问:
Derivative: Application of the Product rule
Suppose we have a function $\pi_1(t|v_1)=b^*(t)^{n-1}(v_1-b^*(t))$

When I differentiate this, I need to use the chain rule and the product ruke,

$(n-1)b^*(t)^{n-2}\frac{db^*}{dt}(v_1-b^*(t)) - b^*(t)^{n-1}\frac{db^*}{dt}$

As you can see I apply the chain rule to the first part of the function and the rest is an application of the product rule? Is this correct?

答:
$\pi_1(t|v_1)=b^*(t)^{n-1}(v_1-b^*(t))$

As $\pi_1(t|v_1) = \color{red}{b^*(t)^{n-1}}\times \color{blue}{(v_1-b^*(t))}$ so according the product rule

$\frac {d\pi_1'(t|v_1)}{dt} = \frac {d\color{red}{b^*(t)^{n-1}}}{dt}[\color{blue}{(v_1-b^*(t))}] + [\color{red}{b^*(t)^{n-1}}]\frac {d\color{blue}{(v_1-b^*(t))}}{dt}$

$\color{red}{b^*(t)^{n-1}} = \color{green}(\color{orange}{b^*(t)}\color{green}{)^{n-1}}$ so according to the chain rule

$\frac {d\color{red}{b^*(t)^{n-1}}}{dt}=\frac {d\color{green}(\color{orange}{b^*(t)}\color{green}{)^{n-1}}}{d\color{orange}{b^*(t)}}\frac {d\color{orange}{b^*(t)}}{dt}=\color{green}{(n-1)(}\color{orange}{b^*(t)}\color{green}{)^{n-2}}\frac {d\color{orange}{b^*(t)}}{dt}$

And $\color{blue}{(v_1-b^*(t))} = \color{purple}{v_1} - \color{orange}{b^*(t)}$ so according to the sum/difference rule

$\frac {d\color{blue}{(v_1-b^*(t))}}{dt} = \frac {d\color{purple}{v_1}}{dt} - \frac {d\color{orange}{b^*(t)}}{dt}= \color{purple}0 - \frac {d\color{orange}{b^*(t)}}{dt}=-\frac {d\color{orange}{b^*(t)}}{dt}$

Putting it all together we get exactly what you got

$\frac {d\pi_1'(t|v_1)}{dt} =  \color{green}{(n-1)(}\color{orange}{b^*(t)}\color{green}{)^{n-2}}\frac {d\color{orange}{b^*(t)}}{dt}\color{blue}{(v_1-b^*(t))} - \color{red}{b^*(t)^{n-1}}\frac {d\color{orange}{b^*(t)}}{dt}$

I'm not comfortable with you language of "parts" but I imagine what you meant to say was correct. The above is more detail then you probably need but it's good practice to really run a comb through things to make sure one really understands.

.....

Oh we can further factor this to get

$\frac {d\pi_1'(t|v_1)}{dt} = [(n-1)b^*(t)^{n-2}(v_1-b^*(t)) - b^*(t)^{n-1}]\frac {db^*(t)}{dt}=[(n-1)(v_1-b^*(t))- b^*(t)]b^*(t)^{n-2}\frac{db^*(t)}{dt}$

======

Alternative as

$\pi_1(t|v_1) = b^*(t)^{n-1}(v_1-b^*(t)) = v_1b^*(t)^{n-1} - b^*(t)^n$ we can calculate the derivative as

$\frac {\pi_1(t|v_1) }{dt} = v_1(n-1)b^*(t)^{n-2}\frac{db^*(t)}{dt} - nb^*(t)^{n-1}\frac{db^*(t)}{dt}$


问:
How to prove $p\land q\to \neg r \vdash r \to p \to \neg q$ using natural deduction?
I'm having a lot of trouble proving the sequent below with Natural Deduction rules. I'm new to this and find it difficult to come up with proof strategies. The first thing I do is to assume $r$ but really confused on where I could go from there.

The sequent:
$$(p\land q)\to \neg r \vdash r \to (p \to \neg q)$$

Thanks so much, any help appreciated!

答:
Your first step seems good, assume $r$ Now you want to show that $p\to\neg q$, so what do you want to do? well you are trying to prove an implication again, so assume $p$ and show $\neg q$!

You are given that $p\wedge q\to \neg r$ if you already have that $(a\to b) \leftrightarrow (\neg b \to \neg a)$ then this would be a good place to use it. If you dont have this, then you can prove $\neg(p\wedge q)$ using contradiction. (using the fact that you are assuming $r$)

once you have $\neg(p\wedge q)$ you can now just assume $p$ and prove $\neg q$ again by contradiction by assuming $q$.

ContentLicense: CC BY-SA 4.0


问:
How can we find the derivative of a circle if a circle is not a function?
If we consider the equation of a circle, $x^2+y^2=r^2$, then I understand that $dy/dx$ can be computed in the following way via implicit differentiation:
\begin{align}
2x + 2y\frac{dy}{dx} &= 0 \\
\frac{dy}{dx} &= -\frac{2x}{2y} = -\frac{x}{y} \, .
\end{align}
Although I feel comfortable deriving this result, I don't really understand how I should interpret it. On an intuitive level, the formula $dy/dx = -x/y$ seems to suggest that the gradient of the tangent to any given point $(x,y)$ is $-x/y$. However, since the curve $x^2+y^2=r^2$ fails the vertical line test, it doesn't look like it is even a function. Usually, $dy/dx$ can be thought of as a shorthand for
$$
\lim_{h \to 0}\frac{y(x+h)-y(x)}{h} \, .
$$
However, in this case each $x$-value maps to two $y$-values, and so the limit definition doesn't seem to apply here. So what does $dy/dx$ actually represent in this context?

答:
While each $x\in(-r,\,r)$ is compatible with two choices of $y$, continuous motion along the circumference well-defines the choice of $y$ at each point, giving a local $y$-as-a-function-of-$x$ behaviour wherever $dy/dx$ is finite and nonzero (i.e. $x,\,y$ are both nonzero), which happens at all but four of the circumference's points.

This local behaviour is more easily described in terms of the polar angle $\theta$, and since $x=r\cos\theta,\,y=r\sin\theta$, by the chain rule $\frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta}=\frac{x}{-y}=-\cot\theta$.


问:
Show that $\exists c \in \mathbb{R} : f(c) = 0$
Let $f : \mathbb{R} \longrightarrow \mathbb{R}$ be a differentiable function such as $\forall x \in \mathbb{R}, f'(x) > 2$.

What could I do to show that $\exists c \in \mathbb{R} : f(c) = 0$ ?

At first I thought I could use the mean value theorem but it did not help me. In fact, I don't really understand why $f(x)$ couldn't be strictly positive $\forall x \in \mathbb{R}$.

Would you know how to proceed ?

Thanks for your help.

答:
From the mean-value theorem you get
$$
 \frac{f(x)-f(0)}{x-0} > 2
$$
for all nonzero $x$. It follows that
$$
 f(x)  \ge f(0) + 2x \text{ for } x > 0
$$
and
$$
 f(x)  \le f(0) + 2x \text{ for } x < 0
$$
so that $f$ necessarily takes both positive and negative values. Now conclude with the intermediate value theorem.


问:
Prove that every group $G$ of order $98,$ having just one element $g$ of order $2$ is an abelian group.
Prove that every group $G$ of order $98,$ having just one element $g$ of order $2$ is an abelian group.

I was thinking about use that the group $G = 7^2·2$ so I have a group $N_2$ and $H_7$ and I need to prove what are the subgroups, watch if i can put G like cyclic groups one $Z_2$ and the other ones with no elements order 2. So I have $G\simeq Z_2$ x $Z_n$ x ... I hope it is enough.

答:
You have the right idea in terms of showing that $G$ is a direct product of abelian groups, but you are missing some details. Moreover, the subgroup of order $7^2$ need not be cyclic (not sure if you are actually trying to say this tbh).

By the first Sylow theorem we have a subgroup $H_7$ of order $7^2$ and a subgroup $N_2$ of order two $2$, which you also seemed to have figured out. The latter is unique by the assumption that there is only one element of order $2$. We will now show that also the first one is unique:

We know by the third Sylow theorem that the number of Sylow-$7$ subgroups needs to divide $2$ and that this number is congruent to $1 \text{ mod } 7$. Hence it must be odd and is therefore $1$. This tells us, that both of these groups are normal subgroups of $G$.

Since the intersection of $H_7$ and $N_2$ is trivial by Lagrange, we know that $H_7 \cdot N_2$ must be $G$ by comparison of the cardinalities. To summarize, we have

$\bullet$ $H_7$ and $N_2$ are normal subgroups of $G$

$\bullet$ $H_7 \cap N_2 = 1$

$\bullet$ $H_7 \cdot N_2 = G$

which means that $G \cong H_7 \times N_2$. Now note, that a group of prime order is cyclic and hence abelian and that a group of order $p^2$ for some prime $p$ is either cyclic or a direct product of two cyclic groups and hence also abelian. Thus $H_7 \times N_2 \cong G$ is abelian, as wanted.


问:
Why this two dynamical Systems are not isomorphic?
Given two dynamical Systems on [0,1) with the Borel $\sigma-Algebra$ and the Lebesgue measure l.

$T_a (x) = x + a$ mod1
$T_2 (x) = 2x$ mod1.
 Show that this two systems are not isomorphic for any choice of a:

We've already showed, that $T_2$ is ergodic. $T_a$ is just ergodic if a is irrational. Now I still have to show, that the two systems are not isomorphic for irrational a. Don't know how to do this? I first tried to assume that there is an isomorphism and to construct a contradiction…I failed. What are other invariants of dynamical systems under isomorphism besides ergodicity? may someone help me?

答:
There are a few possibilities here. I'm gonna use $m$ to refer to Lebesgue measure on the unit interval $[0 , 1)$. These are roughly in ascending order of "sophistication".

(1) The map $T_a$ is invertible, meaning that it not only has the property that $m(A) = m \left(T_a^{-1} A \right)$ for all measurable $A$, which is the definition of a measure-preserving map, but also $m(A) = m(T_a A)$ for all measurable $A$. The map $T_2$ does not have this property. Note that it's not quite enough to just observe that $T_2$ is not invertible as a map from $[0, 1)$ to itself. As others remarked, a lot can happen on a null set.

(2) The map $T_2$ is weakly mixing, strongly mixing, and even Bernoulli. The map $T_a$ is none of these.

(3) The map $T_2$ has positive entropy, whereas $T_a$ has entropy $0$.


问:
A problem about Lévy's continuity theorem

Problem: Use the continuity's theorem for to solve this problem: Let $X_{1},X_{2}, \overset{i.i.d}{\sim}\textbf{Poisson}(\lambda)$.Also, let $S_{n}=\sum_{i=1}^{n}X_{i}$. Prove that $$\frac{S_{n}}{n}\overset{D}{\to}\lambda$$

My attempt: I know that result is inmediate of the Weak Law of Large Numbers (WLLN), by I need to use the continuity's theorem.

Now, I know that the continuity'theorem statement that: $$F_{X_{n}}\overset{D}{\to} F_{X}(x) \iff M_{X_{n}}(t) \to M_{X}(t) \quad \text{as} \quad n\to \infty$$
where $F_{X}(x)$ is the CDF and $M_{X}(t)$ is the generating moment function.

But,
\begin{eqnarray*}
M_{Z_{n}}(t)=\mathbb{E}[e^{tZ_{n}}]=\mathbb{E}[e^{tS_{n}/n}]=\mathbb{E}[e^{\frac{tS_{n}}{n}}]=[\mathbb{E}[e^{tX}]]^{n}=[e^{\lambda(e^{t}-1)}]^{n}
\end{eqnarray*}

But, I don't know how I would to conclude, because $\displaystyle \lim_{n \to \infty}M_{Z_{n}}(t)\not=\lambda$

What is the mystake? What am I not understanding well?

答:
You are missing a $\frac{t}{n}$ instead of $t$ in the expectation: $M_{Z_n}(t) = \big[\mathbb{E}\big(e^{\frac{t}{n}X}\big)\big]^n = \big[e^{\lambda (e^{\frac{t}{n}}-1)}\big]^n$.

Since $n(e^{\frac{t}{n}}-1) \to t$, you get $M_{Z_n}(t) \to e^{\lambda t}$.

Since the MGF of the constant variable $\lambda$ is $t \mapsto e^{\lambda t}$, $Z_n$ converges in distribution to $\lambda$.


问:
Existence Universal goto-programm (turing machine)
May you can help me out with my problems with source codes. Well first of all we proved that for recursive functions $N:\mathbb N^2\rightarrow \mathbb N$ and $A^k: \mathbb N^k\rightarrow \mathbb N$ and for every goto programm (I am not 100% sure, but I think this is identical to a turing machine) there exist a source code $\mathcal{P}=<p_0,...,p_{L-1}>$ where $L$ is the number of lines of the program and $p_i$ is the code of the i-th line.

Question: How to construct a simple universal program $U$? I.e a program which delivers the ouput for the input $(\mathcal{P},n)$, this is what the program $P$ would give for the input $n$

This is nothing more than a goto-interpreter, similar to a java-interpreter, but what about a formal argument, why this program exist?

EDIT: Next to my original question (the construction of $U$), two other questions arises:Let $f(n)$ is the output of the universal program $U$ on the input $(n,n)$. Is $f$ recursiv? Same with $g(n)=f(n)+1$. Is it recursiv?

答: Quote from Wikipedia: "To show that something is Turing complete, it is enough to show that it can be used to simulate some Turing complete system. For example, an imperative language is Turing complete if it has conditional branching (e.g., "if" and "goto" statements, or a "branch if zero" instruction. See OISC) and the ability to change arbitrary memory locations (e.g., the ability to maintain an arbitrary number of variables). Since this is almost always the case, most if not all imperative languages are Turing complete if we ignore any limitations of finite memory"</p>

ContentLicense: CC BY-SA 3.0


问:
Prove that the Galois group $G = Gal(x^5-2; \mathbb{Q})$ is a group metacyclic.
Prove that the Galois group $G = Gal(x^5-2; \mathbb{Q})$ is a group metacyclic, that is, $G$ contains a normal cyclic subgroup $N$ whose quotient $G/N$ is also cyclic.

MY ATTEMPTY: I was trying to follow this idea I found, which is:

You have a group $G$ that has a cyclic normal subgroup $K$ such that its quotient $G/K$ is also cyclic. Now take a quotient group $G/N$ of $G$ by a normal subgroup $N$. The image $KN/N$ of $K$ in $G/N$ is clearly also cyclic, as a homomorphic image of the cyclic group $K$. And for the quotient we have
\begin{equation}
(G / N) / (K N / N) \cong G / K N \cong (G /K) / (KN / K),
\end{equation}
which is then also cyclic, as a homomorphic image of the cyclic group $G/K$.

However, first I need to prove that the mentioned Galois Group is metacyclic.

Would you help me?

答:
An explicit solution. (I started it, and although there are already good answers, i will submit it, sometimes the pedestrian view may be useful for the reader.)

Let $K\subset \Bbb C$ be the field generated by the roots of the poylnomial
$$
X^5-2\in\Bbb Q[X]\ .
$$
Then one root is $a:=\sqrt[5]2\in\Bbb R$. (The usage of analysis is done rather to have clear notations.) The other roots are of the shape
$$
a\zeta^k\ ,\qquad k\in\{0,1,2,3,4\}\ .
$$
Here, $\zeta=\zeta_5=\cos\frac{2\pi}5+i\sin\frac{2\pi}5$ is a primitive root of unity (in $\Bbb C$, in case analysis is not wanted, take $\zeta$ in some theoretical algebraic closure).

Then a subfield of $K$ is the cyclotomic field $\Bbb Q(\zeta)$. And
$$
K=\Bbb Q(\zeta,a)\ .
$$
A Galois substitution $s\in G$ in the Galois group $G:=\operatorname{Gal}(K:\Bbb Q)$ is determined by the images of the generators $\zeta$ and $a$, which are respectively conjugated to $\zeta$ and $a$. So:
$$
\begin{aligned}
s(\zeta) &=\zeta^k&&\text{ for some unique $k$ among $1,2,3,4$, and}\\
s(a) &= a\zeta^j&&\text{ for some unique $j$ among $0,1,2,3,4$ .}
\end{aligned}
$$
Let us denote by $s(k, j)\in G$ the substitution determined by the above relations.
Then $G$ is the set of all $4\cdot 5=20$ substitutions $s(k,j)$, $k,j$ as above. (A vector space basis of $K$ over $\Bbb Q$ has also $20$ elements, $a^j\zeta^k$, $j$ among $0,1,2,3,4$, and $k$ among... $0,1,2,3$ usually, since $1+\zeta+\zeta^2+\zeta^3+\zeta^4=0$, but $1,2,3,4$ instead is also possible.)

Let
$\sigma=s(2,0)$ and $\tau=s(0,1)$. So

$\sigma$ maps $\zeta\to\zeta^2$ and $a\to a$. (It invariates $a$, has order four since $\sigma^4\zeta=\zeta^{2^4}=\zeta$, and this is the minimal $k>0$ with $\zeta^{2^k}=\zeta$, the inverse $\sigma^{-1}=\sigma^3$ also invariates $a$ and maps $\zeta$ to $\zeta^{2^3}=\zeta^3$. It generates a cyclic (abelian) group of order four  $S<G$.

$\tau$ maps $\zeta\to\zeta$ and $a\to a\zeta$. It invariates $\zeta$, and has order five, $\tau^j a=a\zeta^j$. Its inverse is $\tau^4$, also invariates $\zeta$ and maps
$a\to a\zeta^4$. So $\tau$ generates a cyclic (abelian) group of order five $T<G$.

So $\sigma, \tau$ generate $G$.
The two mophisms do not commute, but we have.
$$
\begin{aligned}
\sigma^{-1}\tau\sigma\;a
&=
\sigma^{-1}\tau\;a=
\sigma^{-1}\;a\zeta
=a\cdot\sigma^{-1}\zeta
=a\zeta^3
=
\tau^3\; a\ ,
\\
\sigma^{-1}\tau\sigma\;\zeta
&=
\sigma^{-1}\tau\;\zeta^2=
\sigma^{-1}\;\zeta^2=\zeta
=
\tau^3\; \zeta\ .
\qquad\text{This gives:}\\
\sigma^{-1}\tau\sigma &= \tau^3\ ,\\
\sigma^{-1}T\sigma &= T\ .
\end{aligned}
$$
So $T\triangleleft G$ is a normal subgroup, cyclic, with five elements.

The quotient $G/T$ has four elements,
$T=[1]$,
$\sigma T=[\sigma]$,
$\sigma^2 T=[\sigma^2]=[\sigma]^2$,
$\sigma^3 T=[\sigma^3]=[\sigma]^3$, it is a commutative group with four elements isomorphic to $S\cong \Bbb Z/4$, thus cyclic.


问:
Calculate Delta measure
We are looking at the measure space $(\mathbb{R}^d, \mathcal{P}(\mathbb{R}^d),\delta_p)$, for all $A\subset \mathbb{R}^d$:

$\delta_p(A):=\begin{cases}
1, & \text{if } p \in A \\
0, & \text{otherwise}
\end{cases}$.

Let $A \subset \mathbb{R}^d$ and $f:\mathbb{R}^d \rightarrow   \overline {\mathbb {R}}$ arbitrary.
Show that $f$ is measurable and calculate $\int_A f \space d \delta_p$.

Let $\lambda$ be the Lebesgue measure on $\mathbb{R}^d$. Does an integratable function $h: \mathbb{R}^d \rightarrow \mathbb{R}$ exist such that for $A \subset \mathbb{R}^d: \delta_p(A)=\int_Ah \space d  \lambda$?

$\int_{A}fd\delta_p= \begin{cases} f(p) & \text{if}~ p\in A\\ 0 & \text{otherwise} \end{cases}$ is the integral but how do I show this rigorously?

答:
Use the fact that, for an increasing sequence $(f_n)_n$ of simple functions with $\lim_{n \to +\infty}f_n(x) = f(x)$,
$$\int f~d\delta = \lim_{n \to \infty} \int f_n~d\delta$$
by definition of the integral. (Note that in the definition of the integral we suppose $f \ge 0$, you just have to separate $f^+$ and $f^-$).

Title: Calculate Delta measure


问:
Finding complete set of representatives
For $x, y \in \mathbb N \times \mathbb N$ with $x=(a,b)$ and $y=(c,d)$. Define $ x \sim y$ if $a+d=b+c$ and show that $\{(n,0) : n\in \mathbb N \} \cup \{(0,m) : m \in \mathbb N, \ m\neq 0\}$ is a complete set of representatives for $(\mathbb N \times \mathbb N)/\sim$
(We assume $0 \in \mathbb N$)

I have written the equivalence class of an element $(a,b)$ with $[(a,b)]=\{(n,m) \in \mathbb N \times \mathbb N : n+b=m+a\}$

I can see that if we consider set of real numbers, an equivalence class is a line having slope 1 and $y-axis$ is a complete set of representatives. However I could not see how can I show the given set is a complete set of representatives.

In addition, what if we take equivalence relation as $|a|+|b|=|c|+|d|$ on $\mathbb R^2$? I am in we get the lines having slope 1 in first quadrant and third quadrant and having slope -1 in second quadrant and fourth quadrant. Is it true? How can I describe geometrically equivalence classes better and what is the complete set of representatives for this?

I appreciate any help.

Thank you

答:
If $a+d=b+c$ we also have $a-b=c-d$ so that the equivalence class of $(a,b)$ is characterized by the number $a-b$.  Do you see how to do it now?

In the second part of the question, the equivalence class of $(a,b)$ is characterized by $|a|+|b|$, so $\{(0,a)\mid a\geq0\}$ is a complete set of representatives.

Title: Finding complete set of representatives


问:
Infinitely many $ n \in \mathbb{N} $ such that $ n^2+1 $ has two divisors $ a,b $ such that $a-b=n $
Prove that there is infinitely many $ n \in \mathbb{N} $ such that $ n^2+1 $ has two divisors $ a,b $ such that $a-b=n $.

It is obvious that if $ p\mid n^2+1 $ then $\gcd(p,n)=1$. I tried to use the Chinese remainder theorem, but I got nothing. Please help me.

答:
It works for $b,a$ consecutive terms in the sequence $x_j$ beginning
$$ 1, 2, 5, 13, 34, ..$$
with
$$ x_{j+2} = 3 x_{j+1} - x_j $$
which are (every second) Fibonacci numbers.

Although the problem does not require this, the product is precisely the given number $n^2 + 1.$  Thus the hint that $a^2 - 3ab + b^2 = -1$ really did finish the matter.


问:
Do we need compactness hypothesis in Lemma 6.2 in Lee's "Introduction to Smooth Manifolds"?
The following lemma is from Lee's "Introduction to Smooth Manifolds."

Lemma 6.2. Suppose $A\subseteq\mathbb R^n$ is a compact subset whose intersection with $\{c\}\times\mathbb R^{n-1}$ has $(n-1)$-dimensional measure zero for every $c\in\mathbb R$. Then $A$ has $n$-dimensional measure zero.

Do we need to hypothesize that $A$ is compact? I think the author included the compactness hypothesis so that the elementary proof (no measure theory) in the book will work. But if I prove the lemma using Tonelli's theorem, I think $A$ does not need to be compact:

Let $\chi_A$ be the characteristic function of $A$. We have to prove that $\int\chi_Adm^n=0$. By Tonelli's theorem, $\int\chi_Adm^n=\int\left(\int(\chi_A)_cdm^{n-1}\right)dm(c)$, where $(\chi_A)_c(x_2,\ldots,x_n)=(c,x_2,\ldots,x_n)$. Since we are given that $\{c\}\times\mathbb R^{n-1}$ has $(n-1)$-dimensional Lebesgue measure zero, it follows that $\int(\chi_A)_cdm^{n-1}=0$ for all $c\in\mathbb R$. Thus $\int\chi_Adm^n=\int 0dm(c)=0$.

Am I right, or did I miss something?

答:
You're absolutely right -- the lemma is true without assuming compactness, with an easy proof based on Fubini/Torelli. And you're also right that the reason I assumed compactness was so I could give an elementary proof that doesn't use any measure theory.

My intention for my series of manifolds books (Introduction to Topological/Smooth/Riemannian Manifolds) was to assume only a decent undergraduate math degree as prerequisite, and to make the series otherwise self-contained. Since many undergraduate math programs don't cover Lebesgue integration (or at least that was the case when I started writing the books), I decided to use only Riemann integration and some very basic facts about sets of measure zero.


答:
If you consider the matrix
$$
B_n=\begin{bmatrix}
1&0&\cdots&0\\
1&1&0&\cdots\\
&&\cdots\\
1&1&\cdots &1
\end{bmatrix}
$$
(i.e. the $i,j$ entry of $B_n$ is $1$ is $i\geq j$ and $0$ otherwise), then the $k^{\rm t h}$ column of $A_n$ is obtained by adding the first $k$ columns of $B_n$. So
$$
\det A_n=\det B_n=1.
$$</p>

Title: Evaluation of a specific determinant.

Body:
Evaluate $\det{A}$, where $A$ is the $n \times n$ matrix defined by $a_{ij} = \min\{i, j\}$, for all $i,j\in \{1, \ldots, n\}$.

$$A_2
\begin{pmatrix}  1&  1\\
  1&  2
\end{pmatrix};
A_3 = \begin{pmatrix}  1&  1&  1\\
  1&  2&  2\\
  1&  2&  3
\end{pmatrix};
A_4 = \begin{pmatrix}  1&  1&  1&  1\\
  1&  2&  2&  2\\
  1&  2&  3&  3\\
  1&  2&  3&  4
\end{pmatrix};
A_5 = \begin{pmatrix}  1&  1&  1&  1&  1\\
  1&  2&  2&  2&  2\\
  1&  2&  3&  3&  3\\
  1&  2&  3&  4&  4\\
  1&  2&  3&  4&  5
\end{pmatrix}$$

$$A_6 = \begin{pmatrix}  1&  1&  1&  1&  1&  1\\
  1&  2&  2&  2&  2&  2\\
  1&  2&  3&  3&  3&  3\\
  1&  2&  3&  4&  4&  4\\
  1&  2&  3&  4&  5&  5\\
  1&  2&  3&  4&  5&  6
\end{pmatrix};
A_7 = \begin{pmatrix}  1&  1&  1&  1&  1&  1&  1\\
  1&  2&  2&  2&  2&  2&  2\\
  1&  2&  3&  3&  3&  3&  3\\
  1&  2&  3&  4&  4&  4&  4\\
  1&  2&  3&  4&  5&  5&  5\\
  1&  2&  3&  4&  5&  6&  6\\
  1&  2&  3&  4&  5&  6&  7
\end{pmatrix}
$$


问:
For any field $F$, prove that $\Sigma(2,F)$ is isomorphic to ${\rm Aff}(1,F)$, where $\Sigma(2,F)$ denotes the stochastic group.
$$\Sigma(2,F)=\left\{\begin{bmatrix}a&b\\1-a&1-b\end{bmatrix},a,b\in F\right\}$$

$${\rm Aff}(1,F)=\left\{\begin{bmatrix}a&b\\0&1\end{bmatrix},a\neq 0,a,b\in F\right\}$$

First I tried linear transformation:

$\begin{bmatrix}c&d\\e&f\end{bmatrix}\begin{bmatrix}a&b\\1-a&1-b\end{bmatrix}=\begin{bmatrix}ca+d(1-a)&cb+d(1-b)\\ea+f(1-a)&eb+f(1-b)\end{bmatrix}$

I want $eb+f(1-b)=1$ for any $b\in F$. Then it must be $e=f=1$. But I also want $ea+f(1-a)=0$. This is impossible.

答: Your parametrization of $\Sigma(2,F)$ isn't entirely correct; if $a=b$ then the matrix is singular. And more critically, linear transformation as you've written it (multiplying by some fixed matrix $M$) doesn't preserve the group product; $(MA)(MB)\neq M(AB)$. You probably want to conjugate your matrix $A\in\Sigma(2,F)$ by some matrix M; that is, to look at $MAM^{-1}$. The most straightforward way of doing this is to try and solve $MA=BM$ for  $A=\left(\begin{smallmatrix}\alpha&\beta\\ 1-\alpha&1-\beta\end{smallmatrix}\right)\in\Sigma(2,F)$ and $B=\left(\begin{smallmatrix}a&b\\ 0&1\end{smallmatrix}\right)\in\mathrm{Aff}(1,F)$.


答:
You can still think of it in terms of the slope of a tangent line, and even in terms of a limit.  However, as you've pointed out, $x^2 + y^2 = r^2$ isn't a function because it fails the vertical line test.  However, by the Implicit Function Theorem we can consider $F(x,y) = x^2 + y^2 - r^2$, and for any $(x_{0},y_{0})$ where $\frac{\partial F}{\partial y}\ne 0$ then there exists some neighborhood around the point $(x_{0},y_{0})$ for which we can express $F(x,y) = 0$ as some function $y = f(x)$.  Note that in this case, $$\frac{\partial F}{\partial y} = 2y,$$ which is zero whenever $y = 0$, so at the points $(r,0)$ and $(-r,0)$.  On the circle.  Anywhere else we can define the curve by either $$y = \sqrt{r^{2} - x^{2}}$$ or $$y = -\sqrt{r^{2} - x^{2}}$$ and these functions are differentiable so long as $y\ne 0$.  To verify that the derivative via the limit definition matches that obtained by implicit differentiation, we can compute as follows for the positive semicircle:
\begin{align}
y' &= \lim_{h\to 0}\frac{\sqrt{r^{2} - (x+h)^{2}} - \sqrt{r^2 - x^2}}{h}\\
&=\lim_{h\to 0}\frac{\sqrt{r^{2} - (x+h)^{2}} - \sqrt{r^2 - x^2}}{h} \cdot \frac{\sqrt{r^{2} - (x+h)^{2}} + \sqrt{r^2 - x^2}}{\sqrt{r^{2} - (x+h)^{2}} + \sqrt{r^2 - x^2}}\\
&=\lim_{h\to 0}\frac{r^{2} - (x+h)^{2} - r^{2} - x^{2}}{h(\sqrt{r^{2} - (x+h)^{2}} + \sqrt{r^2 - x^2})}\\
&=\lim_{h\to 0}\frac{-2xh -h^{2}}{h(\sqrt{r^{2} - (x+h)^{2}} + \sqrt{r^2 - x^2})}\\
&=\lim_{h\to 0}\frac{-2x - h}{\sqrt{r^{2} - (x+h)^{2}} + \sqrt{r^2 - x^2}}\\
&=-\frac{2x}{2\sqrt{r^{2} - x^{2}}}\\
&=-\frac{x}{y}.
\end{align}
Where the last equality comes from the equation $y = \sqrt{r^2 - x^2}.$  We could do the same for the negative semicircle.

Title: How can we find the derivative of a circle if a circle is not a function?

Body:
If we consider the equation of a circle, $x^2+y^2=r^2$, then I understand that $dy/dx$ can be computed in the following way via implicit differentiation:
\begin{align}
2x + 2y\frac{dy}{dx} &= 0 \\
\frac{dy}{dx} &= -\frac{2x}{2y} = -\frac{x}{y} \, .
\end{align}
Although I feel comfortable deriving this result, I don't really understand how I should interpret it. On an intuitive level, the formula $dy/dx = -x/y$ seems to suggest that the gradient of the tangent to any given point $(x,y)$ is $-x/y$. However, since the curve $x^2+y^2=r^2$ fails the vertical line test, it doesn't look like it is even a function. Usually, $dy/dx$ can be thought of as a shorthand for
$$
\lim_{h \to 0}\frac{y(x+h)-y(x)}{h} \, .
$$
However, in this case each $x$-value maps to two $y$-values, and so the limit definition doesn't seem to apply here. So what does $dy/dx$ actually represent in this context?


问:
Why are vector fields defined to be sections of the disjoint union of the tangent spaces? Isn't that overly complicated?
It is usual to define the tangent bundle $TM$ as the disjoint union of all tangent space and then to define vector fields as sections of
\begin{align}
TM=\bigsqcup_{p\in M}T_pM=\bigcup_{p\in M}T_pM\times\{p\}&\to M\\
(v,p)&\mapsto p
\end{align}
(I am reading John Lee's book, where tangent vectors are introduced as derivations, such that $T_pM\cap T_qM\neq\emptyset$.)

Why don't we simply define $TM=\bigcup_{p\in M}T_pM$ and say that a vector field is a map $\omega\colon M\to TM$ with $\omega_p\in T_pM$ for all $p\in M$.

Isn't this the essential property of a vector field: That is assigns to all $p\in M$ an element of $T_pM$?

An analogous question arises for Differential forms.

An attempt to clarify the question: I have to prepare a talk about the Maxwell equations in terms of differential forms and I have an audience who does neither know what a manifold is, nor what a tensor ist. I have 30 minutes for a crash course in this stuff. This means that I have to leave out as much information as possible, but I want everything I say to be correct. So one question I asked myself is: Do I miss out on something if I don't explain what disjoint unions and sections are and simply give them my definition above? Is there a good reason to define the tangent bundle as the disjoint union of tangent spaces instead of the ordinary union (besides wanting to introduce the notion of sections)? But this are questions I already asked myself before having to give this talk.

答: There are lots of good answers here, all of which elucidate certain parts of the situation. But there's one important point that hasn't been mentioned -- in the definition of tangent spaces that I use in my Smooth Manifolds book, the zero derivation is an element of $T_pM$ for every $p\in M$, so if you don't use disjoint union in the definition of the tangent bundle, the tangent spaces will all intersect. See also this answer.


问:
Approaches to Prove Every Planar Graph $G$ is $6$-colorable
I have asked similar question in an earlier post, but this time I would like to get better understanding of the difference between using strong induction and regular induction to prove that every planar Graph $G$ is $6$-colorable.

My understanding: by induction hypothesis, for $n \ge 6$ and assume that every simple, connected and planner graph on up to $n$ vertices is $6$-colorable. Then we can not regular induction here because we have the connectivity condition, but if we removed connectivity condition from the assumption, then we could use regular expression. The issue here is if we delete a vertex $u$ from $G$, which might make the graph disconnected and this is why we need the strong induction and connectivity in the assumption.

Please let me know your thought of my understanding about proving every planar graph $G$ is $6$-colorable and why we need connectivity condition in the induction hypothesis.

答:
There is no reason to require connectivity, it just makes your life harder.

There's one reason to try to stick to connected graphs, and that's probably why the source you're reading from does it: Euler's formula. We only have $|V(G)| - |E(G)| + |F(G)| = 2$ for a plane embedding of a connected planar graph, and this is the equation from which we derive $|E(G)| \le 3|V(G)| - 6$ (for graphs on at least $3$ vertices), which in turn lets us deduce that there is a vertex of degree less than $6$.

We can avoid this by observing that every planar graph can be turned into a connected planar graph by adding edges. Therefore if $|E(G)| \le 3|V(G)| - 6$ holds for all connected planar graphs on at least $3$ vertices, it holds for all planar graphs. Having deduced this, we can use this inequality with impunity and no care for whether our graph is connected.

But, if you want to prove the theorem for connected graphs, then we use strong induction, and say:

If the graph is connected, do the usual thing where you remove a vertex, color the remainder by the induction hypothesis, and put the vertex back.

If the graph is not connected, color every connected component by the induction hypothesis. (This is the part where we need strong induction: if the original graph has $n$ vertices, then the components can have any number of vertices between $1$ and $n-1$, so we need the induction hypothesis for all of those cases.)


问:
Solving $\sum_{k=a}^{b-1} k=\sum_{k=b+1}^c k$ for natural numbers $a,b,c$
Find natural numbers $a<b<c$ such that:

$$\sum_{k=a}^{b-1} k=\sum_{k=b+1}^c k$$

I found infinitely many solutions when $a=1$, where $b$ and $c$ can be found in the following way:

Calculate, in the simplest form, the continued fractions of the series: $1+\frac{1}{2}$,$1+\frac{1}{2+\frac{1}{2+\frac{1}{2}}}$, where each term contains an odd number of $2$'s.

Then $b$ is $\frac{1}{2}D$ and $c$ is $\frac{N-1}{2}$ where $D$ and $N$ are the numerator and the denominator of any term of the previous series, respectively.

For example,$1+\frac{1}{2+\frac{1}{2+\frac{1}{2}}}=\frac{17}{12}$, So $b=\frac{1}{2}(12)=6$, and $c=\frac{17-1}{2}=8$.

So, $a=1,b=6,c=8$ is satisfied.

Calculating continued fractions by hand is easy, but tedious!, So is there a general way to calculate $b$ and $c$ given any natural number $a$?

Your help would be appreciated. Thanks!

答:
Not an answer, but rather a long comment.
Hint. Use the well-known formula $$\sum_{i=1}^n i=\frac{n\cdot (n+1)}{2}$$ in order to deduce $$\sum_{k=a}^{b-1} k=\sum_{k=1}^{b-1}k-\sum_{k=1}^{a-1}k=\ldots=\frac{(b-a)(a+b-1)}2$$ Similarly $$\sum_{k=b+1}^c k=\frac{(c-b)(c+b+1)}2$$Thus, your problem comes down to solving the following Diophantine equation $$(b-a)(a+b-1)=(c-b)(c+b+1)\iff2b^2=a^2-a+c^2+c$$ Given some $a\in\mathbb N$, observe that the latter equation is equivalent to $$\begin{align*}8b^2+2=(2a-1)^2+(2c+1)^2&\iff(2a-1)^2-2=8b^2-(2c+1)^2\\&\iff ... \end{align*}$$

I've been playing with Wolfram Alpha for a while and can't find any pattern for the solutions of the respective $a$ values, neither can the OEIS.


答:
$$(a,b) \sim\begin{cases}(a-b,0)&\text{ if }\;a\ge b,\\ (0,b-a) &\text{ if }\;a<b.\end{cases}$$
Note this is the starting point of the construction of $\mathbf Z$ from $\mathbf N$:
$$\mathbf Z\overset{\text{def}}{=} (\mathbf N\times \mathbf N)/\sim$$

Note: This  construction is quite similar to the construction of the field of fractions of $\mathbf Z$:

On $\mathbf Z\times\mathbf Z^*$, one defines the equivalence relation:
$$(p,q)\sim(r,s)\quad (q,s\ne 0)\iff ps=qr$$
and $\:\mathbf Q\overset{\text{def}}{=} (\mathbf Z\times \mathbf Z^*)/\sim$.

Title: Finding complete set of representatives

Body:
For $x, y \in \mathbb N \times \mathbb N$ with $x=(a,b)$ and $y=(c,d)$. Define $ x \sim y$ if $a+d=b+c$ and show that $\{(n,0) : n\in \mathbb N \} \cup \{(0,m) : m \in \mathbb N, \ m\neq 0\}$ is a complete set of representatives for $(\mathbb N \times \mathbb N)/\sim$
(We assume $0 \in \mathbb N$)

I have written the equivalence class of an element $(a,b)$ with $[(a,b)]=\{(n,m) \in \mathbb N \times \mathbb N : n+b=m+a\}$

I can see that if we consider set of real numbers, an equivalence class is a line having slope 1 and $y-axis$ is a complete set of representatives. However I could not see how can I show the given set is a complete set of representatives.

In addition, what if we take equivalence relation as $|a|+|b|=|c|+|d|$ on $\mathbb R^2$? I am in we get the lines having slope 1 in first quadrant and third quadrant and having slope -1 in second quadrant and fourth quadrant. Is it true? How can I describe geometrically equivalence classes better and what is the complete set of representatives for this?

I appreciate any help.

Thank you


问:
Solve the equation in $\mathbb{Z}_{7}$
The problem is taken from my textbook in algebra and is given as:

Solve $x+4=3$ in $\mathbb{Z}_{7}$

From the euclidean algorithm I found that $2=4^{-1}\bmod(7)$

Now I'm not completly sure how to continue but I tried:
$$x+4+4^{-1}=3+4^{-1} $$
$$x=3+4^{-1}=3+2=5   $$

However, the correct answer is $x=6$ and I'm not sure were I went wrong since the book don't have any examples so I would appreciate some help!

Title: Solve the equation in $\mathbb{Z}_{7}$

Body:
Solve $x+4=3$ in $\mathbb{Z}_{7}$

$(x+4)+3=3+3$

$x+(4+3)=6$

$x+0=6\;,\;$ indeed in $\;\mathbb{Z}_{7}\;,\;4+3=0\;.$

$x=6\;.$


问:
Problem regarding the definition of field of rational-functions over $k$.
In the construction of $\operatorname{Frac}(R)$, where $R$ is a domain, we define a partition on $R \times R^\times$ where $R^\times:= R \setminus \{0\}$. which in turn becomes a field containing $R$.

Taking $R:=\mathbb{Q}[x]$, can we tell that $\frac{x^2+1}{x-1} \in \mathbb{Q}(x)$? As $x-1$ is not the zero-polynomial it is a member of $\mathbb{Q}[x]^\times$, hence $\frac{x^2+1}{x-1}$ should be a member of $\mathbb{Q}(x)$. But this is undefined when 1 is plugged in $x$.

Where am I going wrong?

答:
Your understanding has a gap starting at the ring of polynomials $\mathbb Q[X]$, not at the field of fractions. So I'm going to focus on the former, not the latter.

To start with, polynomials are not functions. Memorize this sentence and never forget it: Polynomials are not functions.

What polynomials really are should be understood through first understanding what their purpose is. A polynomial's purpose is to be a template for expressions that involve some element $X$ of a ring using only the operations provided by the ring: addition, subtraction and multiplication. We are allowed to multiply $X$ by an element of the ring or by itself, and we can add or subtract the results of these multiplications. The end result is a polynomial expression in $X$ with coefficients in the underlying ring. This expression is a template into which we want to plug in any object for which the ring operations addition, subtraction and multiplication are meaningful. For instance, real matrices can be multiplied by themselves or by real numbers, and they can be added. So matrices are a kind of object we might want to plug into a polynomial. But then the result is a matrix, so it wouldn't make the least bit of sense to define the polynomial to be a function mapping numbers to numbers. On the other hand, we might as well plug in numbers, and then the results are numbers. So we also can't define polynomials to be functions mapping matrices to matrices. Rather, polynomials should be something that allows us to plug in a variety of objects before we even know what kind of object we want to plug in. What kind of map we get shouldn't be predefined.

The takeaway is that polynomials do not serve as maps, but as templates for maps involving all kinds of objects, where we're free to adjust the domain to our liking: we could make it a matrix space, or the ring itself, or something entirely different. The same goes for the fraction field. Its elements are templates for functions whose domain we can adjust dynamically. Simply don't use an expression like $\frac{1}{X}$ as a template for functions whose domain contains $0$ and you're good.


问:
Notation problem in polynomial rings. What is $\mathfrak{p}R[x]$?
I have seen a lot the notation "$\mathfrak{p}R[x]$", where $\mathfrak{p}$ is a prime ideal and $R[x]$ is a polynomial ring with coefficients in $R$.

My first question is: What is $\mathfrak{p}R[x]$ exactly? I think that it is not the same as $\mathfrak{p}[x]$, whose elements are polynomials with coefficients in $\mathfrak{p}$.

Just to give some context, I am trying to prove that $1 + \dim(R) ≤ \dim(R[X]) ≤ 1 + 2 \dim(R)$. And the first thing that appears in every single proof I've seen is the prime ideals chain $$\mathfrak{p}_0R[x]\subsetneq\mathfrak{p}_1R[x]\subsetneq\dots\subsetneq\mathfrak{p}_nR[x]\subsetneq\mathfrak{p}_nR[x]+(x).$$
Somehow, there is a relationship between the ideals of $R$ and the ideals of $R[x]$, but I am not able to spot it.

答:
That is a notation for the ideal generated by the set $\mathfrak{p}$. Let me specify that:

It gets used when you have a ring extension $R \rightarrow S$ and want to describe the ideal generated by the image of the a prime ideal $\mathfrak{p} \subset R$.

In this case, the extension is the inclusion $R \subset R[x]$.


问:
product of two diffeomorphic maps are diffeomorphic?
Suppose I have a diffeomorphism  $f_i: U_i \subseteq \mathbb{R}^{n_i} \to f_i(U_i) \subseteq \mathbb{R}^{m_i}$ where $U_i$ is an open subset. Does it then follow that
$$
f_1 \times f_2: U_1 \times U_2 \to f_1(U_1) \times f_2(U_2)
$$

is still diffeomorphic? I can see that the map is still bijective, but I was struggling to show that the map and its inverse are continuous. Any explanation would be appreciated!

答:
Notice first that if $f : U \subset \mathbb{R}^{a} \to \mathbb{R}^{b}$ is a diffeomorphism onto its image, and if $p \in U$, then
$$
\mathrm{d}f(p) : \mathbb{R}^{a} \to \mathbb{R}^{b}
$$
is a linear isomorphism. Indeed, the chain rule shows that
$$
\mathrm{d}f(p)\circ\mathrm{d}f^{-1}(f^{-1}(p)) = \mathrm{id}_{\mathbb{R}^b}
$$
and
$$
\mathrm{d}f^{-1}(f(p))\circ\mathrm{d}f(p) = \mathrm{id}_{\mathbb{R}^a}
$$

Hence, $a = b$, and in you question, you have $n_i = m_i$.

Now, suppose $f : U \subset \mathbb{R}^n \to \mathbb{R}^n$ and $g : V\subset \mathbb{R}^m \to \mathbb{R}^m$ are diffeomorphisms onto their image. Define
$$h : U\times V \to \mathbb{R}^n\times \mathbb{R}^m$$
by $h(p,q) = (f(p),g(q))$. Let us first show that $h$ is continuous. As the vector space topology does not depend on the norm (we are here in finite dimension), let us take the norm $\|x\|_{\mathbb{R}^n} = \sup_{1\leqslant i\leqslant n} |x_i|$. This norm is a good choice because with this norm, we have this fact:
$$
B_{\mathbb{R}^n\times \mathbb{R}^m}((p,q),r) = B_{\mathbb{R}^n}(p,r)\times B_{\mathbb{R}^m}(q,r)
$$
that is, an open ball of radius $r$ in $\mathbb{R}^n\times\mathbb{R}^m$ is the direct product of open balls of radius $r$. As balls generate the topology, to show a function is continuous, it is sufficient to show that the preimage of any ball is open in $U\times V$. Let $r>0$ and $(p,q) \in \mathbb{R}^n\times\mathbb{R}^m$. Then
\begin{align}
h^{-1}\left(B((p,q),r) \right) &= h^{-1} \left(B(p,r)\times B(q,r) \right) \\
&= \left(f^{-1}\left(B(p,r)\right) \times V \right)\cap \left(U\times g^{-1}\left(B(q,r) \right) \right)
\end{align}
Now, as $f$ and $g$ are continuous, $f^{-1}\left(B(p,r) \right)$ is open in $U$, and $g^{-1}\left(B(q,r)\right)$ is open in $V$. It follows that $f^{-1}\left(B(p,r) \right)\times V$ and $U\times g^{-1}\left(B(q,r)\right)$ are open in $U\times V$, and so is their intersection.

This shows that $h$ is continuous.

Now, to show that $h$ is differentiable, let us just fix $(p,q) \in U\times V$ and $(u,v) \in \mathbb{R}^n\times \mathbb{R}^m$ with small norms. Then
\begin{align}
h\left((p,q) + (u,v)\right) &= \left(f(p+u),g(q+v) \right) \\
&= \left(f(p) + \mathrm{d}f(p)u + o(\|u\|),g(q)+\mathrm{d}g(q)v + o(\|v\|) \right) \\
&= (f(p),g(q)) + (\mathrm{d}f(p)u,\mathrm{d}g(q)v) + o(\|(u,v)\|)
\end{align}
and this shows that $h$ is differentiable at $(p,q)$ with $\mathrm{d}h(p,q) = \left(\mathrm{d}f(p),\mathrm{d}h(q) \right)$.

Remark that we could just have shown the differentiability and the continuity would have followed. But I think, in a pedagogical point of view, that showing the continuity by its definition was important.

Now, let $f^{-1}$ and $g^{-1}$ be the inverse of $f$ and $g$ defined on $f(U)$ and $g(V)$ respectively. What we have done before is still valid by replacing $f$ by $f^{-1}$, $g$ by $g^{-1}$, $U$ by $f(U)$ and $V$ by $g(V)$. This shows that if $l = (f^{-1},g^{-1})$, then $l$ is also differentiable and continuous.

Moreover, it is a straightforward calculation that $l\circ h = \mathrm{id}_{U\times V}$ and $h \circ l = \mathrm{id}_{f(U)\times g(V)}$.

To conclude, we have shown that:

$h$ is differentiable map

$h$ is bijective (onto its image) with invere $l$

$h^{-1}$ is differentiable

This is exactly saying that $h$ is a diffeomorphism.


问:
Local times and quadratic variation
Suppose we have a continuous local martingale $(M_t)_{t \geq 0}$, where $M_0 = 0$. Let $L_{t}^{0}$ denote the local time of this process at zero. I would like to prove the simple statement that:
\begin{equation}
\inf \{t : L_{t}^{0} > 0 \} = \inf \{t : \langle M \rangle_t > 0 \},
\end{equation}
where the bracket denotes the quadratic variation process.

My initial idea was to use the occupation density formula, which relates local times and the quadratic variation through the integrals involving non-negative Borel measurable functions. However, the formula integrates against the superscript of $L$ rather than the subscript (that is, one measures the change with respect to the position of the process, not with respect to the time for a fixed position). Therefore, I do not hope that this could be useful.

I am not aware of any other (useful) results connecting the quadratic variation and the local times, so I would appreciate any help. I appreciate the fact that (heuristically) the local time measures the time a process spends at a particular point, with the clock given by the quadratic variation rather than the time indexing the process, but I can't quite get  a rigorous proof of this.

答:
By the theorem of Dubins & Schwartz, there is a Brownian motion $B$ such that $M_t=B_{\langle M\rangle_t}$ for all $t\ge 0$, a.s.
The  local times of $B$ and $M$ are related in the same way: $L^x(M)_t = L^x(B)_{\langle M\rangle_t}$.
In particular, taking $x=0$ and defining $A(u):=\inf\{t:\langle M\rangle_t>u\}$, you have
$$
\inf\{t:L^0(M)_t>0\}=A(\inf\{t:L^0_t(B)>0\}).
$$
But as $M_0=0$, you have $B_0=0$ as well, and so $\inf\{t:L^0_t(B)>0\}=0$ a.s. Therefore
$$
\inf\{t:L^0(M)_t>0\}=A(0)=\inf\{t:\langle M\rangle_t>0\},
$$
a.s.

Title: Local times and quadratic variation


问:
Averages clarifications!
If i know an average such as ( catch an average of one rat every 39 minutes ) what is the average number of rats caught in 10 hours ?

is it as simple as dividing 600/39 (600 = 10 hours in minutes) and multiplying this value by number of rats caught in 39 minutes ?

in this case 15.38 rats caught in 10 hours.

答:
Graphical comment. Suppose rat catching is at random so that the number $X$ of rats caught in 10 hours is distributed
as $\mathsf{Pois}(\lambda = 15.3846).$ Then, according to @ATF's computation, the probabilities $P(X = x),$ for $x=0, 1, \dots, 40$ can be plotted in R as follows:

```<code>x = 0:40;  lam = 15.3846
PDF = dpois(x, lam)
plot(x, PDF, type="h", lwd=2, col="blue", main="PDF of POIS(15.3846)")
 abline(h = 0, col="green2");  abline(v = 0, col="green2")
</code>```

Only rarely, would one catch fewer than 5 or more than 30 rats in 20 hours.

Title: Averages clarifications!

Body:
Graphical comment. Suppose rat catching is at random so that the number $X$ of rats caught in 10 hours is distributed
as $\mathsf{Pois}(\lambda = 15.3846).$ Then, according to @ATF's computation, the probabilities $P(X = x),$ for $x=0, 1, \dots, 40$ can be plotted in R as follows:

x = 0:40;  lam = 15.3846
PDF = dpois(x, lam)
plot(x, PDF, type="h", lwd=2, col="blue", main="PDF of POIS(15.3846)")
 abline(h = 0, col="green2");  abline(v = 0, col="green2")

Only rarely, would one catch fewer than 5 or more than 30 rats in 20 hours.


问:
Showing two group presentations are isomorphic to known groups
I'm trying to help a friend through his Maths degree and he gave my a question that I just can't remember how to do.
Show
$$\langle x,y | y^5=x^2=[x,y]=1\rangle \ \text{and}\ \langle u, w | u^{-1} wu=w^2, w^{-1}uw=u^2\rangle$$

are isomorphic and justify.
I recall that there's some work to be done with generators etc but I can't find questions similar enough that I would feel confident helping him without mistakes.

答:
I don't think they are isomorphic... Call the first group $G$ and the second group $H$.

The first group is abelian, since $x$ and $y$ commute; it has an element of order $2$ and an element of order $5$, hence it has an element of order $10$, namely $xy$. The group is, in fact, cyclic of order $10$: if $G=C_{10}$ generated by $z$, then $z^5$ and $z^2$ satisfy the relations given for $x$ and $y$, so we get a map $G\to C_{10}$ given by $x\mapsto z^5$ and $y\mapsto z^2$. The map is surjective, so $G$ has order at least $10$. Conversely, we can map $C_{10}$ to $G$ by sending $z$ to $xy$, which has order $10$ since $[x,y]=1$; the map is surjective, since $(xy)^2 = y^2$ and $\langle y^2\rangle =\langle y\rangle$ (as the order of $y$ divides $5$), so $\langle xy\rangle$ contains $y$ and hence contains $x$. This proves $G$ has order at most $10$, and so $G\cong C_{10}$.

On the other hand, if $H$ were abelian, then $u^{-1}wu=w^2$ would imply $w=w^2$, so $w=e$, and similarly $w^{-1}uw=u^2$ would imply $u=e$, and so we would immediately conclude that if $H$ is abelian, then it is trivial. Thus, either $H$ is not abelian (and so cannot be isomorphic to $G$), or else $H$ is abelian and trivial (and so cannot be isomorphic to $G$).

So, what is $H$ isomorphic to? Note that $[u,w]=u^{-1}w^{-1}uw=u$ and $[w,u]=w^{-1}u^{-1}wu = w$. Since $[u,w]=[w,u]^{-1}$, this means $w=u^{-1}$. But then $u^2 = w^{-1}uw = uuu^{-1}$, so $u^2=u$, hence $u=e$ and similarly $w=e$, so $H$ is in fact trivial.


问:
Is $\operatorname{Aut}(\mathbb{I})$ isomorphic to $\operatorname{Aut}(\mathbb{I}^2)$?
Is $\def\Aut{\operatorname{Aut}}\Aut(\mathbb{I})$ isomorphic to $\Aut(\mathbb{I}^2)$ ? ($\mathbb{I},\mathbb{I}^2$ have their usual meaning as objects in $\mathsf{Top}$).

I show some of one of my attempts. I was trying to show that the two are not isomorphic by considering the elements of order $2$ of both groups. If $\Aut(\mathbb{I})$ has a finite number of elements of order $2$, then $\Aut(\mathbb{I}^2)$ would have more elements of order $2$ (because for each $f\in \Aut(I)$ of order $2$, the functions $f_1,f_2$ given by $f_1(x,y)=(f(x),f(y)),f_2(x,y)=(x,f(y))$ are elements of order $2$ of $\Aut(\mathbb{I}^2)$ ).

Shortly, I found that $\Aut(\mathbb{I})$ has infinitely many elements of order $2$, thus this does not work.

答:
If you knew that there were only a non-zero finite number of elements of $\text{Aut}(\Bbb I)$ of order $2$, then your approach would do the trick. However, since you know that each map $x\mapsto(1-x^n)^{1/n}$ with $n\in\Bbb Z_+$ is an element of $\text{Aut}(\Bbb I)$ of order $2$, then there are infinitely many such elements in both automorphism groups. Without something more--like showing that the respective infinite collections of order $2$ elements are of different cardinality--this won't be enough.

Consider instead the function $g:\Bbb I^2\to\Bbb I^2$ given by $$g(x,y)=(1-y,x).$$ You should be able to show that $g$ is an automorphism of order $4$. Does $\text{Aut}(\Bbb I)$ have any elements of order $4$?

Hints: Each auto-homeomorphism on $\Bbb I$ will either be strictly increasing or strictly decreasing. The only strictly increasing auto-homeomorphism on $\Bbb I$ of finite order is the identity function. Consider the multiplicative subgroup $C_2:=\{1,-1\}$ of $\Bbb C^\times,$ and the function $\phi:\text{Aut}(\Bbb I)\to C_2$ taking strictly increasing functions to $1$ and strictly decreasing functions to $-1$. You should be able to show that $\phi:\text{Aut}(\Bbb I)\to C_2$ is a surjective homomorphism. From this, we see that if $f$ is a strictly decreasing auto-homeomorphism of $\Bbb I$ of finite order, then $f$ has even order. How is the order of $f$ related to the order of $f^2$? Is $f^2$ strictly increasing or strictly decreasing? What can we then conclude about the order of $f$?


问:
Unitary matrix and invertible matrix
Is every unitary matrix invertible?

I think the statement is true? It seems like I can't find a counter example.

答: The definition of "unitary" implies that unitary matrices are invertible: the inverse is its conjugate transpose.</p>

Title: Unitary matrix and invertible matrix


问:
Does the expression $\frac{\mathbf{v}\mathbf{v}^T}{\mathbf{v}^T\mathbf{v}}$ have a name?
I have encountered the matrix $$\frac{\mathbf{v}\mathbf{v}^T}{\mathbf{v}^T\mathbf{v}}$$ (where $\mathbf{v}$ is a column-vector) a several times when differentiating vector expressions and was wondering whether it has a name.

答: This is the matrix representing projection onto $v$.


问:
Möbius transformation change of the variables formula on the unit circle.
Consider the Möbius map $$\varphi_{a}(z)=\frac{a-z}{1-\bar{a}z}$$for $a,z\in\mathbb{D}$.

I'm trying to prove the following change of the variables formula on the unit circle $\mathbb{T}$ $$\int_0^{2\pi} f(\varphi_a(e^{i\theta})) \, d\theta =\int_0^{2\pi} f(e^{it})\frac{1-|a|^2}{|1-ae^{-it}|^2} \, dt$$ for all $f\in L^1(\mathbb{T},d\theta)$. Any help is appreciated.

答:
Let $\varphi_{a}(e^{i\theta})=e^{it}$ hence $\frac{a-e^{i\theta}}{1-\bar{a}e^{i\theta}}=e^{it}$ which leads to $\frac{a-e^{it}}{1-\bar{a}e^{it}}=e^{i\theta}$, hence taking differentials:

$\frac{-1+|a|^{2}}{(1-\bar{a}e^{it})^{2}}ie^{it}dt=ie^{i\theta}d\theta$ and substituting in terms of $t$, we get:

$d\theta=\frac{-1+|a|^{2}}{(1-\bar{a}e^{it})^{2}}\frac{1-\bar{a}e^{it}}{a-e^{it}}e^{it}dt$ or by simplification

$d\theta=\frac{-1+|a|^{2}}{(1-\bar a e^{it})(ae^{-it}-1)}dt=\frac{1-|a|^2}{|1-ae^{-it}|^2}dt$ so the change of variables follows by substituting $d\theta$ (since by defintion $ f(\varphi_a(e^{i\theta}))=f(e^{it})$, while the bounds remain same by periodicity) and we are done!

(note that in the OP, RHS is an integral in $t$ so the denominator must be $|1-ae^{-it}|^2$)


问:
A formula for operator $\frac D{e^D-1}$?
Is there a good unambiguous formula for the linear operator $\frac D{e^D-1}$?

I mean,

$$x^a\to B_a(x)$$
$$\ln x \to \psi(x)$$
$$e^x\to\frac{e^x}{e-1}$$

etc.

答:
Seems, something like this:

$$\frac D{e^D-1}f(x)=\frac1{2 \pi }\int_{-\infty }^{+\infty } \frac{e^{-iwx}}{e^{-i w}-1}\int_{-\infty }^{+\infty } e^{i t w} f'(t) \, dt \, dw$$


问:
Corollary to mean value property for harmonic functions?
For $\Omega \subset\mathbb{R}^n$ open, and $u_i:\Omega \to \mathbb{R}$ a sequence of harmonic functions which are uniformly bounded. Prove that for any multi-index $\alpha$ and for any $K \subset \Omega$ there exists $u_{i_{k}}$ such that $D^\alpha u_{i_{k}}$ converges uniformly in $K$.

The mark scheme apparently uses a corollary to the mean value property which says that $$|u(x) - u(y)| \leq \sup_\xi |\nabla f(\xi)| |x-y|$$ but I can't see where this corollary is coming from? Could someone prove this?

答:
The inequality $$|u(x) - u(y)| \leq \sup_\xi |\nabla f(\xi)| |x-y|\tag1$$ in which the supremum is taken over $\xi$ on the line segment from $x$ to $y$, has nothing to do with harmonic functions, or their mean value property. It is a corollary of the Mean Value Theorem.

As for the original question, the key point is the interior regularity of harmonic functions. Namely, for any multiindex $\alpha$ and every locally uniformly bounded family of harmonic functions $u_j$, the derivatives $u_j^\alpha$ also form a locally uniformly bounded family of harmonic functions. By induction, it suffices to prove this for first order derivative. The quickest way is probably to use the Harnack inequality, which gives an estimate of the form $$|\nabla u(x)|\le \frac{C}{r}\sup_{|y-x|= r} |u(y)-u(x)| \tag2$$ with a universal constant $C$. Note that (2) goes in the direction opposite to (1).


答: Of course it is true. One way to see it that $Q^*Q=I$ implies $detQ \neq 0$. Another way is that the column vectors of $Q$ are orthogonal in complex way (I mean $u^*v=0$), so they have to be linearly independent. So $Q$ is invertible.

Title: Unitary matrix and invertible matrix

Body:
Is every unitary matrix invertible?

I think the statement is true? It seems like I can't find a counter example.


问:
determine the direction of a vector regarding to a plane.
I am trying to study a 3D model, in which I have such details. (concretly, I need to unbed them to export a plate)

In my model I have a list of 3D points, that define the contours of each face. But there is no "convention" regarding order in which information is given.
The order in which the surfaces are given is random, the same as it is clockwise or not.
What I could do now is :

Identify the 2 internal faces

Then find the "formule" to transform 3D lines to a plan

But I see that depending on the direction in Which I receive the points, it generates some symetry problems :

On the end I think I understood why (but couldn't find a way how to solve it :

To determinate the x/y axises, I do the following :

Determinate 2 points common to my 2 faces.

I choose one of them.

Then I define the point next/before that common point.

If equal to second common point, this is Y axis, if not, it is X axis.

But in case 2, my axis "Z" is on the rear of the detail, so I have symetry issues.

I think if I could find the axis is front or behind the plane (x,y), I just would need to change direction of "x", but how can I determinate that?

答:
Well, just found how to solve, was much easier than expected...

The problem was I never knew which detail goes to the left side or to the right side.
I already tried scalar to determine the side, but this didn't answer to 1st question, but didn't expect that scalar was the answer :

Whatever if my "common point" is on bottom or on the top(it is just a rotation of the plate), I just need to : (let's say common point is the one in the bottom corner)

Make cross multiplication of X and Y to find Z axis.

Make scalar product between Z found, and X axis of the opposite plate (always going from common point.

If scalar is positive, I am on the right side of the corner, then if scalar is negative I am on the left side.

As I always take the bottom corner as center of axes, I have all X/Y coordinates from that point. Then just need to join the 2 pieces together.


问:
A problem about positive definite matrices

Given $A\in\mathbb{R}^{n\times n}$, show that all eigenvalues of A has negative real part if and only if for each positive definite matrix $C\in\mathbb{R}^{n\times n}$, there exists an unique positive definite matrix $B\in\mathbb{R}^{n\times n}$ which satisfies $BA+A^TB=-C$

I thought that since $B$ and $C$ are both positive definite, then they can be simultaneously diagonalized, then I cannot go any further.

Can anyone help me? Many thanks

答:
I assume that, in this context, "positive definite" implies "symmetric". Note that the equation $BA + A^TB = -C$ is the continuous Lyapunov equation.

$\implies:$ If $A$ is such that its eigenvalues have negative real part, then to establish existence it suffices to note that the integral
$$
B = \int_0^\infty e^{A^\top\tau} C e^{A\ \tau}\,d\tau
$$
converges. Indeed: if this integral converges, then we find that
$$
\frac d{d\tau} e^{A^\top\tau} C e^{A \tau} =
A^\top e^{A\tau} C e^{A \tau} + e^{A^\top\tau} C e^{A \tau} A,
$$
so that
$$
BA + A^TB = \int_0^\infty [A^\top e^{A^\top\tau} C e^{A \tau} + e^{A^\top\tau} C e^{A \tau} A]d\tau
\\= \int_0^\infty \frac d{d\tau} e^{A^\top \tau} C e^{A \tau}\, d \tau
\\ = 0 - C = -C.
$$
We can see that this matrix must be positive definite because for all $x \in \Bbb C^n$, we have
$$
x^*Bx = \int_0^\infty [e^{A\tau} x]^* C [e^{A\tau} x]\,d\tau > 0,
$$
where $x^*$ denotes the conjugate-transpose of the column-vector $x$.

To establish uniqueness, it suffices to note that the Sylvester equation $BA + A^\top B = -C$ has a unique solution because the eigenvalues of $A$ have negative real part, which means that $A$ and $-A^T$ have no common eigenvalues.

$\Longleftarrow:$ Suppose that $A$ has at least one eigenvalue that does not have a negative real part, and suppose for the purpose of contradiction that $B,C$ are positive definite with
$$
BA + A^TB = -C.
$$
Let $x$ be a (possibly complex) eigenvector of $A$ associated with an eigenvalue $\lambda = a + bi$ for which $a \geq 0$. We note that $x^*Bx = \langle Bx, x \rangle = \langle x,Bx \rangle > 0$, so that
$$
\begin{align}
x^*[BA + A^TB]x &=
(Bx)^*Ax +  (Ax)^*Bx \\
& = 2 \operatorname{Re} \langle Ax, Bx\rangle
\\ & = 2 \operatorname{Re} [(a + bi)\langle x, Bx\rangle]
\\ & = 2a \langle x,Bx \rangle \geq 0.
\end{align}
$$
On the other hand, the fact that $C$ is positive definite means that
$$
x^*[BA + A^TB]x = -x^*Cx < 0,
$$
which contradicts the above.


问:
Hermitian matrix and eigenvalues
It is true that: Every eigenvalue of a Hermitian matrix is real.

But does this mean that : if all of the eigenvalues of a matrix is real, then the matrix is Hermitian?

答: NO. Take, for instance $$A=\begin{pmatrix}1&i\\0&2\end{pmatrix}$$

Title: Hermitian matrix and eigenvalues


问:
$a|p+1$ if and only if $\frac{a}{p} = \frac{1}{m}+\frac{1}{n}$
Let $p$ prime be a number and let $a$ be a natural number so that $a<p$.
prove that $a|p+1$ if and only if there are $M,N$ that are naturals so that $a/p = 1/N + 1/M$.
I have tried a lot of algebraic games to get to something like $p+1 = a\pmod{\text{something}}$ to show that $a|p+1$ but I had no success. I really don't know where to begin.

答:
(For the harder direction.)

Hint: Show that $ \frac{a}{p} = \frac{ n+m}{nm} $ implies that $p \mid nm$.

WLOG, let $ m = pk$.

Hint: Show that $a = \frac{ n + pk}{nk} $ implies that $ k \mid n$ and $ n \mid pk$.

Conclude that $ n =k $ or $ n = pk$.

Hence, either $ \frac{a}{p} = \frac{1}{k} + \frac{1}{pk} = \frac{ p+1}{pk}$ or $ \frac{a}{p} = \frac{1}{pk} + \frac{1}{pk} = \frac{2}{pk}$.

So either $ ak = p+1$ or $ a = 1, 2$, and we are done.


答:
Any real nonsymmetric matrix is not Hermitian. For example, \begin{bmatrix}
0 & 0 \\
1 & 0
\end{bmatrix}
And eigenvalues are only 0. (This does not mean any real valued matrix have only real eigenvalues, I mean we can find very simple examples instead of using complex number containing $i$)

Title: Hermitian matrix and eigenvalues

Body:
It is true that: Every eigenvalue of a Hermitian matrix is real.

But does this mean that : if all of the eigenvalues of a matrix is real, then the matrix is Hermitian?


问:
Find the maximum value of a function $f(x) = |x(x - 1)(x - 2)(x - 3)(x - 4)(x - 5)(x - 6)(x - 7)|$ where $3 \leq x \leq 4$
Find the maximum value of a function $f(x) = |x(x - 1)(x - 2)(x - 3)(x - 4)(x - 5)(x - 6)(x - 7)|$ where $3 \leq x \leq 4$.

This question is supposed to be solved without using derivatives.

答:
The function is symmetric around the center point $x=\frac{7}{2}$. Letting $x = y + \frac{7}{2}$, we have:

$$ f(x) = \left| \left(y + \frac{7}{2}\right) \left(y + \frac{5}{2}\right) \left(y + \frac{3}{2}\right) \left(y + \frac{1}{2}\right) \left(y - \frac{1}{2}\right) \left(y - \frac{3}{2}\right) \left(y - \frac{5}{2}\right) \left(y - \frac{7}{2}\right) \right| $$

$$ f(x) = \left| \left(y^2-\frac{7^2}{4}\right) \left(y^2-\frac{5^2}{4}\right) \left(y^2-\frac{3^2}{4}\right) \left(y^2-\frac{1^2}{4}\right) \right| $$

Since $3 \leq x \leq 4$, $-\frac{1}{2} \leq y \leq \frac{1}{2}$ and $y^2 \leq \frac{1}{4}$, so all four factors are non-positive.

$$ f(x) = \left(\frac{7^2}{4} - y^2\right) \left(\frac{5^2}{4} - y^2\right) \left(\frac{3^2}{4} - y^2\right) \left(\frac{1^2}{4} - y^2\right) $$

$f(x)$ clearly decreases when $y^2$ increases, so the maximum of $f$ is achieved when $y^2$ is as small as possible. The smallest possible is $y^2=0$, at $x=\frac{7}{2}$.

$$ \max_{3 \leq x \leq 4} f(x) = \frac{7^2 \cdot 5^2 \cdot 3^2}{2^8} = \frac{11025}{256} \approx 43.066 $$


答:
The maximum of the product occurs at the maximum of the logarithm of the product.  Write this out to see the maximum must occur at $x = 3.5$.

Specifically, maximize $\log x + \log (x-1) + \cdots + \log (x -7)$

Consider the sum of just the first and last terms.  How would you maximize that sum?  (Convince yourself that it would be for $x = 3.5$.)  Now consider the sum of the second and the second-to-last term.  Again, $x=3.5$.  And so on.

Put these answers together to see the result.

Body:
Find the maximum value of a function $f(x) = |x(x - 1)(x - 2)(x - 3)(x - 4)(x - 5)(x - 6)(x - 7)|$ where $3 \leq x \leq 4$.

This question is supposed to be solved without using derivatives.


问:
Rings | Homomorphisms | Units

Question

Show that if $f :R\rightarrow S$ is a homomorphism, and if $a$ is a unit of $R$, then $f(a)$ is
a unit of $S$. Show, in fact, that $f(a^{−1}) = f(a)^{−1}$ for any unit $a$ of $R$.

Attempt

Let $f:R\rightarrow S$ be a homomorphism. Then

$$ f(a)=f(a\cdot 1_R)=f(a)f(1_R),$$

so

$$ f(1_R)=\frac{f(a)}{f(a)}=1_S, $$

which is me trying to show why $(6)$ holds:

Consequently, we then have that,

$$ f(a\cdot a^{-1})=f(a)f(a^{-1})=f(1_R)=1_S, $$

so it is then easy to see that because $f(a)f(a^{-1})=1_S$, then $f(a)$ is a unit in $S$. Furthermore, by the same equality

$$ f(a^{-1})=\frac{1_S}{f(a)}=f(a)^{-1}, $$

as needed.

答: Hint: if $a\in R$ is a unity then you can write $1_R=a\cdot b$. Use the fact that $f(1_R)=1_S$ and compute $f(a\cdot b)$.

Title: Rings | Homomorphisms | Units


问:
If we take a vector out of a base and put a vector with all co-ordinates equal to $1$ on it's place it is not sure that we will get a base.
A set of vectors: $ \alpha_1$, $ \alpha_2$, ...$ \alpha_n$ is a base of linear space $V$. Vector $\beta \in V$ has all co-ordinates equal to $1$. Is sequence $ \alpha_1$, $ \alpha_2$, ...$ \beta$ also a base of linear space V?

No, it is not. We can't be sure that after we took $\alpha_n$ out of the set of vectors $ \beta$ is not still dependent on any other of them. Particularly, among vectors $ \alpha_1$, $ \alpha_2$, ...$ \alpha_{n-1}$ there could be a vector that is equal to $ \beta$. Am I right?

答:
Yes, what you wrote is correct.

It can even happens that $\{\alpha_1, \alpha_2, \dots, \alpha_{n-1}, \beta\}$ is not a basis while $\beta$ is not one of the $\alpha_1, \alpha_2, \dots, \alpha_{n-1}$.

For example take

$$\alpha_1 = \begin{pmatrix}1\\0\\ 0\end{pmatrix}, \,
\alpha_2 = \begin{pmatrix}0\\1\\ 1\end{pmatrix}, \,
\alpha_3 = \begin{pmatrix}0\\1\\ 0\end{pmatrix}, \,
\beta = \begin{pmatrix}1\\1\\ 1\end{pmatrix}
$$

$\beta \notin \{\alpha_1, \alpha_2, \alpha_3\}$ but if you replace $\alpha_3$ by $\beta$, the new family of vectors is no more a basis.


问:
Generalized eigenvalue problem with symmetrical real matrices
I am investigating the generalized eigenvalue problem

$$(\lambda\,\boldsymbol{A}+\boldsymbol{B})\,\boldsymbol{x}=\boldsymbol{0}$$

where $\boldsymbol{A}$ and $\boldsymbol{B}$ are real-valued symmetrical matrices, $\lambda$ are the eigenvalues and $\boldsymbol{x}$ are the eigenvectors.

According to Wikipedia, the eigenvalues $\lambda$ are all real-valued if $\boldsymbol{A}$ is positiv definite. Is there a simple proof for that?

What happens if $\boldsymbol{A}$ is only positiv semidefinite? Are the eigenvalues are still all real-valued?

I assume, from a mechanical point of view, that for positiv definite matrices $\boldsymbol{A}$ and $\boldsymbol{B}$, the eigenvalues are all negative real-valued. Is there a simple proof for that statement? And again, what happens if the matrices (or one of them) is only positiv semidefinite?

答:
Regarding the specific question about positive definite. Suppose $A$ is positive definite symmetrical matrix. So $x^TAx$ is positive for all non zero vector $x$.

For symmetrical matrices eigenvectors constitute orthogonal basis. Say here the eigen vectors are $x_1,x_2,,,x_n$, we can represent $x$ as $x = a_1x_1+a_2x_2+...+a_nx_n$.

So $Ax = a_1\lambda_1x_1+a_2\lambda_2x_2+...+a_n\lambda_nx_n$. Now as the vectors are orthogonal and $x^TAx$ is the dot product of $x$ and $Ax$. Assume all $\lambda_i$ are positive, so here we get

$x^TAx=  {a_1}^2\lambda_1{(x_1.x_1)}+{a_2}^2\lambda_2{(x_2.x_2)}+...+{a_n}^2\lambda_n{(x_n.x_n)}$

Note that this is always positive as not all $a_i$ are zero and all$\lambda_i$ are positive. So A is positive definite.

Now say $A$ is positive definite so for any non zero vector $x$, $x^TAx$ is positive. We can put eigenvactors in place of $x$ and get $x^TAx = x^T\lambda x = \lambda x^Tx$

Now the sign of this completely depends on $\lambda$ and as $x^TAx$ is positive, $\lambda$ has to be positive (true for all eigenvalues).

edit:

So 1) $x^TAx$ is positive for all non zero vector $x$

2) All eigenvalues of $A$ are positive

Both are equivalent statement to define a positive definite matrix.


问:
Another second order PDE in canonical form
Once again I have to solve a PDE:

$e^{2y} u_{xx} + u_y = u_{yy}$

I have found this is hyperbolic, with canonical form:

$u_{\phi\psi}=\frac{1}{\phi-\psi} u_\psi$

I think this is how to do it: let $z=u_\psi$

Then $z_\phi = \frac{1}{\phi-\psi} z$. This is separable if we treat $\psi$ as a constantso we get $u_\psi=A(\phi-\psi)$ where A is some constant. Then we get $u=\frac{-A\psi^2}{2}+A\psi\phi+B$ for some constants A and B.

My question is, is this allowed? Can I treat $\phi$ as a constant in the first part of this calculation?
Thanks in advance.

答:
$$u_{\phi\psi}=\frac{1}{\phi-\psi} u_\psi$$
Yes substitute $z=u_{\psi}$:
$$(\phi-\psi)z_{\phi}=z$$
$$(\phi-\psi)z_{\phi}-z=0$$
$$\left ( \dfrac {z}{\phi-\psi} \right)'=0$$
$$ \dfrac {z}{\phi-\psi}=C(\psi)$$
Note that $C$ is not a constant but a function of $\psi$:
$$  {u}_{\psi}=C(\psi)(\phi-\psi)$$
Integrating by part gives:
$$  {u}{(\psi,\phi)}=g(\psi)(\phi-\psi)+\int g(\psi)d\psi+f(\phi)$$


答:
Graphical Comment. The function crosses the x-axis at $0, 1, 2, \dots, 7.$ So it seems that the maximum for $x$ in $(3,4)$ must be about halfway between $3$ and $4.$

```<code>curve(abs(x*(x-1)*(x-2)*(x-3)*(x-4)*(x-5)*(x-6)*(x-7)), 3, 4, ylab="y", main="")
</code>```

A numerical search shows that the maximum $y= 43.0664$ is very nearly at $x = 3.5.$

```<code>x= seq(3,4,by=.001)
y = abs(x*(x-1)*(x-2)*(x-3)*(x-4)*(x-5)*(x-6)*(x-7))
x[y==max(y)]
[1] 3.5
max(y)
[1] 43.06641
</code>```

Computations in R.

Title: Find the maximum value of a function $f(x) = |x(x - 1)(x - 2)(x - 3)(x - 4)(x - 5)(x - 6)(x - 7)|$ where $3 \leq x \leq 4$

Body:
Graphical Comment. The function crosses the x-axis at $0, 1, 2, \dots, 7.$ So it seems that the maximum for $x$ in $(3,4)$ must be about halfway between $3$ and $4.$

curve(abs(x*(x-1)*(x-2)*(x-3)*(x-4)*(x-5)*(x-6)*(x-7)), 3, 4, ylab="y", main="")

A numerical search shows that the maximum $y= 43.0664$ is very nearly at $x = 3.5.$

x= seq(3,4,by=.001)
y = abs(x*(x-1)*(x-2)*(x-3)*(x-4)*(x-5)*(x-6)*(x-7))
x[y==max(y)]
[1] 3.5
max(y)
[1] 43.06641

Computations in R.


问:
Show that the following set has measure 0
Let $g:[a,b] \rightarrow \mathbb{R}$ a strictly increasing and absolutly continous function with $g(a)=c$ and $g(b)=d$. Let $H=\{x: g'(x) \neq 0\}$ and $E \subseteq [c,d]$ with $|E|=0$. Show that $|g^{-1}(E) \cap H|=0$.

I want to check if this is fine or not

The previous exercise was to show that if $G \subseteq[c,d]$ was open then $\int_{g^{-1}(G)}g'(x)dx=|G|$ and I've done it.

So, what I was thinking is that given some $\epsilon > 0$ I can take a set $G=\cup_{n\in \mathbb{N}}G_n$ with $G_n=(c_n, d_n)$ such that $E \subseteq G$ and $|G| < \epsilon$, so, since $g$ is absolutely continous and biyective (because is strictly incresing) then exists a $\delta_\epsilon >0$ such that $\sum g^{-1}(d_n)-g^{-1}(c_n)<\delta_\epsilon$ and $\delta_\epsilon \rightarrow 0$ when $\epsilon \rightarrow 0$ because is a strictly incresing function.

Is this ok? I'm bothered by the fact that I don't use the set $H$ at all and a little bit because I don't use the previous part either.

Thanks

答:
What your wrote at your argument its not always true, even if both $g$ and $g^{-1}$ are absolutely continuous functions. For example take $g(x)=x^2$ defined on the interval $[0,1]$. Then both $g$ and $g^{-1}$ are strictly increasing, hence both are absolutely continuous functions on $[0,1]$. Now consider the intervals $$[\alpha_n,\,\beta_n]=\biggl[\frac{1}{2n},\,\frac{1}{n}\biggr]$$

Then, $$\sum_{n=1}^{N}f(\beta_n)-f(\alpha_n)=\sum_{n=1}^{N}f(\frac{1}{n})-f(\frac{1}{2n})=\sum_{n=1}^{N}\frac{1}{n^2}-\frac{1}{4n^2}\sim \sum_{n=1}^{N}\frac{1}{n^2}$$

Hence, the tails, $\sum_{k=N}^{\infty}f(\beta_k)-f(\alpha_k)\sim \sum_{k=N}^{\infty}\frac{1}{k^2}\to 0$ as $N\to \infty$. But on the other hand,
$$\sum_{k=N}^{\infty}\beta_n-\alpha_n=\sum_{k=N}^{\infty}\frac{1}{n}-\frac{1}{2n}=\sum_{k=N}^{\infty}\frac{1}{2n}=\infty$$
Which means that for every $\epsilon>0$ you can find large $N$ such that
$$\sum_{k=N}^{\infty}f(\beta_n)-f(\alpha_n)<\epsilon$$
but you always have $\sum_{k=N}^{\infty}(\beta_n-\alpha_n)=\infty$. Now, to answer your question since you have proved that
$$\tag{1} \int_{g^{-1}(G)}g'(x)\,dx=|G|$$
you can use this to prove that $|g^{-1}(E)\cap H|=0$. First observe that since $g$ is strictly increasing, then $g'(x)\geq 0$ almost everywhere. Using $(1)$ for $G=E$ we get
$$\int_{g^{-1}(E)}g'(x)\,dx=|E|=0$$
On $H^c$ we have $g'(x)=0$, hence
\begin{align}
\int_{g^{-1}(E)}g'(x)\,dx&=\int_{g^{-1}(E)\cap H}g'(x)\,dx+\int_{g^{-1}(E)\cap H^c}g'(x)\,dx\\
&=\int_{g^{-1}(E)\cap H}g'(x)\,dx=0
\end{align}
Now since on $g^{-1}(E)\cap H$ we have that $g'(x)>0$ we deduce that $|g^{-1}(E)\cap H|=0$.


问:
Does $f(x,y)=x^4+y^4$ have an extremum at $(0,0)$?
Let us consider the following function
$$f(x,y)=x^4+y^4.$$

I want to find the local extrema of $f$.

We can verify that
$$(\frac{\partial f}{\partial x}(x,y),\frac{\partial f}{\partial y}(x,y) )=(0,0)\Longleftrightarrow (x,y)=(0,0)$$

Further the hessian matrix of $f$ is given by
$$H_f(x,y)=\begin{pmatrix}
12x^2 &0\\
0 &12y^2
\end{pmatrix}.$$
So the determinant of the matrix $H_f(0,0)$ is equal to $0$.

答:
$f(x,y)=x^4+y^4 \ge 0=f(0,0)$ for all $(x,y)\in\Bbb R^2$.

So YES, it is a minimum (in fact, at $(0,0)$ the function attains its global minimum).

MOREOVER

For all $n\in\Bbb N$, the same argument says that $f_n(x):=x^{2n}+y^{2n}$ as a (global) minimum at $(0,0)$.

For odd exponents, namely, for functions $g_n(x,y):=x^{2n+1}+y^{2n+1}$, the poinbt $(0,0)$ is a Saddle point: Given $\varepsilon>0$,
$$g_n(\varepsilon,0)=\varepsilon^{2n+1}>0>-\varepsilon^{2n+1}=g_n(-\varepsilon,0)$$


答:
Interestingly, even if the determinant is $0$ at $(x,y) = (0,0)$ we can say some things. The matrix is positive semidefinite at this point but is positive definite at any point in a neighborhood of $(0,0)$. Hence it's a local minimum.

However if you want to prove that it's a global minimum, you should consider Tito Eliatron's answer.

Title: Does $f(x,y)=x^4+y^4$ have an extremum at $(0,0)$?

Body:
Let us consider the following function
$$f(x,y)=x^4+y^4.$$

I want to find the local extrema of $f$.

We can verify that
$$(\frac{\partial f}{\partial x}(x,y),\frac{\partial f}{\partial y}(x,y) )=(0,0)\Longleftrightarrow (x,y)=(0,0)$$

Further the hessian matrix of $f$ is given by
$$H_f(x,y)=\begin{pmatrix}
12x^2 &0\\
0 &12y^2
\end{pmatrix}.$$
So the determinant of the matrix $H_f(0,0)$ is equal to $0$.


问:
Intuition for Algebraic Value of Covariant Derivative
Reading through Carmo's Differential Geometry of Curves and Surfaces, the following statement is made on an algebraic value for the covariant derivative (pg. 252, section 4-4):

Proposition 3: Let $x(u,v)$ be an orthogonal parametrization (that is, $F = 0$), of a neighborhood
of an oriented surface $S$, and $w(t)$ be a differentiable field of unit vectors along the curve
$x(u(t), v(t))$. Then,
$$[\frac{Dw}{dt}] = \frac{1}{2\sqrt{EG}}\{G_{u}\frac{dv}{dt} - E_{v}\frac{du}{dt}\} + \frac{d \varphi}{dt},$$
where $\varphi(t)$ is the angle from $x_{u}$ to $w(t)$ in the given orientation

The proof of this statement is given in the text. My question is: what does this mean? My linear algebra is a bit rusty, so I'm not sure if that's the intuition I'm missing, but is there any extra intuition, geometric or otherwise, to be taken away from this expression of the covariant derivative (regarding the fundamental form and what it means for the surface), or is it just a tidy way of expressing the derivative that will make computation for other proofs more manageable? It certainly achieves the latter part, but I feel like I'm missing some important details on the former, more intuitive part.

Any input is much appreciated.

答: You can read more about the intuition in my (free) differential geometry text (linked in my profile).  It's a superposition principle: The first term is telling you at what rate the frame $\{x_u/\sqrt E,x_v/\sqrt G\}$ is rotating, and you add to that $d\phi/dt$, which tells you at what rate your unit-length vector field $w$ is rotating relative to the frame.


问:
How valid is this concept or does this already have a name?
I was going through my school papers and found an interesting question, so I experimented a bit more and found out a pattern, so I made a formula for such matrices.

$$A = \begin{bmatrix}x&-(x-1)\\x+1&-x\end{bmatrix}$$ where $x > 0$ is an integer

$$A^n =
\begin{cases}
I,  & \text{if $n$ is even} \\
A, & \text{if $n$ is odd}
\end{cases}$$
where $I$ is identity matrix of order 2.

I just wanted to know if this has been found before or whether it has a name too or if there are some cases that does not obey this.

Hope someone can format my question properly, I'm new to this community. Hope this is the correct way of putting things together too. Thanks!!

答:
You simply have $A^2=I$, from which it follows that $A^3=AA^2=AI=A$, $A^4= AA^3=AA=A^2=I$ and so on.  Of course $A^2=I$ is the same as saying that $A=A^{-1}$ so that $A$ is its own inverse.

This is sometimes stated as $A$ is self-inverse, or that $A$ is an involution, in more highfalutin language.


答:
Thank you for addressing our comments and clarifying your question!  Any square matrix which satisfies your equation $$A^n =
\begin{cases}
I,  & \text{if $n$ is even} \\
A, & \text{if $n$ is odd}
\end{cases}$$
is its own inverse, i.e. it is an involutory matrix.  This is because if $n = 2$ you have $AA = I$, which by definition means that $A$ is its own inverse.  From that simple statement you can extrapolate your formula, because if $n$ is even then $$A^{n} = A^{2m} = (AA)^{m} = I^{m} = I$$ for some integer $m$, and if $n$ is odd then $$A^{n} = A^{2k+1}=A^{2k}A=(AA)^{k}A = I^{k}A=IA=A$$ for some integer $k$.  In fact, for any $2\times 2$ matrix $$\begin{pmatrix} a & b\\c & -a\end{pmatrix},$$ of which your matrix is an example, such a matrix will be involutory if $a^{2} + bc = 1$.  We can verify this for your matrix $A$: $$x^{2} - (x-1)(x+1) = 1.$$

Title: How valid is this concept or does this already have a name?

Body:
I was going through my school papers and found an interesting question, so I experimented a bit more and found out a pattern, so I made a formula for such matrices.

$$A = \begin{bmatrix}x&-(x-1)\\x+1&-x\end{bmatrix}$$ where $x > 0$ is an integer

$$A^n =
\begin{cases}
I,  & \text{if $n$ is even} \\
A, & \text{if $n$ is odd}
\end{cases}$$
where $I$ is identity matrix of order 2.

I just wanted to know if this has been found before or whether it has a name too or if there are some cases that does not obey this.

Hope someone can format my question properly, I'm new to this community. Hope this is the correct way of putting things together too. Thanks!!


答:
Letting $V=\partial_{\eta} U$ we get
$$\partial_{\xi}V=\frac{1}{\xi-\eta}V$$
This is a straightforward application of characteristics to this semilinear PDE.
$$\begin{bmatrix}
\partial_{t}[ \xi ](t,s)\\
\partial_{t}[ \eta ](t,s)\\
\partial_{t}[ \zeta ](t,s)
\end{bmatrix} =\begin{bmatrix}
1\\
0\\
\zeta\cdot( \xi -\eta )^{-1}
\end{bmatrix}$$
So $\xi(t,s)=t+f_1(s)$, $\eta(t,s)=f_2(s)$ and
$$\zeta(t,s)=V(\xi(t,s),\eta(t,s))=f_3(s)(f_1(s)-f_2(s)+t)=f_3(s)(\xi(t,s)-\eta(t,s))$$
From $\eta(t,s)=f_2(s)$ we can write $s=f_2^{-1}(\eta)$

Hence $V(\xi,\eta)=f_4(\eta)\cdot(\xi-\eta)$, where $f_4=f_3\circ f_2^{-1}$. This can be easily verified. Then,
$$V=\partial_{\eta}U=f_4(\eta)\cdot(\xi-\eta)$$
We can now directly integrate.
$$u(x,y)=U(\xi,\eta)=\int f(\eta)\cdot (\xi-\eta)\mathrm{d}\eta+g(\eta)$$
Which can be integrated by parts from @Aryadeva 's answer. Substitute whatever $\xi,\eta$ are in terms of $x,y$ and you're done.

Title: Another second order PDE in canonical form

Body:
Once again I have to solve a PDE:

$e^{2y} u_{xx} + u_y = u_{yy}$

I have found this is hyperbolic, with canonical form:

$u_{\phi\psi}=\frac{1}{\phi-\psi} u_\psi$

I think this is how to do it: let $z=u_\psi$

Then $z_\phi = \frac{1}{\phi-\psi} z$. This is separable if we treat $\psi$ as a constantso we get $u_\psi=A(\phi-\psi)$ where A is some constant. Then we get $u=\frac{-A\psi^2}{2}+A\psi\phi+B$ for some constants A and B.

My question is, is this allowed? Can I treat $\phi$ as a constant in the first part of this calculation?
Thanks in advance.


问:
Discrete differintegral using Fourier transform?
Using Fourier transform we can give a formula for a differintegral:

$$f^{(a)}(x)=\frac{1}{2\pi}\int_{-\infty}^{+\infty} e^{- i \omega x}{(-i\omega)}^{a} \int_{-\infty}^{+\infty}f(t)e^{i\omega t}dt \, d\omega$$

I wonder, is there a similar formula for a discrete differintegral (finite difference/indefinite sum)?

答:
Well,

$$(e^D-1)^a f(x)=\frac{1}{2\pi}\int_{-\infty}^{+\infty} e^{- i \omega x}(e^{-i\omega}-1)^a \int_{-\infty}^{+\infty}f(t)e^{i\omega t}dt \, d\omega$$


问:
the difference between chains and cochains
I have heard the following definitions, for concreteness I will refer a simplicial complex $K$ and a ring $R$, though the definitions can be extended:

A chain is a formal linear combination of simplices in $K$ with coefficients in $R$.

A cochain is a function on simplices, i.e. a map $K \to R$.

I understand that chains and cochains are dual to each other, but I feel that I do not fully understand how the above two statements meaningfully differ.

The way I am currently thinking of it is something like the following. A $p$-chain is a formal linear combination of $p$-simplices, so I can write that as something like this:
$$\vert f_p \rangle = \sum_{i \in I^p} c_i \vert \sigma^p_i \rangle$$
where $I^p$ is an index set for the $p$-simplices in $K$, $\vert \sigma^p_i \rangle$ is a $p$-simplex indexed by $i$, and $c_i \in R$. Then cochains are the "dual vectors". This is where things a little hazy for me, I would assume we can define "co-simplexes" $\langle \sigma^p_i \vert$ as the basis for the cochains via
$$\langle \sigma^p_i \vert \sigma^p_j \rangle = \delta_{ij}$$
where $\delta_{ij}$ is the Kronecker delta. Then the cochains are the dual vectors to the chains
$$\langle {g_p} \vert = \sum_{i \in I^p} \langle \sigma_i^p \vert c_i $$

Is this an accurate way to think about them? I think my confusion is what the difference is between "a function which maps a simplex to a scalar" (cochain) and "a formal linear combination of simplices" (chain), which to me is the same as assigning a scalar value to each simplex. Is there a clear distinction between these two concepts?

In other words, if a cochain is a map $K\to R$, what is a "formal linear combination of elements of $K$"?

答:
Practically, the difference is that if there are infinitely many simplices, a cochain can take nonzero values on all of them, whereas a chain must be a finite linear combination of simplices. In other words, if we think of a chain in terms of its coefficients, those coefficients, regarded as a function on simplices, must be finitely supported. We say that chains must be "compactly supported" (since each simplex is compact).

This leads to a big difference in functoriality: chains push forward and cochains pull back. This is easiest to explain for just bare sets: given a set $X$, "chains on $X$" consists of the free vector space $C_0(X) = \mathbb{R}[X]$, and "cochains on $X$" is the vector space of functions $C^0(X) = \mathbb{R}^X$. (These are in fact the zeroth homology vs. zeroth cohomology of $X$ regarded as a discrete space.) Given a map $f : X \to Y$ of sets, we get a pushforward on chains

$$f_{\ast} : C_0(X) \ni \sum_x c_x x \mapsto \sum_x c_x f(x) \in C_0(Y)$$

and a pullback on cochains

$$f^{\ast} : C^0(Y) \ni c^y \mapsto c^{f(x)} \in C^0(X).$$

Neither of these directions can be reversed in general: we don't get a pullback on chains or a pushforward on cochains (check this; you can take $X$ to be infinite and $Y$ to be a point), and the reason has to do with the finite support condition on chains. However, chains and cochains can be identified if $X$ is finite, and this is a very simple form of Poincare duality.


问:
Relationship between PDFs and derivatives
There is a well-known formula in probability theory - and by the way I m looking for the name of this formula or theorem? (Because it would be nice to be able to refer to it by a name) - :

$f_Y(y) = f_X(x)|dx/dy|$

The main question is: how to intuitively interpret the role of the derivative of $x$ with respect to $y$? Which by the way is harder to interpret given the kind of "inverse" notion $x$ with respect to $y$ instead of y with respect to x. Aslo y is a function of $x$ e.g. $y = g(x)$

I imagine it is some kind of variation in the output domain with respect to the input domain... but I can't formulate it well...

Reference to formula in Wikipedia entry for pdf function under the section "scalar to scalar".

答:
Start from the definition

$$F_Y(y)=P(Y \leq y)=F_X(g^{-1}(y))$$

Derivating you get

$$f_Y(y)=f_X(g^{-1}(y))\frac{dx}{dy}$$

That is your formula but the absolute value. This is due to the fact that this formula is the same also for $g$ decreasing.

P.S. this formula is known as "fundamental transformation Theorem "


问:
Are Taylor series and power series the same "thing"?
I was just wondering in the lingo of Mathematics, are these two "ideas" the same? I know we have Taylor series, and their specialisation the Maclaurin series, but are power series a more general concept? How does either/all of these ideas relate to generating functions?</p>

答:
Taylor series are a special type of power series. A Taylor series has a very special form, given by $$T_f(x) = \sum_{n = 0}^{\infty}\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n,$$ and a general power series looks like $$P(x) = \sum_{n = 0}^{\infty} a_n (x - x_0)^n,$$ where the $a_k$'s are just the constants associated to this power series in particular. <s>The $a_n$'s may not have the form $f^{(n)}(x_0)/n!$, so that not every power series is a Taylor series (although every Taylor series is a power series)</s>.
Edit: as Matt noted, in fact each power series is a Taylor series, but Taylor series are associated to a particular function, and if the $f$ associated to a given power series is not obvious, you will most likely see the series described as a "power series" rather than a "Taylor series."

Both of these types of series can be generalized to forms involving more variables, and you can also come up with types of series that involve negative powers of $x$.

As for generating functions, these are more formal objects, the analysis of which doesn't really deal with the issue of convergence as much as the analysis a power series or a Taylor series does. In this case, the coefficients are encoding information about some sequence of numbers $\{a_n\}$, and we examine the series formally to gather information about this sequence.


问:
commutes with a nilpotent matrix and invertible
The exercise is:

Given $V$ a vector space, $\dim V<\infty$ and $T, N\in \mathcal{L}(V)$, where $N$ is nilpotent and $TN=NT$. Prove that: $T$ is invertible iff $T+N$ is invertible. Furthermore, $\det(T)=\det(T+N)$ and $p_T(t)=p_{T+N}(t)$.

I need a hint about how can I start this kind of exercise, which involves a nilpotent matrix.

答: Hint: Note that $N$ and $T$ are simultaneously upper triangularizable.


问:
function of a function (composite function) with a function not equal to 3
Hello I have the question
functions g and h are defined by

g(x)=6/3-x ; x ≠ 3

h(x)=5x+2

I have never come across this type of question where a function has this sign in it ≠

I answered the following question not taking this into account.

A. determine function of g(h(x))

replace x with hx

6/3-(5x+2)

expand bracket -(5x+2)

=-5x-2

=3-5x-2

=-5x-1

g(h(x))=6/-5x-1

B. determine function of h(g(x))

replace x with gx

5(6/3-x)+2

expand 5x6=30

=(30/3-x)+2

h(g(x))=(30/3-x)+2

How do I incorporate the x≠3 into my answer? do I just add it onto the end of g(h(x)) answer?

答:
$x \ne 3$ indicates that the domain of $g$ does not include the value $x=3$.

Indeed, $g(x) = \dfrac {6}{3-x}$ is undefined at $x=3$.

Basically, the rest of the question follows from what you have done, but for $g(h(x))$, the function is not defined for $h(x) = 3$, i.e. $x = -\frac15$, while $h(g(x))$ is not defined for $x=3$.


问:
Changing the quality of the transient process in a nonlinear system (Part II)
My question is a continuation of the topic.

Changing the quality of the transient process in a nonlinear system (in Mathematica)

Unfortunately, last time I didn’t get any help, so I decided to reformulate the question.

I have a system of differential equations like this:

$\begin{cases} \frac{dx}{dt} = y \cdot \alpha \cdot sin(\omega t) + \frac{d}{dt}(\alpha \cdot sin(\omega t))
   \\ \frac{dy}{dt} + y = \frac{d}{dt}(f(t)) \end{cases}$

where:
$f(t)$ - Any function that has one extreme (minimum or maximum). For example $f(t) = e^{-(x - x_{*})^2}$

$x$ and $y$ - variables of the system of differential equations.

$x_{*}$ - the point at which the maximum or minimum of the function $f(t)$ is reached.

Let's pay special attention to the variable $x$.

Parameters: $\alpha = 0.3, \omega = 2 \cdot \pi \cdot 0.5, x(0)=1/4, y(0)=0$

The solution is to go from the starting point $x(0)$ to the point $x_{*}$ at which the minimum or maximum of the function $f(t)$ is reached.

For my system with such parameters, the solution will look like this:

It can be seen that the transition process is a transition from the initial point to the final one with a certain character, which is added to the additional sinusoidal signal $\alpha \cdot sin(\omega t)$.

It is necessary to change the nature of the transition from one point to another, making it exponential (naturally retaining an additional sinusoidal signal $\alpha \cdot sin(\omega t)$), i.e. it must be described by law:

$x(t)=(x(0)−x_{*}) \cdot exp(-t) + x_{*} + \alpha \cdot sin(\omega t)$,

Like this:

The problem is complicated by the fact that the function $f(t)$ and the point at which its minimum/maximum $x_{*}$ is located, generally speaking, are not known to us.

How to get out of the situation using the structure of the differential equation? Add additional input signal? Include an additional differential equation in the system? Use adaptive techniques?

Help! Please give at least some idea. I am desperate and my hands give up.

答:
Certainly adaptive control won't help either, your problem is not in the stability of the system but its performance.

Assuming that $f$ has a minimum then you could let your system be
$$
\dot{y}=-\alpha \frac{df}{dy}\\
\dot{x}+\beta x =\beta y+\gamma (\omega \cos{\omega t}+\beta\sin{\omega t})
$$

if $\alpha >> \beta$ then the response would be exponential provided that $f$ is of function of $y$, if $f$ is a function of $t$ explicitly then let $T$ be the time it takes to reach its critical point then if $\beta << \alpha <<\frac{1}{T}$ it will work. The idea behind this is to make $y$ converge to the maximum or minimum much faster than $x$.

If $f$ is dependent on $x$ then you could make $\dot{x}=\beta u'+\gamma \sin \omega t$ where $u'=sign(-\frac{df}{dx})$ for $\frac{df}{dx} \neq0$ and $u'=0$ when $\frac{df}{dx}=0$. This system will converge to $x^*$ in a linear manner when  $\gamma >> \beta$

If $f$ has a maximum then
$$
\dot{y}=\alpha \frac{df}{dy}\\
\dot{x}+\beta x =\beta y+\gamma (\omega \cos{\omega t}+\beta\sin{\omega t})
$$
or
$\dot{x}=\beta u'+\gamma \sin \omega t$ where $ u'=sign(\frac{df}{dx})$ for $\frac{df}{dx} \neq0$ and $u'=0$ when $\frac{df}{dx}=0$.


问:
How can we find the derivative of a circle if a circle is not a function?
If we consider the equation of a circle, $x^2+y^2=r^2$, then I understand that $dy/dx$ can be computed in the following way via implicit differentiation:
\begin{align}
2x + 2y\frac{dy}{dx} &= 0 \\
\frac{dy}{dx} &= -\frac{2x}{2y} = -\frac{x}{y} \, .
\end{align}
Although I feel comfortable deriving this result, I don't really understand how I should interpret it. On an intuitive level, the formula $dy/dx = -x/y$ seems to suggest that the gradient of the tangent to any given point $(x,y)$ is $-x/y$. However, since the curve $x^2+y^2=r^2$ fails the vertical line test, it doesn't look like it is even a function. Usually, $dy/dx$ can be thought of as a shorthand for
$$
\lim_{h \to 0}\frac{y(x+h)-y(x)}{h} \, .
$$
However, in this case each $x$-value maps to two $y$-values, and so the limit definition doesn't seem to apply here. So what does $dy/dx$ actually represent in this context?

答:
The curve is indeed not the graph of a function. At any point $(x,y)$ on the curve, if an open disk about that point is small enough, then that portion of the curve that is within that neighborhood is the graph of a function, and the slope of the tangent line to the graph of that function is $-x/y.$

Derivatives are local, that is the slope of a curve at a point is determined by the behavior of that curve within a small open neighborhood of the point, no matter how small it is.

Also, the concept of a tangent line to a curve is not limited to curves that are the graph of a function, so


问:
Lemma used to prove $\left|HK\right|=\frac{\left|H\right|\left|K\right|}{\left|H \cap K\right|}$
Given a group $G$ and $H,K \le G$,then :

$$\left|HK\right|=\frac{\left|H\right|\left|K\right|}{\left|H \cap K\right|}$$

Where $HK:=\left\{hk:h \in H ,k \in K\right\}$

Lemma:

For $h_1,h_2 \in H$

$$hK=h'K \iff h(H \cap K)=h'(H \cap K)$$

We have:

$$HK=\bigcup_{h \in H}hK$$

Not every such left cosets of $K$ in $H$ are distinct, on the other hand the function $\phi:hK \to K$ with $hk \mapsto k$ is a bijection, so the number of elements in $hK$ is the same as that $K$'s  I showed that the set of left cosets (equivalently right cosets) partitions the group.

By this we see that :$$\left|HK\right|=\left|\color{blue}{\text{the set consiting of all distinct left cosets }}hK\right|\left|K\right|$$

One concludes from the lemma that the number of such distinct left cosets is the same as $\left|H: H \cap K\right|$ but I don't know how such a conclusion is possible, how the lemma helps us?

It looks that $hK \ne h^{'} K$ iff $h(H \cap K) \ne h^{'}(H \cap K)$ and the order of the set of all such distinct $h(H \cap K)$ for $h \in H$  is $\left|H: H \cap K\right|$...

Also, it would be appreciated if someone gives me an example where such left cosets $hk$ are identical.

答:
The equivalence relation $(h,k)\sim (h',k')\stackrel{(def.)}{\iff} hk=h'k'$ induces a partition of $H\times K$ into equivalence classes each of cardinality $|H\cap K|$, and the quotient set $(H\times K)/\sim$ has cardinality $|HK|$. Therefore, $|H\times K|=|H||K|=|H\cap K| |HK|$, whence (if $H$ and $K$ are finite, in particular if they are subgroups of a finite group) the formula in the OP. Hereafter the details.

(Note that the formula holds irrespective of $HK$ being a subgroup.)

Let's define in $H\times K$ the equivalence relation: $(h,k)\sim (h',k')\stackrel{(def.)}{\iff} hk=h'k'$. The equivalence class of $(h,k)$ is given by:

$$[(h,k)]_\sim=\{(h',k')\in H\times K\mid h'k'=hk\} \tag 1$$

Now define the following map from any equivalence class:

\begin{alignat*}{1}
  f_{(h,k)}:[(h,k)]_\sim &\longrightarrow&  H\cap K \\
  (h',k')&\longmapsto& f_{(h,k)}((h',k')):=k'k^{-1} \\
\tag 2
 \end{alignat*}

Note that $k'k^{-1}\in K$ by closure of $K$, and $k'k^{-1}\in H$ because $k'k^{-1}=h'^{-1}h$ (being $(h',k')\in [(h,k)]_\sim$) and by closure of $H$. Therefore, indeed $k'k^{-1}\in H\cap K$.

Lemma 1. $f_{(h,k)}$ is bijective.

Proof.

\begin{alignat}{2}
f_{(h,k)}((h',k'))=f_{(h,k)}((h'',k'')) &\space\space\space\Longrightarrow &&k'k^{-1}=k''k^{-1} \\
&\space\space\space\Longrightarrow &&k'=k'' \\
&\stackrel{h'k'=h''k''}{\Longrightarrow} &&h'=h'' \\
&\space\space\space\Longrightarrow &&(h',k')=(h'',k'') \\
\end{alignat}

and the map is injective. Then, for every $a\in H\cap K$, we  get $ak\in K$ and $a=f_{(h,k)}((h',ak))$, and the map is surjective. $\space\space\Box$

Now define the following map from the quotient set:

\begin{alignat}{1}
  f:(H\times K)/\sim &\longrightarrow&  HK \\
  [(h,k)]_\sim &\longmapsto& f([(h,k)]_\sim):=hk \\
\tag 3
 \end{alignat}

Lemma 2. $f$ is well-defined and bijective.

Proof.

Good definition: $(h',k')\in [(h,k)]_\sim \Rightarrow f([(h',k')]_\sim)=h'k'=hk=f([(h,k)]_\sim)$;

Injectivity: $f([(h',k')]_\sim)=f([(h,k)]_\sim) \Rightarrow h'k'=hk \Rightarrow (h',k')\in [(h,k)]_\sim \Rightarrow [(h',k')]_\sim=[(h,k)]_\sim$;

Surjectivity: for every $ab\in HK$ , we get $ab=f([(a,b)]_\sim)$. $\space\space\Box$

Finally, the formula holds irrespective of $HK$ being a subgroup, which was never used in the proof.


问:
length of common cycle in two k-connected subgraphs of a graph
In a given graph $G$, $G_1$ and $G_2$ are maximal k-connected subgraphs, I have to prove that the length of the common cycle is atmost $K$ i.e. they have atmost $k-1$ common nodes.

I am unsure on how to get started in the proof, any help and related links are appreciated.

答:
I would try proving this by contradiction. Suppose $V(G_1)\cap
V(G_2) \geq k$. Then I would show that $G_1\cup G_2$ is $k$-connected. This would contradict the maximality of $G_1$ and $G_2$. I would look at the proof that any two distinct
blocks in a graph contain at most one common vertex for some more help.


答:
You have looked for the critical points of $f$. There is only $(0,0)$, and the Hessian there is $0$. This means that you have to study qualitatively what happens at $(0,0)$; it does not mean that things cannot be decided.

We have $f(0,0)=0$, and when $x\ne0$ or $y\ne0$ then $f(x,y)>0$. This already proves that $f$ has a global (a fortiori: a local) minimum at $(0,0)$. Furthermore
$$\nabla f(x,y)=(4x^3,4y^3)\ne(0,0)\qquad\bigl((x,y)\ne(0,0)\bigr)$$
shows that $f$ has no other local minima or maxima in ${\mathbb R}^2$.


答:
A power series is just a series whose terms are monomials in some number of variables, such as
$$
\sum_{n=0}^\infty a_n x^n
\qquad \text{or} \qquad
\sum_{m,n=0}^\infty a_{m,n} x^my^n.
$$
These are sometimes formal algebraic objects that encode sequences in their coefficients.

In my experience, the term Taylor series is used when the power series is built from a function.  In this context, issues of convergence are central.

Title: Are Taylor series and power series the same "thing"?

Body: I was just wondering in the lingo of Mathematics, are these two "ideas" the same? I know we have Taylor series, and their specialisation the Maclaurin series, but are power series a more general concept? How does either/all of these ideas relate to generating functions?</p>


问:
Help with a proof on subgroups
I'm trying to prove that if ($\pmb H$, *) and ($\pmb K$, *) are subgroups of ($\pmb G$, *) then ($\pmb H \cup \pmb K$, *) is a subgroup of ($\pmb G$, *) iff ($\pmb K$, *) is a subgroup of ($\pmb H$, *) (or the other way around).

Here is my attempt:

Pick any 4 elements $x_1,x_2 \in \pmb H \setminus \pmb K \neq \emptyset$ and $y_1,y_2 \in \pmb K \setminus \pmb H \neq \emptyset$ and assume
$$x_1 *y_1 =x_2$$
this means that
$$y_1=x_1^{-1} * x_2$$
but because ($\pmb H$, *) is closed we now know that
$$y_1 \in \pmb H$$
but that contradicts with
$$y_1 \in \pmb K \setminus \pmb H$$
We can see that this argument is symmetric (i.e. it gives the same result if we consider $x_1 * y_1=y_2$) so we can conclude that if $x \in \pmb H \setminus \pmb K$ and $y \in \pmb K \setminus \pmb H$ then
$$x*y \in (\pmb H \cap \pmb K)$$

This is where I get stuck. I know that I have to show that no elements can satisfy the last equation so I think that all $x*y \not \in(\pmb H \cup \pmb K)$. Like how in $(3\Bbb Z, +)$ and $(5\Bbb Z, +)$ $3 + 5 = 8 \not \in (3\Bbb Z \cup 5\Bbb Z)$.

Maybe this is a dead end but I'm not sure. I also don't want to just look up the awnser in the book.

Any hints on what I need to do?

For context: Im a freshman applied math major but because I like pure math I'm doing some self studying. I'm self studying Dan Sacrino's a first course in Abstract Algebra.

Thanks in advance

答:
$\Leftarrow:$ If we have $K \le H$ or $H \le K$, then $H \cup K = H$ or $H \cup K = K$ so the left implication is clear.

$\Rightarrow:$ Now, suppose $H \cup K \le G$ but $H \nleq K$ or $K \nleq H$. Then, we have $h \in H \backslash K$ and $k \in K \backslash H$. Now, since $H \cup K \le G$ and $hk \in H \cup K$, we must have $hk \in H$ or $hk \in K$. If $hk \in H$, then $hk = h'$ for some $h' \in H$. Thus, $k = h^{-1}h' \in H$ since $H$ is a group, a contradiction. Similarly, $hk \in K$ gives a contradiction so the right implication follows.

Title: Help with a proof on subgroups