问:
Calculating the probability of reaching each absorbing state in Markov Chain
I'm starting with a Matrix that looks like this:

```<code>[[ 0 , 1/2, 0,  0 ,  0 , 1/2],
 [4/9,  0 , 0, 3/9, 2/9,  0 ],
 [ 0 ,  0 , 1,  0 ,  0 ,  0 ],
 [ 0 ,  0 , 0,  1 ,  0 ,  0 ],
 [ 0 ,  0 , 0,  0 ,  1 ,  0 ],
 [ 0 ,  0 , 0,  0 ,  0 ,  1 ]]
</code>```

And I'm being asked to calculate the probability of reaching each of the absorbing states C, D, E, and F starting from A. Obviously C is 0, and the other probabilities are given as:

A to D -> 3/14

A to E -> 1/7

A to F -> 9/14

but I don't know what steps or formulas I need to arrive at those values.
I don't really know where to start with this so any help would be greatly appreciated!

答:
If you have access to mathematical software like Python, you can solve it using the following trick.

The Eigendecomposition lets us split the matrix into

V x D x  V^-1 where V is the eigenvector, and d is the eigenvalues.

the terminal states are then V x D ^ inf x V^-1, and we can find D ^ inf by looking at whether the values are >= or < 1.

Here is code that solves your problem.

```<code>import numpy as np
m = np.array(
    [
    [0, 1 / 2, 0, 0, 0, 1 / 2],
    [4 / 9, 0, 0, 3 / 9, 2 / 9, 0],
    [0, 0, 1, 0, 0, 0],
    [0, 0, 0, 1, 0, 0],
    [0, 0, 0, 0, 1, 0],
    [0, 0, 0, 0, 0, 1],
    ]
)
d, v = np.linalg.eig(m)
v @ np.diag(d >= 1).astype(int) @ np.linalg.inv(v)

[[0.        , 0.        , 0.        , 0.21428571, 0.14285714, 0.64285714],

   [0.        , 0.        , 0.        , 0.42857143, 0.28571429,   0.28571429],

   [0.        , 0.        , 1.        , 0.        , 0.        ,  0.        ],

   [0.        , 0.        , 0.        , 1.        , 0.        ,  0.        ],

   [0.        , 0.        , 0.        , 0.        , 1.        ,    0.       ],

   [0.        , 0.        , 0.        , 0.        , 0.        ,1.        ]]
</code>```

Body:
If you have access to mathematical software like Python, you can solve it using the following trick.

The Eigendecomposition lets us split the matrix into

V x D x  V^-1 where V is the eigenvector, and d is the eigenvalues.

the terminal states are then V x D ^ inf x V^-1, and we can find D ^ inf by looking at whether the values are >= or < 1.

Here is code that solves your problem.

import numpy as np
m = np.array(
    [
    [0, 1 / 2, 0, 0, 0, 1 / 2],
    [4 / 9, 0, 0, 3 / 9, 2 / 9, 0],
    [0, 0, 1, 0, 0, 0],
    [0, 0, 0, 1, 0, 0],
    [0, 0, 0, 0, 1, 0],
    [0, 0, 0, 0, 0, 1],
    ]
)
d, v = np.linalg.eig(m)
v @ np.diag(d >= 1).astype(int) @ np.linalg.inv(v)

[[0.        , 0.        , 0.        , 0.21428571, 0.14285714, 0.64285714],

   [0.        , 0.        , 0.        , 0.42857143, 0.28571429,   0.28571429],

   [0.        , 0.        , 1.        , 0.        , 0.        ,  0.        ],

   [0.        , 0.        , 0.        , 1.        , 0.        ,  0.        ],

   [0.        , 0.        , 0.        , 0.        , 1.        ,    0.       ],

   [0.        , 0.        , 0.        , 0.        , 0.        ,1.        ]]


问:
Expectation of a sum of independent and identical exponential variables
Suppose $c>0$. Let $X_i$ be independently and identically distributed exponential variables with parameter $\lambda >0$ and $$S_n=\sum_{i=1}^n X_i.$$ Let $I=\min \{i: X_i >c\}$. How do I show that
$\mathbb{E} [S_{I-1}+c] = \dfrac{e^{\lambda c} -1}{\lambda}$?

I started as:

$S_{I-1}$ follows a Gamma distribution with parameters $(I-1,\lambda)$, where if a random variable $X$ follows a Gamma distribution with parameters $n$ and rate $\lambda$, then
$$f(x)= e^{-\lambda x}\dfrac{{\lambda}^n x^{n-1}}{(n-1)!}.$$

Then I conditioned as follows:

$$\mathbb{E}[S_{I-1}]=\sum_{i=1}^{\infty} \mathbb{E}[S_{I-1}\mid I=i]\mathbb{P}(I=i).$$

Here I'm wondering whether I should sum from $i=2$ instead!

Anyway, I obtained $\mathbb{P}(I=i)=(1-e^{-\lambda c})^{i-1}e^{-\lambda c}$ as $$\mathbb{P}(I=i)=\mathbb{P}(X_1<c, X_2<c, \dots, X_{i-1}<c, X_i>c).$$

I have doubts here also because the way I did it I obtained $\mathbb{E}[S_{I-1}]=\dfrac{e^{\lambda c} -1}{\lambda}$, which is wrong.

答:
You are correct in your assertion that $I$ is a geometric random variable with parameter $e^{-\lambda c}$ and hence expected value $e^{\lambda c}$.
But, given the value of $I$, all the $X_i, 1 \leq i < I$, are known to have value $\leq c$, and therefore each such $X_i$ has conditional marginal density
function
$$f_{X_i \mid I}(x) = \begin{cases}\frac{\lambda e^{-\lambda x}}{1-e^{-\lambda c}},
&0 \leq x \leq c,\\\quad\\
0,&\text{otherwise}\end{cases}$$
which is just the truncation (and normalization) of the exponential density to the
interval $[0,c]$. Hence, the conditional mean is
$$\begin{align}
E[X_i\mid I] &= \int_0^c \frac{x\lambda e^{-\lambda x}}{1-e^{-\lambda c}}\,\mathrm dx
= \frac{1}{1-e^{-\lambda c}}\int_0^cx\lambda e^{-\lambda x} \,\mathrm dx\\
&= \left.\left.\frac{1}{1-e^{-\lambda c}}\right[-xe^{-\lambda x}\biggr|_0^c
+ \int_0^c e^{-\lambda x} \,\mathrm dx\right]\\
&= \left.\left.\frac{1}{1-e^{-\lambda c}}\right[-ce^{-\lambda c}
+ \left.\left.\frac{1}{\lambda}\right(1- e^{-\lambda c} \right)\right]\\
&= \frac{1}{\lambda}-\frac{c}{e^{\lambda c}-1}
\end{align}$$
which gives
$$E[S_{I-1}\mid I] = E\left[\sum_{i=1}^{I-1}X_i\right] = (I-1)\left[\frac{1}{\lambda}-\frac{c}{e^{\lambda c}-1}\right]$$
and
$$E[S_{I-1}] = E\left[E[S_{I-1}\mid I]\right]
= (e^{\lambda c}-1)\left[\frac{1}{\lambda}-\frac{c}{e^{\lambda c}-1}\right]
= \frac{e^{\lambda c}-1}{\lambda} - c.$$

ContentLicense: CC BY-SA 3.0


问:
How to evaluate inverse trig functions without a calculator or trig tables?
I am trying to understand how to evaluate the following equation without using a calculator or trig tables:

arcsin(4/5) - arccos(12/13) = arccos(56/65)

答:
Hint Compute $\sin(\arccos(12/13) + \arccos(56/65))$ and prove that the result is $4/5$. You can use
\begin{equation}
\sin(a+b) = \sin a \cos b + \cos a \sin b
\end{equation}
Also note that $65^2-56^2 = 33^2$ and $13^2-12^2=5^2$.

ContentLicense: CC BY-SA 4.0


问:
Problem of Hahn-Banach theorem
in $(l_\infty)'$ space there is a $f$ functional that
$$\forall x \in c_{00};f(x)=0$$
$$f((1,1,1,...))=1$$
Provide these properties. I find this functional and showed by using Hahn-Banach theorem please help

答:
$Y=\overline{c_{00}}$ ,$x=(1,1,...)$ and $V=Y \oplus span\{ {x} \}$ $f:V \to \Bbb{F} $  ,$v=y+a.x \in V$,$y\in Y , a \in \Bbb{F}$ $f(v)=a.dis(x,Y)=a.inf_{y \in Y} ||x-y||$ f is a functional. Also  $x=(1,1,1,...) \notin Y $ so $dis(x,Y)\neq 0 $ let $f(x)=f(0.y+1.x)=1.dis(x,Y)=m>0$ There is a $F \in (l_\infty)'$that $f=F|_V$ from Hahn-Bannach Theorem.$g(t)=\frac{F(t)}{m}$ $g((1,1,...))=\frac{F(1,1,1,...)}{m}=\frac{m}{m}=1$ and $\forall y \in c_{00}; g(y)=0$
$$$$
Let see f is a lineer functional .Take $v.w∈V$then $v=y1+a.x,w=y2+b.x,y1,y2∈Ya,b∈F $and take $λ∈F$.
$$f(λ.v+w)=λ.f(v)+f(w)(?)$$
$$f(λ.v+w)=f(λy1+\lambda.a.x+y2+b.x)=f([λy1+y2]+[λa+b].x)$$
and  $ λy1+y2∈Y \ then \ ∃y3∈Y  \ that \  :y3=λy1+y2$ so we can write
$$f(y3+[λa+b].x)=[λa+b]+dis(x,Y)=λ.a.dis(x,Y)+b.dist(x,Y)=λ.f(v)+f(w)$$

Title: Problem of Hahn-Banach theorem


问:
Evaluate $\lim_{n \rightarrow \infty } \int_0^{\pi/2} \sqrt{n \sin{\frac{x}{n}} }dx$
Evaluate $\lim_{n \rightarrow \infty }  \int_0 ^{\pi/2} \sqrt{n \sin{\frac{x}{n}}} dx$ .

I already have that $\lim_{n \rightarrow \infty} \sqrt{n \sin{\frac{x}{n}}}= \sqrt{x}$.

So I'm trying to use DCT so that $\lim_{n \rightarrow \infty} \int f_n = \int \lim_{n \rightarrow \infty} f_n$.

But I'm having a hard timer finding a bound.

Also, I'm a little concerned about the range of $x$, can we still use the DCT with the range?

Thanks

答: In fact $0\le\sin\tfrac{x}{n}\le\tfrac{x}{n}$ for $x\in[0,\,\pi/2]$, so you can use $\sqrt{x}$ as your upper bound on the integrand in the DCT.


问:
Relation between cubes of Fibonacci numbers and even Fibonacci numbers
While recently revisiting an old Project Euler problem, which states:

Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be:

1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...

By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms.

I noticed the following.

Let $F_n = 1, 2, 3, 5, 8, \ldots$ as defined in the problem. Then, assuming zero indexing, for all $n \ge 2$,

$$F_n^3 + F_{n-1}^3 - F_{n-2}^3 = F_{3(n-1)}.$$

This evidently holds for the first few $n$:

$$F_2^3 + F_1^3 - F_0^3 = 27 + 8 - 1 = 34 = F_7$$
$$F_3^3 + F_2^3 - F_1^3 = 125 + 27 - 8 = 144 = F_{10}$$

and so forth.

I have tried using Cassini's Identity and its generalizations to prove this, but to no avail.

I've also found this answer, but cannot find the derivation of this solution, and cannot quite reconcile it with my observation.

My question: does my observation hold true, and if so, how can I go about proving this. What am I missing?

答:
We have Binet's formula,
$$ F_n=\frac{\phi^n-\psi^n}{\sqrt 5}$$
where $\phi=\frac{1+\sqrt 5}2$ and $\psi=\frac{1-\sqrt 5}2$ are the two roots of $X^2-X-1$.
You can plug this into your conjectured equality, expand and simplify (using in particular that $\phi+\psi=1$, $\phi-\psi=\sqrt 5$, $\phi\psi=-1$), and check.


问:
Proof whether or not it is Hausdorff (T_2) space $X = \mathbb{R} \cup \{\alpha\} $ with $\alpha \notin \mathbb{R}$
Let $X = \mathbb{R} \cup \{\alpha\} $ with $\alpha \notin \mathbb{R}$. In $X$ we have the topology $\tau$ with the basis $$B = \{(a,b): a, b \in \mathbb{R}, a  0 \}$$

I want to show if $(X, \tau)$ is Hausdorff ($T_2$) or not, I tried to show that the open sets do not collide but I am confused with the $\alpha$. Any help is appreciated!

答:
Space will be Hausdorff if for all $x,y\in X$ with $x \ne y$ there are $U, V \in \tau$ with $x \in U, y \in V$ with empty intersection, $U \cap V = \emptyset$, but we can see that $x= 0, y = \alpha$ can't be separated with open disjoint sets.

$U, V\in \tau$, such that $0 \in U, \alpha \in V$, for the definition of $B$, $0 \in B_1 = (a,b)$, $\alpha \in B_2 = (-\epsilon,0) \cup \{\alpha\} \cup (0,\epsilon)$ with $a<0<b$ and $\epsilon > 0$, because $((a,b) \cap (-\epsilon,0)) \ne \emptyset$, then $ \emptyset \ne B_1 \cap B_2 \subset U \cap V$ and the space is not Hausdorff.


问:
The numbers in these patterns go together following the same rule. Find the missing number.
I don't see the pattern of the numbers, can someone tell me what the pattern is?

答: I think the second number is the smallest Primes after the first number in pair with an extra condition that it has to at least 40 + 6 ( significance of 6 is the AP starting from 0 in pair (0,2) ) . Eg- $29\geq 25+4$


问:
How does the absence of a pivot in a given column make a variable 'free'?
Why is it that, if a given variable does not have a pivot variable within the column of a given REF / RREF matrix, it can be arbitrarily chosen?</p>

答:
When you rewrite the rref matrix in equation form, you backsolve. Any row that contains a pivot element gives you a dependent variable, for example:
$$x_1 + 2x_2 - x_3 = 4$$
$$3x_3 = 2$$
The variable $x_2$ is a free variable, and $x_1$ and $x_3$ are pivot variables. Basically, there is no equation that starts with a multiple of $x_2$, so it is impossible to solve for it. Rather, we can solve for $x_1$ in terms of $x_2$ and $x_3$, so $x_2$ is free.


问:
Routine question about derivatives of automorphic forms being L^2
I consider Automorphic forms on $G = SL_2 (\mathbb{R})$, which are $\Gamma$-invariant, $K$-finite, $Z(g)$ finite, and of moderate growth. If I have such an automorphic form, which happens to be in $L^2(\Gamma \backslash G)$, is it true that if we will act on it by some element of the Lie algebra, the result will also be in $L^2$?

Thank you,
Sasha

答:
If you allow yourself to use the fact that $L^2$, $\mathfrak z$-finite $K$-finite automorphic forms are finite linear combinations of $L^2$ eigenfunctions, here cuspforms and constants, then on-the-right derivatives map such finite linear combinations to finite sums of the same type, which are again in $L^2$.

Thus, one would want to prove that no $L^2$ continuous spectrum can be $\mathfrak z$-finite and $K$-finite, which is a corollary of Plancherel.


问:
Not understanding this notation
What is the magnitude (length) of vector µv for the following µ?
a) µ = 0 b) µ = length of v c) µ = 1/length of v µ = 1/length of v squared

What is µv? what is  µ?

答:
Let $||v||$ represent the length/norm of a vector $v$. $\mu$ is a scalar, and $\mu v$ is the vector multiplied by the scalar. Generally, we have that $$||\mu v||=|\mu|\cdot||v||$$

So, the question is asking you what $||\mu v||$ is when:
$$a) \ \mu= 0$$
$$b) \ \mu=||v||$$
$$c) \ \mu=\frac{1}{||v||}$$
$$d) \ \mu=\frac{1}{||v||^2}$$

Title: Not understanding this notation


问:
How many nonprincipal ultrafilters exists on $X$ such that $|X|^2=|X|$?
Let $X$ be an infinite set, idemmultiple ($|X|^2 = |X|$) if that helps.  I am looking for a proof that $X$ has more than $|X|$-many ultrafilters.  It has $|X|$-many principal ultrafilters. Without at least some AC I know of no proof that $X$ must have any non-principal ultrafilters.  So let us assume BPI, the Prime ideal theorem.  If we assume BPI can we prove $|\beta X| > |X|$?   ($\beta X$ is of course the set of ultrafilters on $X$).  If we have AC then it is standard that $|\beta X| = 2^{2^{|X|}}$.  I don't much mind which set theory we work in (the question actually arose in NF, with $X = V$) co's it's the construction (if any) that I am after.

Summary: i am interested in a proof (no theory specified) that BPI implies that an idemmultiple (infinite) set $X$ has more than $|X|$-many ultrafilters

答:
Recall the usual proof of $|\beta X|=2^{2^{|X|}}$ using AC.  To sketch the argument, you replace $X$ with the set $Y$ of pairs $(A,S)$ where $A$ is a finite subset of $X$ and $S$ is a finite set of finite subsets of $X$.  Then, you explicitly construct a family of $2^{2^{|X|}}$ pairwise incompatible filters on $Y$, and extend them each to ultrafilters.  This gives $2^{2^{|X|}}$ different ultrafilters on $Y$, and hence on $X$ since $|X|=|Y|$.

With a bit of care, you can make this argument still work using only BPI and $|X|^2=|X|$.  First, we still have $|X|=|Y|$.  To prove this, note that we can totally order $X$ by BPI, and so from $|X|^2=|X|$ we can obtain a family of injections $[X]^n\to X$ for each finite $n$.  Also, $|X|^2=|X|$ implies $\aleph_0\leq |X|$ so $|X|\times\aleph_0\leq |X|^2=|X|$.  Thus $|X|\leq [X]^{<\omega}\leq|X|\times\aleph_0=|X|$, and so $|Y|=|[X]^{<\omega}\times[[X]^{<\omega}]^{<\omega}|=|X|^2=|X|$.

Now, using BPI, we can extend each of our filters on $Y$ to an ultrafilter.  In fact, we can actually extend them simultaneously to get a family of $2^{2^{|X|}}$ different ultrafilters on $Y$, proving that $|\beta X|=|\beta Y|=2^{2^{|X|}}$.  More generally, suppose you have a family $(B_i)_{i\in I}$ of nonzero Boolean algebras and you wish to find ultrafilters on all of them simultaneously.  Let $B$ be the free product of the $B_i$.  Then $B$ is a nonzero Boolean algebra: if we had $1=0$ in $B$, then we would have $1=0$ in the free product of some finitely many $B_i$ (since $B$ is the direct limit of the finite free products), but that is impossible (for instance, by picking ultrafilters on those finitely many $B_i$ to get a homomorphism from their free product to $\{0,1\}$).  Thus by BPI, $B$ has an ultrafilter, and it can be pulled back along the canonical homomorphisms $B_i\to B$ to get a family of ultrafilters on each $B_i$. (Thanks to Andreas Blass for providing a variant of this argument in the comments.)


问:
How to show two properties about the Cantor Set
Define $C_0=[0,1]$ and for $n\in\mathbb{N}$, define $$C_n=C_{n-1}\setminus\bigg(\bigcup_{k=0}^{3^{n-1}-1}\bigg(\frac{1+3k}{3^n},\frac{2+3k}{3^n}\bigg)\bigg) $$ Then the Cantor set is defined as $$C=\bigcap_{n\in\mathbb{N}}C_n$$

Things I need to show:

(1) Each $C_n$ is the disjoint union of $2^n$ closed sub-intervals of length $3^{-n}$ and that the endpoints of each $C_n$ are in $C$.

(2) For any distinct $x,y\in C$, there exists non-empty, disjoint, open sets (open in in $C$) $A,B\subset C$ such that $A\cup B=C$ with $x\in A$ and $y\in B$.

For (1), using induction seemed to me to be the best way to go about proving (1), since I am unsure how to go about proving it for an arbitrary $n\in\mathbb{N}$. However, using the inductive hypothesis with this definition turned out to be unwieldy and awkward, and did not yield any results with the methods I tried.

Proof outline for (2):

Let $x,y\in C$ and WLOG, let $x<y$. Choose $N\in\mathbb{N}$ such that $$3^{N-1}(y-x)>2$$ Now $$3^{N-1}x<3^{N-1}x+2<3^{N-1}y $$ so there is an integer $k\in\mathbb{N}$ such that $$3^{N-1}x<k<k+1<y3^{N-1}$$ Then $$x<\frac{3k}{3^{N}}<\frac{3(k+1)}{3^{N}}<y $$ and finally, we have $$x<\frac{3k}{3^{N}}<\frac{1+3k}{3^N}<\frac{2+3k}{3^N}<\frac{3(k+1)}{3^N}<y $$

The idea here was to find one of the "deleted" intervals which separated $x$ and $y$. If we set $A=(-1,z)$ and $B=(z,2)$ for some $z$ between $\frac{1+3k}{3^N}$ and $\frac{2+3k}{3^N}$, then $A\cap C$ and $B\cap C$ will be open in $C$, they will be disjoint, and their union will equal $C$.

Questions:

How should I proceed in proving (1)? Is this proof for (2) sufficient?

答:
Here we look at the induction proof of the claim

Let
\begin{align*}
C_n&=C_{n-1}\setminus\bigcup_{k=0}^{3^{n-1}-1}\bigg(\frac{1+3k}{3^n},\frac{2+3k}{3^n}\bigg)\qquad\qquad n\geq 1\tag{1}\\
C_0&=[0,1]\\
\\
C&=\bigcap_{n\in\mathbb{N}}C_n
\end{align*}
then each $C_n$ is the disjoint union of $2^n$ closed sub-intervals of length $3^{-n}$ and that the endpoints of each $C_n$ are in $C$.

Base step: $n=0$

$C_0=[0,1]$ is the disjoint union of $2^0=1$ closed sub-intervals of length $3^{0}=1$. We also see the endpoints $0$ and $1$ are in $C$, since the open intervals which are subtracted in the definition of $C_n$ do not contain these points.

Induction hypothesis: $n=N$

We assume the claim is valid for $n=N$. In order to apply the hypothesis we look somewhat closer at $C_n$ and derive a further representation. We see (more or less clearly) that in each step the middle third of intervals is subtracted. This indicates to use a representation of the reals at base $b=3$. We start with small $n=0,1,2$ and get
\begin{align*}
\color{blue}{C_0}&=[0,1] \color{blue}{= [0_3,0.\dot{2}_3]}\tag{2.0}\\
\color{blue}{C_1}&=[0,1]\setminus\left(\frac{1}{3},\frac{2}{3}\right)
=\left[0,\frac{1}{3}\right]\cup\left[\frac{2}{3},1\right]\color{blue}{=[0_3,0.0\dot{2}_3]\cup[0.2_3,0.\dot{2}_3]}\tag{2.1}\\
\color{blue}{C_2}&=C_1\setminus
\left(\left(\frac{1}{9},\frac{2}{9}\right)\cup\left(\frac{4}{9},\frac{5}{9}\right)
\cup\left(\frac{7}{9},\frac{8}{9}\right)\right)\\
\\
&\,\,\color{blue}{=\left[0_3,0.00\dot{2}_3\right]\cup\left[0.02_3,0.0\dot{2}_3\right]
\cup\left[0.2_3,0.20\dot{2}_3\right]\cup\left[0.22_3,0.\dot{2}_3\right]}\tag{2.2}
\end{align*}
We observe in (2.0) - (2.2) the numbers in $C_n$ can be written in ternary representation (indicated with an index $3$) without the digit $1$ at the first $n$ positions after the comma. We can generally write
\begin{align*}
C_n=\left\{z\in[0,1]\bigg|z=\sum_{j=1}^\infty a_j\cdot 3^{-j}, a_k\in\{0,2\}, 1\leq k\leq n\right\}\tag{3}
\end{align*}

We are now well prepared for the induction step.

Induction step: $n=N+1$

We obtain
\begin{align*}
C_{N+1}&=C_N\setminus\bigcup_{k=0}^{3^{N}-1}\left(\frac{1+3k}{3^{N+1}},\frac{2+3k}{3^{N+1}}\right)\\
&=C_N\setminus\bigcup_{k=0}^{3^{N}-1}\left(\frac{k}{3^N}+\frac{1}{3^{N+1}},\frac{k}{3^N}+\frac{2}{3^{N+1}}\right)\tag{4.1}\\
&=\left\{z\in[0,1]\bigg|z=\sum_{j=1}^\infty a_j\cdot 3^{-j}, a_k\in\{0,2\}, 1\leq k\leq N+1\right\}\tag{4.2}
\end{align*}
and the claim follows.

Comment:

In (4.1) we observe the intervals $\left(\frac{k}{3^N}+\frac{1}{3^{N+1}},\frac{k}{3^N}+\frac{2}{3^{N+1}}\right)$ which are subtracted from $C_N$ have length $\frac{1}{3^{N+1}}$. Furthermore these are intervals which are precisely the numbers which can be written with a $1$ at position $N+1$ in ternary representation.

According to the induction hypothesis $C_N$ is the union of $2^N$ disjoint intervals. Since we subtract from each of these intervals the middle part, the number of disjoint intervals is doubled, giving $2^{N+1}$ intervals.

In (4.2) we use the result from (4.1) together with the induction hypothesis to obtain this representation. We observe the endpoints of the intervals in $C_{N+1}$ can be written without using the digit $1$ and are this way justified as points in $C$.


问:
Intuition on joint conditional
For the equation:

$p(a, b \mid c)=p(a \mid b, c) p(b \mid c)$

I was able to prove it but can you help with some easy intuition.

$p(a,b \mid c) = p(a,b,c)/p(c) \\
p(c) =p(b,c)/p(b \mid c)$

$\therefore$

$p(a,b \mid c) =p(a,b,c)p(b \mid c)/p(b,c) \\
p(a,b \mid c) =p(b \mid c) p(a,b,c)/p(b,c) \\
p(a,b \mid c) =p(b \mid c) p(a \mid b,c)$

Q1. Can you help me understand the intuition on previous equation.

Q2. Is there some site online that can do symbolic math on probability such as: $p(a, b \mid c)$

答: I think of the condition on conditional probability as "environment."  Such as, rolling two dice, the numbers are independent...conditional on the dice being fair, being balanced the same, not being marked differently, etc.  You just take the usual independence relation and add "given z" to every probability.</p>

Title: Intuition on joint conditional


问:
Simplifying Differentiation
I'm working off my textbook and I've followed the steps easily enough until it gets to this

$ \dfrac {dy}{dx} = \dfrac{(x^2 + 1)^3}{2 \sqrt{x - 1}} + \sqrt{x-1}~(6x)~(x^2 + 1)^2$

$= \dfrac{(x^2 + 1)^2}{2\sqrt{x - 1})}[(x^2 + 1) + 12x(x - 1)]$

$= \dfrac{(x^2 + 1)^2(13x^2 - 12x + 1)}{2\sqrt{x - 1}}$

How do they go from the first line onwards?

Thanks in advance (sorry about formatting)

答:
I will assume that we are looking at
$$\frac{(x^2+1)^3}{2\sqrt{x-1}} +\sqrt{x-1}(6x)(x^2+1)^2.$$
Take out the common factor $(x^2+1)^2$, and multiply and divide by $2\sqrt{x-1}$. We get
$$\frac{(x^2+1)^2}{2\sqrt{x-1}}\left((x^2+1)+(2)(x+1)(6x)     \right).$$
Now simplify the expression on the right.

That seems to be the way it was thought of. I would prefer to first bring the expression to the common denominator $2\sqrt{x-1}$. We then get
$$\frac{(x^2+1)^3 +2(x-1)(6x)(x^2+1)^2}{2\sqrt{x-1}}.$$
Now take out the common factor
$\dfrac{(x^2+1)^2}{2\sqrt{x-1}}$. Perhaps that was the way it was done, with a step skipped.

Remark: If you have seen the (natural) logarithm already, the following approach, called logarithmic differentiation, may appeal to you. It is of at most marginal utility in this problem, but could be useful when differentiating a longer product.

We were differentiating $(x^2+1)^3 \sqrt{x-1}$. Let
$$y=(x^2+1)^3\sqrt{x-1}.$$
Take the natural logarithm of both sides. Using "laws of logarithms," we get
$$\ln y=3\ln(x^2+1)+\frac{1}{2}\ln(x-1).$$
Differentiate, using the Chain Rule. We get
$$\frac{1}{y}\frac{dy}{dx}=\frac{6x}{x^2+1} +\frac{1}{2(x-1)}.$$
Finally, multiply by $y$ to get $\frac{dy}{dx}$.

Title: Simplifying Differentiation


问:
Categorical semantics explained – what is an interpretation?
I’ve been really having a hard time trying to understand categorical semantics. In fact, I am confused to the point I am afraid I don't know how to ask this question!

I’ve been reading textbooks like Makkai and Reyes’ book, Jacobs’ Categorical logic and type theory (1992) or Categorical Logic by Andrew Pitts, but I simply cannot get the feeling of what are they trying to do. This entry and this one in the nlab do not seem to be very helpful for my understanding. In fact, it seems that the more I read the more confused I get.

Can those questions be answered crystal clearly?

How we define a categorical interpretation?

If it a functor, what is its domain? That is, in model theory, we interpret languages. What are we interpreting here, languages or theories .. or something else?

Let me take an example: I know, very informally, that intuitionistic propositional logic (IPL) correspond to bicartesian closed categories (BCC).

But how to make this statement mathematically precise?
How do we define such an interpretation?

If it is indeed a functor, what is its domain? The category of intuitionistic propositional logics or the category of intuitionistic propositional languages? Or even a particular intuitionistic propositional logic understood as a categry (possibly given by the category of proofs)?

I would really appreciate any answer that could shed some light not only into this example, but also into the big picture of it.

Background: I believe to have all the required background in category theory (UPs, natural transformations, equivalences, adjoints etc.) and type theory (polymorphism, dependent types etc.), having attended courses in both disciplines.

答:
The idea of formulating everything in terms of natural bijections makes things simpler; but in order for the natural bijections for the quantifiers $(C,∀A) ⇔ ^T(C,A)$ and $(∃B,D) ⇔ ^T(B,D)$ to be made consistent, the occurrences of $C$ and $D$ on the right have to be lifted to the category $^T$.

There are actually adjunctions we're talking about here!

The lifting is by the $K$ combinator, represented here as a functor $_T:  → ^T$, given on objects, like $C ∈ ||$ as $_TC(t) = C$, for $t∈T$, and on morphisms $f: A → B$ in $$ as $_Tf: _TA → _TB$ as $_Tf(t) = f$, for $t ∈ T$ and $A, B ∈ ||$.

So, the natural bijections are actually $(C,∀A) ⇔ ^T(_TC,A)$ and $(∃B,D) ⇔ ^T(B,_TD)$.

Of course, once you do that, the category-theory people are going to have a field day and go "$T$ is only a set or a family of sets? Oh, we can fix that!" and off we go, launch sequence activated. They're a modest crowd and generally don't sit well with bare sets out in the open. A way will always be found to "categorify" it yet further.


答:
Hint:

Use
Why it's true? $\arcsin(x) +\arccos(x) = \frac{\pi}{2}$  to convert all the inverse ratios to arcsin

and then utilize Proof for the formula of sum of arcsine functions $ \arcsin x + \arcsin y $

Title: How to evaluate inverse trig functions without a calculator or trig tables?

Body:
I am trying to understand how to evaluate the following equation without using a calculator or trig tables:

arcsin(4/5) - arccos(12/13) = arccos(56/65)


问:
Proving $\sum_{k=0} \frac{(-1)^k}{1+2k} = \frac{\pi}{4} $ taking $4 \int_0^1 \sqrt{1 - x^2} \; \text{d}x$ as the definition of $\pi$.
I'm trying to prove the Madhava-Leibniz formula for $\pi$ using one quarter of the area of the unit circle as the definition of $\frac{\pi}{4}$.

$$ \int_{0}^{1} \sqrt{1-xx}\; \text{d}x = \sum_{k=0}^{\infty} \frac{(-1)^k}{2k+1} $$

I have a proof of this identity, but it uses a lot of trigonometric substitutions and makes direct use of $\pi$ as a constant.

This feels a bit circular since the left hand side is supposed to be the definition of $\frac{\pi}{4}$.

Is there a way to prove this identity directly, instead of proving that both sides are equal to $\frac{\pi}{4}$?

All integrals are over the interval $[0,1]$ in the variable $x$. Let $\theta$ refer to $\frac{\pi}{4} x$. $\theta$ is a constant multiple of $x$, I'm using it so I don't need to change the bound of integration.

$$ \sum_{k=0} \frac{(-1)^k}{1+2k} $$

Using Lemma 101.

$$ \sum_{k=0} \int_{0}^{1} (-1)^k x^{2k} \text{d}x $$
$$ \sum_{k=0} \int_{0}^{1} (-xx)^k \text{d}x $$

Exchanging the sum and integral.

$$ \int_{0}^{1} \sum_{k=0} (-xx)^k \text{d}x $$

Applying Lemma 102

$$ \int_{0}^{1} \frac{1}{1+xx} \text{d}x $$

Applying Lemma 103, with $g$ equal to $\tan(\theta)$ and $g'$ equal to $\frac{\pi}{4} \sec^2(\theta)$.

$$ \int_{0}^{1} \frac{\pi}{4} \cdot \sec^2(\theta) \cdot \frac{1}{1+\tan^2(\theta)} \text{d}x $$

$$  \frac{\pi}{4} \int_{0}^{1} \sec^2(\theta) \cdot \frac{1}{\sec^2(\theta)} \text{d}x $$

$$  \frac{\pi}{4} \int_{0}^{1} 1\; \text{d}x $$

$$ \frac{\pi}{4} $$

And for the other side.

$$ \int_0^1 \sqrt{1-xx} \; \text{d}x $$

Apply lemma 103, with $g = \sin(2\theta)$ and $g' = \frac{\pi}{2}\cos(2\theta)$.

$$ \int_0^1 \frac{\pi}{2} \cdot \cos(2 \theta) \sqrt{1-\sin^2(2 \theta)} \; \text{d}x $$

$$ \frac{\pi}{2} \int_0^1 \cos^2(2 \theta) \; \text{d}x $$

Apply lemma 104.

$$ \frac{\pi}{2} \cdot \frac{1}{2} $$

$$ \frac{\pi}{4} $$

Lemmas

Lemma 101: integral of monomials

$$ \int_0^1 x^n \text{d}x = \frac{1}{1+n} $$

Lemma 102: sum of geometric series

$$ \sum_{k=0} r^k = \frac{1}{1-r} $$

Lemma 103: Special case of change of variables for integration

Suppose $g(0)=0$ and $g(1)=1$, then the following holds.

$$ \int_0^1 f(x) \text{d}x = \int_0^1 g'(x) \cdot f'\circ g(x) \text{d}x $$

Lemma 104: Integral of $\cos^2 + \sin^2$

For all values of $x$, $\cos^2(x) + \sin^2(x) = 1$, therefore the following holds

$$ \int_0^1 \cos^2(2\theta) + \sin^2(2\theta) \text{d}x = 1 $$

However, $\cos\left(\frac{\pi}{2}-x\right) = \sin(x) $, therefore, over the interval $[0, 1]$, $\sin$ and $\cos$ take on the same values at the same offsets, but in the opposite order. Thus

$$ \int_0^1 \sin^2(2\theta) \text{d}x = \int_0^1 \cos^2(2\theta) \text{d}x $$

Thus

$$ \int_0^1 \cos^2(2\theta) \text{d}x = \frac{1}{2} $$

答: The right hand side of your formula is the Maclaurin series expansion of $\arctan 1$.


问:
Why is the notation for the inner product of two columns of the Vandermonde matrix expressed this way?

I don't see how the above is a dot product for a Vandermonde matrix.

答: It is a dot product for two columns of a Vandermonde matrix, not the entire matrix. If one column is $(1, \omega^j, \omega^{2j}, \omega^{3j}...$, and say x = $\omega^j$, then that column is the more familiar (1, x, $x^2, x^3, ...)$ and another column likewise with $\omega^k$, then taking the complex transpose of the second vector gives a row vector (1, $\omega^{-k}, \omega^{-2k}$, ...) and then you take the dot product to get the result.


问:
Being neither a minimum element nor a successor of any element implies being a limit point
I am proving the following theorem

Suppose $<_{X}$ strictly order $X$ and $c \in X$ is neither a minimum element of $X$ nor a successor of any $x \in X$. Then $c$ is a limit point.

I think I have constructed a reasonable proof for this statement. However, the proof seems verbose, and it looks more like a logic game rather than a common proof in mathematics. I am seeking opinions on how to write this proof more succinctly.

Proof. Assume that $c$ is not a minimum element. Then there exists some $y \in X$ such that $y <_{X} c$.

Next assume that $c$ is not a successor of any $x \in X$. According to the definition of a successor, we know that for any $x \in X$, either $c \leq_{X} x$, or there exists some $y$, $x <_{X} y$ and $y <_{X} c$. This statement is equivalent to the following statement: for any $x \in X$, either $\lnot\left(x <_{X} c\right)$, or there exists some $y$, $x <_{X} y$ and $y <_{X} c$. This is further equivalent to the following statement: for any $x \in X$, if $x <_{X} c$, then there exists some $y$, $x <_{X} y$ and $y <_{X} c$.

It is now clear that $c$ is a limit point, given the definition of a limit point. As a result, $c \in X$ being neither a minimum element of $X$ nor a successor of any $x \in X$ implies that $c$ is a limit point.

答:
It’s okay, but it really is a bit wordy. You could safely cut it down to something like this:

Suppose that $c$ is neither a minimum element nor a successor. Let $P=\{x\in X:x<_Xc\}$; by hypothesis $P\ne\varnothing$. Let $x\in P$; $c$ is not the successor of $x$, so there is a $y\in X$ such that $x<_Xy<_Xc$. This is the case for each $x\in P$, so by definition $c$ is a limit point.

That really is all that you need to say.


问:
What is the exact definition of polynomial functions?
I'm trying to understand the difference between polynomial functions and the evaluation homomorphisms. I noticed that I don't know what's the exact definition of a polynomial function, although I've used it since I was in school.

I'm confused.

Thanks in advance

答:
A polynomial with coefficients in a commutative ring $R$ is a formal sum $r_n x^n + \ldots + r_1 x + r_0$. It is not a function. It is merely a finite sum of terms of the form $r_i x^i$, where $x$ is some abstract symbol. The ring structure of $R$ lets us make these symbols into a ring, that we denote $R[x]$.

For each $r \in R$, there is a ring homomorphism $e_r:R[x] \rightarrow R$. These are the evaluation maps, defined by replacing each instance of $x$ in the formal sum of a polynomial with $r$ and considering this as an element of $R$. We sometimes denote $e_r(f(x))$ as $f(r)$, but this is not the same as the formal polynomial $f(x)$. $f(r)$ for $r \in R$ is actually an element of the ring, while the formal polynomial $f(x)$ is not.

For each polynomial $f(x) \in R[x]$, there is a map $R \rightarrow R$ defined by $r \mapsto e_r(f(x)) = f(r)$. This is not a ring homomorphism! (For instance, the polynomial function on $\mathbb{R}$ defined by $f(x) = x^2$ is not a homomorphism of rings.) It is a function determined by a polynomial, and so we call such a function a polynomial function.

All of these things can be generalized to polynomials in multiple variables, as well.

A note: Why the fuss about formal polynomials and the functions they define? Typically, you're safe not making a distinction between the formal sum and the function it defines. However, some issues can arise and you need to be aware of them. For instance, consider the polynomial $x^2 - x = (x-1)x$ with coefficients in the ring $\mathbb{F}_2$, the field with two elements. This polynomial, formally, is not zero... it has nonzero coefficients! However, the function it defines on $\mathbb{F}_2$ is the zero function, since there are only two points in $\mathbb{F}_2$ and the evaluation homomorphism for each point sends the function to zero.


问:
Prove $a(n, k) = t(n, k + 1)$ with permutations and cycles where 1 and 2 are/are not in the same cycle?
Let $a(n, k)$ be the number of permutations of length n with k cycles in which the entries 1 and 2 are
in the same cycle. Let $t(n, k) = c(n, k) − a(n, k)$ be the number of permutations of length
n with k cycles in which the entries 1 and 2 are not in the same cycle. Prove that $a(n, k) = t(n, k + 1)$
for all $k ≤ n − 1.$

EDIT: I've tried directly summing the generating functions for (,),(,), and we got $x^2(+1)...(+−1)$, but we don't know where to go from there. We also tried using the fact that $t(n, k) = c(n, k) − a(n, k)$ to perhaps get some sort of relation that makes more sense, but the only relations that we can make rely on using the fact we're trying to prove, which is sort of counterproductive.

答:
Hint: Let $\sigma$ be one of the permutations counted by $a(n,k),$ then $$\sigma = (1\, a_1\, \cdots\, a_i\, 2\, b_1\, \cdots \, b_j)c_2\cdots c_k,$$ where $c_i$ are cycles and $\{a_m,b_m\}$ are numbers. What if you define a function
$$\varphi (\sigma )=(1\, a_1\, \cdots\, a_i)(2\, b_1\, \cdots \, b_j)c_2\cdots c_k.$$
Is this a bijection?


问:
How to evaluate inverse trig functions without a calculator or trig tables?
I am trying to understand how to evaluate the following equation without using a calculator or trig tables:

arcsin(4/5) - arccos(12/13) = arccos(56/65)

答:
Let $\theta=\arcsin(4/5)$, $\:\varphi =\arccos(12/13)$, and compute $\cos(\theta-\varphi)$.

First note that $\theta\in \bigl[\frac \pi 3,\frac\pi 2\bigr]$, so $\cos\theta >0$ and $\varphi\in\bigl[0,\frac\pi 6\bigr]$, so $\sin \varphi \ge 0$.

Therefore $\cos\theta=\sqrt{1-\frac{16}{25}}=\frac 35$ and $\sin\varphi=\sqrt{1-\frac{144}{169}}=\frac 5{13}$, and consequently, by the addition formula,
$$\cos(\theta-\varphi)=\frac35\frac{12}{13}+\frac45\frac5{13}=\frac{56}{65}. $$
Furthermore, $\theta-\varphi\:$ lives in  the interval $\bigl[\frac\pi3-\frac\pi 6,\frac\pi 2\bigr]\subset[0,\pi]$, so it is exactly $\:\arccos(56/65)$.


答:
Yea the earlier answer basically said it all. More explicitly, obviously we have (because of Bayes theorem, or drawing the relevant Venn diagram, or whatever):

$$P(a,b)=P(a|b)P(b)$$

Then we can add the condition on $c$ on both sides to get

$$P(a,b|c)=P(a|b,c)P(b|c)$$

Title: Intuition on joint conditional

Body:
For the equation:

$p(a, b \mid c)=p(a \mid b, c) p(b \mid c)$

I was able to prove it but can you help with some easy intuition.

$p(a,b \mid c) = p(a,b,c)/p(c) \\
p(c) =p(b,c)/p(b \mid c)$

$\therefore$

$p(a,b \mid c) =p(a,b,c)p(b \mid c)/p(b,c) \\
p(a,b \mid c) =p(b \mid c) p(a,b,c)/p(b,c) \\
p(a,b \mid c) =p(b \mid c) p(a \mid b,c)$

Q1. Can you help me understand the intuition on previous equation.

Q2. Is there some site online that can do symbolic math on probability such as: $p(a, b \mid c)$


问:
What is the meaning of inner product of two circles?
I was reading this post about orthogonal circles (here) and in the question body something about inner product of circles was used.. it seemed quite mysterious but still interesting. I know of inner products of vector which is simply the dot product but what exactly is the inner product of circles? Does it having an associated geometric meaning?</p>

答:
I checked Iversen's book "Hyperbolic Geometry". This formula is indeed the Lorentzian inner product restricted to points of the 1-sheeted hyperboloid $M$. Here is how the map from circles to points in the  1-sheeted hyperboloid works. Consider a circle $C$ in the Euclidean plane. If you think of the upper half-space model $U$ of the hyperbolic 3-space ${\mathbb H}^3$:
$$
U=\{(x_1, x_2, x_3): x_3> 0\}
$$
then $C$ is the intersection of the closure of a unique hyperbolic plane $H_C\subset U$ with the Euclidean plane $\{x_3=0\}$. Now, you switch to the Lorentzian model of the hyperbolic 3-space. Then ${\mathbb H}^3$  becomes the upper sheet $S$ of the 2-sheeted hyperboloid
$$
\{v: \langle v, v\rangle=-1\},
$$
where $\langle v, w\rangle$ denotes the Lorentzian inner product in the Lorentzian space $R^{3,1}$. Then hyperbolic planes in ${\mathbb H}^3$ correspond to the intersection of 3-dimensional linear subspaces $V\subset R^{3,1}$ with $S$. The assumption that the intersection $V\cap S$ is nonempty translates into the condition that the Lorentzian perpendicular vector $u$ to $V$ is space-like, i.e.
$$
\langle u, u\rangle > 0
$$
WLOG, we can assume then that $u$ is a Lorentzian unit vector.
I will consider the circle $C$ oriented, then $H_C$ is oriented as well, hence $V$ is also oriented. This determines the direction of $u$, hence, determines $u$ uniquely. The vector $u$ then belongs to the 1-sheeted hyperboloid $M$ in $R^{3,1}$.

Thus, we obtain the map $C\mapsto u=u_C$, sending oriented Euclidean circles to points of the 1-sheeted hyperboloid $M$. (This map is 1-1 but not onto: The mossing points correspond to Euclidean lines which can be regarded as circles of infinite radius.) Next, each circle $C$ in the plane is given by the quadratic equation
$$
b (x\cdot x) - 2(x\cdot f) + a=0,
$$
where $\cdot$ is the Euclidean dot product, $f\in {\mathbb R}^2$ and $a, b\in {\mathbb R}$. This equation is not quite unique since we can multiply $a, b, f$ by the same scalar without changing the zero set of the quadratic polynomial. So, one normalizes
$$
|f|^2 -ab=1.
$$
Then the "inner product" of two circles becomes
$$
\langle C_1, C_2\rangle = (f_1\cdot f_2) - \frac{1}{2}( a_1 b_2 + b_1 a_2).
$$

This is the formula from the linked MSE question.
Now, one has to do a computation which yields:
$$
\langle u_{C_1}, u_{C_2}\rangle= \langle C_1, C_2\rangle,
$$
where the inner product on the left is Lorentzian.

Another geometric interpretation of this inner product is the negative of the cosine of the angle between the circles.

Needless to say, all this also works in higher dimensional Euclidean and hyperbolic spaces, just one has to replace circles by spheres.


问:
A $2021$ problem: $20\sim 21$ and $43\times 47$
Notice that $2021$ is a concatenation of consecutive integers:  $20\sim 21$

Also $2021$ is a product of consecutive primes: $43\times 47$.

What is the next number with both of these properties?

$24502451$ is close, $4943\times 4957$  but $4951$ is in between.

$2484224843$ is close, $49831\times 49853$ but $49843$ is in between.

$715353612\sim 715353607$ isn't quite there but is $845785793\times 845785799$

Are there any other numbers with the $2021$ property?

Body:
I found more twin prime solutions. For $n = 396$, $d = 2$ I found four; the smallest is $x =$
$${\small
434219772837481616940726338933362452273916097301492635627291443512051997247961750150908315825206335025681182252954355463934702107964728269885096262688644034853639395077615105799339919571601786192665103151170947581386505769710635828116973131647706180301379657813220432413064536727826883282252811784121256116486385454859849292857777667719016881251956101283374338871503089431823276421261037909974359}$$

For $n = 420, d= 2$ I found $x =$

$${\small797632045320122442922746848370123218495083238021539262463793196558953236606673684986633691999805212011226776035396567202878987657452660659814874176096052099397205337415190293836597252999499063444222715671072742933640380698030812583452694556061309433747142801949662038045758937332990610602377203706320829077950947557183280793831442171456664369310820656043310649088552521802135329739094023210425721483954126295715977820771}$$

I wrote a Java program to find modular square roots using known factorizations of $10^n+1$ at https://stdkmd.net/nrr/repunit/10001.htm

The $n = 396$ solution shown may not be the second smallest twin prime solution because I did not analyse $10^n+1$ with unknown factorization (eg: $323, 392$).

Note that all these solutions involve only probable primes.

UPDATE:

I analyzed all $n \le$ 500 for which complete factorizations are known,
using Shank's $(ln\ 10^n)^2$ estimate for prime gaps: $2 \le d < 5.3n^2$. The following table shows the number of solutions of $n$ for each $d$.

\begin{array}{r|l|r} d & n & total  \\
\hline 2 & 36 \ 396(4) \ 420 \ 468 & 7
\\ 4 & 2 \  70 \ 150 \ 154 \ 210(27) \ 270(8) \ 298 \ 306(2) \ 330(216) \ 350(4) \ 390(10) \ 450(519) & 791
\\ 12 & 370(3)  & 3
\\ 14 & 336(2) \ 396(3)  & 5
\\ 54 & 72  & 1
\\ 76 & 150(2) \  210(16) \ 238 \ 270(4) \ 306(2) \ 330(209) \ 350(3) \ 390(15) \ 450(464) & 716
\\ 82 & 288 \ 336 \ 396(5) & 7
\\ 136 & 370(2) & 2
\\ 478 & 336(2) \ 420(2) & 4
\\ 1364 & 350 \ 370 & 2
\\ \hline
\end{array}

I analyzed many of the remaining known factorizations ($10^{2021}+1$ is not one of them!). I found another twin prime solution: $n=1008$ with $d=2$, $x=$

$${\small415119285335713138107859159895921470120127901771012286705147863784377859054383205862088683121228843823644091024472578295715781891723489781341595609399956375971523325157129204031439607287157565203433091945798569106366453441681028958853009984899996962841286597261664058243897757317082387302022827796188228872367615940280481644558354868664735437056336615963301573812328050683875074515986286867031968871583403569829949341102058719072290611202615565955752049854880027669507042067569064851453754109081689314779674580036777493429088593987592215948631397812098103849006767378312184473327978598863061300578414744304808368322583101196999821431867760847600988156486022394149378510713398269801057016361947970128875578649362095899180615435692074486807012156095811778896443290518016088027948808773134635025220295845062105375107159001836188571569967805721691214024809680987944490924904786048603374165125520526174101682333287056313013054613391604106858759925794745594124906616469238525225716143856670923606283395248479131507}$$

I found three large solutions: $n=1530$ with $d=4$, $n=2442$ with $d=4$, and $n=3011$ (odd!) with $d=20690$ and $x=$

$${\small88690623974342445320706035164907895149666724533149745483024674929494064077161748069463728371848588950263928708885107056062087458873420997129551966175554795722180833514256206712208756791349423181129064370039115228768196330069003307522127855902035611295636718786724552834173077606816947565351851981240290592919330234435002605544818898704383050138705647019342509295291603847356454590383853903649034831518476190799174284971745361604565835545543751726529610775367997499873686078754015523170389723175091842178101899712480880871125974133817901518538674999703972616649632379652346395327832021456141800194701647932112291489233775928062470099635623529564608650310241193184759187003057364275380404785676729319256559060638811849744657325135099995176231775532397706618363314457193916318557002601459452270771049765689157482464270159966495073998419806308753511281083712198273207669118838966447628803262451478908503097089308456131000369654881518026667709557445311154935973975062432608109650859738547758359753161811236281693386411516599337580682444733887155249147703101490725039867545594773022646082904694935305608529996539856673800894879204151359288642251946537962882362542598385355877258435885255720463281832318996714749127620565928295871807381765069313332957959393036520560765777218866425812066806566968512664298253194097198110254106398916193178093323285510247757090049514423671929016532305341758972853095343072371415631421656596311059421635680366649419960217875499119738356663626923434263846557987005717753122760707664468827714474707911315796646846223047293968656604803860055531395007086544292734839551265327070525560544860595807571049301946018895129458512514361858289027536454436450967551887360811792658089837039529387070983777359590036217218745184175803793616945894809150630864349323615934382263207217535053701872322473625965756309360763380729005686068625698261693375747392352007831638590079855262236534947866527970435414431816193335704235728803592301557775355412534721854441649050543489915852763945114633640120859359425753981309635846954784400203166716706835000886558080563709914367339817642986577630512066137497993335131126095180539966000395250657557890723647715298981683035616063901855227374869260175148423588397623462842825972640026055333967889877654729507728851982076103225242572504527429707508826564462178655965426323365276645348503046435607191738892135058428362640796305615358118855629551341911741579489817482134866791604658301235483655504562219338888352571792484327753544399478907161167075036271824602275963945345494267916881834140919492205225308097367770069805455016950009471893642408322180438063500438940054744323617310086174876710854406021502917708656972234075774384689936433268632860789883740306973585819078135116827900294591791739792869006503475384136963506082899865375050762137583035828837915859970271078112397338135202345323724701287354928723862525723289692127491314630211880612187498071770083275673021172070970359804795314571223002210253991420896839561827296295193582310969508838235157649439819871296875917}$$

UPDATE (May 28, 2021)

I attempted to find solutions bigger than $n = 3011$, but found none!
I rewrote my program in C using GMP and ran it on AWS with 96 cores.
The following table summarises my results for all known complete factorizations
of $10^n+1$ for $n > 3011$.
For each $n$ the search was up to  $d$ for the given ratio $\frac{d}{n}$.
Note that a ratio of $\mathbf{2.3}$ corresponds to the expected number of
primes in the range $[10^n,10^n+2.3n] = 1$,
a ratio of $\mathbf{96}$ corresponds to the current record prime gap
(of $8350$ for an $87$ digit prime number),
and a ratio of $\mathbf{6.9}$ corresponds to the biggest solution I found: $n=3011, d=20690$.

\begin{array}{r|l} n & \frac{d}{n}   \\
\hline <22292 & 100
\\ 22292 & 21
\\ 22303 & 10
\\ 23592 & 53
\\ 23734 & 3
\\ 26014 & 10
\\ 32962 & 10
\\ 46957 & 3
\\ 47248 & 12
\\ 64439 & 1.0
\\ 80363 & 1.0
\\ 95594 & 1.0
\\ 103624 & 1.0
\\ 132586 & 1.0
\\ ^*180178 & 0.10
\\ ^*268207 & 0.10
\\ ^*1600787 &  -
\\ \hline
\end{array}

$^*$For the last  three values of $n$ the largest prime factor is a probable prime. I did not analyse the last value, because I estimate that a single calculation of $x^y$ mod $z$ with 1600000 digit numbers would take about 70 hours.

UPDATE (June 11, 2021)

For the largest known solution $(n=3011, d=20690)$ the primes $x$ and $x+d$ have been certified using Primo and recorded in factordb.com - search for primes with 3011 digits.

UPDATE (July 8, 2021)

I searched for all twin prime solutions $(d = 2)$ for $ n = 1008$ and found a total of 18. The largest, a new record, is $x=$

$${\small959366284693377033651389735182738458526276121193598553484439760604241337237451497452789960716688792639495153272015332792541955520436759663588757045933885829578173457125721695213851669376206161760331060025909588497751253712340327354707310139498467167235392925523839181171301134866240381669769835527505988247393867108561444781188402425283990574068760196925230932242847785476986116528280435669208099164158860477631087793039498552963323814612240444137080246220725308557204315700968442132019195862098047534857069372757341953852762143251682584867890954165891418336661720575672057626771116234845905889545915860780107445581724924411155250266226552503036267220337257336647317889646751823745301888424349108532006368071312231688002693274043002649314803790153000719816292711634980344850814644852432359001479449220906110949158858923721913836167918792291736846984222096022979367999907004169558381802601034072450079249418137015546418369167885051996435963305528417728081298205640144099322592183826280769897515421235519976901}$$

No more were found for $1008 < n \le 268207$.


问:
prove $ F(x)=\int_0^\infty {\sin(tx)\over(t+1)\sqrt t} dt \in C^\infty(\mathbb R^*) $
prove that :

$$ F(x)=\int_0^\infty {\sin(tx)\over(t+1)\sqrt t} \, dt \in C^\infty(\mathbb R^*) $$

i end up proving that $F(x)\in C^ \infty(\mathbb R^{*+})$ not $\mathbb R^*$ , and i studied the case with :

$F(x)= \sqrt x G(x)$ where $G(x)=\int_0^\infty {\sin(t)\over(t+x)\sqrt t} \, dt$ ($\sqrt x , G(x) \in C^\infty(R^{*+}))$

is my answer wrong ? , wich is the correct interval for $C^\infty$ ?

答: One would think that proving $F(x)\in C^ \infty(\mathbb R^{*+})$ would be sufficient since $F(-x) = -F(x)$.</p>


问:
Homology of Alexander Horned Sphere
I am taking a course in homology this semester, and so far we have only examined spaces/surfaces that the simplicial structures are rather easy to find. I was curious about the Alexander Horned Sphere, and how one would approach finding its homology groups.

I figure $H_{0}(X) \cong \mathbb{Z}$, as the horned sphere is path connected (I think), and I know the fundamental group is infinitely generated, but what explicitly would be $H_{1}(X)$?

And then that begs the question of how to find $H_{2} (X)$.

Looking more for a conversation than a proof, just to satisfy my curiosity.

答: The Alexander horned sphere $A$ is the image of an "exotic" embedding $\phi : S^2 \to \mathbb R^3$. Hence $A \approx S^2$ and $H_n(A) = \mathbb Z$ for $n = 0,2$ and $H_n(A) = 0$ else.

Title: Homology of Alexander Horned Sphere


问:
Using shell method, find the volume of the solid generated when $y=3-x^2$ is revolved around x-axis from $y=0$ to $y=3$

Question

Via the Shell Method, what is the volume of the solid of revolution formed by revolving a portion of $y=3-x^2$ about the x-axis bounded by $y=0$ and $y=3$.

Context

The answer from Wolfram Alpha leads me to believe the solution is $V=\frac{48\sqrt{3}\pi}{5}$. While I believe that this problem can be solved via the Washer Method, I am interested in how to do solve this problem via the Shell Method.

Work So Far

Shell method formula for finding volume

The formula for volume via the shell method is $V=2\pi\int_a^{b}[r(x)h(x)]\mathrm dx$, where:

$V$ represents volume;

$a$ represents the lower bound;

$b$ represents the upper bound;

$r(x)$ represent the distance from the axis of rotation to $x$; and

$h(x)$ represents the height of the solid at $x$.

Given that the revolution occurs around the x-axis, I know I need to set the given equation from terms of $x$ into terms of $y$: $V=2\pi\int_a^{b}[r(y)h(y)]\mathrm dy$. When I rewrite the formula to account for this problem, I end up with $V=2\pi\int_0^{3}[y*\sqrt{3-y}]\mathrm dy$.

Use of u-substitution

Via u-substitution, I would set $u=3-y$ and determine two things: $du=-dy$ and $y=3-u$. Rewriting the volume formula with this information reveals:
$$V=2\pi\int_0^{3}[(u-3)*\sqrt{u}]\mathrm du$$

This simplifies to:

$$V=2\pi\int_0^{3}[u^\frac{3}{2}-3u^\frac{1}{2}]\mathrm du$$

Which further simplifies to:

$$V=2\pi*[\frac{2u^\frac{5}{2}}{5}-2u^\frac{3}{2}]\Big|_0^3$$

When I plug $3-y$ for $u$, the formula reaches it final form:

$$V=2\pi*[\frac{2(3-y)^\frac{5}{2}}{5}-2(3-y)^\frac{3}{2}]\Big|_0^3$$

At this point, I'm getting a volume that is nowhere close to $V=\frac{48\sqrt{3}\pi}{5}$. Any help would be much appreciated!

Appendix

This is my first time encountering an equation where $x=\pm\sqrt(3-y)$ with these types of volume problems. For now, I'm blindly ignoring the negative and am not sure if that's a good idea.

答:
Approach 1: the total volume is the sum of the volume of revolution of the region to the right of $x=0$ and the volume of revolution of the region to the left of $x=0$. By symmetry, the two volumes are the same. On the right, you have $x=\sqrt{3-y}$, so the volume of the part on the right is $2 \pi \int_0^3 y \sqrt{3-y} \, dy$ and then the total volume is twice that.

Approach 2: in general the height $h(y)$ is really the distance from the left side of the region to the right side of the region for each fixed value of $y$, so $h(y)=\sqrt{3-y}-(-\sqrt{3-y})=2\sqrt{3-y}$. (Note that this needs to be adjusted if you have a curve determined by a function that isn't even two-to-one, like the region located above the $x$ axis and below $y=\sin(x)$ for $x$ between $0$ and $4\pi$. But there is no issue here.)


答:
There are of course several easier ways to solve this question, but I'll stick to your shell method. You have close to the right answer, but I found two mistakes.

When you change variables from $y$ to $u$, you need to change your bounds (so you would have the integral from 3 to 0, rather than from 0 to 3).

Your height function is only half of what it should be. This is because you ignored the negative square root; you are only counting the height (width in this problem) from the y-axis to the right, rather than from the left side of the rotated surface (the negative square root) to the right.

After those changes, I got the same answer $\dfrac{48\sqrt{3}\pi}{5}$.

Title: Using shell method, find the volume of the solid generated when $y=3-x^2$ is revolved around x-axis from $y=0$ to $y=3$

Body:
Question

Via the Shell Method, what is the volume of the solid of revolution formed by revolving a portion of $y=3-x^2$ about the x-axis bounded by $y=0$ and $y=3$.

Context

The answer from Wolfram Alpha leads me to believe the solution is $V=\frac{48\sqrt{3}\pi}{5}$. While I believe that this problem can be solved via the Washer Method, I am interested in how to do solve this problem via the Shell Method.

Work So Far

Shell method formula for finding volume

The formula for volume via the shell method is $V=2\pi\int_a^{b}[r(x)h(x)]\mathrm dx$, where:

$V$ represents volume;

$a$ represents the lower bound;

$b$ represents the upper bound;

$r(x)$ represent the distance from the axis of rotation to $x$; and

$h(x)$ represents the height of the solid at $x$.

Given that the revolution occurs around the x-axis, I know I need to set the given equation from terms of $x$ into terms of $y$: $V=2\pi\int_a^{b}[r(y)h(y)]\mathrm dy$. When I rewrite the formula to account for this problem, I end up with $V=2\pi\int_0^{3}[y*\sqrt{3-y}]\mathrm dy$.

Use of u-substitution

Via u-substitution, I would set $u=3-y$ and determine two things: $du=-dy$ and $y=3-u$. Rewriting the volume formula with this information reveals:
$$V=2\pi\int_0^{3}[(u-3)*\sqrt{u}]\mathrm du$$

This simplifies to:

$$V=2\pi\int_0^{3}[u^\frac{3}{2}-3u^\frac{1}{2}]\mathrm du$$

Which further simplifies to:

$$V=2\pi*[\frac{2u^\frac{5}{2}}{5}-2u^\frac{3}{2}]\Big|_0^3$$

When I plug $3-y$ for $u$, the formula reaches it final form:

$$V=2\pi*[\frac{2(3-y)^\frac{5}{2}}{5}-2(3-y)^\frac{3}{2}]\Big|_0^3$$

At this point, I'm getting a volume that is nowhere close to $V=\frac{48\sqrt{3}\pi}{5}$. Any help would be much appreciated!

Appendix

This is my first time encountering an equation where $x=\pm\sqrt(3-y)$ with these types of volume problems. For now, I'm blindly ignoring the negative and am not sure if that's a good idea.


问:
categorical interpretation of quantification
Many constructions in intuitionistic and classical logic have relatively simple counterparts in category theory.  For instance, conjunctions, disjunctions, and conditionals have analogues in products, coproducts, and exponential objects.

More esoteric concepts still permit reasonably simple categorical interpretations.  For instance, the modal operator $\Box$ (necessity) can be described as a monoidal endofunctor (see, for instance, Paiva & Ritter's Basic Constructive Modality (PDF)).

What are the similar constructions that are suitable interpretations for universal and existential quantification in first-order logic?

答:
It's a good exercise to see the natural bijections worked out in some detail, but as I noted in your other reply (Categorical semantics explained – what is an interpretation?), padawan, there is an additional minor detail that you left out, which should be made clear.

While the natural bijection for the $∧$ instantiates to bijections between the hom sets $(C,A∧B)$ and $(C,A)×(C,B)$ for $A,B,C ∈ |××^+|$, for consistency the natural bijection for $∀$ can't instantiate to bijections between $(C,∀A)$ and $^T(C,A)$ for $A ∈ |^T| = ||^T$ and $C ∈ |^+| = ||$: the $C$ on the right has to be lifted to $||^T$ as a constant function. This is implemented by Curry's $K$ combinator, represented here as a functor $_T:  → ^T$ that embeds the objects and morphisms of $$ as constant functions in $^T$ (i.e. $_T C: t∈T ↦ C$ for $C∈||$ and $_T f: t∈T ↦ f$ for $f ∈ (A,B)$ and $A,B ∈ ||$.

Therefore, the correct statement of the natural bijection ... which produces the details you spelled out ... is actually $(C,∀A) ⇔ ^T(_TC,A)$, for $C ∈ ||$ and $A ∈ |^T|$. This forms an adjunction pair $(∀_T,_T)$ for each set $T$ that is being considered: be they the finite sets $T = ∅$ or $T = 2 = \{0,1\}$ or the infinite sets $T$ of terms in a language or families of such sets if the language is multi-sorted.

This is a natural launching point for considering possible generalizations, where the set $T$, or family of such sets, is replaced by other kinds of objects, perhaps even by categories.


问:
Is there any situation where $x^2 \not= \sqrt{x^4}$
This is new for me so sorry if i am missing something, thanks for any helpful pointers.

So I was thinking about this classic equation $E=mc^2$

Then I was, why is there $c^2$. I found it is just expression of units in the equation. And from this another question popped in my mind (maybe unrelated).

I found that we do not know if speed of c is same in all directions and i started thinking, if there is any possibility if this $c^2$ can be something like $\sqrt{c^4}$ with something hidden we do not know yet.

Like if in more dimensional numbers (complex, quaternions), or some other situation when you can add some "hidden number" or be in different numeric "world" when the equation $x^2 = \sqrt{x^4}$ will be not true.

I am really interested if there is any possibility of this and I would like to study it.

答:
If $x = i$ then

$x^2 = -1$

$x^4 = 1$

$\sqrt{x^4} = 1$

$x^2 \ne \sqrt{x^4}$

This answer assumes that the $\sqrt{}$ symbol always yields a positive number if such a number is available.


问:
Drake, Seven Axioms of the Algebra of Events
On page 3 of Drake Fundamentals of Applied Probability he lists The Seven Axioms of the Algebra of Events.

$1. A \cup B = B \cup A \\
2. A \cup (B \cup C) = (A \cup B) \cup C \\
3. A\cap( B \cup C) = A \cap B \cup A \cap C \\
4. (A')' = A \\
5. (A\cap B)' = A' \cup B'\\
6. A\cap A' = \phi \\
7.A \cap U = A  $

I have two questions.

First while Drake states that his choice is not unique is there a reason that he selected these seven and is seven the minimum possible number of axioms with which the Algebra of events can be expressed.

Second he further states that any other relation in the algebra of events can be proved directly from these seven axioms with no additional information and in fact in the chapter problem section asks the student to do so. Does the author mean that all the other set relations in the Algebra of Events can be proved by direct manipulation of the seven axioms with out resorting to the method of $x \in A$

For example, proofs of $A \cup A' = U$ are given in this posting
Proof of union of a set and its complement is equivalent to a universe

but these proofs use the method of $x \in A$ and not direct manipulation of Axiomatic statements as required by Drake. How would I prove $A \cup A' = U$ or $A \cup A = A$ using only direct manipulation of Drake's Seven Axioms of the Algebra of Events?

答:
Let us answer first the second part of your question and the first part of your question.

Let us prove $A \cup A' = U$ and $A \cup A = A$, using only direct manipulation of Drake's Seven Axioms of the Algebra of Events (as it is asked in the question).

Consider the seven axioms:

$1. A \cup B = B \cup A \\
2. A \cup (B \cup C) = (A \cup B) \cup C \\
3. A\cap( B \cup C) = (A \cap B) \cup (A \cap C) \\
4. (A')' = A \\
5. (A\cap B)' = A' \cup B'\\
6. A\cap A' = \phi \\
7.A \cap U = A  $

Let us go result by result .

a. From axioms $1$, $5$  and $4$, we have $A\cap B = B\cap A $.

$ A' \cup B' = B' \cup A' \textrm{ by } 1 \\
(A\cap B)' = (B\cap A)'  \textrm{ by } 5 \\
A\cap B = B\cap A     \textrm{ by } 4 $

b. We also have $U'=\phi$ and $\phi'=U$.

$  U\cap U' = \phi   \textrm{ by } 6 \\
U'\cap U = \phi   \textrm{ by item a. above}  \\
U'= \phi    \textrm{ by } 7  \\
U= \phi'    \textrm { by } 4 $

c. From axioms $6$, $5$, $4$ and $1$ you can now easily get  $A\cup A' = U$:

$A\cap A' = \phi  \textrm{ start in  } 6 \\
A'\cup (A')'= \phi' \textrm{ by } 5 \\
A'\cup A  = \phi'  \textrm{ by } 4 \\
A \cup A'  = \phi'  \textrm{ by } 1 \\
A \cup A'  = U  \textrm{ by item b. above} $

d. We have that $A \cup \phi = A$.

$ A' \cap U = A' \textrm{ by } 7 \\
(A'\cap U)' = A   \textrm{ by } 4 \\
(A')' \cup U' = A  \textrm{ by } 5 \\
A \cup U' = A   \textrm{ by } 4 \\
A \cup \phi = A  \textrm{ by item b. above} $

e. Now let us prove that $A \cap A =A$.

$ A \cap U = A \textrm{ by } 7 \\
A \cap (A\cup A') = A \textrm {by item c. above} \\
(A \cap A) \cup (A \cap A') = A \textrm{ by } 3 \\
(A \cap A) \cup \phi = A \textrm{ by } 6 \\
A \cap A = A \textrm{ by item d. above }  $

f. Let us prove now that $A\cup A  = A$

$ A' \cap A' = A' \textrm{ by item e. above }  \\
( A' \cap A' ) ' = A \textrm{ by } 4 \\
( A')' \cup ( A' ) ' = A \textrm{ by } 5 \\
A \cup A = A  \textrm{ by } 4 $

Now regarding the first part of your question.

The choice is not unique, of course. Consider, for instance,

$1. A \cup B = B \cup A \\
2. A \cup (B \cup C) = (A \cup B) \cup C \\
3. A\cap( B \cup C) = (A \cap B) \cup (A \cap C) \\
4. (A')' = A \\
5. (A\cap B)' = A' \cup B'\\
6'. A \cup A' = U  \\
7'. A\cup \phi = A $

From axioms $1$ to $7$ we have already proved $6'$ (see item c. above) and $7'$ (see item d. above).

Now using axioms $1$ to $5$, $6'$ and $7'$, the proof of item a. above remains unchanged because it uses only axioms $1$, $4$ and $5$. Then

b'.  Let us prove $\phi'=U$  and $U'=\phi$.

$  \phi\cup \phi' = U   \textrm{ by } 6' \\
\phi'\cup \phi = U   \textrm{ by item a. above}  \\
\phi'= U    \textrm{ by } 7'  \\
\phi= U'    \textrm { by } 4 $

Using the item b'. above and axiom $4$, it is easy to recover axiom $6$ from axiom $6'$ and axiom $7$ from axiom $7'$. So we can conclude that the set of axioms $1$ to $7$  is equivalent to the set of axioms $1$ to $5$, $6'$ and $7'$.

Which set to choose is more a matter of personal choice. Both sets are fine.

Remark: Regarding your question if the number of axioms is the minimum. Well, we can always reduce the number of axioms simply using "and", so we could replace, for instance, axiom $4$ and axiom $5$ by an axiom $4+5$ saying:

$ 4+5. \:\: (A')' = A \textrm{ and } (A\cap B)' = A' \cup B' $

Actually, any finite set of axioms is equivalent to a single (long) axiom.

On the other hand, disregarding the simple concatenation by "and", I would say that seven is probably the minimum number of axioms. The argument here is not a "rigorous proof", it is more an argument of intuition. Let us see it.

We have three operation ($\cup$, $\cap$, $'$)  and two constants ($\phi$, $U$). We should expect to have at least one axiom to define each of the operations and each of the constants. So we should have at least five axioms. But we also need axioms to define the interaction of the operations and that would bring the total of axioms to six or seven axioms.

Note that in Drake's axioms, he uses two axioms to describe union (axioms $1$ and $2$), but just one axiom to define intersection and how complement interacts with unions and intersections (axiom $5$).


问:
Can two eigenvectors share one eigenvalue?
I apologize in advance for the mess. This is my attempt.

The matrix in question is

\begin{align} \textbf{A} = \frac{1}{100}\cdot\begin{pmatrix}
92 & 0 & -144 \\
0 & 100 & 0 \\
-144 & 0 & 8
\end{pmatrix}.\end{align}
We have to find the normalised eigenvector $v = \begin{pmatrix} v_{1} \\ v_{2} \\ v_{3} \end{pmatrix}$ for the eigenvalue $\lambda = -1$, such that $\lvert v \rvert = 1 $.

I solved the equation $$(\textbf{A} - \lambda \textbf{I})\cdot v = 0$$
i.e. $$\begin{bmatrix} 192 & 0 & -144 \\ 0 & 200 & 0 \\ -144 & 0 & 108 \end{bmatrix}\cdot\begin{bmatrix}v_{1} \\ v_{2} \\ v_{3}\end{bmatrix} = 0$$
I used Gaussian elimination and found that $$v = \begin{bmatrix}\frac{3}{4}v_{3} \\ 0 \\ v_{3} \end{bmatrix}.$$
Using the fact that $\lvert v \rvert = 1$, I found that $$ 1 = \Big(\frac{3}{4}v_{3}\Big)^{2} + \Big(v_{3}\Big)^{2} \implies v_{3} = \pm \frac{4}{5}.$$
I tried plugging both $\begin{bmatrix}\frac{3}{5} \\ 0 \\ \frac{4}{5} \end{bmatrix}$ and $\begin{bmatrix} -\frac{3}{5} \\ 0 \\ -\frac{4}{5} \end{bmatrix}$ in the original equation and they both work. So I'm wondering if it's possible that matrix $\textbf{A}$ has two eigenvectors for the eigenvalue $\lambda = -1$, or is the sign irrelevant? If so, which one should I consider?

 Any help or suggestions is appreciated.

答:
Great observation! Yes, it is absolutely possible for a matrix to have two eigenvectors corresponding to the same eigenvalue. We know that for each distinct eigenvalue there will be at least one eigenvector, but there could certainly be more. For example, if our matrix is the $n\times n$ identity matrix then every vector in $\mathbb{R}^n$ is an eigenvector corresponding to the eigenvalue 1.

Why does this matter? Well one important result is that if an $n\times n$ matrix has $n$ distinct eigenvalues then it is diagonalizable.


问:
Matrix Powers A^k - Eigenvalues and Eigenvectors - Chapter 6 Gilbert Strang
I confused about the formula below:

It's different from his lecture note. Why it's different? It doesn't matter the order?

答:
The second one in his lecture notes must be a typo; it should be $S\Lambda^kc$. In general, $S\Lambda^k= \Lambda^kS$ is not true.

Here is a counterexample:
$$
\begin{pmatrix}
1&2\\3&4
\end{pmatrix}
\begin{pmatrix}
1&0\\0&2
\end{pmatrix}
\ne
\begin{pmatrix}
1&0\\0&2
\end{pmatrix}
\begin{pmatrix}
1&2\\3&4
\end{pmatrix}
$$


问:
How do you rotate a vector by a unit quaternion?
Given a 3-variable right-handed vector v that is a translation measured in local space and a unit quaternion representing an orientation from local to world space, how do you use the quaternion to rotate the vector from local space to world space?

For ease of use, the values are:

Vector v = $[1.0, 0.0, 0.0]$

Quaternion $q = [W: 0.7071068, X: 0, Y: 0.7071068, Z: 0]$, which I understand to be a rotation $90^\circ (\frac{\pi}{2})$ around the $Y$-axis and which converts from the local space to the world space. (That is, the resulting vector is $[0.0, 0.0, 1.0]$, and if this was the nose of a spaceship, it'd be pointing to the right in world coordinates)

Thanks.

答:
You seem to be having a good deal of trouble with this, over several questions. At the same time, I am confident that you will get no satisfying answers as long as you stick with the terminology you are using. Maybe on the original stack overflow site, aimed at programmers.

Please read this:
http://en.wikipedia.org/wiki/Quaternions_and_spatial_rotation

and this: http://en.wikipedia.org/wiki/Quaternions

To answer not much more than your question, any quaternion is an expression
$$ q = w + x \; \mathbf{i} + y \; \mathbf{j} + z \; \mathbf{k}.$$
where the multiplication rules use
$$ \mathbf{i}^2 =  \mathbf{j}^2 = \mathbf{k}^2 = \mathbf{i} \mathbf{j}\mathbf{k} = -1, $$ and consequences of those, see wikipedia as I said.
Any quaternion $ q = w + x \; \mathbf{i} + y \; \mathbf{j} + z \; \mathbf{k}$ has a conjugate, that on wikipedia is written $q^\ast,$ given by
$$ q^\ast = w - x \; \mathbf{i} - y \; \mathbf{j} - z \; \mathbf{k}.$$

The "norm" of the quaternion $q$ is exactly $$ \parallel q \parallel^2 = w^2 + x^2 + y^2 + z^2 = q q^\ast = q^\ast q$$

A quaternion $q$ is called a "unit" quaternion when $$ w^2 + x^2 + y^2 + z^2 = 1. $$

A quaternion is called "pure" or a vector in 3-space when $ w = 0,$ so a vector in 3-space is
$$ v =   v_1 \; \mathbf{i} + v_2 \; \mathbf{j} + v_3 \; \mathbf{k} $$
I have no idea what engineers and programmers call these concepts. You are asking mathematicians.

Given two quaternions, the norm of the product is the product of the norms.

The "real part" (the $w$) of the product of two quaternions $pq$ is the same as the "real part of $qp.$

So, what happens when I take a unit quaternion $q$ and a "pure" quaternion $v,$ and calculate
$$ p = q^\ast v q.$$

Well, we have $$\parallel p \parallel   = 1 \cdot \parallel v \parallel \cdot 1 = \parallel v \parallel $$

But as to the "real part," we begin with
$$ \Re v = 0,$$

then
$$ \Re q^\ast (v q) = \Re  (v q) q^\ast = \Re  v (q q^\ast) = \Re v = 0. $$

So $ p = q^\ast v q$ is another pure quaternion, another "vector," the same length as $v,$ but rotated from where it was.

That's enough for a start.


问:
$\liminf, \limsup$, Measure Theory, show: $\lim \int n \ln(1+(f/n)^{1/2})\mathrm{d}\mu=\infty$

Let $(X,\Omega,\mu)$ be a measure space and $M^+(X,\Omega)$ denote the set of all non-negative real valued measurable functions.

If $f \in M^+(X,\Omega)$ and $0< \int f \mathrm{d}\mu < \infty$ then $$\lim_{n\rightarrow\infty} \int n \ln\left(1+\left(\frac{f}{n}\right)^{\frac{1}{2}}\right)\mathrm{d}\mu = \infty.$$

The hint given says use Fatou's Lemma. I am not really too sure exactly what Fatou's Lemma will do to help me solve this:

If I let $$f_n = n \ln\left(1+\left(\frac{f}{n}\right)^{\frac{1}{2}}\right)$$ then $f_n \in M^+(X,\Omega)$ and so Fatou's Lemma says:

$$\int \liminf f_n \mathrm{d}\mu \leq \liminf \left( \int f_n \mathrm{d}\mu \right).$$

To be honest, $\liminf f_n$ looks daunting to me as I know it as
$$\sup_{m \in \mathbb{N}}\left( \inf_{n \geq m} f_n \right)=\sup_{m \in \mathbb{N}}\left( \inf_{n \geq m}\left[ n \ln\left(1+\left(\frac{f}{n}\right)^{\frac{1}{2}}\right)\right] \right).$$

My actual attempt at solving the problem was looking at the set $E = \{ x \in X : 1+\left(\frac{f}{n}\right)^{\frac{1}{2}} \geq e \}$ and then breaking
$$\int f_n \mathrm{d}\mu = \int_E f_n \mathrm{d}\mu + \int_{X\setminus E}f_n\mathrm{d}\mu$$
and then we have
$$\int_E n \ln\left(1+\left(\frac{f}{n}\right)^{\frac{1}{2}}\right) \mathrm{d}\mu \geq \int_{E} n \mathrm{d}\mu=n\mu(X)$$
and assuming we have a space with non-zero measure (Do some axioms include just looking at measures that give the space positive measure?) then we'd have $\int f_n \mathrm{d}\mu \geq n\mu(X)$ and by taking limits of both sides (and as long as $\mu(E) \neq 0$) we'd get $\lim \int f_n \mathrm{d}\mu = \infty$ as needed. Problem is I wasn't sure what to do in the case that $\mu(E)=0$.

I would like to add I am interested in exercises to help work on improving my comfort and understanding of $\limsup$ and $\liminf$ if possible.

Thanks very much!

答:
You can check that for $a>0$ the function
$$
\varphi(t)=t\log\left(1+\left(\frac{a}{t}\right)^{1/2}\right)
$$
is strictly increasing on $[1,+\infty)$ and tends to infinity. Unfortunately you had to dig up to second derivative to prove this. Thus for each $x\in X$ you have strictly increasing to infinity sequence $\{f_n(x):n\in\mathbb{N}\}$, so
$$
\liminf\limits_{n\to\infty} f_n(x)=\lim\limits_{n\to\infty} f_n(x)=+\infty
$$
The rest is clear.</p>


答: A real matrix has infinite eigenvectors for an eigenvalue. In fact the set of all eigenvectors relative to an eigenvalue is a vector subspace(if $v$ and $w$ are eigenvector w.r.t an eigenvalue $\lambda$, $A(w+v)=Aw+Av=\lambda w+ \lambda v=\lambda (w+v)$ and $A(kw)=\lambda kw$, so $w+v$ and $kw$ are also eigenvectors w.r.t $\lambda$).

Title: Can two eigenvectors share one eigenvalue?

Body:
I apologize in advance for the mess. This is my attempt.

The matrix in question is

\begin{align} \textbf{A} = \frac{1}{100}\cdot\begin{pmatrix}
92 & 0 & -144 \\
0 & 100 & 0 \\
-144 & 0 & 8
\end{pmatrix}.\end{align}
We have to find the normalised eigenvector $v = \begin{pmatrix} v_{1} \\ v_{2} \\ v_{3} \end{pmatrix}$ for the eigenvalue $\lambda = -1$, such that $\lvert v \rvert = 1 $.

I solved the equation $$(\textbf{A} - \lambda \textbf{I})\cdot v = 0$$
i.e. $$\begin{bmatrix} 192 & 0 & -144 \\ 0 & 200 & 0 \\ -144 & 0 & 108 \end{bmatrix}\cdot\begin{bmatrix}v_{1} \\ v_{2} \\ v_{3}\end{bmatrix} = 0$$
I used Gaussian elimination and found that $$v = \begin{bmatrix}\frac{3}{4}v_{3} \\ 0 \\ v_{3} \end{bmatrix}.$$
Using the fact that $\lvert v \rvert = 1$, I found that $$ 1 = \Big(\frac{3}{4}v_{3}\Big)^{2} + \Big(v_{3}\Big)^{2} \implies v_{3} = \pm \frac{4}{5}.$$
I tried plugging both $\begin{bmatrix}\frac{3}{5} \\ 0 \\ \frac{4}{5} \end{bmatrix}$ and $\begin{bmatrix} -\frac{3}{5} \\ 0 \\ -\frac{4}{5} \end{bmatrix}$ in the original equation and they both work. So I'm wondering if it's possible that matrix $\textbf{A}$ has two eigenvectors for the eigenvalue $\lambda = -1$, or is the sign irrelevant? If so, which one should I consider?

 Any help or suggestions is appreciated.


问:
Big list: Free textbooks and resources
We have had multiple book-recommendation, but I have not found a question where the many free (yet legal) books are available under one umbrella. Therefore I propose a thread where all the best, free books can be found.

Please post the links  with a slight explanation on what the reader may find (topic/ requirements).

答:
Openstax has Math (anywhere from pre-algebra to 3 volumes of calculus books), science, humanities, and business textbooks all for free.

-all you have to do is click the link, click the book you want to read, and click “view online” and start reading.

Title: Big list: Free textbooks and resources


问:
Countable sub-order $C$ to infinite totally ordered set $T$ such that $\forall t\in T\ :\exists \: c\in C,\:t<c$.
Given any infinite totally ordered set $T$, does there exist a countable subset $C$ such that $C$ is not upper bounded in $T$?

Edit: $T$ also is unbounded from above. My bad for not including this first.

This statement is non-trivial for large $T$, and I don't really have a grasp on how to prove it (my inclination says it is probably true since every simple example that comes to mind satisfies the condition).

Any pointers would be much appreciated. I'm sure this is far simpler than I'm making it out to be.

答: It depends on the cofinality of $T$, i.e., the smallest cardinality of a cofinal subset of $T$. If $\operatorname{cf}T>\omega$, then every countable subset of $T$ is bounded above; $\omega_1$, the first uncountable ordinal, is the simplest example. If $\operatorname{cf}T=\omega$, let $\langle x_n:n\in\omega\rangle$ be a cofinal sequence; clearly $\{x_n:n\in\omega\}$ is a countable subset of $T$ with no upper bound. The only other possibility is that $T$ has a maximum element, in which of course every non-empty subset of $T$ is bounded above, but that possibility is excluded by hypothesis.


问:
Lebesgue integration and Riemann integration of a characteristic function
Let $E$ be a closed nowhere dense set on $[0,1]$.

a) Is it true that $\chi_E$ is Lebesgue integrable?

b) is it true that $\chi_E$ is Riemann integrable if and only if $m(E)=0$?

I know that $f$ is Riemann integrable if and only if $f$ is continuous except on a set of measure 0.

By definition of nowhere dense - the interior of the closure  is empty - I think that b) is true. But not sure about a).

Also, please let me know if you think my reasoning for b) is incorrect.

Thanks.

答: You did not give details for b). Here is a proof: If $m(E)=0$ then $\chi_E$ is continuous a..e because it is continuous on $E^{c}$ (It is the constant function $1$ on this open set). Hence, it is RI. Conversely, if it is RI then it is continuous a.e.. If $x \in E$ then no interval around $x$ can be conatined in $E$ (becasue $E$ has no interior) and hence there is  a sequence $x_n$ in $E^{c}$ converging to $x$. But then $\chi_E (x_n) =0$ for all $n$ and $\chi_E(x)=1$. So $\chi_E$ is not continuous at any point $x$ of $E$. Since it is continuous a.e we must have $m(E)=0$.


答:
Yes, there can be more than one eigenvector associated with a certain eigenvalue; we call the number of eigenvalues of a eigenvalue its geometric multiplicity $m_g(\lambda)$.

The eigenvectors need to be linearly independent however for them to count as two (or more) different ones. In your case, one is a scalar multiple of $-1$ of the other and thus not linearly independent so they are they 'same' eigenvector, i.e. there is only one for the eigenvalue $\lambda = -1$.

In general, the number of eigenvectors $m_g(\lambda)$ corresponding to a eigenvalue is the dimension of the eigenspace $E_\lambda$ which is spanned by the eigenvectors $\xi_1, \xi_2, \ldots$. If you get two eigenvectors which are linearly dependent, the dimension of $E_\lambda$ does not change so we do not consider it as a different one.

As an example of a matrix with 2 eigenvectors belonging to one eigenvalue, consider the matrix
$$A = \begin{pmatrix}
2 &0\\
0&2
\end{pmatrix}.$$
Since $A$ is upper triangular, its eigenvalues $\lambda$ are given by the elements on the diagonal which are both $2$. If we calculate the eigenvectors corresponding to $2$ we get two linearly independent ones (check this!)
\begin{align*}
\left\{\begin{pmatrix}
1\\
0
\end{pmatrix},
\begin{pmatrix}
0\\
1
\end{pmatrix}\right\}.
\end{align*}


问:
Uncountable chain of negligible sets
Let $(E,\mathcal{A},m)$ be a (non empty) measure space, with $m$ a complete measure, it's easy to prove that an uncountable union of negligible sets does not need to be negligible or even measurable. But, now, let $(I,<)$ be a totally ordered set ($I\neq \emptyset$ and I not countable) and $(A_i)_{i\in I}$ be an increasing family of negligible sets (for $m$) (ie $\forall i\in I \ \ m(A_i)=0$ and $\forall (i,j)\in I² \ \ i<j \implies A_i \subset A_j$). Is $A=\bigcup_{i\in I} A_i$ measurable and $m(A)=0$?
Thanks.

答:
Let $\omega _1$ be the smallest uncountable ordinal equipped with the measure $\mu $  defined on the $\sigma $-algebra of all subsets by
$\mu (E) = 0$, if $E$ is countable,  and   $\mu (E) = \infty $,  otherwise.

Then the set
$$
  \omega _1 = \bigcup_{x\in \omega _1} [0, x]
  $$
has infinite measure despite $\mu ([0, x])$ being zero for all $x$.

EDIT.
Assuming the continuum hypothesis,  here is an adaptation of the above example for Lebesgue's measure on $\mathbb R$.

Choose any uncountable subset $A\subseteq \mathbb R$, and observe that  the cardinality $|A|$ satisfies
$$
  \aleph_0<|A|\leq 2^{\aleph_0}.
  $$
As we are assuming the continuum hypothesis, we deduce that
$$
  |A|=\aleph_1=|\omega _1|,
  $$
so there exists a bijective mapping
$$
  \varphi :\omega _1\to A.
  $$
With an eye on the terminology introduced in the question we take the totally ordered index set $I$ to be $\omega _1$, and for
every $i$ in $\omega _1$, we put
$$
  A_i = \varphi ([0,i]).
  $$
It is then clear that each $A_i$ is countable and hence  negligible relative to Lebesgue measure.

On the other hand,  we have
$$
  \bigcup_{i\in \omega _1}A_i =
  \bigcup_{i\in \omega _1}\varphi ([0, i]) =
  \varphi \left(\bigcup_{i\in \omega _1}[0, i]\right) =
  \varphi (\omega _1) = A.
  $$

Choosing the apropriate $A$ we may now  answer the last two questions posed by the OP.

Is $A=\bigcup_{i\in I} A_i$ measurable?

No.  Just choose any non-Lebesgue measurable $A$.

Must  $A=\bigcup_{i\in I} A_i$ have Lebesgue measure zero?

No.  Just choose $A=\mathbb R$.

Title: Uncountable chain of negligible sets


问:
Coupon Collecting Problem
There are $c$ different types of coupon, and each coupon obtained
is equally likely to be any one of the $c$ types. Find the probability that the first $n$ coupons which you collect do not form a complete set.

It is clear that the probability of getting a new coupon on the first draw is 1, and that the probability of getting a new coupon on the second is $(c-1)/c$, or $(c-i)/c$ for the $i$-th new coupon after the initial draw (or $(c-(i-1))=c$ if you include the first draw as a new $i$-th coupon) I don't know where to go from here.

答:
Just because you accepted a wrong answer.

There are in total $c^n$ possible collections among the first $n$ collected coupons. Among them there are
$$
c!{n \brace c}
$$
combinations with each coupon collected at least once. Here ${n \brace c}$ is the Stirling number of second kind.

Therefore the probability in question is:
$$
1-\frac{c!}{c^n}{n \brace c}.
$$

Title: Coupon Collecting Problem


问:
What is $P(X=x)$ when $X\sim \text{Unif}(a,b)$?
I'm trying to solve this excursive A stick of length 1 is broken in a random place that is U(0, 1). Let X be the length of the longer piece and let Y be the length of the shorter piece. in another way. The question:

A stick of length 1 is broken in a random place that is $\text{Unif}(0,1)$. Let $X$ be the point chosen on the steak and $Y$ be the length of the longer piece. Find $E(Y)$.

So we get:
$$
Y=\begin{cases}
1-X & X<0.5\\
X & X\geq0.5
\end{cases}
$$
Then we can say:
$$
\begin{align*}
E\left(Y\right)&=\int_{-\infty}^{\infty}y\cdot P\left(Y=y\right)dy\\&=\int_{0}^{1}y\cdot P\left(Y=y\right)dy=\int_{0}^{0.5}y\cdot P\left(Y=y\right)dy+\int_{0.5}^{1}y\cdot P\left(Y=y\right)dy\\&=\int_{0}^{0.5}y\cdot P\left(Y=y\right)dy+\int_{0.5}^{1}y\cdot P\left(Y=y\right)dy\\&=\int_{0}^{0.5}(1-x)\cdot P\left(X=x\right)dx+\int_{0.5}^{1}x\cdot P\left(X=x\right)dx\\&=\int_{0}^{0.5}(1-x)\cdot P\left(X=x\right)dx+\int_{0.5}^{1}x\cdot P\left(X=x\right)dx
\end{align*}
$$
I want to say $P(X=x)=1$ when $x\in[0,1]$. But is it true to say that $P(X=x)=\frac{1}{b-a}$ when $x\in[a,b]$?

答:
See the other answer for a direct method to get the expectation. Since you asked how to compute the pdf of y, here is one way, though not very intuitive.

$$\begin{split}Y=\begin{cases}1-X&X<.5\\X&X\ge.5\end{cases}\end{split}$$

For X<.5, you can write

$$\begin{split}Y&=|1-X-\frac 1 2|+\frac 1 2\\
&=|\frac 1 2 -X|+\frac 1 2\\
&=\left|\frac{1-2X}{2}\right|+\frac 1 2\\
&=\frac{|1-2X|+1}{2}\end{split}$$

and for X>.5 you can write

$$\begin{split}Y&=|X-\frac 1 2|+\frac 1 2\\
&=\left|\frac{2X-1}{2}\right|+\frac 1 2\\
&=\frac{|2X-1|+1}{2}\end{split}$$

Thus $$Y=\frac{|2X-1|+1}{2}\text{ for }0\le X\le1$$

and can take on the values $1 \ge Y\ge \frac 1 2$. The upper bound is achieved at X=0 or 1, and the lower bound is achieved at X=1/2.

$$F_Y(y)=P(Y\le y)\\...\\f_Y(y)=...$$


问:
What are the implications of a surjection between free modules?
My professor said the following in his lecture:

Suppose $M$ and $P$ are free modules and $\alpha: M\rightarrow P$ is a surjection. Then $\alpha$ splits and the kernel of $\alpha$ is a summand of $M$.

I know that $M$ and $P$ being free means that they have a basis and hence, any element can be written as a linear combo of the basis elements. I also know that if $\alpha$ being split means that there exists a $\beta: P\rightarrow M$ such that $\alpha\beta=1_{P}$. I am having trouble understanding why this surjection of free modules implies that the surjection splits. How do I get started?

答:
The key property of the free module $F$ with basis $\{f_i\}_{i\in I}$ is the following:

For every module $N$ and every choice of elements $n_i\in N$, $i\in I$, there exists a unique module homomorphism $\phi\colon F\to N$ such that $\phi(f_i)=n_i$.

Let $\{p_i\}_{i\in I}$ be a basis for $P$. For each $i$, there exists $m_i\in M$ such that $\alpha(m_i)=p_i$. Define $\beta$ to be the (unique) morphism induced by mapping $\beta(p_i)=m_i$. Then $\alpha\beta(p_i) = \alpha(m_i)=p_i$. Thus, $\alpha\beta$ is the identity on the basis, hence is the identity morphism.

Now let $A=\mathrm{ker}(\alpha)$, and $B=\mathrm{Im}(\beta)$. If $x\in A\cap B$, then $x=\beta(p)$ for some $p\in P$, hence $0 = \alpha(x) = \alpha\beta(p) = p$. Thus, $x=\beta(p)=\beta(0)=0$, so $A\cap B=\{0\}$. And if $m\in M$, then $\beta(\alpha(m))\in B$, and $m-\beta(\alpha(m))\in A$, with $m=\beta(\alpha(m)) + (m-\beta(\alpha(m))$, proving that $A+B=M$. Thus, $M=A\oplus B$.

Note the last part does not use the facts that $M$ and $P$ are free. In general, if $N$ and $N’$ are any modules, and $\alpha\colon N\to N’$ splits (has a right inverse), then $\mathrm{ker}(\alpha)$ is a direct summand of $N$, and $N$ is isomorphic to a direct sum $N’\oplus N’’$.

The first part shows that free modules are projective. A module $P$ is projective if and only if, for every pair of modules $M$ and $M’$, and surjective module morphism $f\colon M\to M’$, if there is a morphism $\alpha\colon P\to M’$, then there is a morphism $\beta\colon P\to M$ such that $f\beta=\alpha$. That is: if you have a diagram
$$\begin{array}{ccccc}
&& P \\
&&\downarrow\alpha\\
M&\stackrel{f}{\longrightarrow}&M’&\rightarrow &0
\end{array}$$
with bottom row exact, then there exists a morphism $\beta$ such that
$$\begin{array}{ccccc}
&& P \\
&{\scriptstyle\beta}\swarrow&\downarrow\alpha\\
M&\stackrel{f}{\longrightarrow}&M’&\rightarrow &0
\end{array}$$
that makes the diagram commutative.

Free modules are always projective, and for some rings, projective modules are necessarily free; but for other rings, you may have projective modules that are not free. A module is projective if and only if it is a direct summand of a free module.


问:
Is there an easy way to see that all derivatives are bounded?

Show that all derivatives of $f:\mathbb{R}\to\mathbb{R}$ given by $$f(x):=\frac{1}{\sqrt{x^2+1}+1}$$ are bounded.

It's easy to see that all derivatives are continuous. So the only potential problem is that a derivative might blow up at $\infty$. Is there an easy way to see that this does not happen? I guess deriving an explicit form of the n-th derivative is not the way to do it (I'd doubt there is an easy closed form).

答:
If you know some complex analysis, you can just say something like

The function is analytic and bounded in the strip $|\Im z|\le 1/2$. Apply the Cauchy estimate in each circle of radius $1/2$ centered on the real line and conclude.

If you don't, but still strongly prefer to avoid any talks about "the general form of the expression for the $n$-th derivative", you can just say that the function $[0,\infty)\ni y\to \frac{1}{1+y}$ is bounded with all derivatives (explicit formula), so, by the chain rule, it will suffice to show that all derivatives of $F(x)=\sqrt{x^2+1}$ are bounded. However, $F(x)^2=x^2+1$, $F'(x)$ is bounded by an explicit formula, and, by the product rule for the $n$-th derivative with $n\ge 2$, we have
$$
2F^{(n)}F+\sum_{k=1}^{n-1}{n\choose k}F^{(k)}F^{(n-k)}=\begin{cases}2,&n=2\\0,&n>2\end{cases}
$$
from where the conclusion follows by induction using the inequality $F\ge 1$.


问:
How to evaluate $_2F_1(\frac{1}{2},-\frac{1}{2};\frac{1}{2}; \sin^{2}(x))=\cos(x)$
$$_2F_1(\frac{1}{2},-\frac{1}{2};\frac{1}{2}; \sin^{2}(x))=\cos(x)$$

I plug these values into the definition of the hypergeometric function:

$$_2F_1(a,b,c,x)=\displaystyle\sum_{n=0}^{+\infty}\dfrac{a^{\overline{n}}b^{\overline{n}}}{c^{\overline{n}}}\dfrac{x^n}{n!}$$

And I obtain:

$$\sum_{n=0}^{+\infty}\dfrac{\frac{1}{2}^{\overline{n}}\left(-\frac{1^{\overline{n}}}{2}\right)}{\frac{1^{\overline{n}}}{2}}\frac{(\sin^{2}(x))^n}{n!}$$

I don't know how to simplify the $-\dfrac{1^{\overline{n}}}{2}$ and $n!$.

$$-\dfrac{1^{\overline{n}}}{2}=\left(-\dfrac{1}{2}\right)\left(\dfrac{1}{2}\right)\left(\dfrac{3}{2}\right)...\left(\dfrac{2n-3}{n}\right)$$

$$n!=(n-3)(n-2)(n-1)n...$$

Also, what should I do with the $n$ power of $\sin^2(x)$

答:
First of all we can cancel one parameter:
$$\mbox{$_2$}F_1(a,b;a; x)=\mbox{$_1$}F_0(b;;x)$$

Further:
$$
\mbox{$_1$}F_0\left(-\frac12;;x\right)=\sum_{n\ge0}\frac{\left(-\frac12\right)^\overline{n}}{n!}x^n=\sum_{n\ge0}\binom{\frac12}n(-x)^n=\sqrt{1-x}.
$$


问:
Show that for every linear map $T:V\rightarrow V$ there exists $a\in F$ so that $Tv=av$ for all $v\in V$ if $V$ is $1$-dimensional
Attempting my first problem on linear maps, and I'm fairly certain it's incorrect, but I'm not sure what I'm missing. Here's the problem:

Suppose that $V$ is 1-dimensional. Show that for every linear map $T:V\rightarrow V$ there exists $a\in F$ so that $Tv=av$ for all $v\in V$.

Suppose that $V$ is $1$-dimensional and choose an arbitrary $v \in V$. We know that $T(v)$ must output an element in $V$ and that output can be represented as $av$ for an appropriate $a \in F$ because any vector can be represented as a scalar multiple of $v$. Since we chose an arbitrary $v$, we can just apply this thinking for all vectors.

答:
Take some non-zero vector $v_0$. This vector forms a (trivial) basis for $V$ as $V$ is $1$-dimensional by assumption. Hence every vector $v\in V$ is of the form $v=cv_0$ for some $c\in F$ as this are the linear combinations in a $1$-dimensional vector space.

Now, $T(v_0)\in V$ and hence $T(v_0)=av_0$ for some $a\in F$. Now take $v\in V$ arbitrary. As noted earlier $v=cv_0$ for some $c\in V$. But then

$$T(v)=T(cv_0)=c\, T(v_0)=b(av_0)=a(bv_0)=av$$


答: Actually $_2F_1(a,b;a;z)=(1-z)^{-b}$ for $|z|<1$ and any $a,b$ (well, with $a\notin\mathbb{Z}_{\leqslant 0}$). Indeed, both are equal to $\sum_{n=0}^\infty b(b+1)\dots(b+n-1)z^n/n!$, by definition of $_2F_1$ for the LHS, and by the binomial series for the RHS.

Title: How to evaluate $_2F_1(\frac{1}{2},-\frac{1}{2};\frac{1}{2}; \sin^{2}(x))=\cos(x)$

Body:
$$_2F_1(\frac{1}{2},-\frac{1}{2};\frac{1}{2}; \sin^{2}(x))=\cos(x)$$

I plug these values into the definition of the hypergeometric function:

$$_2F_1(a,b,c,x)=\displaystyle\sum_{n=0}^{+\infty}\dfrac{a^{\overline{n}}b^{\overline{n}}}{c^{\overline{n}}}\dfrac{x^n}{n!}$$

And I obtain:

$$\sum_{n=0}^{+\infty}\dfrac{\frac{1}{2}^{\overline{n}}\left(-\frac{1^{\overline{n}}}{2}\right)}{\frac{1^{\overline{n}}}{2}}\frac{(\sin^{2}(x))^n}{n!}$$

I don't know how to simplify the $-\dfrac{1^{\overline{n}}}{2}$ and $n!$.

$$-\dfrac{1^{\overline{n}}}{2}=\left(-\dfrac{1}{2}\right)\left(\dfrac{1}{2}\right)\left(\dfrac{3}{2}\right)...\left(\dfrac{2n-3}{n}\right)$$

$$n!=(n-3)(n-2)(n-1)n...$$

Also, what should I do with the $n$ power of $\sin^2(x)$


问:
find $a,b,c$ such that $e^{2\pi A}=I$

Given $A$ is real valued matrix of the form $\begin{pmatrix}a&b\\c&0\end{pmatrix}$ find all $a,b,c$ such that $e^{2\pi A}=I$

My attempt is the following, Calculating the eigenvalues I get $\frac{a\pm\sqrt{a^2+4bc}}{2}$ and I said that $a=0$ otherwise we won't get the desired result and $bc<0$ to get the right form of diagonlize matrix of $A$ and afterwards that $bc\in\mathbb{Z}$ but not sure if I've done this right

答:
You already got $a=0$ and $bc=-\lambda^2$. Suppose $bc\neq0$. Note that
$$ x(t)=e^{tA}x_0 $$
is the solution to $$x'=Ax $$
or
$$ x_1'=bx_2,x_2'=cx_1,x_1(0)=x^0_1. \tag 1$$
From (1), one has
$$ x_1=c_1\cos(\lambda t)+c_2\sin(\lambda t),x_2=\frac{1}{b}\bigg[-c_1\lambda\sin(\lambda t)+c_2\lambda\cos(\lambda t)\bigg] $$
and hence
$$ x(t)=\binom{x_1}{x_2}=\left(\begin{matrix}\cos(\lambda t)&\sin(\lambda t)\\
-\frac{1}{b}\lambda\sin(\lambda t)&\frac{\lambda}{b}\cos(\lambda t)\end{matrix}\right)\binom{c_1}{c_2}.$$
So
$$ e^{At}=\left(\begin{matrix}\cos(\lambda t)&\sin(\lambda t)\\
-\frac{1}{b}\lambda\sin(\lambda t)&\frac{\lambda}{b}\cos(\lambda t)\end{matrix}\right). $$
Since $e^{2\pi A}=I$, one has
$$ \sin(2\pi\lambda)=0, \cos(2\pi\lambda)=1,b=\lambda $$
which implies
$$ \lambda=k, b=k, c=-k, k\in N. $$

Title: find $a,b,c$ such that $e^{2\pi A}=I$


问:
Find the height of the following rectangle
We have a rectangle ABCD, and a point P on the diagonal AC. From P we see BC at an angle $\alpha$. Knowing $\alpha$, AP and AB, find BC.

Here's a probably unnecessary illustration of the problem, with known data in red:

I am lost with this, certainly, extremely simple problem. I might use the cosine law to define BC as a function of $\alpha$, PC and PB, but I find no way of finding those latter. I thought of finding PB constructing a parallelogram including sides PB and AP, having BC as diagonal, but I'm stuck there since I, in my poor knowledge on solving parallelograms, I would need the other diagonal of the parallelogram to find PB. On the other hand, I looked where could I transpose $\alpha$ as to express its trigonometric functions using known information, but didn't found anything useful.

Sorry for coming to you with high-school problems (I feel ridiculous to be asking this after three semesters of real analysis, one of complex, and one of linear algebra). I'd appreciate, in addition to a method of resolution, ressources that might help me get a comprehensive approach to planar, high-school like geometry.

答:
Find $\angle APB = 180º - \alpha$.

Use the sine rule in $\Delta APB$ to find $\angle PBA$.

Find $\angle PAB = 180º - \angle APB - \angle PBA$.

then since $\Delta CAB$ is right-angled, $\tan \angle PAB = \frac{BC}{AB}$, so you now have $BC$.


问:
Given a triangle with points in $\mathbb{R}^3$, find the coordinates of a point perpendicular to a side
Consider the triangle ABC in $\mathbb{R}^3$ formed by the point $A(3,2,1)$, $B(4,4,2)$, $C(6,1,0)$.

Find the coordinates of the point $D$ on $BC$ such that $AD$ is perpendicular to $BC$.

I believe this uses projections, but I can't seem to get started. I tried the projection of $AC$ onto $BD$ and $AB$ onto $BC$, but to no avail.

Any help is loved! Thanks.

答:
Here is one way, presumably unsuitable.

The direction vector of $BC$ is $(2,-3,-2)$.

A generic point $D_t$ on $BC$ is given by $(4,4,2)+t(2,-3,-2)$.
The direction vector of $AD_t$ is $(1,2,1)+t(2,-3,-2)$. The dot product of this with $(2,-3,-2)$ is $-6+17t$. This dot product must be $0$.

We end up with $D=\left(\frac{80}{17},\frac{50}{17},\frac{22}{17}  \right)$.


问:
$X$ is a normal scheme iff $O_X (U)$ is normal integral domain.
This is a proposition in Liu's book.(proposition 4.1.5)
I can't understand the proof....

Proposition 1.5. Let $X$ be an irreducible scheme. The following properties are equivalent:
(i) The scheme $X$ is normal.
(ii) For every open subset $U$ of$X$, $O_X(U)$ is a normal integral domain.
Proof: Let us suppose $X$ is normal. Let $U$ be an open subset of $X$ and $α ∈ Frac(O_X(U))$ integral over $O_X(U)$. Let $V$ be an affine open subset of $U$. Then $α$ is integral over $A := O_X(V)$ and $α ∈ Frac(A)$. Let us set $I = Ann((αA+A)/A)$. If $I \neq A$, then $I ⊆ p$ for a prime ideal $p$ of $A$.As $A_p$ is normal, there exists an $s ∈ A \setminus p$ such that $sα ∈ A$. It follows that $s ∈ I$, whence a contradiction. Hence $1 ∈ I$ and $α ∈O_X(V)$. As this is true for every affine open subset $V$ of $U$,we have $α ∈O_X(U)$ (Proposition 2.4.18). This shows that (i) implies (ii).

In proof why $\alpha$ is integral over $A := O_X(V)$ and $\alpha ∈ Frac(A)$
And why is $I$ defined like that?
I really need help with this proof....

答:
Here by abuse of notation, Liu is identifying $\alpha\in Frac(O_X(U))$ with its image under the restriction map $O_X(U)\to O_X(V)$ which induces a map $Frac(O_X(U))\to Frac(O_X(V))$.  So, this restriction of $\alpha$ is an element of $Frac(O_X(V))$ by definition.  Since $\alpha$ is integral over $O_X(U)$, it is a root of some monic polynomial with coefficients in $O_X(U)$; applying the restriction map to the coefficients of this polynomial, we conclude that the restriction of $\alpha$ is a root of a monic polynomial with coefficients in $O_X(V)$, i.e. it is integral over $O_X(V)$.

The ideal $I$ is just the ideal of possible denominators of the fraction $\alpha$: that is, it is the set of elements $s\in A$ such that $\alpha=\frac{t}{s}$ for some $t\in A$.  This is equivalently the set of $s\in A$ such that $s\alpha\in A$, which is equivalent to saying that $s\alpha A\subseteq A$ or that $s$ annihilates $(\alpha A+A)/A$ (which is the somewhat opaque definition of $I$ that Liu uses).  So, if you can prove that $1\in I$, that means that $s$ can be written with a denominator of $1$, so it is just an element of $A$.  So to prove this, you suppose that $1\not\in I$, so $I$ is a proper ideal that is contained in some prime ideal, and then you use the normality of the localization at that prime ideal to reach a contradiction.


问:
Describe image under mapping $w=\frac{z-i}{z+i}$ of set $\{z=x+iy : x>0, y>0 \}$
I'm new to complex analysis. In polar form, when I plug in a general $z=re^{i\theta}$ to $w$, I end up with $$w(z)=\dfrac{r^2-2ir\cos\theta-1}{r^2+2r\sin\theta+1}$$

I know since $x,y>0$, I have $0 < \theta < \frac{\pi}{2}$, but that's all I can tell.

Where do I go from here?

答:
Rewrite $w=\frac{z-i}{z+i}$ as $z = i\frac{1+w}{1-w} =re^{i\theta}, \theta\in (0,\frac\pi2)$. Then
$$e^{i2\theta} =\frac{z}{\bar z} =  \frac{i\frac{1+w}{1-w}}{-i\frac{1+\bar w}{1-\bar w}}
= -\frac{(1-\bar w)(1+w)}{(1+\bar w)(1-w)}\implies
|w - i\tan\theta |^2 = \sec^2\theta$$
which represents a family of circles with center $i\tan\theta$ and radius $\sec\theta$. Thus, along with $\text{Im}(w)=\frac{-2r\cos\theta}{r^2+2r\sin\theta+1}<0$, the image is a half unit-disk, the shaded area below.


问:
Find the highest natural number which is divisible by $30$ and has exactly $30$ different positive divisors.

Find the highest natural number which is divisible by $30$ and have exactly $30$ different positive divisors.

What I Tried: I am not sure about any specific approach to this problem. Of course as the number is divisible by $30$ , some of the factors of the number would be $1,2,3,5,6,10,20,30$ , but that only makes $8$ divisors and there will be $12$ more which I have no idea of .

Can anyone help me?

答:
Because it must be a multiple of $30 = 2 \cdot 3 \cdot 5$, then the number must be of the form $$N = 2^{\alpha+1} \cdot 3^{\beta+1} \cdot 5^{\gamma+1} \cdot M $$ where $M$ is $1$ or a product of primes that are greater than $5$ and $\alpha, \beta$, and $\gamma$ are non negative integers.

Let $d(n)$ represent the number of different positive divisors of $n$. Then

$$d(N) = (\alpha+2)(\beta +2)(\gamma + 2)d(M)$$

and $d(N)$ is a multiple of $30 = 2 \cdot 3 \cdot 5$. It follows that
$\{\alpha+2, \beta+2, \gamma+2\} = \{2,3,5\}$ and $d(M)=1$. The largest value of $N$ will occur when $\alpha = 0, \beta = 1$, and $\gamma = 3$. The value of $N$ is therefore

$$N = 2^1 \cdot 3^2 \cdot 5^4 = 11250$$