N^*$ because $OA = OB = OC$ with $C$ has integer coordinate on the axis $\vec{Ox}$

Hint 3

The solution of Pythagorean triple $a^2+b^2 = c^2$ (link) is, for example, $(a,b,c) = (m^2-n^2,2mn, m^2+n^2)$ with $m,n \in \Bbb N^*, m>n$.

Hint 4

Calculate the $\cos(\widehat{AOx}) = \frac{m^2-n^2}{m^2+n^2}$ which must be in the interval $(v_1,v_2)=\left(\cos(\frac{\pi}{3}+\epsilon), \cos(\frac{\pi}{3}-\epsilon)   \right)$.

Hint 5

Prove there always exists a rational number  $t\in \Bbb Q^+$ such that $$\sqrt{\frac{1+v_1}{1-v_1}} <t < \sqrt{\frac{1+v_2}{1-v_2}}$$ with $(v_1,v_2)$ defined in the hint 4. The couple $(m,n)$ will be defined by $t = \frac{m}{n}$. Hence, you deduce the value of $(a,b,c)$ from $(m,n)$


问:
minimal sufficient statistic
Let X ~ Ber(n1; p), Y ~ Ber(n2; p^2), where X and Y are independent. Find a minimal
sufficient statistic T and, using a nontrivial function, show that it is not complete.

I get confused by having two different distribution! is it true to simply multiply these two distribution and use factorization theory?

答:
First write the likelihood function: $$f(x,y|p) = [p^x(1-p)^{1-x}][p^{2y}(1-p^2)^{1-y}]$$
From this use the Lehmann-Scheffe Theorem. Need to find $T$ such that $$T(x_1, y_1) = T(x_2, y_2) \Leftrightarrow \frac{f(x_1,y_1|p)}{f(x_2,y_2|p)} = c(x_1, y_1, x_2, y_2)$$
To this end, let's compute the likelihood ratio:
\begin{align}
\frac{f(x_1,y_1|p)}{f(x_2,y_2|p)} &=
\frac{p^{x_1}(1-p)^{1-x_1}p^{2y_1}(1-p^2)^{1-y_1}}{p^{x_2}(1-p)^{1-x_2}p^{2y_2}(1-p^2)^{1-y_2}} \\
&= p^{(x_1 - x_2)} (1-p)^{x_2 - x_1}p^{2(y_1 - y_2)} (1-p^2)^{y_2-y_1}
\end{align}
It can be seen then that the ratio does not depend on $p$ if and only if $x_1 = x_2, y_1 = y_2$. By the Lehmann-Scheffe, this implies that $T(X, Y) = (X, Y)$ is the minimal sufficient statistic.

Title: minimal sufficient statistic


问:
Using prior conjugates to find posterior distribution
If we have a negative binomial like
$N \sim NegBin(r, p)$.

Then
$(N = i \mid p) = \binom{i - 1}{r - 1} p^r (1 - p)^{i - r} \mathbb{1}(\{i \in \{r, r + 1, \dots)$

We are given that $p$  has a beta prior with distribution $a$ and $b$.
I am trying to find the posterior distribution $[p∣N,r,a,b]$. Do I need to find the density of the negative binomial distribution first?

答:
The negative binomial distribution is a discrete distribution so has a probability mass function (as you have stated - note there are other parametrisations of the negative binomial) rather than a density.  It is the Beta distribution which has a density, proportional to $p^a(1-p)^b$.

If you have a single observation $x$ from your negative binomial distribution with $x \in \{r,r+1,\ldots\}$ then the posterior density for $p$ given $x$ will be (ignoring multiplicative constants) proportional to $p^a (1-p)^b p^r (1-p)^{x-r}$ when $0 <p<1$, with the first part coming from the Beta prior and the second part from the negative binomial likelihood.

Tidying this up gives a posterior density proportional to $p^{a+r}(1-p)^{b+x-r}$, which is the density of a Beta distribution with parameters $a+r$ and $b+x-r$.

If instead you had $n$ observations, the posterior distribution would be a  Beta distribution with parameters $a+nr$ and $b+(\sum x_j)-nr$.


问:
Help with this trigonometric identity
Can anyone help me? Why is

$$\frac{1}{\sin(x) + \cos(x) + 1} = \frac{\sec^{2}\big(\frac{x}{2}\big)}{2\tan\big(\frac{x}{2}\big) + 2}$$

Title: Help with this trigonometric identity

Body:
$$\sin x + \cos x + 1 = 2 \sin \frac{x}{2} \cos \frac{x}{2} + \cos^2 \frac{x}{2} - \sin^2 \frac{x}{2} + \sin^2 \frac{x}{2} + \cos^2 \frac{x}{2}$$

$$= 2 \cos \frac{x}{2} \left(\cos \frac{x}{2} + \sin \frac{x}{2}\right) = 2 \cos^2 \frac{x}{2} \left(1 + \tan \frac{x}{2}\right) $$

Hence $$\frac{1}{\sin x + \cos x + 1} = \frac{1}{2 \cos^2 \frac{x}{2} \left(1 + \tan \frac{x}{2}\right)} = \frac{\sec^2 \frac{x}{2}}{2 \tan \frac{x}{2} + 2}$$


Title: Help with this trigonometric identity

Body:
Can anyone help me? Why is

$$\frac{1}{\sin(x) + \cos(x) + 1} = \frac{\sec^{2}\big(\frac{x}{2}\big)}{2\tan\big(\frac{x}{2}\big) + 2}$$

Body:
Let $\displaystyle y = \frac{x}{2}$.  Then:

$$\frac{1}{\sin(x) + \cos(x) + 1} = \frac{1}{\sin(2y) + \cos(2y) + 1}$$

Using the double-angle identities:

$$ = \frac{1}{2\sin(y)\cos(y) + 2\cos^{2}(y) - 1 + 1} = \frac{1}{2\sin(y)\cos(y) + 2\cos^{2}(y)}$$

$$ = \frac{\frac{1}{\cos^{2}(y)}}{2\cdot\frac{\sin(y)}{\cos(y)} + 2} = \frac{\sec^{2}(y)}{2\tan(y) + 2} =\boxed{\frac{\sec^{2}\big(\frac{x}{2}\big)}{2\tan\big(\frac{x}{2}\big)
 + 2}}\ \ \blacksquare$$


问:
Dice Roll Probability of winning
Dice A sides: 4, 4, 4, 4, 13, 13

Dice B sides: 6, 6, 6, 9, 9, 9

If I roll both dice

Which dice is more likely to win here? How can I approach this?

答: Presumably winning means getting the larger number.  As stated dice B numbers are all in between both dice A numbers.  2/3 of A rolls are 4 (A loses), while 1/3 are 13 (A wins).  So A wins 1/3 of the time.</p>

Title: Dice Roll Probability of winning


问:
Is there a tree of Pythagorean quadruples?
It is well-known that there is a trinary tree that contains every primitive Pythagorean triple exactly once. It even has a Wikipedia page.

Is there a similar tree that contains each primitive Pythagorean quadruple exactly once?

答:
You can find the answer in the following paper:
http://math.colgate.edu/~integers/u73/u73.pdf</p>


问:
Which one result in mathematics has surprised you the most?
A large part of my fascination in mathematics is because of some very surprising results that I have seen there.

I remember one I found very hard to swallow when I first encountered it, was what is known as the Banach Tarski Paradox. It states that you can separate a ball $x^2+y^2+z^2 \le 1$ into finitely many disjoint parts, rotate and translate them and rejoin (by taking disjoint union), and you end up with exactly two complete balls of the same radius!

So I ask you which are your most surprising moments in maths?

Chances are you will have more than one. May I request post multiple answers in that case, so the voting system will bring the ones most people think as surprising up. Thanks!

答:
Mamikon’s Theorem: The area of a tangent sweep is equal to the area of its tangent cluster, regardless of the original shape of the curve.

This theorem allows you to, among other things, easily obtain results that were obtained before only with difficulty, such as the area under one arch of a cycloid. This theorem is the basis of what has come to be known as Visual Calculus. Here’s the link to Tom Apostol’s account of this awesome insight:

http://eands.caltech.edu/articles/Apostol%20Feature.pdf

LastEditorDisplayName: anon


问:
Area of intersection between two circles
Suppose you have 2 circles that intersect each other in such a way that each circle passes through the other's center. What is the area between the circle(or common area) i.e. area between the centres of the circles?

答:
The angle between points of intersection is $2 \pi / 3$, so it is:
$$
2 r^2 \cdot \left( \frac{\pi}{3} - \frac{\sqrt{3}}{2} \right)
$$</p>


问:
Poisson-gamma mixture when likelihood terms are raised to powers
Consider a Poisson model with gamma prior:

$$
x \sim \text{Poisson}(\theta), \quad \theta \sim \text{gamma}(\alpha, \beta).
$$

I am interested in deriving the posterior predictive for this model with a power $\zeta_i$ raised to each term in the likelihood:

$$
\mathcal{L}(\theta) \propto p(\theta) \prod_{i=1}^t p_{\theta}(x)^{\zeta_i}
$$

Without these powers, it's straightforward to show that the posterior predictive is negative binomially distributed, e.g.

https://en.wikipedia.org/wiki/Negative_binomial_distribution#Gamma%E2%80%93Poisson_mixture

https://people.stat.sc.edu/Hitchcock/stat535slidesday18.pdf.

I'm fairly certain based on prior work (e.g. see equation 2.4 here: https://jwmi.github.io/publications/C-posterior.pdf) that this power-based likelihood will still be tractable. However, I can't push the powers through the derivation cleanly.

Here's what I have. The posterior is tractable:

$$
\begin{aligned}
p(\theta \mid X_{1:t})
&\propto \pi(\theta) \prod_{i=1}^t p_{\theta}(x)^{\zeta_i}
\\
&= \frac{\beta^{\alpha}}{\Gamma(\alpha)} \theta^{\alpha-1} e^{-\beta\theta} \prod_{i=1}^t \left(\frac{\theta^{x_i} e^{-\theta}}{x_i!} \right)^{\zeta_i}
\\
&\propto \theta^{\alpha-1} e^{-\beta\theta} \prod_{i=1}^t \left( \theta^{x_i} e^{-\theta} \right)^{\zeta_i}
\\
&\propto \theta^{\sum_i \zeta_i x_i + \alpha-1} e^{-(\sum_i \zeta_i + \beta) \theta}
\end{aligned}
$$

It's just gamma distributed:

$$
\pi(\theta \mid X_{1:t}) = \text{gamma}\!\left(\sum_{i=1}^t \zeta_i x_i + \alpha, \sum_{i=1}^t \zeta_i + \beta \right).
$$

The posterior predictive is therefore

$$
\begin{aligned}
p(x_{t+1} \mid X_{1:t})
&= \int_{\Theta} p(x_{t+1} \mid \theta) p(\theta \mid X_{1:t}) \text{d}\theta
\\
&= \int_{\Theta} \left[\frac{\theta^{x_{t+1}} e^{-\theta}}{x_{t+1}!}\right]^{\zeta_{t+1}} \left[ \frac{(\sum_i \zeta_i + \beta)^{\sum_i \zeta_i x_i + \alpha}}{\Gamma(\sum_i \zeta_i x_i + \alpha)} \theta^{\sum_i \zeta_i x_i + \alpha - 1} e^{-(\sum_i \zeta_i + \beta) \theta} \right]\text{d}\theta
\\
&= \frac{(\sum_i \zeta_i + \beta)^{\sum_i \zeta_i x_i + \alpha}}{\Gamma(\sum_i \zeta_i x_i + \alpha)(x_{t+1}!)^{\zeta_{t+1}}} \int_{\Theta} \theta^{\zeta_{t+1} x_{t+1} + \sum_i \zeta_i x_i + \alpha - 1} e^{-(\zeta_{t+1} + \sum_i \zeta_i + \beta) \theta} \text{d}\theta
\\
&= \frac{(\sum_i \zeta_i + \beta)^{\sum_i \zeta_i x_i + \alpha}}{\Gamma\left(\sum_i \zeta_i x_i + \alpha\right) (x_{t+1}!)^{\zeta_{t+1}}} \frac{\Gamma(\zeta_{t+1} x_{t+1} + \sum_i \zeta_i x_i + \alpha)}{\left(\zeta_{t+1} + \sum_i \zeta_i + \beta\right)^{\zeta_{t+1} x_{t+1} + \sum_i \zeta_i x_i + \alpha}}
\\
&= \frac{\Gamma(\zeta_{t+1} x_{t+1} + \sum_i \zeta_i x_i + \alpha)}{\Gamma\left(\sum_i \zeta_i x_i + \alpha\right) (x_{t+1}!)^{\zeta_{t+1}}} \left(\frac{\sum_i \zeta_i + \beta}{\zeta_{t+1} + \sum_i \zeta_i + \beta} \right)^{\sum_i \zeta_i x_i + \alpha} \left(\frac{1}{\zeta_{t+1} + \sum_i \zeta_i + \beta} \right)^{\zeta_{t+1} x_{t+1}}
\end{aligned}
$$

This is almost negative binomally distributed. However, I don't know what to do with $(x_{t+1}!)^{\zeta_{t+1}}$. It seems like with a nice trick, maybe a property of the gamma distribution raised to a power, this can be resolved to induce a negative binomial distribution with parameters that are functions of $x_{t+1}$, $\zeta_{t+1}$, and the posterior parameters.

答:
I want to clean up the notation.  Let the posterior gamma hyperparameters be $$\alpha' = \sum_{i=1}^t \zeta_i x_i + \alpha, \quad \beta' = \sum_{i=1}^t \zeta_i + \beta.$$  And for convenience, let us write $y = x_{t+1}$ and $\xi = \zeta_{t+1}$.  Then your question basically boils down to evaluating
$$\begin{align}
p(y \mid x_{1:t}, \xi)
&\propto \int_{\theta=0}^\infty \left(\frac{e^{-\theta} \theta^y}{y!}\right)^\xi \frac{(\beta')^{\alpha'} \theta^{\alpha'-1} e^{-\beta' \theta}}{\Gamma(\alpha')} d\theta \\
&= \frac{(\beta')^{\alpha'}}{(y!)^\xi \Gamma(\alpha')} \int_{\theta=0}^\infty \theta^{y\xi + \alpha' - 1} e^{-(\xi + \beta')\theta} \, d\theta \\
&= \frac{(\beta')^{\alpha'}}{(y!)^\xi \Gamma(\alpha')} \cdot \frac{\Gamma(y\xi + \alpha')}{(\xi + \beta')^{y \xi + \alpha'}}.
\end{align}$$
This is equivalent to what you obtained, just written more compactly.  Note that this posterior predictive likelihood is not in general a density unless $\xi = 1$.  Hence we are motivated to consider the kernel of the posterior predictive likelihood:
$$\kappa(y \mid x_{1:t}, \xi) = \frac{\Gamma(y\xi + \alpha')}{\Gamma(y+1)^\xi (\xi + \beta')^{\xi y}}.$$  It is at this point it becomes very plainly obvious that there is no way for the posterior predictive to be negative binomial, whose kernel is $$\kappa(y) = \frac{\Gamma(y+r)}{\Gamma(y+1)} p^y,$$ unless $\xi = 1$.  A generalization could be considered for positive integer $\xi$, such that the ratio $$\frac{\Gamma(y\xi + \alpha')}{\Gamma(y+1)^\xi} \propto \binom{y\xi + \alpha' - 1}{y, y,  \ldots, y}$$ is proportional to a $\xi+1$-nomial coefficient, but to my knowledge, there is no name for such a distribution.  And if $\xi$ is not an integer, I should think there is no hope for a simple model; the integrating factor to make it a density is likely to not have a closed form.


问:
What happens in the long run?
Suppose we have a round table with 4 boys, each having a bowl with cereals. There is an adult in the same room. When the adult leaves the room, the boys start stealing each other's cereal, but they can $\textbf{only}$ steal from its neighbor as they $\textbf{cannot}$ leave their seat. When the adult comes back, they stop stealing each other's cereal.

${\bf question:}$ If the adult leaves and comes back $N$ times, can we say about the number of cereals in each boy's bowl? Are they eventually all have the same number of cereals?

${\bf thought}:$ We can start by letting $x_i$ be the number of cereal in ith bowl. So, initially we can encode our situation in a vector ${\bf x} = (x_1,x_2,x_3,x_4)$. After the adult leaves the room, now $x_1$ becomes $x_1 + k_1 x_2 + k_2 x_4$ ;
$x_2$ becomes $x_2 + k_3 x_1 + k_4 x_3$; $x_3$ becomes $x_3 + k_5 x_2 + k_6 x_4$ and $x_4$ becomes $x_4 + k_7 x_3 + k_8 x_1$, so after one step we obtain new vector

$$ {\bf y }  = \left( \begin{matrix}  x_1 + k_1 x_2 + k_2 x_4 \\ x_2 + k_3x_1 + k_4 x_3 \\  x_3 + k_5 x_2 + k_6 x_4 \\ x_4 + k_7 x_3 + k_8 x_1 \end{matrix} \right) = \underbrace{ \left( \begin{matrix} 1 & k_1 & 0 & k_2 \\ k_3 & 1 & k_4 & 0 \\ 0 & k_5 & 1 & k_6 \\ k_8 & 0 & k_7 & 1 \end{matrix} \right) }_{A} {\bf x}$$

So, in order to solve this problem, we may have to compute $A^N {\bf x}$ which seems pretty laborious.

Is this a correct approach to the problem?

${\bf Update:}$
Let us assume that at each round, each boy takes half the number of cereals from the neighbor. In this case, then

$$ {\bf y} = \left( \begin{matrix} (x_2+x_4)/2 \\ (x_1+x_3)/2 \\ (x_2+x_4)/2 \\ (x_1+x_3)/2 \end{matrix} \right) $$

答:
Assume that the stealings, made and suffered, when summed over the period of teacher's absence might turn out to be proportional to the starting amounts on which they are perpetrated.

Then for the guy in e.g. position $2$ we will have the following incoming and outgoing flux

That is
$$
\left\{ \matrix{
  \Delta x_{\,1}  = k_{\,1,\,2} x_{\,2}  + k_{\,1,\,4} x_{\,4}
  - \left( {k_{\,2,\,1}  + k_{\,4,\,1} } \right)x_{\,1}  \hfill \cr
  \Delta x_{\,2}  = k_{\,2,\,1} x_{\,1}  + k_{\,2,\,3} x_{\,3}
  - \left( {k_{\,1,\,2}  + k_{\,3,\,2} } \right)x_{\,2}  \hfill \cr
  \Delta x_{\,3}  = k_{\,3,\,2} x_{\,2}  + k_{\,3,\,4} x_{\,4}
  - \left( {k_{\,2,\,3}  + k_{\,4,\,3} } \right)x_{\,3}  \hfill \cr
  \Delta x_{\,4}  = k_{\,4,\,1} x_{\,1}  + k_{\,4,\,3} x_{\,3}
  - \left( {k_{\,1,\,4}  + k_{\,3,\,4} } \right)x_{\,4}  \hfill \cr}  \right.
$$
where the factors will presumably be non-negative (no donations).

We shall also assume that the out-flux at each step will not be such as to leave the affetcted cup empty, independently from the in-flux.
So it is a reasonable model that the factors, at least those in parenthesis,  will lay in the range $0 \le k_{\, n\, m} \le 1/2$.
It is also assumed that the cereals, between the various mutual stealings, are not eaten, so that the total quantity
remains fixed, and the sum of the net incomings is null.

In matricial notation
$$
\left( {\matrix{ {\Delta x_{\,1} }  \cr {\Delta x_{\,2} }  \cr {\Delta x_{\,3} }  \cr {\Delta x_{\,4} }  \cr } } \right)
 = \left( {\matrix{
   { - \left( {k_{\,2,\,1}  + k_{\,4,\,1} } \right)} & {k_{\,1,\,2} } & 0 & {k_{\,1,\,4} }  \cr
   {k_{\,2,\,1} } & { - \left( {k_{\,1,\,2}  + k_{\,3,\,2} } \right)} & {k_{\,2,\,3} } & 0  \cr
   0 & {k_{\,3,\,2} } & { - \left( {k_{\,2,\,3}  + k_{\,4,\,3} } \right)} & {k_{\,3,\,4} }  \cr
   {k_{\,4,\,1} } & 0 & {k_{\,4,\,3} } & { - \left( {k_{\,1,\,4}  + k_{\,3,\,4} } \right)}  \cr
 } } \right)
\left( {\matrix{{x_{\,1} }  \cr {x_{\,2} } \cr {x_{\,3} }  \cr {x_{\,4} }  \cr  } } \right)
$$
i.e.
$$
\Delta {\bf x} = {\bf M}\,{\bf x}
$$

Designating the unit vector and its transpose as
$$
{\bf u} = \left( {\matrix{ 1  \cr  1  \cr  1  \cr   1  \cr } } \right)
\quad \overline {\bf u}
  = \left( {1,1,1,1} \right)
$$
we have infact
$$
\overline {\bf u} \;{\bf M} = {\bf 0}\quad  \Rightarrow \quad \overline {\bf u} \;\Delta {\bf x}
 = \Delta \left( {\overline {\bf u} \;{\bf x}} \right) = \Delta N = 0
$$
So $\bf M$ has determinant null, and it has rank $3$ in general:  for instance the minor in the upper left corner
has determinant
$$
\left| {{\bf M}_3 } \right| =  - k_{\,1,\,2} k_{\,4,\,1} \left( {k_{\,2,\,3}  + k_{\,4,\,3} } \right)
 - k_{\,3,\,2} k_{\,4,\,3} \left( {k_{\,2,\,1}  + k_{\,4,\,1} } \right)
$$
The rank will be lower than $3$ for particular values of the $k$'s.

If the rank is $3$ then the null-space of $\bf H$ has dimension $1$.

At equilibrium we shall have
$$
\Delta {\bf x}_\infty   = 0\quad
  \Rightarrow \quad {\bf x}_\infty   \in {\rm null - space}\left( {\bf M} \right)
$$
as well as that the total of the bowls content shall remain constant

$$
\overline {\bf u} \;{\bf x}_\infty   = \overline {\bf u} \;{\bf x}_0
$$
We can then solve the homogeneous system to get
$$ \bbox[lightyellow] {
{\bf x}_\infty   = \lambda \;\left( {\matrix{
   {k_{\,1,\,2} k_{\,2,\,3} \left( {k_{\,1,\,4}  + k_{\,3,\,4} } \right)
 + k_{\,1,\,4} k_{\,4,\,3} \left( {k_{\,1,\,2}  + k_{\,3,\,2} } \right)}  \cr
   {k_{\,2,\,1} k_{\,3,\,2} \left( {k_{\,1,\,4}  + k_{\,3,\,4} } \right)
 + k_{\,4,\,1} k_{\,3,\,4} \left( {k_{\,1,\,2}  + k_{\,3,\,2} } \right)}  \cr
   {k_{\,2,\,1} k_{\,1,\,4} \left( {k_{\,2,\,3}  + k_{\,4,\,3} } \right)
 + k_{\,2,\,3} k_{\,3,\,4} \left( {k_{\,2,\,1}  + k_{\,4,\,1} } \right)}  \cr
   {k_{\,1,\,2} k_{\,4,\,1} \left( {k_{\,2,\,3}  + k_{\,4,\,3} } \right)
 + k_{\,3,\,2} k_{\,4,\,3} \left( {k_{\,2,\,1}  + k_{\,4,\,1} } \right)}  \cr
 } } \right)
}$$
with $\lambda$ being the constant that provides the above normalization.

Then in the example in which the "rapacities" are unbiased , i.e. when $k_{1,2} = k_{1,4} = k_1$ and so on, we have
$$
{\bf M} = \left( {\matrix{
   { - \left( {k_{\,2}  + k_{\,\,4} } \right)} & {k_{\,1} } & 0 & {k_{\,1} }  \cr
   {k_{\,2} } & { - \left( {k_{\,1}  + k_{\,3} } \right)} & {k_{\,2} } & 0  \cr
   0 & {k_{\,3} } & { - \left( {k_{\,2}  + k_{\,\,4} } \right)} & {k_{\,3} }  \cr
   {k_{\,\,4} } & 0 & {k_{\,\,4} } & { - \left( {k_{\,1}  + k_{\,3} } \right)}  \cr
 } } \right)
$$
and
$$
{{{\bf x}_\infty  } \over {\overline {\bf u} \;{\bf x}_0 }}
 = {1 \over {k_{\,1}  + k_{\,2}  + k_{\,3}  + k_{\,\,4} }}\;
\left( {\matrix{{k_{\,1} }  \cr{k_{\,2} }  \cr{k_{\,3} }  \cr{k_{\,\,4} }  \cr } } \right)
$$
which means that the contents of the bowls will tend to stabilize in proportion to the rapacities
independently from the starting point.

But for the sequence of the  ${\bf x}_n$ to actually converge to ${\bf x}_\infty$ we need that
$$
{\bf x}_n  = \left( {{\bf I} + {\bf M}} \right)^{\,{\bf n}} {\bf x}_0 \; \to \;{\bf x}_\infty
  \quad  \Leftrightarrow \quad \left( {{\bf I} + {\bf M}} \right)^{\,{\bf n}} \; \to \;{\bf L}
$$
the sequence $\left( {{\bf I} + {\bf M}} \right)^{\,{\bf n}}$ be also convergent.

For that,  it is necessary that the matrix in question has a preminent eigen-value, so that
its reiterated application will bring the ${\bf x}_n$ towards the corresponding eigen-vector.

That is clear once the matrix is diagonalized.

And since the null-vector of $\bf M$ is an
eigenvector with $\lambda = 1$ of $\left( {{\bf I} + {\bf M}} \right)$, then for the convergence
the other three eigen-values shall have a lower absolute value .

If $\left( {{\bf I} + {\bf M}} \right)^{\,{\bf n}} $ diverges, since we cannot have  quantities negative nor
higher than the starting total amount, ${\bf x}_n$ will perennially oscillate.

The eigen-values are the roots of a fourth degree polynomial, so a general expression is quite involute.

But considering that the vectors ${\bf x}_n$ move over the diagonal (hyper)plane $\overline {\bf u} \;{\bf x}_n /\overline {\bf u} \;{\bf x}_0  = 1$,
then we can tell that a necessary condition for the convergence is that the matrix $\left( {{\bf I} + {\bf M}} \right)$ brings a vector from the border of the plane to the interior of it.

This is the assumption that $0 < k_{n,m} < 1/2$.

Title: What happens in the long run?


问:
Finding probability distribution from given moments
Some quantum mechanics derivation gave me the expectation values of some operator. I got some clean analytical result which makes me think that there is an analytical answer for the probability distribution.The odd moments are all $0$ and the even moments are given by:

$$\langle p^{2n}\rangle=2^{-2n}\binom{2n}{n}(k)^{2n}$$

$ghk$ is just some parameter. The most significant result for me will happen if this will result in some probability distribution based on Bessel $J$ functions but anything is good.

答:
Let us denote $q := ghk$ and your variable by $X.$ You can consider the moment-generating function of the variable $X,$ $M_X(t) = \mathbb{E}[e^{tX}] = \sum_{0 \leq k} \mathbb{E}[X^k] \frac{t^k}{k!},$ which in your case would be $$M_X(t) = \sum_{0 \leq k} 2^{-2k} \binom{2k}{k} \frac{(tq)^{2k}}{(2 k)!} = \sum_{k \geq 0} 2^{-2 k} \frac{(2 k)!}{(k!)^2} \frac{(t q)^{2k}}{(2k)!} = \sum_{k \geq 0} (\frac{t^k q^k}{2^k k!})^2.$$ Note that we already get that $M_X(t) \leq e^{tq},$ so this series is actually well defined in this case (it is not always the case with moment generating functions),. In fact, this is just a modified Bessel function of the first kind. More precisely, $M_X(t) = I_0(t q),$ so your intuition was right that we would get some kind of Bessel function somehow.

P.S.: We can determine the PDF of $X$ by using the inversion formula. Note that by the same reasoning as above, we deduce that the characteristic function of $X$ is given by $\varphi_X(t) = J_0(q t),$ a Bessel function of the first kind. It follows from the Inversion Theorem that $X$ admits a probability density function, which is given simply as the Fourier Transform of the characteristic function of $X.$ This has been computed before on this very site (see for example here or here). We have that $$\mathcal{F}(J_0)(\xi) = \chi_{(-1, 1)}(\xi)\frac1\pi \frac{1}{\sqrt{1 - \xi^2}}.$$ However, we are interested in a scaling of this function, so performing a simple substitution in the Fourier Transform integral, we get the expression of the PDF of $X:$ $$f_X(\xi) = \chi_{(-q, q)}(\xi) \frac{1}{\pi} \frac{1}{\sqrt{q^2 - \xi^2}}.$$


问:
SUPPOSE versus IF
What is the difference between

suppose .  .  ., then .  .  .

and

if .  .  ., then .  .  .

?

For instance, in this...

"Let P(n) be a statement that is defined for all nEZ and let a be a fixed integer.
Suppose that both of the following statements are true:

P(a) is true

for every integer, k ≥ a, if p(k) is true then p(k+1) is true.

Then the statement: "for all integers n ≥ a, p(n)" is true."

Why is it suppose that as opposed to if

答:
As single sentences, "Suppose X, then Y" and "If X, then Y" are synonymous.

We say "Suppose X." when we want to keep X around as an assumption for the next while. This might just be because we have a statement of a theorem that's too long to fit into a single sentence. That's all that is going on in your example.

In proofs, we might prove an implication of the form "If X, then Y" by saying "Suppose X", continuing with several steps which are free to use X as an assumption, and concluding Y.

Title: SUPPOSE versus IF


问:
Expression for the sum of two iid bernoulli random variables
The Bernoulli distribution pmf can be written $f(x)=p^x(1-p)^{1-x}$; when x is 0, this evaluates to 1-p, and when x is 1, it is p.

Suppose $X, Y\sim Ber(p)$ and are iid. Can the pmf of their sum $f(x,y)=\left[(1-p)^2\right]^{1\{x+y=0\}}\left[2p(1-p)\right]^{1\{x+y=1\}}\left[p^2\right]^{1\{x_1+x_2=2\}}$ be written with exponents that are functions of x, y (that aren't indicators)? In other words, three separate expressions are needed

$$(1)1 \text{ when x=y=0 and 0 for x or y=1}\\
(2)1 \text{when x=1 or y=1, and 0 for x=y=0 or x=y=1}\\
(3)1 \text {when x=y=1 and 0 for x or y=0}$$

E.g. for (1) you need $g(0,0)=1, g(0,1)=g(1,0)=g(1,1)=0$

答:
First of all, I think you started using $x_1$ and $x_2$ when you were using $x$ and $y$.

Note that the PMF you wrote is expressible in terms of a third variable, say $z = x+y$, that takes on values in $\{0, 1, 2\}$.  As such you do not need to handle separate cases for the outcomes of $x$ and $y$, only whether the sum $z$ equals $0$, $1$, or $2$.  So your mass function reduces to

$$f(z) = ((1-p)^2)^{\mathbb 1(z = 0)} (2p(1-p))^{\mathbb 1 (z = 1)} (p^2)^{\mathbb 1(z = 2)}.$$  You could certainly leave it this way, but it is worth noticing that there is a nice relationship between the exponents of $p$ in each case, and the value of $z$.  Specifically, the exponent of $p$ is equal to the value of $z$, and the exponent of $1-p$ is equal to $2-z$.  So we may write this as $$f(z) = (\mathbb 1 (z = 1) + 1) p^z (1-p)^{2-z}.$$  Notice that we had to use a little trick to make sure that the factor of $2$ is correctly accounted for when $z = 1$, but everything else works out neatly.

Is there a better way to write this constant factor?  After all, when there are more than $2$ IID Bernoulli random variables, the number of distinct outcomes to handle becomes larger and this approach is not illuminating.  You will find that the PMF for the sum of three IID Bernoulli random variables, written your way, looks like this:
$$f(z) = \begin{cases}(1-p)^3, & z = 0 \\ 3p(1-p)^2, & z = 1 \\ 3p^2(1-p), & z = 2 \\ p^3, & z = 3. \end{cases}$$  So what we really need instead of $\mathbb 1 (z = 1) + 1$ is the binomial coefficient  $$\binom{2}{z} = \frac{2!}{z! (2-z)!} = \begin{cases}1, & z = 0 \\ 2, & z = 1 \\ 1, & z = 2. \end{cases}.$$  This gives us the PMF $$f(z) = \binom{2}{z} p^z (1-p)^{2-z}, \quad z \in \{0, 1, 2\}.$$


问:
Why does a convex polyhedron being vertex-, edge-, and face-transitive imply that it is a Platonic solid?
Suppose that we have a convex polyhedron $P$, such that the symmetry group of $P$ acts transitively on its vertices, edges, and faces (that is, it is isogonal, isotoxal, and isohedral). It then follows that $P$ is one of the Platonic solids, which have a number of equivalent definitions but usually includes some stipulation that the faces are regular. (Once you have regular faces plus these symmetry conditions, most other properties follow easily.)

This can be shown via a rather gross casework argument on the number of sides at each face. By face-transitivity, all faces are congruent (in fact, we only need that the polyhedron be monohedral for this). By edge-transitivity, said faces are equilateral polygons. Then we break into cases by the number of sides of the polygons:

If the faces are triangles, being equilateral implies being regular.

If the faces are non-regular equilateral quadrilaterals, they are rhombi. Then, at every vertex there must be an equal number of "skinny" and "wide" angles meeting, and there must be at least 3 faces meeting at a vertex. But then there are at least two skinny and two wide angles (since we assume the angles are different), so the angles at each vertex add up to at least 360 degrees, which is not possible in a convex polyhedron.

If the faces are non-regular equilateral pentagons, then they have angles which are not all the same; however, each vertex must have the same angles around it, so each vertex has a proportional fraction of the angles of the pentagons. But for any distribution other than the uniform one, the only way to have a proportional amount of every angle is to have some $n$ copies of every angle with multiplicity, since $5$ is prime. (E.g. if there are $2$ angles of one size and $3$ of another, we can't get a $2:3$ ratio without having at least $5$ angles at every vertex.) But the angles of a pentagon add to more than $360^\circ$, so we can't have them all at once.

If the faces are hexagons or higher, they cannot form a topological sphere by Euler's formula.

However, I don't like this proof very much; I walk away from it with no understanding of why this behavior occurs, and it certainly doesn't suggest any generalizations to other dimensions (where I believe the analogous statement - that transitivity on every degree of cell in a convex polytope implies regularity - is true, though I would appreciate a reference for this fact).

Is there a natural or more elegant proof of this statement?

One thing to note is that any proof must somehow use the fact that we are working on the sphere. In the case of planar tilings, the corresponding implication is false: being transitive on faces, vertices, and edges does not imply that one is looking at a tiling of regular polygons! As a counterexample, consider a rhombic tiling like this one:

As another aside, relaxing any one of these conditions allows for polyhedra whose faces are not congruent regular polygons: the rhombic dodecahedron is face-transitive and edge-transitive, an isosceles tetrahedron is face-transitive and vertex-transitive, and the cuboctahedron is edge-transitive and vertex-transitive.

答:
Let a vertex have degree $\rho$ and a face have $s$ edges and note that $$\rho V=2E=sF\tag 1$$

Let $G$ be the group of symmetries. Let $G_v,G_e\text { and }G_f$ be the stabiliser of a vertex $v$, edge $e$ and face $f$, respectively. Then $$|G|=|G_v||V|=|G_e||E|=|G_f||F|\tag 2$$
If $G_e$ has a non-identity symmetry then it is a rotation of $180^o$ about the mid-point of the edge.  Comparing (1) and (2) we then have $|G_f|=s$ and the face is regular.

Otherwise $|g_e|=1$. Again comparing (1) and (2) we have $|G_v|=\frac{\rho}{2} \text { and } |G_f|=\frac{s}{2}$. Therefore $\rho\ge  4$ and $s\ge 4$ and then Euler's formula gives $$\frac{E}{2}+\frac{E}{2}\ge E+2.$$

If $G$ includes reflections

The only 'extra' case is $|G_v|=\rho, |G_e|=2, |G_f|=s$, where both $G_v$ and $G_f$ contain reflections. However, the orders of the two groups, $\rho$ and $s$, are then both even and we still have $\rho\ge  4$ and $s\ge 4$.


问:
Solving $-y' +y^2 = 1$
I must solve the ODE $-y' + y^2 = 1$. Rewriting this I can get: $$y' - y^2 + 1 = 0$$ However, because of the nonlinear term $y^2$ I do not know how to solve this. We cannot use the characteristic equation I think  and I cannot see that this is seperable.

Can someone please lend a hand?

I get the inverse of what WolframAlpha says the answer is. I get $y = \frac{e^{2x+2c} + 1 }{1 - e^{2x + 2c}}$.

From $\frac{dy}{dx} = y^2 - 1$ I write $\frac{dy}{y^2 - 1} = dx$ and then integrate: $$\int \frac{1}{y^2 - 1} dy = \int 1 dx = x + c_1$$

The integral on the left I decompose into partial fractions: $$\frac{1}{y^2 -1} = \frac{1}{(y-1)(y+1)} = \frac{A}{y-1} + \frac{B}{y+1}$$ and I get $A = \frac{1}{2}$ and $B = -\frac{1}{2}$

So then I have $\frac{1}{2}\int \frac{1}{y-1} - \frac{1}{2} \int \frac{1}{y+1} = x + c_1$ which is $$\frac{1}{2}ln(y-1)-\frac{1}{2}ln(y+1) = x + c$$ This simplifies to $$\frac{1}{2}ln(\frac{y-1}{y+1}) = x+c$$ Raising both sides by $e$: $$(\frac{y-1}{y+1})^{\frac{1}{2}} = e^{x+c}$$
Now squaring... $$\frac{y-1}{y+1} = e^{2x+2c}$$
Then: $$y - 1 = e^{2x+2c}y + e^{2x+2c} \implies y(1-e^{2x+2c}) = e^{2x+2c}+1$$
and finally $$y = \frac{e^{2x+2c} + 1}{1 - e^{2x+2c}}$$

答:
Try $\displaystyle y' = \frac{dy} {dx} = y^2 - 1$ which is immediately separable and solvable by partial fraction decomposition. A hyperbolic trigonometric substitution will likely work as well, it's less elementary though.

Adding on, there is a very important subtlety here. The integral of $\displaystyle \frac 1x$ is not simply $\displaystyle \ln x + c$, it is actually $\displaystyle \ln |x| + c$. That absolute value sign is actually critical. It means that for the domain $\displaystyle x<0$, you should take $\displaystyle \ln (-x) + c_1$, while for $\displaystyle x>0$, you should take $\displaystyle \ln x + c_2$.

After integration, you wrote: $\displaystyle \frac 12\ln(y-1) - \frac 12\ln(y+1) = x + c$

This is imprecise (and not correct without domain restriction). For the general solution, consider three cases (note that $\displaystyle y$ cannot be equal to either $\displaystyle -1$ or $\displaystyle 1$):

Case 1: $\displaystyle y<-1$:

In this case, integration yields $\displaystyle \frac 12\ln(-(y-1)) - \frac 12\ln(-(y+1)) = x + c$  which, after you do the remaining working, will get you the exact solution you got, namely $\displaystyle y = \frac{1+e^{2(x+c)}}{1- e^{2(x+c)}}$

Case 2: $\displaystyle -1<y<1$:

In this case, integration yields $\displaystyle \frac 12\ln(-(y-1)) - \frac 12\ln(+(y+1)) = x + c$

Working through this would get you the same form as the Wolfram Alpha solution:

$\displaystyle y = \frac{1-e^{2(x+c)}}{1+e^{2(x+c)}}$

And finally,

Case 3: $\displaystyle y>1$ would get you, after integration:

$\displaystyle \frac 12\ln(+(y-1)) - \frac 12\ln(+(y+1)) = x + c$

Working through which you would get the exact same form as for case 1:

$\displaystyle y = \frac{1+e^{2(x+c)}}{1- e^{2(x+c)}}$

Splitting it up into cases gives you the full picture. You can visualise exactly what is going on by graphing.

Title: Solving $-y' +y^2 = 1$


问:
Definition of coroots in a semisimple Lie algebra
For $\mathfrak{g}$ a semisimple Lie algebra, $\mathfrak{h}$ a Cartan subalgebra, $\Phi$ a root system with respect to $\mathfrak{h}$, the coroot $\check{\alpha} $ associated to a root $\alpha \in \Phi$ is usually defined as $$\check{\alpha} = \frac{2}{(\alpha,\alpha)}\alpha,$$ where $(\cdot,\cdot)$ is a bilinear form defined to be the dual of the Killing form $\kappa$, which I understand to be defined on $\mathfrak{h}^*$ by setting $(\gamma, \delta) = \kappa (t_\gamma, t_\delta)$, where $t_\gamma \in \mathfrak{h}$ for which $\gamma(X) = \kappa(t_\gamma, X)$ for all $X \in \mathfrak{h}$.

In my textbook (Goodman-Wallach) the definition of coroot is slightly different: $$ \check{\alpha} = \kappa(e_\alpha, f_\alpha) \alpha,$$ where $\{e_\alpha, f_\alpha, h_\alpha\}$ is an $\mathfrak{sl}_2$-triple in $\mathfrak{g}$.
However from my working out, I have that $$ \kappa(h_\alpha, h_\alpha) = 2 \kappa (e_\alpha, f_\alpha),$$ so that on rearranging the given definition I obtain $$ \check{\alpha} = \frac{\kappa(h_\alpha,h_\alpha)}{2} \alpha = \frac{(\alpha,\alpha)}{2}\alpha.$$

So is the definition in my textbook wrong, and should we instead define $$ \check{\alpha} = \frac{1}{\kappa(e_\alpha, f_\alpha)} \alpha,$$ or have I misunderstood something in the theory?

答:
Half of my confusion stems from the fact that Goodman-Wallach (unlike e.g. Humphreys) do not explicitly write out what form $h_\alpha$ should take: $$ h_\alpha = \frac{2}{\kappa(t_\alpha, t_\alpha)}t_\alpha,$$ and finding appropriate $e_\alpha$ and $f_\alpha$ will still give an $\mathfrak{sl}_2$-triple.
Using this explicit formula, it is easy to check that my identity $\kappa(h_\alpha, h_\alpha) = 2 \kappa (e_\alpha, f_\alpha)$ implies $$ \check{\alpha} = \frac{2}{(\alpha, \alpha)}\alpha.$$
The other half of my confusion was that in my second-last line, $\kappa(h_\alpha, h_\alpha) \neq (\alpha, \alpha)$.
Once there is an explicit form for $h_\alpha$, everything else is clarified.


问:
Conditional Probability using an urn.
An urn contains $5$ white balls and $6$ black balls. $3$ balls are drawn sequentially without replacement. What are the probabilities...

a. $P(\text{2nd is black} | \text{1st is white})$?

For this one I did the

$$ P(\text{1st is white})=\frac{5}{11} , $$

and

$$ P(\text{second is black}) = \frac{3}{11}*\frac{3}{11} $$

(that is the probability of WB and BB added together). I multiplied the probability of

$$ \frac{P(\text{2nd black} \cap \text{1st white})}{P(\text{1st is white})} , $$

for this I got

$$ \frac{\frac{5}{11}*\frac{3}{11}}{\frac{5}{11}} = \frac{3}{11} . $$

However since this is without replacement I did not think that the answer could be independent, some feedback would help. Alright so I have this figured out that it should be $\frac{6}{10}$.

b. If I am looking now for $P(\text{3rd is black} | \text{1st is white and 2nd is black})$, I ended up with

$$ \frac{\frac{5}{9}*\frac{3}{11}}{\frac{3}{11}} , $$

this is again an independent answer is this correct?

答:
There's no need to use any formulas here, just look at what happens when you follow the conditions and remove the relevant balls from the urn step by step.

If we start with $5$ white balls and $6$ black balls, then we're given that the first ball drawn is white, then after that first draw, $6$ black balls remain and $5-1=4$ white balls remain. So, we have $6$ black balls out of $10$ total balls, and the answer to the first question is

$$ P[\text{2nd is black} | \text{1st is white}] = \frac{6}{10} . $$

For the second question, we're given that the first is white and the second is black. So, after these two draws, we know that $5-1=4$ white balls remain and $6-1=5$ black balls remain. Out of a total of $9$ remaining balls, $5$ are black, so we have

$$ P[\text{3rd is black} | \text{1st is white and 2nd is black}] = \frac{5}{9} . $$

Title: Conditional Probability using an urn.


问:
The inverse of $1+x$ in $\frac{\mathbb{Q}[x]}{\langle x^2+x+1 \rangle}$ and elegant methods of this computation
I would like to find the inverse of $1+x$ in $\frac{\mathbb{Q}[x]}{\langle x^2+x+1 \rangle}$ . I argued that the remainder of the division of any polynomial by ${x^2+x+1 }$ is of the form $ax+b$ and after doing the long division and comparing like terms, I concluded that the inverse of $1+x$ is $-x$ in this field.

Now my question is if there are more elegant ways to go about this that preferably avoids long divisions.

答:
You know that in this field we have $x^2 + x + 1 = 0 \implies 1 = -x^2 - x = -x(x+1)$.

Thus $-x$ is the multiplicative inverse of $x+1$.


问:
Numerically evaluating $\iiint_{C} \frac{ {\rm d}x \, {\rm d}y \, {\rm d}z}{x^2 +y^2 + z^2}$

Numerically evaluate $$\iiint_{C} \frac{ {\rm d}x \, {\rm d}y \, {\rm d}z}{x^2 +y^2 + z^2}$$ where $$C = [-1,1] \times [-1,1] \times [-1,1] = [-1,1]^3$$

I do not expect an analytic solution. I would be satisfied with a numeric value. But even this is not so easy.

How to deal with the singularity at the origin? One idea is to first integrate over the unit ball, which can be done analytically, then use monte carlo to integrate over the rest irregular region. But I am not sure whether this will converge fast enough.

答:
We can use divergence theorem to say

$$\iiint_{[-1,1]^3} \frac{dV}{r^2} = \iint_{\partial [-1,1]^3} \frac{\hat{r}}{r}\cdot\vec{dS}$$

By symmetry we can choose to only integrate over the square in the $z=1$ plane and evaluate

$$\iint_{[-1,1]^2}\frac{6}{\sqrt{x^2+y^2+1}}\cos\theta\:dA = \iint_{[-1,1]^2}\frac{6}{x^2+y^2+1}\:dA$$

By even more symmetry we can reduce this to $8$ times an integral over a triangle in the first quadrant

$$\int_0^1\int_0^x \frac{48}{x^2+y^2+1}dydx$$

You can show this is equal to

$$\int_0^{\sinh^{-1}(1)}48\tan^{-1}(\tanh t)\:dt$$

by integrating directly or

$$\int_0^{\frac{\pi}{4}}24\log(\sec^2\theta+1)\:d\theta$$

by integrating in polar coordinates. Numerically these integrals both evaluate to $\approx 15.3482$


问:
Notation for optional addition sign
Is there a way to notate an optional addition? I'm thinking of something similar to the plus-minus sign $\pm$. For reference, I have two angles $\theta$ and $\theta+\pi/2$ and want to include them both in one line.

Hopefully, there is something simpler than $\theta + \pi/4 \pm \pi/4$.

答: I suggest $\alpha = \theta + \epsilon\cdot \pi/2$ for $\epsilon \in \{0,1\}$.

Title: Notation for optional addition sign


问:
Quaternions +Geometric (Clifford) Algebra: What Is the Proper Prerequisite Sequence Before Learning These Subjects
What is the systematic prerequisite sequences of learning that must be mastered before approaching the subject of learning Quaternions, and then Clifford Algebra? My ultimate goal is to learn and study Maxwell’s equations in their original Quaternion form, plus other obscure scientific papers that utilize Clifford Algebra and Octonians. But before I can approach that, I need the solid mathematical background that comes before studying and understanding Quaternions and Clifford (Geometric) Algebra, beginning with a recommended syllabus that starts with pre-algebra mathematics.

I know that there is a systematic approach to learning mathematics in general. For example: Multiplication and division at the foundation, then fractions, then pre-algebra, then algebra, then geometry, then trigonometry, and so on. With each division of mathematics itself having a systematic sequence of learning the subject from its fundamentals to their ultimate advanced topics. So to state my question differently, what I want to know is this:

What is the best sequence of mathematical prerequisites (the syllabus) to master before I begin studying Quaternions, and then advance to Clifford (Geometric) Algebra?

Where does Quaternion and Clifford (Geometric) algebra fit in the sequence of learning algebra, and what will be their prerequisites necessary to have been already mastered before entering these subjects?

Also, can someone recommend a track of material that would properly prepare me for the topic of Quaternions and Clifford (Geometric) Algebra from a pre-algebra starting point? i.e. Completed knowledge of addition/subtraction, multiplication/division, fractions and percentages.

答:
You learn arithmetic, basic geometry, then what is called "prealgebra", then algebra 1 and algebra 2.  Once you have gotten through the chapter on complex numbers, you want to learn at least a little bit about matrices. You might get a taste in your algebra 2 class or precalculus, but the subject is called "linear algebra" and the more of it you know the better--it doesn't require calculus.

Once you have complex numbers and matrices, you are good to go. I actually would not recommend Conway; I own it and found it frustrating.  I have hunted for good materials to learn quaternions and there mostly aren't any books that explain them well.  But YouTube videos fill the gap; since quaternions get used in computer programming to handle rotations in three dimensions, there is some demand.  Good luck!


问:
Proof of convergence of Kaprekar's Constant
I've tried googling this one a bit but nothing seems to come up, even though its considered to be a well known fact.

Why does the kaprekar process of taking a 4 digit number: L, generating L' and L'' such that L' is the digits of L in ascending order and L'' is the digits in descending order and subtracting L' - L'' always converge to the Kaprekar constant of 6174?

Clearly the only value for which this process is constant is 6174 but that doesn't explain why there should be convergence.

One attempt at proof is to determine all the possible numbers that converge to 6174 after a single iteration, and then attempt to reason that each too can be reached by the convergence of even more numbers in such a way that if I continue this reasoning I should cover ALL 4 digit numbers.

But to turn my strategy into induction from brute force casework has become futile. What is the actual proof?

答:
Somewhere I have a complete proof written out, but I'll just share a couple of random things here.  First, with some work you can show that if two digits of a fixed point are equal, then three digits have to be equal, and in turn all four digits have to be equal, and hence you must have $0000$.  So you can assume all the digits of a non-trivial fixed point are different.

Second, notice that after one iteration of the process, you must have a multiple of $9$.  This considerably reduces the number of cases to check.

Of these multiples of $9$, if you do two iterations to all of them, you end up with a small set.  Some clever things help you get a small system of equations to solve.  If the snow ever melts in Austin, I might post the whole proof, if I can find it.  I hope my hints help.


问:
Finding $Pr(Y < X < Z)$ where $X, Y, Z$ are random variables with known CDF's
I'm interested in the problem of computing $Pr(Y < X < Z)$ where $X, Y, Z$ are random variables (either discrete or continuous), not necessarily defined over the same support. I found one other thread which concluded that, for the continuous case, we have

$Pr(Y < X < Z) = \int_{-\infty}^\infty Pr\left( x \in [Y, Z)\right)f_X(x)dx = \int_{-\infty}^\infty \left( \int \int_{(x, \infty) \times (-\infty, x)}f_{YZ}(y, z)dy dz\right)$

However, I do not understand the bounds for the RHS. Is $y$ integrated out in $(x, \infty)$ or $(-\infty, x)$?

答:
You should have:

$$\begin{align}\mathsf P(Y<X<Z)&=\int_\Bbb R \mathsf P(Y<x<Z\mid X=x)~f_{\small X}(x)~\mathrm d x\\&=\int_\Bbb R\iint_{(-\infty..x)\times(x..\infty)} f_{\small X}(x)\ f_{\small Y,Z\mid X}(y,z\mid x)\,\mathrm d\langle y,z\rangle\,\mathrm d x\\&=\int_\Bbb R\iint_{(-\infty..x)\times(x..\infty)} f_{\small X,Y,Z}(x,y,z)\,\mathrm d\langle y,z\rangle\,\mathrm d x\end{align}$$


问:
Is there a precise way to state that a structure does not satisfy a theory because its signature is different?
A structure $\mathcal{S}$ is a triple $<\sigma, D, I>$ containing a signature $\sigma$, which determines what functions and relations the structure is able to interpret, a domain $D$, which specifies what objects are in consideration, and an interpretation $I$, which assigns truth values to predicates, functions, and relations.

Yesterday I learnt that the $\vDash$ and $\not\vDash$ symbols are only defined for structures when the signature of the structure (on the LHS), at least contains the symbols of the theory or sentence (on the RHS).

$$ \mathcal{S}\vDash T$$

This got me wondering: It is a true statement to say that a structure $\mathcal{S}$ does not satisfy a theory $T$, if the structure does not contain in its signature at least the syntax used by the theory $T$ (for instance, it may not contain a predicate the theory $T$ uses).

However, with the way $\vDash$ is defined, we cannot write $\mathcal{S}\not\vDash T$ from this observation.

A question then arises: Is there any way to express that a structure does not satisfy a sentence or theory, because its signature doesn't match, in more precise terminology than natural language?

答:
For what it's worth I disagree with your initial claim, although of course this will come down to a convention: I would say that "$\not\models$" is exactly the complement of "$\models$," so that if $\varphi$ is not in the signature of $\mathcal{A}$ we have both $\mathcal{A}\not\models\varphi$ and $\mathcal{A}\not\models\neg\varphi$. In particular, under my convention we have the following equivalence: $\varphi$ uses a symbol not in the signature of $\mathcal{A}$ iff $\mathcal{A}\not\models\varphi\vee\neg\varphi$.

However, that's a bit messy - in general, I think that even though interpreting "$\not\models$" as the literal complement of "$\models$" is the right call, I also think that expositions ought not to crucially exploit pathologies when there's a natural way to avoid this. In this case, I think the right call is something like "$\varphi\not\in Sent(Sig(\mathcal{A}))$," where "$Sig(\cdot)$" sends each structure to its signature and "$Sent(\cdot)$" sends each signature to the set of sentences in that signature. These two component functions are quite useful to have lying around, so using them here is a reasonable thing to do.


答:
The arcsine distribution stretched to have support on $[-ghk, ghk]$ has the moments you are looking for.   So, with $-ghk < x < ghk$, the probability density function is $$f(x)=\frac{1}{\pi  \sqrt{{{(ghk)}^{2}}-{{x}^{2}}}}$$ and the cumulative distribution function is $$F(x)=\mathbb P(X\le x) = \frac1\pi \cos^{-1}\left(\frac{-x}{ghk}\right)$$

As a quick check in R of the first twelve moments, setting $ghk=2$ gives the central binomial coefficients:

```<code>qstretchedarcsine <- function(prob, ghk){ - ghk * cos(prob * pi) }
thisdata <- qstretchedarcsine(ppoints(10^6), ghk=2)
round(colMeans(outer(thisdata, 1:12, "^")), 12)
#    0   2   0   6   0  20   0  70   0 252   0 924
</code>```

See https://math.stackexchange.com/a/2629048/6460 for the relationship between the arcsine distribution and Bessel functions

Title: Finding probability distribution from given moments

Body:
Some quantum mechanics derivation gave me the expectation values of some operator. I got some clean analytical result which makes me think that there is an analytical answer for the probability distribution.The odd moments are all $0$ and the even moments are given by:

$$\langle p^{2n}\rangle=2^{-2n}\binom{2n}{n}(k)^{2n}$$

$ghk$ is just some parameter. The most significant result for me will happen if this will result in some probability distribution based on Bessel $J$ functions but anything is good.

Body:
The arcsine distribution stretched to have support on $[-ghk, ghk]$ has the moments you are looking for.   So, with $-ghk < x < ghk$, the probability density function is $$f(x)=\frac{1}{\pi  \sqrt{{{(ghk)}^{2}}-{{x}^{2}}}}$$ and the cumulative distribution function is $$F(x)=\mathbb P(X\le x) = \frac1\pi \cos^{-1}\left(\frac{-x}{ghk}\right)$$

As a quick check in R of the first twelve moments, setting $ghk=2$ gives the central binomial coefficients:

qstretchedarcsine <- function(prob, ghk){ - ghk * cos(prob * pi) }
thisdata <- qstretchedarcsine(ppoints(10^6), ghk=2)
round(colMeans(outer(thisdata, 1:12, "^")), 12)
#    0   2   0   6   0  20   0  70   0 252   0 924

See https://math.stackexchange.com/a/2629048/6460 for the relationship between the arcsine distribution and Bessel functions


问:
Dice Roll Probability of winning
Dice A sides: 4, 4, 4, 4, 13, 13

Dice B sides: 6, 6, 6, 9, 9, 9

If I roll both dice

Which dice is more likely to win here? How can I approach this?

答: If A rolls a 13, then A wins.  If A rolls a 4, then B wins.  The probability A rolls a 13 is 2/6 and the probability A rolls a 4 is 4/6.</p>

Title: Dice Roll Probability of winning


问:
Number of elements of prime order $p$ in $\mathrm{SL}_2(\mathbb{F}_q)$
Let $p,q$ be prime numbers.
It is known that there is one element of order $2$ in $\mathrm{SL}_2(\mathbb{F}_q)$.

How many elements of order $p>2$ are there in $\mathrm{SL}_2(\mathbb{F}_q)$?

答:
The answer is going to vary with $p$ and the factorization of the order of $Sl_2(F_q)$, which is $q^3-q$.

For example with $q=101$, we have $|Sl_2(F_{101})|=2^3\cdot3\cdot5^2\cdot 17\cdot101$.

So you can see, there won't be any elements with order a prime not appearing here, and if $p$ is a prime in the factorization, then we'd need to count how many elements are in the $p$-Sylow subgroups of $Sl_2(F_{101})$. It seems like that would need to be done on a case-by-case basis.