问:
Find the limit $\lim_{x\to 0}\frac{\sqrt[5]{1+\sin x}-1}{\ln(1+\tan x)}$
Find the limit $\lim_{x\to 0}\frac{\sqrt[5]{1+\sin x}-1}{\ln(1+\tan x)}$.

I tried subtracting $1$ for using $\lim$ of $e$ but I realized that this is not a problem that can be solved in that way.

Using only algebra, famous limits without L'Hôpital.

答:
Hint:

$$\lim_{x\to0}\dfrac{\sqrt[n]{1+\sin x}-1}{\ln(1+\tan x)}=\lim_{x\to0}\dfrac{\sqrt[n]{1+\sin x}-1}{\sin x}\cdot\lim_{x\to0}\dfrac{\sin x}{\tan x}\cdot\dfrac1{\lim_{x\to0}\dfrac{\ln(1+\tan x)}{\tan x}}$$

For the first limit, set $$\sqrt[n]{1+\sin x}-1=y\implies\sin x=(1+y)^n-1$$

$$\lim_{x\to0}\dfrac{\sqrt[n]{1+\sin x}-1}{\sin x}=\lim_{y\to0}\dfrac y{(1+y)^n-1}=\lim_{y\to0}\dfrac y{ny+O(y^2)}=\cdots=\dfrac1n$$


问:
Prove that a certain sum of binomial coefficients is divisible by a power of 2
As many will recognize the following expression is a closed form for the Fibonacci numbers. Can it be proved that

$$\frac{1}{2^{n-1}} \sum_{j=0}^{\lfloor n/2 \rfloor} \binom{n}{2j+1} 5^{j} \quad \text{where} \quad n \in\mathbb{N}$$
are integers without using the fact that these are the Fibonacci numbers?

答:
Hint:

$$S_n=\frac{1}{2^{n-1}} \sum_{j=0}^{\lfloor n/2 \rfloor} \binom{n}{2j+1} (\sqrt{5})^{2j} \quad \text{where} \quad n \in\mathbb{N}$$
Use ${n \choose k}={n-1 \choose k}+{n-1 \choose k-1}$
$$\implies S_n=2^{n-1} \sum_{j=0}^{[n/2]}\left({n-1 \choose 2j} (\sqrt{5})^{2j}+{n-1 \choose 2j-1} \sqrt{5}^{2j}\right)$$
Next, these two parts can yield $S_n=S_{n-2}+S_{n-1}$ sfter interesting adjustments.

I hope to come back.

ContentLicense: CC BY-SA 4.0


问:
Double Integrals
$(a)$ Sketch the region of integration in the integral

$$\int_{y=-2}^{2} \int_{x=0}^{\sqrt{4-y^2}} x e^{{(4-x^{2})}^{3/2}} dx dy$$

By changing the order of integration, or otherwise, evaluate the integral.

$(b)$ Let $R$ be the region in the $x-y$ plane defined by $0 \leq x \leq y \leq 2x$, $1 \leq x+2y \leq 4$. Evaluate: $$\mathop{\int\int}_{R} \frac{1}{x} dx dy$$

I understand how to draw these but I am not sure how to caluculate the limts in either case (especially part $b$).

Can someone explain how we calculate the limits for integration? Once I know that I am sure I can integrate the function myself. Thanks!!

答: The first thing, as stated, is to sketch the region.  Only then calculate the limits of integration.  So, what does your sketch look like?</p>

ContentLicense: CC BY-SA 3.0

Title: Double Integrals


问:
Number Of Solutions $X^{2}=X$
The equation $x^{2}=x$ $,x\in \mathbb{R}$ has only the solutions $x=0$ and $x=1$.

For a $2 \times 2$ matrix X, how many solutions does $X^{2}=X$ have?

What if $X$ is symmetric?

答:
Compute the square of a symmetric matrix:
$$\begin{bmatrix}
a & b\\
b & d
\end{bmatrix}^2=
\begin{bmatrix}
a^2 + b^2 & ab + bd \\
ab + bd   & b^2 + d^2
\end{bmatrix}
$$
Now impose the conditions
$$\begin{cases}
a^2 + b^2 = a\\
(a+d)b = b\\
b^2 + d^2 = d
\end{cases}$$
What can you say?

What if you do the same with a non symmetric matrix?

Title: Number Of Solutions $X^{2}=X$


问:
Nonlinear System of ODES
I have been trying to solve this nonlinear system of ODEs analytically with no luck:

$$\frac{d\psi}{dt}=-cot \theta cos \psi$$

$$\frac{d\theta}{dt}=-sin\psi$$

$$\frac{d\phi}{dt}=\frac{cos\psi}{sin\theta}$$

Where of course $\psi$, $\theta$ and $\phi$ are functions of $t$.

Any hints would be appreciated!

答:
Assuming that $\psi$ can be written in some interval as a function of $\theta$, one can write
$$
\frac{d\psi}{d\theta}=\frac{d\psi}{dt}\Big/\frac{d\theta}{dt}=\cot \theta \cot \psi.
$$
This gives
$$
\frac{d}{d\theta}\log\cos\psi=-\frac1{\cot\psi}\frac{d\psi}{d\theta}=-\cot\theta
$$
and so
$$
\log\cos\psi=-\log\sin\theta+C\quad\Leftrightarrow\quad
\cos\psi=\frac1{\sin\theta}e^C.
$$
Now you can potentially integrate explicitly all the equations. For example, the first equation becomes
$$
\frac{d\psi}{dt}=-\cot \theta \cos \psi=-e^{-C}\sqrt{1-e^C/\cos^2\psi}\cos^2\psi.
$$

Title: Nonlinear System of ODES


问:
Prove that $\angle{BEC}=\angle{DAC}$.
A convex quadrilateral $\quad{ABCD}$ has $AD=CD$ and$\angle{DAB}=\angle{ABC}<90$. The line through $D$ and the midpoint of $BC$ meets line $AB$ in $E$. Prove that $\angle{BEC}=\angle{DAC}$.

I have to approaches: Either we can prove that $EB.EA=EC^2$ or we can prove that line $AD$ is the tangent to the circumcircle of $\triangle{ACE}$ as shown, but I am not getting how to prove them. Pure geometrical method is preferred.

答:
I have an Angle Chasing based approach to this problem. It is as follows;

Let $M$ be the midpoint of Line Segment $\overline{BC}$, we construct a line parallel to $BC$ through $D$, and let it intersect $AB$ and $EC$ at $F$ and $G$.

We know that $\angle AFD=\angle ABC=\angle BAD=\angle FAD$. Hence, $\triangle ADF$ is isosceles, and we must have $DA=DF$. We let $DA=DF=DC=x$.

We know by the Basic Proportionality Theorem that;
$$\frac{FD}{DG}=\frac{BM}{MC}=1$$
$$\implies FD=DG.$$

Since $FD=x$, we must have $DG=x$, as well. Now, we know that $DA=DF=DC=DF=x$, and thus $AFCG$ is a lies on a circle centered at $D.$ Hence, $AFCG$ is a Cyclic Quadrilateral.

Consider $\angle ACG$ and $\angle AFG$, since the two angles are inscribed in the same arc ${AG}$  of circle $AFCG,$ we know that $\angle ACG=\angle AFG.$

Hence, we know that;
$$180^{\circ}-\angle ACG = 180^{\circ}-\angle AFG$$
$$\implies \angle ACE= \angle GFE=\angle CBE$$
$$\implies \angle ACE=\angle CBE.$$

Consider triangles $\triangle ACE$ and $\triangle CBE,$ we know that they share an angle at $E,$ and $\angle ACE=\angle CBE.$ Hence, the two triangles are similar by $\text{AA Similarity}$ and we thus have;
$$\frac{EB}{EC}=\frac{EC}{EA}$$
$$\implies EB\cdot EA=EC^2,$$
which is what you needed. But just to continue, we know by similarity that $\angle BCE=\angle CAE,$ let this angle be $y.$

We know that $\angle BAD=\angle ABC,$ but since $\angle BAD=\angle BAC+\angle CAD,$ and $\angle ABC=\angle BCE+\angle BEC,$ we must have;
$$\angle BAC+\angle CAD=\angle BCE+\angle BEC$$
$$\implies y+\angle CAD=y+\angle BEC$$
$$\implies \angle CAD=\angle BEC$$
as required.

Title: Prove that $\angle{BEC}=\angle{DAC}$.


问:
Homotopy-coherent replacement of diagrams in quasi-categories
In Emily Riehl's book Categorical Homotopy Theory, a proposition (16.3.1) attributed to Cordier-Porter states that if $\underline{\mathcal{C}}$ is a fibrant simplicial category, and if $F : \mathcal{A} \to \underline{\mathcal{C}}$ is a so-called homotopy coherent diagram, then given a family of equivalences $Fa \to Ga$ in $\underline{\mathcal{C}}$, then there is some way to extend all of this data to a new homotopy coherent diagram $G : \mathcal{A} \to \underline{\mathcal{C}}$, as well as a (homotopy-coherent) natural transformation $F \Rightarrow G$ extending the family of maps given. There is a similar statement (16.3.2) for modyfiyinh natural transformations.

I am looking for similar statements in the setting of $(\infty,1)$-categories à la Joyal-Lurie. I guess the kind of statement I am looking for would be that given a simplicial set $K$, an $\infty$-category $\mathcal{C}$, a functor $F : K \to \mathcal{C}$ and given a family of equivalences $Fa \to Ga$ for all $0$-simplex $a$ of $K$, then there is a functor $G : K \to \mathcal{C}$ as well as a natural transformation $F \Rightarrow G$ extending this data.

I have never seen in the litterature such a kind of statement in this setting so that I fear that it might be "too good to be true", and maybe there needs to be some additional assumptions to be made on $K$ or on $F$, I'm fine with this.

Given the equivalence between quasi-categories and simplicial categories, I guess the statement holds if $F$ arises as the homotopy-coherent nerve of some homotopy-coherent diagram. This pdf seems to suggest (page 10) that this would hold for instance if $K$ is the nerve of a $1$-category. It would be good to see it stated precisely somewhere.

Any reference for that kind of statement (maybe with additionnal assumptions on K) would be appreciated.

答:
It is true.

For every simplicial set $K$, the exponential object $[K, \mathcal{C}]$ is a quasicategory if $\mathcal{C}$ is.
Let $K_0$ be the set of vertices of $K$, considered as a discrete simplicial set.
The inclusion $K_0 \hookrightarrow K$ is a monomorphism, so the induced functor $[K, \mathcal{C}] \to [K_0, \mathcal{C}]$ is an isofibration of quasicategories.
But an equivalence in $[K_0, \mathcal{C}]$ is precisely a $K_0$-indexed family of equivalences, so this says that for any functor $F : K \to \mathcal{C}$ and any $K_0$-indexed family of equivalences $F a \to G a$ ($a \in K_0$), there is a functor $G : K \to \mathcal{C}$ whose objects are the specified objects and a natural equivalence $F \Rightarrow G$ whose components are the specified equivalences.

The hard work is in the statement that $[-, \mathcal{C}]$ sends monomorphisms to isofibrations.
This is basically the fact that the Joyal model structure is cartesian, plus the explicit identification of fibrations between fibrant objects as isofibrations of quasicategories.


答:
Let $\textbf{X} = \left( {\matrix{
   a & b  \cr
   c & d  \cr
 } } \right)$, then  ${\textbf{X}}^2-\textbf{X} =\textbf{X}(\textbf{X} - \textbf{I})= 0$ is equivalent to

$$ \left( {\matrix{
   a & b  \cr
   c & d  \cr
 } } \right)\left( {\matrix{
   a-1 & b  \cr
   c & d-1  \cr
 } } \right) = 0$$
Solve this equation for $a,b,c,d$, I get that

$\left\{\left\{c=\frac{a-a^2}{b},d=1-a\right\},\{a=0,b=0,d=1\},\{a=1,b=0,d=0\},\{a=0,b=0,c=0,d=0\},\{a=0,b=0,c=0,d=1\},\{a=1,b=0,c=0,d=0\},\{a=1,b=0,c=0,d=1\}\right\}$

Obviously, there are infinite solutions.

Title: Number Of Solutions $X^{2}=X$

Body:
The equation $x^{2}=x$ $,x\in \mathbb{R}$ has only the solutions $x=0$ and $x=1$.

For a $2 \times 2$ matrix X, how many solutions does $X^{2}=X$ have?

What if $X$ is symmetric?


问:
Find the minimum of $\sqrt{4y^2-12y+10}+\sqrt{18x^2-18x+5}+\sqrt{18x^2+4y^2-12xy+6x-4y+1}$
Find the minimum of $$f(x,y)=\sqrt{4y^2-12y+10}+\sqrt{18x^2-18x+5}+\sqrt{18x^2+4y^2-12xy+6x-4y+1}$$ It seems that $f_x=f_y=0$ is very hard to compute. Is there any easier idea?

答:
Since
$$f(x,y)= \sqrt{(2y-3)^2+1}+\dfrac{\sqrt2}2\sqrt{(6x-3)^2+1}+\dfrac12\sqrt{(6x-4y+2)^2+(6x)^2},$$
then the substitutions
$$2y-3 = \sinh s,\quad 6x-3=\sinh t,\quad s,t\in\mathbb R,\tag1$$
transform $\;f(x,y)\;$ to
$$g(s,t)= \cosh s + \dfrac{\sqrt2}{2} \cosh t+\dfrac12\,r(s,t),\tag2$$
where
$$r(s,t)=\sqrt{\left(\sinh t-2\sinh s-1\right)^2+\left(\sinh t+3\right)^2}.\tag3$$

The stationary points of $\;g(s,t)\;$ correspond to solutions of the system $\;g'_s=g'_t=0,\;$ or
\begin{cases}
\sinh s-\dfrac1{r(s,t)}\,\cosh s\,(\sinh t-2\sinh s-1)=0\\[4pt]
\dfrac{\sqrt2}2\sinh t-\dfrac1{r(s,t)}\,\cosh t\,(\sinh t-\sinh s+1)=0,\\[4pt]
\end{cases}

\begin{cases}
r(s,t) = \coth s\,(\sinh t-2\sinh s-1)\\[4pt]
r(s,t) = \sqrt2\,\coth t\,(\sinh t-\sinh s+1),\\[4pt]
\end{cases}

\begin{cases}
\big(\left(\sinh t-2\sinh s-1\right)^2+\left(\sinh t+3\right)^2\big)\sinh ^2s\\[4pt]
=(\sinh t-2\sinh s-1)^2(1+\sinh^2s)\\[4pt]
\big(\left(\sinh t-2\sinh s-1\right)^2+\left(\sinh t+3\right)^2\big)\sinh ^2t\\[4pt]
=2(\sinh t-\sinh s+1)^2(1+\sinh^2t)\\[4pt]
\coth s\,(\sinh t-2\sinh s-1)\ge0\\[4pt]
\coth t\,(\sinh t-\sinh s+1)\ge0,
\end{cases}

\begin{cases}
\left(\sinh t+3\right)\sinh s=\sinh t-2\sinh s-1\\[4pt]
\big(2\sinh^2t-4\sinh t \sinh s +4\sinh^2s+4\sinh t+4\sinh s+10\big)\sinh ^2t\\[4pt]
=2(\sinh^2t-2\sinh t \sinh s+\sinh^2s+2\sinh t-2\sinh s+1)(1+\sinh^2t)\\[4pt]
\sinh t+3\ge0,\\[4pt]
\coth t\,(\sinh t-\sinh s+1)\ge0,
\end{cases}

\begin{cases}
\sinh t+3 = \dfrac{2\sinh s+4}{1-\sinh s}\\[4pt]
(\sinh s+2)^2\sinh ^2t =(\sinh t-\sinh s+1)^2\\[4pt]
\sinh t\ge -3,\quad 1 > \sinh s \ge -2\\[4pt]
\sinh t\,(\sinh t-\sinh s+1)\ge0,
\end{cases}

\begin{cases}
\sinh t = \dfrac{5\sinh s+1}{1-\sinh s}\\[4pt]
(\sinh s+2)^2(5\sinh s+1)^2 =(5\sinh s+1+(\sinh s-1)^2)^2\\[4pt]
\sinh t\ge -3,\quad 1 > \sinh s \ge -2\\[4pt]
\sinh t\,(\sinh t-\sinh s+1)\ge0,
\end{cases}

\begin{cases}
\sinh t = \dfrac{5\sinh s+1}{1-\sinh s}\\[4pt]
(\sinh s+2)^2(5\sinh s+1)^2 =(\sin s+1)^2(\sinh s+2)^2\\[4pt]
\sinh t\ge -3,\quad 1 > \sinh s \ge -2\\[4pt]
\sinh t\,(\sinh t-\sinh s+1)\ge0,
\end{cases}

$$\begin{pmatrix}\sinh s\\ \sinh t\\ g(s,t)\end{pmatrix} =
\left\{\begin{pmatrix}-2\\ -3\\ 2\sqrt5\end{pmatrix},
\begin{pmatrix}-\frac13\\ -\frac12\\ \sqrt10 \end{pmatrix},
\begin{pmatrix}0 \\ 1\\ 4\end{pmatrix}\right\}.$$

Therefore, the least value of $\;f(x,y)\;$ is $\;\color{brown}{\mathbf{\sqrt{10}}}\;$ at $\;\color{brown}{\mathbf{(x,y)=\left(\dfrac5{12},\dfrac43\right)}}.\;$


问:
Problem proving that there exists an irrational number between any two real numbers
I want to show that: for $a,b \in \mathbb R$ with $a<b, \exists x \in (a , b)$ such that $x \notin \mathbb Q$ .
I start with $0< \frac 1{\sqrt 2} < 1 \Rightarrow 0< \frac{ b-a}{\sqrt 2} < b-a \Rightarrow a < \frac{b-a}{\sqrt 2} +a < b$ .

I know I am close to completing the proof but I do not know how to prove that the number $\frac{b-a}{\sqrt 2} +a \notin \mathbb Q$

Any help would be great!

答: If $\;\cfrac{b-a}{\sqrt2}+a\in\Bbb Q\;$ and since $\;b-\left(\cfrac{b-a}{\sqrt2}+a\right)=\epsilon>0\;$ , take either $\;\cfrac{b-a}{\sqrt2}+a+\cfrac\epsilon2\;$ , if $\;\epsilon\notin\Bbb Q\;$ , or else $\;\cfrac{b-a}{\sqrt2}+a+\cfrac\epsilon\pi\;$ if $\;\epsilon\in\Bbb Q\;$ . Either way, you get an irrational number.


问:
SMO problem: Sequence and series.
Problem on Series and Sequences (SMO Test):

For each positive integer $n \ge 1$ , we define the recursive relation given by $a_{n+1}=\cfrac{1}{1+a_n}$.

Suppose that $a_1=a_{2012}$.Find the sum of the squares of all possible values of $a_1$.

How to do this? I do not know how to approach it at all. What does recursive relation mean?

答:
Hint :

Lets calculate first few terms.

$$a_2=\frac{1}{1+a_1}$$
$$a_3=\frac{1}{1+a_2}=\frac{1+a_1}{2+a_1}$$
$$a_4=\frac{1}{1+a_3}=\frac{2+a_1}{3+2a_1}$$
$$a_5=\frac{1}{1+a_4}=\frac{3+2a_1}{5+3a_1}$$
$$a_6=\frac{1}{1+a_5}=\frac{5+3a_1}{8+5a_1}$$

It can be shown by induction that
$$a_n=\frac{F_n+F_{n-1}\, a_1}{F_{n+1}+F_n \, a_1} \tag{$n \ge 2$}$$

where $F_n=0,1,1,2,3,5,8,13,\ldots$ are Fibonacci numbers.

Then $a_{2012}=a_1$ is a quadratic in $a_1$ and using Vieta's, sum of squares of its roots can be found.

Title: SMO problem: Sequence and series.


答:
Assumed that $0 < a < b.$

Case 1

$a$ rational.

Since $a < b$, you should be able to find $n \in \mathbb{Z^+}$ such that $\left(a + \frac{\pi}{10^n}\right)$, which is clearly irrational, is less than $b$.

Case 2

$a$ irrational.

Let $c = \left(a + \frac{b-a}{2}\right) \implies a < c < b.$

If $c$ is irrational, then you are done.

If $c$ is not irrational, then invoke Case 1.

Title: Problem proving that there exists an irrational number between any two real numbers

Body:
I want to show that: for $a,b \in \mathbb R$ with $a<b, \exists x \in (a , b)$ such that $x \notin \mathbb Q$ .
I start with $0< \frac 1{\sqrt 2} < 1 \Rightarrow 0< \frac{ b-a}{\sqrt 2} < b-a \Rightarrow a < \frac{b-a}{\sqrt 2} +a < b$ .

I know I am close to completing the proof but I do not know how to prove that the number $\frac{b-a}{\sqrt 2} +a \notin \mathbb Q$

Any help would be great!


问:
is $\mathrm{depth}(A)=0$ equivalent to $\mathfrak{m}\in\mathrm{Ass}(A)$?
Let $(A,\mathfrak{m})$ a noetherian local ring. If $\mathrm{depth}(A)=0$ then all non unit elements are 0 divisors so $\mathfrak{m}$ is a subset of the union of all the prime divisor of $(0)$ (the associated primes of $A$) so $\mathfrak{m}$ is one of these associated prime. I'm intersted of the converse: if $\mathfrak{m}$ is an associated prime of $A$, is it true the $\mathrm{depth}(A)=0$?

I had the idea of taking $k[x,y]/(x^2,xy)_{(x,y)}$, here $\mathfrak{m}$ is an embedded prime ideal (so it is an associated ideal), but I don't have any idea of the depth of such an ring.

答: If $\mathfrak m$ is an associated prime, then there is $x \in \mathfrak m - \{0\}$ such that $x\mathfrak m = 0$, so every element in $\mathfrak m$ is a zero-divisor, so $\operatorname{depth}(A) = 0$.


问:
Why do we prove that $x_1$ = $x_2$ when looking if a function is injective?
For a function to be considered injective, we consider that $\forall x_1, x_2 \in X, x_1 \neq x_2 \Rightarrow f(x_1) \neq f(x_2)$.

Now, when it comes to proving this, we prove that $\forall x_1, x_2 \in X, x_1 = x_2 \Rightarrow f(x_1) = f(x_2)$, and if it turns out that $x_1 = x_2$, that means that the function is injective, but injectivity means that every $x$ has maps to only one, unique $y$, and we proved that it is not. (?)

So, why do we try to prove the opposite, and if it proves correct we consider it injective?

答:
To prove $1-1$, you are right that we consider that $\forall x_1, x_2 \in X, x_1 \neq x_2 \Rightarrow f(x_1) \neq f(x_2)$

For the other part, just take contrapositive of above statement to get:

$\forall x_1, x_2 \in X,   f(x_1) = f(x_2)\Rightarrow x_1=x_2$.


问:
Multiplicity of parabolic fixed point of iterated holomorphic function
I'm trying to prove Lemma 10.4 of Milnor's Dynamics in One Complex Variable, Third Edition, using a power series argument rather than the geometric argument proposed in the text.

Let $f$ be a holomorphic function $D \in \mathbb{C} \rightarrow \mathbb{C}$ fixing $0 \in D$. Denote by $\lambda := f'(0)$ the derivative of $f$ at $0$ and let it be a primitive $q$-th root of unity,  $\lambda^q = 1$.

Consider the $q$-fold iterate of $f$, $f^{\circ q} := f \circ f \circ \dots \circ f$ $q$ times, and consider the case it's not the identity map. In this situation $0$ is called parabolic fixed point of $f$.

Clearly $f^{\circ q}$ fixes $0$, and by chain rule its derivative at $0$ is $1$: $ \left(f^{\circ q} \right)'(0) = \left(f'(0) \right)^q = \lambda^q = 1$. The origin is hence a parabolic fixed point also for the $q$-fold iterate.

Expand the (holomorphic) $q$-fold iterate around the origin:
$$
\begin{split}
f^{\circ q}(z) &= 0 + 1 \, z + a z^{n+1} + \text{higher terms} \\
& = z+a z^{n+1} + \text{h.t.} \\
& = z \,  (1+az^n) + \text{h.t.}
\end{split}
$$
for some $a \neq 0 \in \mathbb{C}$ and integer $n\geq 1$. Per definition the integer $n+1$ is the multiplicity of $0$ as a fixed point of $f^{\circ q}$.

Lemma The integer $n$ is a multiple of $q$.1

The proof in the text uses a geometric argument making use of previous results 2, but the lemma can also be proved by a direct power series computation observing that $f \circ f^{\circ q} = f^{\circ q} \circ f $ - or so I have been told.

All I could think of is to expand $f$ around $0$ as $f(z) = \lambda z + \dots$, but also using the hint this doesn't seem to lead me anywhere.

Hints?

1 Or, as phrased in the text, the multiplicity $(n+1)$ is congruent to $1$ modulo $q$, which indeed means that $n+1-1$ is a multiple of $q$.

2 The integer $n$ corresponds to the number of attracting vectors of the iterate function at the fixed point. It can be shown that multiplication by $\lambda = e^{\frac{2\pi i j}{q}}$ for some $j = 0, \dots, q-1$ permutes these vectors, whose number must hence be a multiple of $q$.

答:
Expand around the fixed point and compare the coefficients of $f \circ f^{\circ q}$ and $f^{\circ q} \circ f$. Let $f^{\circ q} (z) = z + a z ^{n+1} + \dots$ and $f(z) = \lambda z + \sum_{m=2}^{n+1}b_m z^m + \dots$, where $a$ and $b_m$ are the standars Taylor coefficients. Up to order $n+1$ one has

$$f \circ f^{\circ q} = \lambda z + \sum_{m=2}^{n}b_m z^m + (b_{n+1}+ a \, \lambda \, ) \, z^{n+1} + \dots$$

$$f^{\circ q} \circ f = \lambda z + \sum_{m=2}^{n}b_m z^m + (b_{n+1}+a \,  \lambda^{n+1} \, ) \, z^{n+1} + \dots$$

Since $f \circ f^{\circ q} = f^{\circ q} \circ f$ coefficients must agree at all orders, so $\lambda = \lambda^{n+1}$ i.e. $\lambda^n = 1$. Since $\lambda$ is a primitive $q$-th root of unity, $n$ must be a multiple of $q$. $\quad \square$


问:
The Laplace operator.
What will be the value of $\Delta\left(\frac 1{r^2}\right)$ if $r=|x|=\sqrt{x_1^2+x_2^2+x_3^2}$?
May be this can be determined using Green's formula.</p>

Title: The Laplace operator.

Body:
If we don't wanna use coordinate transform, it can be also done in a "hands-on" way:
$$\Delta u= \sum_{i}\frac{\partial^2 u}{\partial x_i^2},$$
where $$u = \frac{1}{r^2} = \frac{1}{\sum\limits_i x_i^2}= \frac{1}{x_1^2 + x_2^2 + x_3^2}.$$
We can compute it term by term:
$$
\frac{\partial u}{\partial x_1} = -\frac{1}{(\sum_i x_i^2)^2}\cdot 2 x_1 = -\frac{ 2 x_1}{(x_1^2 + x_2^2 + x_3^2)^2} ,
$$
and
$$
\frac{\partial^2 u}{\partial x_1^2} = -\frac{2 (\sum_i x_i^2) ^2 - 2 x_1 \cdot2(\sum_i x_i^2) 2 x_1 }{(\sum_i x_i^2)^4} =  -\frac{2 (x_1^2 + x_2^2 + x_3^2)^2- 8 x_1^2 (x_1^2 + x_2^2 + x_3^2) }{(x_1^2 + x_2^2 + x_3^2)^4} ,
$$
Hence:
$$
\Delta u = -\frac{6 (x_1^2 + x_2^2 + x_3^2)^2- 8  (x_1^2 + x_2^2 + x_3^2)^2 }{(x_1^2 + x_2^2 + x_3^2)^4} = \frac{2}{(x_1^2 + x_2^2 + x_3^2)^2} = \frac{2}{r^4}.
$$


问:
How to turn $|a_n| < 1 + |L|$ into $L -1 < a_n < L + 1$?
So in class, we went over the proof of convergent sequences are bounded, I get pretty much everything except: $|a_n| < 1 + |L|$ part. How to turn $|a_n| < 1 + |L|$ into $L -1 < a_n < L + 1$? Where did the absolute value go?

答: Presumably $n$ is large enough that $|a_n - L| < 1$ is true, since $a_n$ converges to $L$. Then $$-1 < a_n - L < 1$$ so $$L - 1 < a_n < L + 1.$$ But in general, given two arbitrary quantities $x$ and $y$ it is not true that $$|x| < 1 + |y|$$ implies $$y-1 < x < y+1.$$ (To see why, just take $x = -y > 1$.)


问:
Simpler proof of the Hardy-Littlewood-Sobolev inequality in the inhomogeneous case
The Hardy-Littlewood-Sobolev inequality is the statement that there is a $C>0$ such that
$$
\tag{HLS}
\lVert f\ast \lvert\cdot\rvert^{-\alpha}\rVert_p\le C\lVert f\rVert_q, $$
for all $f\in L^q(\mathbb R^d)$, where the convolution is defined as
$$(f\ast \lvert\cdot\rvert^{-\alpha} )(x)= \int_{\mathbb R^d} \frac{f(y)}{\lvert x-y\rvert^{\alpha}}\, dy, $$
and the parameters satisfy the conditions
$$\tag{1} 0<\alpha<d, \quad \frac1p-\frac1 q+1=\frac\alpha d.$$

Now let us define the Japanese bracket $$\langle x \rangle:= \sqrt{1+ \lvert x\rvert^2},\qquad x\in\mathbb R^d,$$
which is a non-homogeneous version of $\lvert\cdot\rvert$. The function $\langle x\rangle^{-\alpha}$ has the exact same decay at $\lvert x \rvert \to \infty$ of its homogeneous counterpart $\lvert x \rvert^{-\alpha}$, but $\langle x \rangle^{-\alpha}$ is not singular at $x=0$. By the obvious estimate $\langle x\rangle^{-\alpha}\le \lvert x\rvert^{-\alpha}$, (HLS) immediately implies its non-homogeneous version
$$\tag{n-HLS}
\lVert f\ast \langle\cdot\rangle^{-\alpha}\rVert_p\le C\lVert f\rVert_q, $$
under the assumptions (1).

Question. Is there a direct proof of (n-HLS) that is simpler than a proof of (HLS)?

Remark 1. The inequality (n-HLS) actually holds for $\frac1 p - \frac1q +1 \le \frac{\alpha}{d}$. However, the non-endpoint case $\frac1 p - \frac1q +1 < \frac{\alpha}{d}$ can be immediately proved by an application of the Young inequality for convolutions. Thus, the present question is concerned solely with the endpoint case (1).

In the following, the letter $C$ will always denote an irrelevant positive constant, whose value may change from line to line.

Remark 2. There are many proofs of (HLS). One that I know uses the Hardy-Littlewood maximal function. By decomposing the ball $B(x, r)$ into dyadic annuli, we obtain the local estimate
$$\tag{2}\left\lvert \int_{B(x, r)}\frac{f(y)}{\lvert x-y \rvert^\alpha}dy\right\rvert \le r^{d-\alpha}\sum_{j=0}^\infty 2^{\alpha j - dj} Mf(x)= C_{d, \alpha}r^{d-\alpha}Mf(x).$$
This constant $C_{d,\alpha}$ equals $\sum 2^{(\alpha  - d)j}$, which is finite because $\alpha<d$. Here $Mf$ denotes the Hardy-Littlewood maximal function
$$
Mf(x):=\sup \frac1{\lvert B\rvert} \int_B \lvert f(y)\rvert\, dy,$$ where the sup is taken over all balls $B$ containing $x$.

We then estimate the tail of the convolution via the Hölder inequality;
$$\tag{3}
\left\lvert \int_{\mathbb R^d\setminus B(x, r)}\frac{f(y)}{\lvert x-y \rvert^\alpha}dy\right\rvert\le C r^{\frac{d}{q'}-\alpha} \lVert f\rVert_q.$$

Combining (2) and (3) gives the pointwise bound
$$
\tag{4} \lvert f\ast \lvert\cdot\rvert^{-\alpha}(x)\rvert \le C\left( r^{d-\alpha} Mf(x) + r^{\frac{d}{q'}-\alpha} \lVert f\rVert_q\right).$$
Choosing the $r$ that minimises the right-hand side, then integrating and applying the Hardy-Littewood maximal estimate $\lVert Mf\rVert_q\le C\lVert f \rVert_q$, we obtain (HLS).

This procedure will of course work if $\lvert\cdot\rvert^{-\alpha}$ is replaced by $\langle\cdot\rangle^{-\alpha}$. However, I think that there shouldn't be the need for the careful dyadic analysis of equation (2). Indeed, $\langle x-y\rangle^{-\alpha}$ is not singular at $y=x$. This is why I am asking for a simpler proof.

答:
In the given range
$$\tag{1}\frac1p + 1 = \frac1q+\frac\alpha d, $$
there is no hope for a simpler proof in the non-homogeneous case.

To explain, let me consider the weak Young inequality
$$\tag{2}
\lVert f\ast g \rVert_p \le C\lVert f\rVert_{q_1} \lVert g\rVert_{q_2,\infty}, \quad \frac1p+1 = \frac1{q_1}+\frac1{q_2},$$
where
$$\tag{3}\lVert g\rVert_{q_2,\infty} = \sup_{t>0} t \lvert \{ \lvert g\rvert >t\}\rvert^\frac1{q_2}.$$
We want to prove (n-HLS) under the assumption (1), which, according to (2) with $g=\langle\cdot\rangle^{-\alpha}$, follows from
$$\tag{4}
\lVert \langle \cdot \rangle^{-\alpha}\rVert_{\frac d \alpha, \infty}<\infty.$$
Now, it is clear that there is no simpler way to prove (4) than to estimate pointwise $\langle x \rangle^{-\alpha}\le \lvert x \rvert^{-\alpha}$, so that $$\lVert \langle\cdot \rangle^{-\alpha}\rVert_{\frac d \alpha, \infty} \le \ \lVert \lvert\cdot \rvert^{-\alpha}\rVert_{\frac d \alpha, \infty}<\infty,$$
which completes the proof. Thus the main step is the proof that $\lVert \lvert\cdot \rvert^{-\alpha}\rVert_{\frac d \alpha, \infty}<\infty$, which yields (HLS) and (n-HLS) essentially at the same time.

No proof can rely on anything simpler than this.

Let me remark that the mathematics behind all this is essentially the Marcinkiewicz interpolation theorem. I have consulted the blog post of Terry Tao https://terrytao.wordpress.com/2009/03/30/245c-notes-1-interpolation-of-lp-spaces/; the weak Young inequality is Exercise 44.

In the proof that I gave in the main question above, the weak-Lp computations are there. They are just hidden behind the Hardy-Littlewood maximal estimate $\lVert Mf\rVert_p\le C \lVert f\rVert_p$, which indeed is proven via the interpolation theorem of Marcinkiewicz.


答:
$$A_{n+1}=\frac{1}{1+A_n}\implies A_{n+1} A_n+A_{n+1}=1~~~(1)$$
Let $A_n=\frac{B_{n-1}}{B_{n}}$, in (1) then
$$B_{n+1}=B_n+B_{n-1},$$
which is like Fibonacci sequence.
Next, let $B_n =x^n$ to get $x=1+x^{-1} \implies x=(1\pm\sqrt{5})
/2=a,b$. So $$B_n= P a^n +Q b^n \implies A_n=\frac{a^{n-1}+R~ b^{n-1}}{a^{n}+R ~ b^{n}}.$$ The constant $R$ can be determined by a given condition.

Title: SMO problem: Sequence and series.

Body:
Problem on Series and Sequences (SMO Test):

For each positive integer $n \ge 1$ , we define the recursive relation given by $a_{n+1}=\cfrac{1}{1+a_n}$.

Suppose that $a_1=a_{2012}$.Find the sum of the squares of all possible values of $a_1$.

How to do this? I do not know how to approach it at all. What does recursive relation mean?


问:
Let $G$ be a group and $H\unlhd G$ with $|H|=25$. Also $G/H \cong A_4$. How many subgroups of order $75$ in $G$? And order $100$? How many are normal?
Let $G$ be a group and $H$ a normal subgroup with $|H| = 25$. Also we know that $G/H \cong A_4$.

I want to find out answers for these 2 questions:

How many subgroups of order 75 in $G$?

How many subgroups of order 100 in $G$? And how many of them are normal?

Here's what I found about that:

By the Lagrange theorem we can say that $|G| = 300$ (Because $|G| = |H||G/H|)$.

By the same reason for every $H'$ such that $|H'| = 75$ we can say that $|G/H| = 4$ (May be we can say that there's not many groups or order 4 for answer the 1'st question).

Second fact I think we can use is that order of every $g \in G$ is divisor of 300. And may be exist some pretty easy proof based on number theory.

答: Looking at this problem without thinking much, probably from Sylow theorems you'll get that $H$ is the only subgroup of order 25, so any subgroup $K$ of $G$ with order divisible by 25 will contain $H$, thus will correspond to $K/H$, which is a subgroup of $G/H$. So we just need the number of respective subgroups of $A_4$.


问:
square cake with raisins
Alice bakes a square cake, with $n$ raisins (= points).

Bob cuts $p$ square pieces. They are axis-aligned, interior-disjoint, and each piece must contain at least $2$ raisins.

Note that a single raisin can be shared by two pieces (if it is on their boundary), or even four pieces (if it is on their corner).

Bob tries to maximize the number of pieces $p$, and Alice tries to minimize $p$ by a sophisticated placement of the raisins.

For a given $n$ (number of raisins), how should Alice place the raisins such as to minimize $p$, and what is the minimum?

Here are some simple cases:

For n=0, 1, obviously p=0.

For n=2, 3, 4, Alice can make p=1, by placing the raisins on the corners of the cake, since the only square piece that can contain 2 raisins is the entire cake.

For n=5, the minimum p is 2: there is at least one quarter of the cake that contains 2 raisins. Bob can cut a piece that contains these 2 raisins on its boundary, and is entirely contained within that quarter of the cake. Now there is enough room for another square, that contains one of these raisins and another raisin.

What is the minimum $p$ for a given $n$?

-

Some insights:

Given two raisins, the minimum length of a square that contains both of them is the chessboard distance between them.

Therefore, it seems that Alice's best strategy is to maximize the  chessboard distances, and so force Bob to use large squares. This makes sense, but I cannot prove this.

If n is a square number ($n = k^2$), then Alice can place the raisins in a k-by-k grid, such that the smallest chessboard distance between raisins is $1/(k-1)$. In this case, Bob can cut at most $(k-1)^2$ squares. But is this really Alice's best strategy?

Bob can always do best with pieces that have raisins only on their boundary (He never has to cut a piece which has a raisin in the interior). PROOF: For any two raisins, there is a square that contains both of them in the boundary. If a piece contains two raisins in the interior, or one in the interior and one in the boundary, Bob can shrink that piece to get a square that has these two raisins in the boundary. This square will be entirely contained in the original square, and so will not affect the other pieces.

答:
Too long for a comment, so I am posting it here.

i can tell the strategy that i think is the best for Alice to place the raisins:

first, put 4 raisins at the four corners, and call any two diaginal raisins A and B;

then,put 2 raisins C and D inside the square, or in the boundary,each raisin A or B can form a square with one of the two raisins that is nearer to it,and take the segment CD as the diagonal, you can form a rectangle inside the square;

then, put another two raisins E and F inside the rectangle or in the boundary, each raisin of C and D can form a square with E or F, and take EF as the diagonal, you can form a smaller rectangle inside the rectangle,continue in this way, untill Alice used all the n raisins.

Title: square cake with raisins


答:
(Modified from the proof of Theorem 4.1 of Set Theory by Thomas Jech, which states that the reals are uncountable):

Let $c_1, c_2, c_3, \cdots$ be the rational numbers in $(a, b)$. We shall constructing an oscillating subsequence $s_n$, so $s_1 < s_3 < s_5 < \cdots < s_6 < s_4 < s_2$. We keep track of the indices $k_n$, so $s_n = c_{k_n}$. Note that here we do not require the indices to be increasing.

Let $s_1 = c_1$, so $k_1 = 1$.

Then, let $s_2 = c_{k_2}$ where $k_2$ is the minimum index such that $c_{k_2} > s_1$. (If none exists, then $\frac{c_1 + b}2$ is irrational.) So $s_1 < s_2$.

After that, let $s_3 = c_{k_3}$ where $k_3$ is the minimum index such that $s_1 < c_{k_3} < s_2$. (It exists because $\frac{s_1+s_2}2$ is rational.) So $s_1 < s_3 < s_2$.

Next, let $s_4 = c_{k_4}$ where $k_4 > k_3$ is the minimum index such that $s_3 < c_{k_4} < s_2$. (It exists for the same reason.) So $s_1 < s_3 < s_4 < s_2$.

Similarly, we let each $s_n$ be $c_{k_n}$ where $k_n$ is the minimum index such that $s_n$ has the appropriate relationship with the $s_r$ that comes before it.

Recall that we have $s_1 < s_3 < s_5 < \cdots < s_6 < s_4 < s_2$.

Then, I claim that $s_1, s_3, s_5, \cdots$ is a Cauchy sequence, so it defines a real number $s$. (If you already proved that suprema exist, then you can just take the supremum and skip this part.) Assume that it is not Cauchy for a contradiction, so there is $\varepsilon$ such that for arbitrarily large $n$ there is $m$ such that $s_{n+2m} - s_n > \varepsilon$, but that would mean that sequence would increase without bound, which contradicts the fact that $s_2$ is (constructed to be) an upper bound.

I now claim that the real number $s$ just constructed cannot be rational. First note that $a < s_1 < s < s_2 < b$, so $s \in (a, b)$. So, if $s$ were to be rational, then it would have to be $c_N$ for some $N$, so $s$ would have been chosen after at most $N$ steps (noting that the relation $s_1 < s_3 < \cdots < s_4 < s_2$ makes the indices all distinct), contradiction.


问:
Prove that the sum of distances from any point on an ellipse to its foci is constant in this framework
A conic section with eccentricity $e$ is defined as :

Given a line L, a point $F$ not on $L$, and a positive number $e$. Let $d(X,L)$ denote the distance from a point $X$ to $L$. The set of all $X$ satisfying the relation $$\|X-F\| = e \cdot d(X,L)$$
is called a conic section with eccentricity $e$. The conic is called an ellipse if $e < 1$.

Now, taking a vector $N$ normal to $L$ with $\|N\| = 1$, we have $d(X,L) = |(X-P) \cdot N|$ with $P$ being any point on $L$, so the equation becomes $$\|X-F\| = e  |(X-P) \cdot N|$$

Now, for an ellipse $(X-P) \cdot N$ is always negative or positive according to the choice of $N$.

Setting $P$ to be that point on $L$ nearest to $F$,and positioning $F$ in the negative half plane (in the sense that $(F-P) \cdot N < 0$,  we get that $P-F = dN$ with $d = \|P-F\|$. So we can rewrite the basic relation as

$$\|X-F\| = e  |(X-F) \cdot N - d|$$

This was the background, I wrote it to make it clear to you and to me :D

From here, I want to show that, if a conic is an ellipse with center at the origin then

$$\|X-F\| + \|X+F\|= 2a$$

with $a = ed+eF \cdot N$. Namely that the sum of distances from any point on an ellipse to its foci is constant.

Attempt :

if $F$ is on the negative half plane then $(X-F) \cdot N - d < 0$, so the basic relation can be written as

$$\|X-F\| = e  (d - (X-F) \cdot N ) = ed+eF\cdot N -eX\cdot N = a -eX\cdot N$$

Now we can see that $$eX\cdot N + a = e((X+F)\cdot N + d)  \stackrel{?}{=} \|X+F\|$$

If I was able to establish the last equality then it would be done, but I'm not able to do it.

Now, I saw that the problem involved a simmetric relation about the origin so if it works for $X$ it works also for $-X$, then substituting in the basic relation we get

$$\|X+F\| = e  |(X+F) \cdot N + d| = e  ((X+F) \cdot N + d) $$

because $(-X-F) \cdot N - d < 0$ . So , we have proven the theorem. Is this right?

答:
There are two parallel line directrices which are separated by constant distance $D_1 D_2 = u_1+ u_2$ symmetrically placed about origin on both sides.

Multiply the relation by $e$ so that for a new constant $a$

$$ 2 a = e u_1+ e u_2 $$

which proves the given proposition.


问:
If $A \cap B = \{x \in \mathbf{R^3} : 2x_1+2x_2-2x_3=0\}$. Then find $\dim A\cap W.$

If $A \cap B = \{x \in \mathbb{R^3} : 2x_1+2x_2-2x_3=0\}$. Then find $\dim A\cap B.$

This implies that $x_3 = x_1+x_2$ from where I get that $2x_1+2x_2-2(x_1+x_2)=0$ and $x_1(2,0,-2)+x_2(2,0,-2)$ spans $A \cap W$, but these vectors are clearly linearly dependent and don't form a basis form $A\cap B$ from where I could deduce the dimension. Is the another basis that I'm not seeing here? Is it the case that $A\cap B = \{0\}$?

答:
Perhaps you need a clearer algorithm for finding vectors that span a given subspace.

Consider that:

$$
x_1= x_1 \\
\;\;\;\;\;\;\,x_2= \,\;\;\;\;\;\;\; x_2\\
\;\;\;\;\;\;\,x_3=x_1+x_2
$$

So two vectors spanning the set are $\begin{bmatrix}1 \\0 \\ 1 \end{bmatrix}$ and $\begin{bmatrix}0 \\1 \\ 1 \end{bmatrix}$. They're not parallel so they must be linearly independent but you could offer a formal argument by assuming that $$c_1 \begin{bmatrix} 1 \\0 \\ 1 \end{bmatrix} + c_2 \begin{bmatrix} 0 \\1 \\ 1 \end{bmatrix} = \mathbf 0$$

And then proving that the two coordinates must be zero.

PS.

Given that the set represents a plane of the vectors that are orthogonal to $\begin{bmatrix}1 \\1 \\ -1 \end{bmatrix}$ it couldn't have had $\text{dim} =1$, planes have two degrees of freedom.


问:
Are trigonometric functions with already substituted arguments (integral) treated with the chain rule, when derivating?
So I am left with this integral after already having done(!) one substitution before (from $dx$ to $dz$):
$$ \int \frac{32\tan(z)}{3\pi} \,dz = \int \frac{32\sin(z)}{3\pi\cos(z)} \,dz $$

and wanted to apply the substitution rule where I choose that

$$ u=\cos(z) .$$

Now I'm a little confused as to whether

$$ u'= -\sin(z)\cdot z' \ \text{or} \ u'=-\sin(z) .$$

Wolframalpha tells me the integral is

$$ -\frac{32\log(\cos(z))}{3\pi} , $$

which makes sense, IF $u'=-\sin(z)$. However, if I'd choose that $u'=-\sin(z) \cdot z'$, I would get (after cancelling out $-\sin(z)$)

$$ \int -\frac{32}{3\pi u} \,\frac{du}{-z'} = \int -\frac{32}{3\pi } \cdot \frac{-1}{uz'} \,du  $$

which would lead to a whole different integral, because the integral of $-\frac{1}{uz'}$ clearly can't be the simple $\log$ from the wolfram alpha solution... if anyone could help me out on the question which one of the derivatives is right I'd be super super thankful!!

答:
$$ \int \frac{32\tan(z)}{3\pi} \,dz = \int \frac{32\left[\sin(z)\,dz\right]}{3\pi\cos(z)} $$

not $u=\cos x$
$$ u=\cos(z) \to du=-\sin z dz\to \left[\sin z \,dz\right]=-du$$
$$\frac{32}{3\pi}\int\frac{-du}{u}=-\frac{32}{3\pi}\log u+C=-\frac{32}{3\pi}\log \cos z+C$$


问:
Checking Proof: Given $\boldsymbol{x}$ there exists a $\boldsymbol{y}$ where $\boldsymbol{x} \cdot \boldsymbol{y} = 0 $
Let bold type-face denote a k-tuple in $\mathbb{R}^k$

Proposition: Suppose there is a, $\boldsymbol{x} \in \mathbb{R}^k$ where $k\ge2$. Then there exists a $\boldsymbol{y} \in \mathbb{R}^k$ where $\boldsymbol{y} \ne \boldsymbol{0}$ such that $\boldsymbol{x} \cdot \boldsymbol{y} = 0 $

Proof: Suppose $k=2$, then $\boldsymbol{x} \cdot \boldsymbol{y} = \sum_{i=1}^2 x_iy_i=x_1y_1+x_2y_2$. Thus, by allowing $y_1=\dfrac{1}{x_1}$ and $y_2 = \dfrac{1}{-x_2}$ the expression goes to zero. Thus there is at least one $\boldsymbol{y}$ that satisfies the expression.

Suppose that the proposition is true for $k=n$, then it must also be true for $k=n+1$.

Suppose $k=n+1$, then $\boldsymbol{x} \cdot \boldsymbol{y} = \sum_{i=1}^{n+1} x_iy_i = \sum_{i=1}^{n} x_iy_i+x_{n+1}y_{n+1}$. Based upon the inductive assumption, the first $n$ values of $\boldsymbol{y}$ can be set such that $\sum_{i=1}^{n} x_iy_i=0$ where $y_i \ne 0$ for at least one $i=1,2,...,n$. Set $y_{n+1}$ to zero, thus, the whole expression is zero and there is at least one $\boldsymbol{y} \ne \boldsymbol{0}$ where $\boldsymbol{x} \cdot \boldsymbol{y} = 0$ $\square$

Question: I would like to request some feedback on my proof. In what ways may it be improved?

Source: Rudin, W (1976). Principles of Mathematical Analysis (ed. 3).

答:
Okay I see OP has a typo in the proposition.

By geometry, $\newcommand{\b}{\boldsymbol}\b{x}\in \mathbb{R}^n$ that is not zero uniquely defines a plane with codimension 1 by being its normal vector. Any nonzero vector $\b{y}$ living on this plane satisfies $\b{x}\cdot \b{y}= 0$.

For $\mathbb{R}^2$, a rotation by $\pi/2$ will do, i.e., let $\b{y} = (x_2,-x_1)$.


问:
Show there exists no diffeomorphism between positive real line and a spiral
I'm given the following two subset of $\mathbb{R}^2$:
$$
M = \{(t,0) \in \mathbb{R}^2 \mid t>0\},
\quad
N = \{(t \cos(1/t),\, t \sin(1/t)) \in \mathbb{R}^2 \mid t>0 \}
$$
The goal of the exercise is to show that there exists no diffeomorphism
$\varphi: \mathbb{R}^2 \to \mathbb{R}^2 $ such that $\varphi (M) = N$.

In a previous exercise I showed that there exists a homeomorphism such that the same conditions are fullfiled.

I'm also given the hint that one should start by showing that the partial derivatives with respect to x of such a diffeomorphism
$\varphi: \mathbb{R}^2 \to \mathbb{R}^2$
(if he would exist) would vanish in $(0,0)$.

I know how to finish the exercise if I find a way of showing that the partial derivatives with respect to x vanish at $(0,0)$. Then the Jacobian determinant is zero at $(0,0)$ so the inverse function is not differentiable at $(0,0)$ and I find a contradiction to the existence of a differeomorphic $\varphi$.

So far I tried to use the difference quotient
$$
\lim_{h\to 0} \frac{\varphi_1(x+h,y) - \varphi_1(x,y)}{h}.
$$
For this Limit to be zero the numerator has to decrease faster than the linear denominator. I figure that I will have to use that the line segment around $(0,0)$ in $N$ is infintely long while the length of the line segment of the preimage is finite but so far I'm not sure how to combine this with the difference quotient.

The concept of differentiating an unknown function is not something I encountered before so I would be thankful for any help.

答:
Assume there exists a diffeomorphism $\varphi$ on $\mathbb R^2$ such that $\varphi(M) = N$. Since $0$ is the only limit point of $M$ not belonging to $M$, also $\varphi(0)$ must be the only limit point of $N$ not belonging to $N$. We conclude $\varphi(0) = 0$.

The curve $\gamma : [0,1] \to \mathbb R^2, \gamma(t) = (t,0)$, is a smooth embedding with derivative $\gamma'(t) = (1,0) \ne 0$. Hence the curve $\varphi \circ \gamma : [0,1] \to \mathbb R^2$ is a smooth embedding with continuous derivative $\bar \gamma'(t) \ne 0$. There is a reparameterization $\bar \gamma : [0,b] \to \mathbb R^2$  of $\varphi \circ \gamma$ such that $\lVert \bar \gamma'(t) \rVert = 1$. By construction $\bar \gamma(0) = 0$ and  $\bar \gamma((0,b]) \subset N$. Moreover $\bar \gamma(t) + \bar \gamma'(t)$ is contained in the tangent at $N$ in the point $\bar \gamma(t)$. But in any neigborhood of $0$ the spiral $N$ has both horizontal and vertical tangents which means that $\lim_{n \to \infty} \bar \gamma'(t)$ does not exist (recall $\lVert \bar \gamma'(t) \rVert = 1$). This is a contradiction because $\bar \gamma'$ is continuous and thus $\lim_{n \to \infty} \bar \gamma'(t) = \bar \gamma'(0)$.


答:
As mentioned, you are missing the possibility that $x_1 = 0$, since we can't divide by 0.

Hint: There is a very easy vector to guess.

Title: Checking Proof: Given $\boldsymbol{x}$ there exists a $\boldsymbol{y}$ where $\boldsymbol{x} \cdot \boldsymbol{y} = 0 $

Body:
Let bold type-face denote a k-tuple in $\mathbb{R}^k$

Proposition: Suppose there is a, $\boldsymbol{x} \in \mathbb{R}^k$ where $k\ge2$. Then there exists a $\boldsymbol{y} \in \mathbb{R}^k$ where $\boldsymbol{y} \ne \boldsymbol{0}$ such that $\boldsymbol{x} \cdot \boldsymbol{y} = 0 $

Proof: Suppose $k=2$, then $\boldsymbol{x} \cdot \boldsymbol{y} = \sum_{i=1}^2 x_iy_i=x_1y_1+x_2y_2$. Thus, by allowing $y_1=\dfrac{1}{x_1}$ and $y_2 = \dfrac{1}{-x_2}$ the expression goes to zero. Thus there is at least one $\boldsymbol{y}$ that satisfies the expression.

Suppose that the proposition is true for $k=n$, then it must also be true for $k=n+1$.

Suppose $k=n+1$, then $\boldsymbol{x} \cdot \boldsymbol{y} = \sum_{i=1}^{n+1} x_iy_i = \sum_{i=1}^{n} x_iy_i+x_{n+1}y_{n+1}$. Based upon the inductive assumption, the first $n$ values of $\boldsymbol{y}$ can be set such that $\sum_{i=1}^{n} x_iy_i=0$ where $y_i \ne 0$ for at least one $i=1,2,...,n$. Set $y_{n+1}$ to zero, thus, the whole expression is zero and there is at least one $\boldsymbol{y} \ne \boldsymbol{0}$ where $\boldsymbol{x} \cdot \boldsymbol{y} = 0$ $\square$

Question: I would like to request some feedback on my proof. In what ways may it be improved?

Source: Rudin, W (1976). Principles of Mathematical Analysis (ed. 3).


问:
Counting the number of passwords of length 6 with the given conditions
A password of length 6 with upper case letters, lower case letters and digits. Password must contain at least 1 upper case, 1 lower case and 1 digit. How many possible passwords are there ?

What I did. Total are $62^{6}$.

Passwords in which upper and lower are mixed(no digits) = $ 52^{6} - 2(26^{6})$.

Passwords in which upper and digits are mixed(no lower) = $ 36^{6} - 10^{6} - 26^{6}$.

Passwords in which lower and digits are mixed(no upper) = $ 36^{6} - 10^{6} - 26^{6}$.

I subtracted all these from the total so get $ 62^{6} - 52^{6} - 2(36^{6}) + 2(26^{6}) + 10^{6}$.
Is this the correct way.

What other ways could one think about ?

答:
I just checked my final answer with that of the OP: are answers match.  I simply had some trouble following the OP's logic, so I arrived at the same answer by a moderately different route:

Also, the alternative to Inclusion-Exclusion, which I will briefly describe, but do not advise using, would be a direct approach:

The number of digits used will either be $1,2,3,$ or $4$.
In each of the 4 cases, you then have to enumerate how many of the remaining characters are uppercase vs lower case.

As an illustration:

In the case of 3 digits used, you will either have 1 uppercase or 2 uppercase.  Exploring the specific case of 3 digits, 2 uppercase, and 1 lowercase, the enumeration would be:

There is $A = \binom{6}{1} \times \binom{5}{2}$ ways of positioning the digits, uppercase letters and lowercase letters.

Then, once the positions are set, there are $B = (10)^3 \times (26)^2 \times (26)^1$ different character assignments that can go in the corresponding positions in the password.

Therefore, the enumeration for this subcase is $A \times B.$

There may be a more elegant approach for the direct approach alternative to Inclusion Exclusion that is escaping me.

Anyway, in any event, Inclusion-Exclusion is much easier.

From Inclusion-Exclusion:

Let $T_k$ denote the number of ways that at least $k$ constraints are violated.

Then, you want $T_0 - T_1 + T_2 - T_3.$

$T_0 = (62)^6.$

To calculate $T_1$, there are 3 possibilities:

No uppercase, no lowercase, or no digits.

Therefore,

$T_1 = (36)^6 + (36)^6 + (52)^6.$

To calculate $T_2$, there are 3 possibilities:

No uppercase or lowercase, no uppercase or digits, no lowercase or digits.

Therefore

$T_2 = (10)^6 + (26)^6 + (26)^6.$

In considering $T_3$, it is impossible for all three constraints to be violated, because no characters are allowed except the 62 letters/numbers.  The assumption that all 3 constraints are violated would mean that there were no characters available.

Therefore

$T_3 = 0.$


问:
How to show that $\exp(x+y) = \exp(x)\times\exp(y)$
I was reviewing some things I had learned on $e$ (or actually $\exp(x)$, which for $x \in \mathbb{R}$ makes sense to write as $e^x$) a few years back, and decided to try and show that  $\exp(x+y) = \exp(x)\times\exp(y)$. I got really far, but don't know what to do now. This is what I have:

$$\exp(x)=\sum_{n=0}^{\infty}{\frac{x^n}{n!}}$$
$$\exp(x)\times\exp(y)=\left(\sum_{n=0}^{\infty}{\frac{x^n}{n!}}\right)\left(\sum_{m=0}^{\infty}{\frac{y^m}{m!}}\right)=\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}{\frac{x^ny^m}{n!m!}}$$
$$\exp(x+y) = \sum_{n=0}^{\infty} {\frac{(x+y)^n}{n!m!}} = \sum_{n=0}^{\infty} \frac{1}{n!} \sum_{k=0}^n \binom{n}{k}x^ky^{n-k} = \sum_{n=0}^{\infty} \frac{1}{n!} \sum_{k=0}^n \frac{n!}{k!(n-k)!} \times x^ky^{n-k} = \sum_{n=0}^{\infty} \sum_{k=0}^n \frac{x^ky^{n-k}}{k!(n-k)!}$$

And that was about it. Now I'm left with showing that:
$$\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}{\frac{x^ny^m}{n!m!}} = \sum_{n=0}^{\infty} \sum_{k=0}^n \frac{x^ky^{n-k}}{k!(n-k)!}$$
And I don't know how to do that. I thought that doing establishing something like $j=n-k$ and rewriting would help give me some insight, but it didn't... I don't know how to handle that relationship between an infinite sum of infinite sums and an infinite sum of a simple sum, even if they both look remarkably similar. I also know that what I wrote this far is true by plugging that last sentence into WolframAlpha (unless I transcribed it into this post incorrectly).

My study of this kind of math is pretty autonomous via research, so I apologize if there was something that could have been phrased better. I'm also not really sure what tags to put in this.

EDIT: A few comments proposed posts that shared the same problem, but they both had this jump:

$$
\sum_{n=0}^{\infty}{\frac{x^n}{n!}} \sum_{m=0}^{\infty}{\frac{y^m}{m!}} =
\sum_{n=0}^{\infty} \sum_{k=0}^n \frac{x^ky^{n-k}}{k!(n-k)!}
$$

And didn't really explain it. I don't know why this is true... In fact, if this I understand this, then the rest of the way is self-explanatory.

The other post did something similar, but used some notation I'm not used to. I think it translates to:

$$
\sum_{n=0}^{\infty} \frac{n!}{n!} \sum_{k=0}^n \frac{x^ky^j}{k!j!} =
\sum_{k=0}^{\infty} \sum_{j=0}^{\infty} \frac{x^ky^j}{k!j!}
$$

And here, there are two problems: one, I'm not sure that's what's written... And if it is, then again, I don't know why that would be true.

EDIT 2: Following the suggestions in the comments, I read up on Cauchy's product and finally see why all of this happens. It is easier with the first example, I thought:

$$
\sum_{n=0}^{\infty}{\frac{x^n}{n!}} \sum_{m=0}^{\infty}{\frac{y^m}{n!}} =
\sum_{n=0}^{\infty} \sum_{k=0}^n \frac{x^ky^{n-k}}{k!(n-k)!}
$$

This is very much the standard form, if you consider $a_n=\frac{x^n}{n!}$ and $b_m=\frac{y^m}{m!}$:

$$
\sum_{n=0}^{\infty}a_n \sum_{m=0}^{\infty}b_m =
\sum_{o=0}^{\infty}c_k =
\sum_{o=0}^{\infty} \sum_{k=0}^{o} a_kb_{o-k} =
\sum_{o=0}^{\infty} \sum_{k=0}^{o} \frac{x^k}{k!}\times\frac{y^{o-k}}{(o-k)!}
$$

The letters are mixed up, but that's because I should have done this a bit better. The point is that I understand it know, because of the Cauchy product. Thank you all!

答:
Theorem: Given two convergent series $X=\sum x_n$ and $Y=\sum y_n$, you define $\sum z_n$ with partial sums $Z_n=\sum_{r=0}^n x_ry_{n-r}$=coefficient of $t^n$ in $(\sum_{r=0} x_rt^r)(\sum_{s=0} y_s t^s)$, then it can be proven that if $\lim Z_n=\sum z_n$ exists (let it be equal to $Z$), then $Z=XY$. (Abel's Theorem).

In your case, $e^x=\sum\frac{x^m}{m!}, e^y=\sum\frac{y^n}{n!}$, both are absolutely convergent series that is their Cauchy product will converge. (Mertens theorem)

Now define Cauchy product, $Z_n=\sum_{r=0}^n \frac{x^ry^{n-r}}{r!(n-r)!}$

This product will converge by Mertens theorem. And by Abel's theorem, $\lim Z_n= \sum_{n=0}^\infty\sum_{r=0}^n \frac{x^ry^{n-r}}{r!(n-r)!}=
(e^x)(e^y)$

$\begin{align}
\implies (e^x)(e^y)&=\sum_{n=0}^\infty\sum_{r=0}^n \frac{n! x^ry^{n-r}}{r!(n-r)! n!}\\              &= \sum_{n=0}^\infty\sum_{r=0}^n \frac{^nC_r x^ry^{n-r}}{n!}\\
                   &=\sum_{n=0}^\infty \frac{(x+y)^n}{n!}=e^{x+y}\end{align}$.

Note that:

$(x+y)^n=\sum_{r=0} ^n$ $^nC_r x^ry^{n-r}$ by Binomial Theorem.

Wherever no superscript is added to $\sum$ in my answer above, please treat the superscript as $\infty$ and in case of missing subscript to $\sum$, please treat the same as index variable $=0$.


问:
Prove that a certain sum of binomial coefficients is divisible by a power of 2
As many will recognize the following expression is a closed form for the Fibonacci numbers. Can it be proved that

$$\frac{1}{2^{n-1}} \sum_{j=0}^{\lfloor n/2 \rfloor} \binom{n}{2j+1} 5^{j} \quad \text{where} \quad n \in\mathbb{N}$$
are integers without using the fact that these are the Fibonacci numbers?

答:
By binomial theorem the expression is equivalent to:
$$
S_n=\frac1 {2^{n}}\frac{(1+x)^n-(1-x)^n}x,
$$
where $x=\sqrt5$. Since the latter fraction which we denote $Z_n (x)$ is clearly integer ($x $ cancels leaving in the numerator only even powers of $x $), we need only to prove that it  is divisible by $2^{n}$.
To prove this we proceed as follows:
$$
2 Z_n (x)=[(1+x)+(1-x)]Z_n (x)=Z_{n+1}(x)+(1-x^2)Z_{n-1}(x)\tag1
$$

Dividing (1) by $2^{n+1} $ and rearranging terms one obtains:
$$
S_{n+1}=S_n+\frac {x^2-1}4S_{n-1}\tag2
$$

Since $S_0=0$ and $S_1=1$ the statement $S_n\in\mathbb Z$ follows by induction for all $n\ge0 $ provided that $4|(x^2-1) $ which is indeed true.


答:
You need to be very careful about notation here. It helps on $u$-substitution not to think about it in terms of $u'$ or $z'$. The prime notation typically incorporates two differentials, which you want to be able to separate. Normally, $u' = \frac{du}{dz}$, but you're treating it like $du$. The integral uses differential notation, so it helps to use differential notation throughout the $u$-substitution process.

Instead of setting $u' = -\sin(z) \cdot z'$, think of it like this:

$$ u = \cos(z) $$
$$ \frac{du}{dz} = -\sin(z) . $$

Then, to transform it from a derivative to differentials, you have

$$ du = \frac{du}{dz} dz $$
$$ du = -\sin(z) dz $$
$$ dz = -\frac{1}{\sin(z)} du . $$

Going back to the integral, we can factor out $\frac{32}{3\pi}$:

$$ \int {\frac{32\sin(z)}{3\pi \cos(z)} \ dz} = \frac{32}{3\pi} \int {\frac{\sin(z)}{\cos(z)} \ dz} $$

Substitute in $u = \cos(z)$:

$$ = \frac{32}{3\pi} \int {\frac{\sin(z)}{u} \ dz} $$

Substitute in $dz = -\frac{1}{\sin(z)} du$:

$$ = \frac{32}{3\pi} \int {\frac{\sin(z)}{u} \frac{du}{-\sin(z)}} $$

Cancel out $\sin(z)$ and factor out the $-1$:

$$ = -\frac{32}{3\pi} \int {\frac{du}{u}} $$

And integrate then substitute back in $u = \cos(z)$:

$$ = -\frac{32}{3\pi} \ln(u) $$
$$ = -\frac{32}{3\pi} \ln(\cos(z)) . $$

With the right notation, the substitution process becomes a lot simpler. With slightly messier notation, it's very easy to get confused.

Title: Are trigonometric functions with already substituted arguments (integral) treated with the chain rule, when derivating?

Body:
So I am left with this integral after already having done(!) one substitution before (from $dx$ to $dz$):
$$ \int \frac{32\tan(z)}{3\pi} \,dz = \int \frac{32\sin(z)}{3\pi\cos(z)} \,dz $$

and wanted to apply the substitution rule where I choose that

$$ u=\cos(z) .$$

Now I'm a little confused as to whether

$$ u'= -\sin(z)\cdot z' \ \text{or} \ u'=-\sin(z) .$$

Wolframalpha tells me the integral is

$$ -\frac{32\log(\cos(z))}{3\pi} , $$

which makes sense, IF $u'=-\sin(z)$. However, if I'd choose that $u'=-\sin(z) \cdot z'$, I would get (after cancelling out $-\sin(z)$)

$$ \int -\frac{32}{3\pi u} \,\frac{du}{-z'} = \int -\frac{32}{3\pi } \cdot \frac{-1}{uz'} \,du  $$

which would lead to a whole different integral, because the integral of $-\frac{1}{uz'}$ clearly can't be the simple $\log$ from the wolfram alpha solution... if anyone could help me out on the question which one of the derivatives is right I'd be super super thankful!!


问:
What is a monoid in simple terms?
I encountered the term "monoid" but I didn't really understand what is it useful for or what's it about.

If I understand correctly a "monoid" is something defined in the context of a set, an element of the set and a binary operation defined on the specific set.

The binary operation has as left and right operators the same element.

So I think it is something like: $1\cdot1=1\cdot1$ where $\cdot$ is the relevant operation on the set $\mathbb{N}$ .

But what does this mean in layman's terms?

Is this some formal way to define a symmetric operation? Then why is it about one element of the set instead of two? I think $1 + 2 = 2 + 1$ on $\mathbb{N}$ is not a "monoid".

Can someone please help me understand this concept?

答:
A monoid is a set $M$ together with an associative binary operation $\ast: M\times M\to M$ with an identity element $1_M\in M$; that is to say that for any $a,b,c\in M$, we have:

Closure: $a\ast b\in M$;

Associativity: $$a\ast(b\ast c)=(a\ast b)\ast c;$$

Identity: There exists a $1_M\in M$ such that $1_M\ast a=a\ast 1_M=a$.

Title: What is a monoid in simple terms?


问:
Suppose that $A$ is an $n \times m$ matrix and $col(A)$ is a subspace of $\mathbb{R}^5$, then what is $n$?
this is a question that appeared on webassign for my linear algebra course. I believe it's a bad question, and want to know what you guys think.

Question:
Suppose that $A$ is an $n \times m$ matrix with $rank(A)=4$, $nullity(A)=3$ and that $col(A)$ is a subspace of $\mathbb{R}^5$. What are the dimensions for $A$?

By the rank nullity theorem, we know that $rank(A) + nullity(A) = \text{# of columns of A}$. Therefore $A$ has $7$ columns.

I believe this is a bad question because they want us to deduce that $n=5$ from the fact that $col(A)$ is a subspace of $\mathbb{R}^5$. The following matrix has $n=6$ and $col(A)$ is still a subspace of $\mathbb{R}^5$.

$$ A = \left( \begin{matrix} 2 & 3 & 2 & 2 & 1 & 3 & 7 \\ 4 & 3 & 2 & 4 & 1 & 4 & 3 \\ 1 & 2 & 3 & 3 & 1 & 1 & 2 \\ 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 6 & 5 & 4 & 3 & 0 & 1 & 9 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{matrix} \right) $$

Here, $col(A)$ will be a subspace of $\mathbb{R}^6$, as the corresponding linear transformation $T(x)=Ax$ has codomain $\mathbb{R}^6$. However, $col(A)$ is also a subspace of $\mathbb{R}^5$ as the last row of the matrix is all zero.

Can somebody verify that my thoughts about this being a bad question are valid, or if not, point out the flaw in my thinking? Thank you.

答:
In your example, the columnspace is not a subspace of $\Bbb{R}^5$, but rather $\Bbb{R}^5 \times \{0\}$ (a subspace of $\Bbb{R}^6$). While this subspace of $\Bbb{R}^6$ is isomorphic to $\Bbb{R}^5$, it is not the same thing.

The columnspace of an $n \times m$ real matrix is always a subset of $\Bbb{R}^n$. If $a \neq b$, then $\Bbb{R}^a \cap \Bbb{R}^b = \emptyset$ ($a$-tuples of real numbers are only $b$-tuples if $a = b$). So, if the columnspace contains even a single element of $\Bbb{R}^5$, then the matrix must have $5$ rows.


问:
Intuition behind definition for limits at infinity using epsilons and deltas

f is said to have a limit at infinity, if there exists a real number L such that for all $\epsilon>0, \exists N>0$ such that  $|f(x) -L|< \epsilon $,$\forall x > N$. In otherwords:
$$ \lim_{x \to \infty } f(x) = L$$
(from)

I don't understand here why we say $\exists N>0$, why not refer to real numbers ?  I don't think we lose any information doing it this way and it is more intuitive because when we discuss concepts of continuity and differentiability in real analysis, we use the set $\mathbb{R}$.

答:
Nothing in the definition of limit you quote says $N\in \mathbb{N}$. In fact, I've seen $N\in \mathbb{R}$ used in the definition before. That's what Wikipedia uses.

I would imagine the notation of using the letter $N$ (which is typically used in reference to natural numbers) comes from limits of sequences, as Vercassivelaunos suggests, where we want to show that it holds beyond some $n$th element of the sequence, and where $N\in \mathbb{N}$ arises naturally since you can think of a real-valued sequence $\{a_n\}$ as a function mapping between $n\in \mathbb{N} \to a_n \in \mathbb{R}$.


问:
How to solve following multidimensional equation?
Let's have an equation
$$
u_{t} = \Delta u ,\quad \mathbf x є R^{n},\quad t > 0, \quad u(\mathbf r , 0) = e^{-(x_{1} + ... + x_{n})^{2}}.
$$
How to solve it? I tried to reduce the equation to the form
$$
u_{t} = \tilde {\Delta} u ,\quad \mathbf x є R^{n},\quad t > 0, \quad u(\mathbf r , 0) = \prod_{i = 1}^{n}f_{k}(x_{i}),
$$
which I solved, but I failed.

Maybe, there be a good way to find solution in form
$$
u(\mathbf r, t) = g(t)e^{-(x_{1} + ... + x_{n})^{2}f(t)}, \quad f(0) = g(0) = 1?
$$
After substituting it to the equation I got
$$
\partial_{t}u = \left( \dot {g} - g\dot {f}(\sum_{i, j}x_{i}x_{j})\right) exp(...)
$$
for time derivation and
$$
(-2nfg + 4ngf^{2}(\sum_{i, j}x_{i}x_{j}))exp(...)
$$
for the Laplacian.

So, I have a system
$$
\dot {g}(t) = -2nf(t)g(t), \quad -g(t)\dot {f}(t) = 4ng(t)f^{2}(t), \quad f(0) = g(0) = 1,
$$
and I'll get a solution fast.

Is this method correct?

How can I solve this equation more "strictly"?

答:
You are solving the heat equation in a half-space. The initial condition has a peculiar form: it depends only on $x_1+\dots+x_n$. This suggests changing the system of coordinates to make $x_1+\dots+x_n$ a new variable. For example:

$$\begin{split}
y_1&=n^{-1/2}(x_1+\dots+x_n) \\
y_2&=2^{-1/2}(x_1-x_2) \\
y_3&=2^{-1/2}(x_1-x_3) \\
\dots&\dots \\
y_n&=2^{-1/2}(x_1-x_n) \\
\end{split}
\tag1$$
Since (1)  is an orthogonal transformation, it commutes with taking the Laplacian. In the new coordinate system you are solving

$$u_{t} = \Delta u ,\quad \mathbf y \in \mathbb R^{n},\quad t > 0, \quad u(\mathbf r , 0) = e^{-n^2y_1^2}. \tag2$$
The initial data in (2) is independent of $y_2,\dots,y_n$. Therefore, the solution will not depend on them either. (Indeed, translating the solution in direction of $y_j$, $j>1$, you get another solution of the same problem; by uniqueness theorem it must be the same function).

So, you end up solving the one-dimensional problem
$$u_{t} = \Delta u ,\quad   y \in \mathbb R,\quad t > 0, \quad u(y , 0) = e^{-n^2y^2}. \tag3$$
which is easy because the Gaussian function becomes a more diffuse Gaussian function under the heat flow. Then return to the original coordinates.

Or you can take a shortcut: obtain a solution using the ansatz $$u(\mathbf r, t) = g(t)e^{-(x_{1} + ... + x_{n})^{2}f(t)}, \quad f(0) = g(0) = 1$$ and appeal to the uniqueness theorem.

(Uniqueness for the heat equation applies to solutions that do not grow too rapidly; there are other, "unphysical" solutions, but nobody ever wants to look for them.)


问:
Computing $\int_{-\infty}^{+\infty} e^{-iky} e^{-c|k|^{\alpha}} |k|^{\beta} dk$
I want to solve the above integral, where $c >0$, $\beta = 0,2$, $0<\alpha<2 $, $y \in \mathbb{R}$, $k \in \mathbb{R}$. I tried to use the Residue Theorem adding the semi-circle contour of infinite radius in the half-plane with negative immaginary numbers.

I don't see singularity inside this half-plane to use for the residue computation; what am I missing?

Any comment is welcome

[EDIT]

As pointed by metamorphy I have a branch cut due to $|k|^{\alpha}$ at exponent, so I can't apply the residue theorem. How should I proceed to solve the integral?

答:
The best you can get is a Fox-Wright-type power series.

Substitute $k=c^{-1/\alpha}t$ (essentially reducing to $c=1$) and split $\int_{-\infty}^\infty=\int_{-\infty}^0+\int_0^\infty$ (only $\alpha>0$, $\alpha\neq 1$ and $\Re\beta>-1$ are assumed below). The problem is then reduced to $$f(\alpha,\beta,z)=\int_0^\infty t^\beta\exp(-zt-t^\alpha)\,dt,$$ where $z\in\mathbb{C}$, and $\Re z\geqslant 0$ if $\alpha<1$ (to have a convergent integral). Now, if $\alpha>1$, then $$f(\alpha,\beta,z)=\sum_{n=0}^\infty\frac{(-z)^n}{n!}\int_0^\infty t^{n+\beta}e^{-t^\alpha}dt=\frac1\alpha\sum_{n=0}^\infty\frac{(-z)^n}{n!}\Gamma\left(\frac{n+\beta+1}{\alpha}\right).$$ If $0<\alpha<1$ and $\Re z>0$, we expand $e^{-t^\alpha}$ this time, and get $\int_0^\infty t^{\eta-1}e^{-zt}\,dt$, which is equal to $\Gamma(\eta)/z^\eta$ even in our case of complex $z$ (this is where contour integration comes into play): $$f(\alpha,\beta,z)=\sum_{n=0}^\infty\frac{(-1)^n}{n!}\int_0^\infty t^{\alpha n+\beta}e^{-zt}\,dt=\sum_{n=0}^\infty\frac{(-1)^n}{n!}\frac{\Gamma(\alpha n+\beta+1)}{z^{\alpha n+\beta+1}}.$$ (The case $\Re z=0$ is covered by taking a limit. The case $\alpha=1$ is trivial.)


问:
Residual vectors and Range
I wanted to understand the question here a little better: Let $r = b - Ax$ be the corresponding residual vector. Which of the following three vectors is a possible value for $r$? Why?

If I understand correctly, the hint posted by the one answerer there implies that I should find the range of $A$ and then check whether the vectors from (a), (b), and (c) are orthogonal to see if they are truly residuals in this case. Lets call these vectors $r_1, r_2, r_3$ respectively.

I know the range of $A$ can be defined as the span of its columns. My question here is, should I start finding all possible dot products of $r_1, r_2, r_3$ with the columns of $A$, and check if any of them come out to zero? I'm not sure that would work. How do I check if  $r_1, r_2, r_3$ is orthogonal to the range?

答:
While it's true that the residual vector needs to be orthogonal to every linear combination of the columns (i.e. every vector in the columnspace), you don't need to compute the dot product with every such vector. It suffices to just check that the possible residual vector is orthogonal with a basis (or even just a spanning set) for the columnspace.

Since every column is in the columnspace, certainly every residual vector must be perpendicular to every column. Conversely, if $r$ is perpendicular to every column, then by linearity of the map $x \mapsto x \cdot r$, then $x$ is perpendicular to every linear combination of columns.

So, you really don't have to worry about checking every vector in the columnspace.

Title: Residual vectors and Range


问:
Confusion regarding the proof of the Multinomial theorem
We saw the following theorem in class:

If $n \in \mathbb{N}$ and $z_1, \dots, z_m \in \mathbb{C}$ we have:

$
(z_1 + \dots + z_m)^n= \sum_{k_1+\dots+k_m=n} \binom{n}{k_1, \dots, k_m} z_1^{k_1} \dots z_m^{k_m}
$

And I tried to understand the combinatorial proof of this theorem, which is as following:

We first consider a term in the expansion of $(z_1 + \dots + z_m)^n$. This must be of the form $\alpha \prod_{i=1}^k x_i^{\beta_i}$ for some integer $\alpha $ and some integers $\beta_i$.

Now it says that: Since each term must come from choosing one summand from $z_1 + \dots + z_m$ we must have $\beta_1+\beta_2+ \dots + \beta_k =n$. That what confuses me here is, what does it mean when one says that each term must come from choosing one summand? For example $(x_1 +x_2)^2 = x_1^2+2x_1x_2+x_2^2$. Then the term $2x_1x_2$ doesn't come from choosing one summand, right? We choose $x_1$ and $x_2$.

Further they say from that it follows that $\beta_1+\beta_2+ \dots + \beta_k =n$. I see that this is true, but I don't see why this has to follow from the statement "choosing one summand from $z_1 + \dots + z_m$".

In the end the argument is like: THe number of different ways that we can get this term will be the number of ways to choose $\beta_1$ copies of $x_1$, $\beta_2$ copies of $x_2$ and so on. How can I conclude from that the claim?

答:
For example, $(x+y)^2=x^2+2xy+y^2$ comes from choosing
$$
\begin{array}{c|c|c}
&(x+y)&(x+y)\\
\hline
x^2&x&x\\
xy&x&y\\
yx&y&x\\
y^2&y&y\\
\end{array}
$$
and $(x+y)^3=x^3+3x^2y+3y^2x+y^3$ comes from choosing
$$
\begin{array}{c|c|c|c}
&(x+y)&(x+y)&(x+y)\\
\hline
x^3&x&x&x\\
x^2y&x&x&y\\
x^2y&x&y&x\\
x^2y&y&x&x\\
xy^2&x&y&y\\
xy^2&y&x&y\\
xy^2&y&y&x\\
y^3&y&y&y\\
\end{array}
$$


问:
Finding the value of $c$ that makes this system of equations have a solution

Find the value of $c$ that makes the system of equations below have a solution.
  $$\begin{align*}
u + v + 2w &= 2 \\
2u + 3v - w &= 5  \\
3u + 4v + w &= c
\end{align*}$$

I have taken a suggestion from a similar question asked 2 months ago.  After row reduction, I have:
$$\begin{bmatrix}
1 & 1 &  \hphantom{-}2 & 2  \\
0 & 1 & -3 & 1  \\
0 & 0 &  \hphantom{-}1 & (3-c)/2
\end{bmatrix}$$
From here I am confused on how to actually solve for $c$.

Any help would be greatly appreciated!

答: Adding the left hand sides of the first two equations gives the left hand side of the third equation, so $c=7$ is the only possibility. </p>


问:
How should I prove that $c^n/n!$, where c is a constant, tends to zero for a large n.
I want to prove that
$$\frac{c^n}{n!}$$ tends to zero when $n$, a positive integer, is quite large.
I do not think I can use L'Hospital's rule here as $n$ is discrete. Should we assume $n$ to be continuous to sove this problem. Furthermore, I do not find approximating $n!$ with gamma function of much help.

答:
Suppose $c=2$ for convenience.  Then
$$
\frac{2^n}{n!}=\frac{2}{n}\cdot\frac{2}{n-1}\cdot\ldots\cdot\frac{2}{3}\left(\frac{2}{2}\cdot\frac{2}{1}\right)\leq\left(\frac{2}{3}\right)^{n-2}\left(\frac{2}{2}\cdot\frac{2}{1}\right)\xrightarrow{n\to\infty}0.
$$
This can easily be generalized to any $c>0$.


问:
Combination of two sets of basis
Given two sets of basis $V = \{v_1,...,v_n\}$ and $U = \{u_1,...,u_n\}$, each spanning $\mathbb{R}^n$, is it possible to select a subset of the two sets $T$ such that T = $\{v_1\, ...,v_m\} \cup \{u_1\, ...,v_o\}$ with $m+o = n$ such that $T$ spans $\mathbb{R}^n$?

答: You can do this many ways. Start with any subset of $U$. Then go through the elements of $V$ one at a time in any order you like. Add each to your growing basis whenever it is independent of what you have so far. Since $V$ by itself spans the space, when you are done you will have a basis of the space consisting of some elements of $U$ and some of $V$.

Title: Combination of two sets of basis


问:
Find $D \subset \mathbb{R}^3$ such that $D \oplus (A\cap B)$

If $A \cap B = \{x \in \mathbb{R^3} : 2x_1+2x_2-2x_3=0\}$. Find $D \subset \mathbb{R}^3$ such that $D \oplus (A\cap B)= \{x \in \mathbb{R^3} : x_1+2x_2-x_3=0\}.$

If I have $\oplus$ then I would have to have that $D \oplus(A\cap B) = \{0\}$? Forgetting about this condition for a minute I have that $D + (A\cap B) = \{(v +(x_1,x_2,x_1+x_2), v \in D, (x_1,x_2,x_1+x_2) \in A\cap B \}$. Now from the equality I have $$\{(v +(x_1,x_2,x_1+x_2), v \in D, (x_1,x_2,x_1+x_2) \in A\cap B \}=\{x \in \mathbb{R^3} : x_1+2x_2-x_3=0\} = \{(x_1,x_2,x_1+2x_2)\}$$ so $(v_1,v_2,v_3)+(x_1,x_2,x_1+x_2) = (x_1+v_1, x_2+v_2, x_1+x_2+v_3) = (x_1,x_2,x_1+2x_2)$. I get $$x_1=x_1+v_1 \\ x_2=x_2+v_2 \\x_1+2x_2=x_1+x_2+v_3$$ I'm not sure if I'm getting in the right direction here?

答:
As this problem is stated, there is no solution, i.e. there does not exist any $D$ as asked for.

The reason is that for any subspaces $D$ and $C$, if the direct sum $D \oplus C$ exists then $C \subset D \oplus C$, because $D \oplus C = D + C$ and for each $x \in C$ we have $x = 0 + c \in D + C$.

However, the set $C = \{x \in \mathbb R^3 \mid 2x_1 + 2x_2 - 2x_3 = 0\}$ is not a subset of the set $\{x \in \mathbb R^3 \mid x_1 + 2 x_2 - x_3 = 0\}$: take $x = (1,1,2)$ as a counterexample.


问:
Metric spaces - sets proof
Let $(X,\rho)$ be a metric space and $A\subset X$. Prove that $\partial A = \overline A \cap \overline{X \setminus A} $.
I have no idea how to prove that. Please help.</p>

答:
To show two sets are equal, first assume we have an element of the first set, and show it is in the second set.  Then, assume we have an element of the second set, and show it is in the first.

Proof outline:

Let $x \in \partial A$ arbitrary.
Show that $x \in \overline{A}$ (this should be easy from your definition).
Then, show that $x \in \overline{X \setminus A}$ using the fact that $\partial A = \partial (X \setminus A)$.

Let $y \in \overline{A} \cap \overline{X \setminus A}$ arbitrary.
Show that $y \in \partial A$ by showing that $y \in \overline{A}$ but $y \not \in \text{Int } A$.

Title: Metric spaces - sets proof


问:
Count number of simple undirected Graphs with exactly one vertex degree of $N - 1$
Let us only be interested in simple undirected graphs.

Find the number of all graphs of $N$ vertices, where exactly one vertex has degree $N - 1$. For example $f(3) = 3, f(4) = 16$.

My idea is: let us know the answer for a fixed vertex number one $V_1$, provided that it is connected to all the others $p_1(N)$, then $f(N) = N * p_1(N)$. So now we should add a new one vertex and count $p_1(N + 1)$. We should connect $V_{N + 1}$ and $V_1$ to hold the task condition, now it's possible to connect new vertex with another N - 1 in different combinations, except only one case, when $V_{N + 1}$ connected with all others, total number is $\sum_{k=0}^{n-2} \binom{n - 1}{k}$. But with the addition of a new vertex, there are cases when some of the vertices $V_2, V_3, ..., V_N$ can now be connected to each other and not included in $p_1(N)$, I don't know how to count such cases.

Are there any other ideas?

P.S. time complexity should be not greater than $O(N^2)$

答:
As you said, you choose the vertex with degree $N-1$ in $N$ ways, and then choose the other connections. If you ignore the vertex with degree $N-1$, what remains is a graph on $N-1$ vertices where no vertex is connected to every other vertex. Taking the complement, this is equivalent to a graph with on $N-1$ vertices with no isolated vertices.

The problem of counting graphs with no isolated vertices has been discussed multiple times on this site. It can be solved fairly easily in $O(N)$ time using the principle of inclusion-exclusion. See

Number of graphs without isolated vertices, which also points to the relevant OEIS sequence: https://oeis.org/A006129.

Number of directed graphs without isolated vertices (The answer here is for directed graphs, but the logic can be modified to undirected graphs).


问:
Sigma notation but each result is stored in a new sequence of equal length (no adding)
My background is Computer Science, so I recognise $\Sigma$ as a for loop that exclusively sums, returning one element/ answer.

What is the notation used to loop over a finite sequence, to perform any calculation, and store the output into a new sequence of equal length?

i.e. I don't want to add each iteration's answer together. One element should yield one unique answer.

All of your comments and answers have given me brilliant insight. I will still be working on this problem as good old fashioned practice. Thank you all.

答:
General Comment

Outside of a block of pseudocode describing an algorithm, mathematical notation is basically stateless. Rather than an imperative programming style which would have things like "for loops", mathematics is usually written like a declarative style. Because of this, the verb "store" (as in "store the output") doesn't really match up with any standard mathematical notation.

Main Answer

If the operation only involves the current entry in the sequence, then variations of the the idea in Blue's comment would be common. Define $f$ to be the function that outputs the "any calculation", and define something like $a_0,a_1,\ldots,a_n$ to be the initial entries of the sequence. Then the new sequence could be written like $\left(\;f(a_k)\;\right)_{k=0}^n$ or $\left\langle f\left(a_k\right)\right\rangle_{k}$, etc., depending on the author's personal style.

To answer a follow-up comment, the sequence can be indexed in whatever way suits your needs in context. For example $a_7,a_8,a_9$ and $\left(\;f\left(a_k\right)\;\right)_{k=7}^9$ would be fine. If the original sequence is a subsequence of some sequence discussed earlier, you could even write things like $\left\langle f\left(a_{k^2-k}\right)\right\rangle_{k=1}^{17}$. In practice, you often might not have need to bundle the sequence together like this, and might just write things like "Let $b_n=f(a_n)$ for all $n\in\left\{7,\ldots,93\right\}$."

Also, this notation has no trouble with infinite sequences. For example:

Define $b_n$ to be $n!$ (the factorial), and define $f(k)=k^3$. Then define $c_n=f(b_n)$ for all $n$. Note that the infinite sequence $\left\langle c_n\right\rangle_{n=1}^{\infty}$ grows even faster than the $a_n$ do.

Sums

As mentioned before, mathematics isn't really written in an imperative way. So mathematical sums with $\sum$ are not best thought of as a for loop. Instead, I would say that the approach in mathematics is often similar to functional programming. With that in mind, $\sum$ is probably most analogous to a fold/reduce with the addition operation and $0$ as the empty sum.

${\displaystyle \sum_{i=1}^{1000}} 2i$ could correspond to things like foldr (+) 0 [2*i | i<-[1..10^3]] in Haskell (Try it online!). In Python, the equivalent (inspired by this SO answer) would be reduce(add, [2*i for i in range(1,10**3+1)], 0) (Try it online!) using the functools and operator modules (and ignoring the existence of sum).


问:
Prove that $\angle{BEC}=\angle{DAC}$.
A convex quadrilateral $\quad{ABCD}$ has $AD=CD$ and$\angle{DAB}=\angle{ABC}<90$. The line through $D$ and the midpoint of $BC$ meets line $AB$ in $E$. Prove that $\angle{BEC}=\angle{DAC}$.

I have to approaches: Either we can prove that $EB.EA=EC^2$ or we can prove that line $AD$ is the tangent to the circumcircle of $\triangle{ACE}$ as shown, but I am not getting how to prove them. Pure geometrical method is preferred.

答:
A shorter approach using symmedians:

It suffices to show that if $\angle BEC = \angle DAC$, then $ED$ passes through the midpoint of $BC$.

From $\angle BEC = \angle DAC$ we obtain that $DA$ (and hence also $DC$) is tangent to the circle through $A$, $C$ and $E$. It follows that $ED$ is the $E$-symmedian in triangle $ACE$. Because $\angle BCE = \angle BAC$, the line $BC$ is an antiparallel in triangle $ACE$, so its midpoint lies on the symmedian.

(The $A$-symmedian in triangle $ABC$ is the reflection of the median through $A$ in the angle bisector of angle $\angle BAC$. I used the known facts that the $A$-symmedian passes through the pole of $BC$ with respect to the circumcircle (i.e., the point of the intersection of the tangents at $B$ and $C$) and that the $A$-symmedian pases through the midpoint of every antiparallel (i.e. a line segment $XY$ with $X$ on $B$ and $Y$ on $AC$ so that $B$, $C$, $X$, $Y$ are concyclic).

Title: Prove that $\angle{BEC}=\angle{DAC}$.


问:
Find all $z\in\mathbb{C}$ such that $i\sin\left(z\right)+\cosh\left(iz\right)=\left(1+i\sqrt{3}\right)^i$.
Firstly, I tried using the identity $\cos\left(ix\right)=\cosh\left(x\right)$ to arrive at this expression: $$i\sin\left(z\right)+\cos\left(z\right)=\left(1+i\sqrt{3}\right)^i$$
Then, using Euler's formula, I got: $$e^{iz}=\left(1+i\sqrt{3}\right)^i$$

That's where I'm at. I thought about using logarithms but I'm not comfortable using them in the complex realm. Thanks.

答: Since $1+j\sqrt{3}=e^{\ln2+\pi j/3+2n\pi j}$ with $n\in\Bbb Z$, $(1+j\sqrt{3})^j$ is multivalued as $e^{j[\ln 2+\pi j/3+2n\pi j]}$, so $z=\ln 2+\pi j/3+2n\pi j$.


答:
Using $\;z=x+iy\;$ (eventually):

$$i\sin z+\cos z=i\frac{e^{iz}-e^{-iz}}{2i}+\frac{e^{iz}+e^{-iz}}2=e^{iz}=e^{i(x+iy)}=e^{-y}(\cos x+i\sin x)$$

On the other hand ( choosing say the standard value of the argument in $\;(-\pi,\pi)\;$ )

$$1+\sqrt3\,i=2e^{\pi i/3}\implies\left(1+\sqrt3\,i\right)^i=2^i e^{-\pi/3}$$

Finally,

$$2^i=e^{i\left(\ln 2\right)}=\cos\ln 2+i\sin\ln 2$$

Now put the above together...

Title: Find all $z\in\mathbb{C}$ such that $i\sin\left(z\right)+\cosh\left(iz\right)=\left(1+i\sqrt{3}\right)^i$.

Body:
Firstly, I tried using the identity $\cos\left(ix\right)=\cosh\left(x\right)$ to arrive at this expression: $$i\sin\left(z\right)+\cos\left(z\right)=\left(1+i\sqrt{3}\right)^i$$
Then, using Euler's formula, I got: $$e^{iz}=\left(1+i\sqrt{3}\right)^i$$

That's where I'm at. I thought about using logarithms but I'm not comfortable using them in the complex realm. Thanks.


问:
Matrix monotonicity $A \ge B$ then $B^{-1} \ge A^{-1}$?
I am wondering if it is true that if in the sense of matrices we have $A \ge B$ that in case both of them are invertible, we also have that $B^{-1} \ge A^{-1}$?

答:
I am supposing by $A \geq B$, you meant $A$ and $B$ are symmetric, and $A - B$ are positive semi-definite.

Your conjecture is not true. For example, take
\begin{align*}
A = I_{(2)}, \quad B = -I_{(2)},
\end{align*}
then $A - B = 2I_{(2)} \geq 0$, but $B^{-1} - A^{-1} = -2I_{(2)} \leq 0$.

On the other hand, given an additional condition $A \geq B \color{red}{> 0}$, then $B^{-1} \geq A^{-1}$ is indeed true. You can try to prove it.


问:
Probability for dummies
I am the least mathematical person around, so apologies if this question is really dumb, but I'm trying to improve! I've been reading loads of examples everywhere but I'm having a hard time applying the logic/rules of probability to new problems.

Let's say you have to win three out of three rounds of a game in order to win a prize. It is a single player game.

The probability of a boy winning a round is $.25$, and the probability of a girl winning a round is .4. Winning one round doesn't influence the result of the next round. So if I haven't misunderstood, the probability of a girl winning a prize is $.4 \cdot .4 \cdot .4 = 0.064$ and the probability of a boy winning a prize is $.25 \cdot .25 \cdot .25 = 0.016$

Now, this is where I'm stuck. What's the overall probability of a person winning a prize if $50\%$ of the players are girls and $50\%$ of the players are boys? Is it just $.016 + 0.064 = 0.08$? Or should I be dividing by $2$ here somewhere given that it's $50$ percent boys and $50$ percent girls.

Thanks in advance for your help.

答: Hint: Would it be reasonable for the probability of a random person winning be higher than both the 'girl' probability and the 'boy' probability?</p>

Title: Probability for dummies


问:
What's the cardinality of the infinity introduced in Riemann sphere model?
In the Riemann sphere model, people extend the complex plane with a point at infinity: $\mathbb {C} \cup \{\infty \}$ and claim this expression: $\frac{1}{0}=\infty$.

I'm confused with what is this $\infty$. Is it a member of $\{\aleph_n\}$ numbers?
When we study infinities, we usually discuss cardinality. Does this $\infty$ corresponds to something's cardinality?

答:
Does it has a cardinality?

No. The $\infty$ there doesn't represent a particular set (though you might have to encode it as a set if you're using set theory as your foundation), nor the cardinality of a particular set.

What's the property of this infinity?

The properties are discussed throughout the English Wikipedia article for "Riemann sphere". Is there a particular property you have a question about?


答:
The sum $\sum_{k=0}^{n-1}a[k]$ over a length-$n$ iterable (let's say an array) $a$ can be achieved in Python as functools.reduce(operator.add, a). Your question is effectively how a mathematician would denote functools.reduce(f, a) for an arbitrary binary function $f$ (unless you only care about how it's said in programming instead, in which case I've already answered that, albeit somewhat stepping on @MarkS.'s toes in doing so).

I don't think there is a general notation. Some special cases have an equivalent of $\sum$, such as $\prod$ for multiplication, $\bigcup$ ($\bigcap$) for sets' unions (intersections), and so on. The best option would be to explicitly define something analogous to partial sums. So let's define partial calculations $C_n$ by$$C_1:=a_0,\,C_{k+1}:=f(C_k,\,a_k)$$(forgive the $+1$s; they result from me using Python, which is $0$-indexed, in the above example, but then many (most?) programming languages are $0$-indexed). In practice, you'd have to present this definition to your readers.

The sum I gave in Python could be written as functools.reduce(operator.add, a, 0). This allows a length-$0$ choice of a to be supported. The equivalent for multiplication is functools.reduce(operator.mul, a, 1). The extra argument checks because $0+x=1\times x=x$. General $f$ won't have a left-identity, which is why I started with $C_1$ rather than $C_0$, which would have to be the left-identity. (We call $0$ an empty sum; we call $1$ an empty product.)

Title: Sigma notation but each result is stored in a new sequence of equal length (no adding)

Body:
My background is Computer Science, so I recognise $\Sigma$ as a for loop that exclusively sums, returning one element/ answer.

What is the notation used to loop over a finite sequence, to perform any calculation, and store the output into a new sequence of equal length?

i.e. I don't want to add each iteration's answer together. One element should yield one unique answer.

All of your comments and answers have given me brilliant insight. I will still be working on this problem as good old fashioned practice. Thank you all.


问:
Maximum likelihood of log-normal distribution
I am trying to find the maximum likelihood function of log-normal distribution for both parameters $\mu$ and $\sigma^2$. I've gotten the derivative of the log-likelihood for $\mu$ to be

$$\frac{\partial l}{\partial\mu}=\frac{\sum^n_{i=1}\ln x_i}{\sigma^2}-\frac{n\mu}{\sigma^2}$$

and found the maximum likelihood estimator by setting the derivative to zero which came about to be

$$\frac{\sum^n_{i=1}\ln x_i}{\sigma^2}-\frac{n\mu}{\sigma^2} = 0 \iff \hat\mu=\frac{\sum^n_{i=1}\ln x_i}{n}.$$

However, the teaching assistant of the course told me that I should explain why the maximum likelihood estimator is found when the derivative is zero and I should've set the derivative to greater or equal to instead so

$$\frac{\sum^n_{i=1}\ln x_i}{\sigma^2}-\frac{n\mu}{\sigma^2} \ge 0.$$

I thought I could figure out this by myself but to be honest, I don't understand why I should've set it to greater or equal instead.

答:
However, the teaching assistant of the course told me that I should explain why the maximum likelihood estimator is found when the derivative is zero

Because when you are looking for a maximum of a function one possible procedure is to find a stationary point (where its first derivative is zero) and then test its second derivative (if it is negative the point is an argmax; it's a point whose image is a maximum)

Remember that is not always true that the MLE is found where the derivative is zero. As a simple example, find MLE for $\theta$ considering a $n$-size random sample from a uniform population $U(0;\theta)$


问:
Finding eigenvectors of a $2 \times 2$ matrix
For the following matrix,

\begin{bmatrix}1&h\\0&1\end{bmatrix}

I was able to find the eigenvalues to be $1$ with multiplicity $2$. How to find the eigenvectors?

答:
Using Python + SymPy:

```<code>>>> from sympy import *
>>> h = Symbol("h")
>>> A = Matrix([[1, h],
                [0, 1]])
</code>```

Let us try to diagonalize the matrix:

```<code>>>> A.diagonalize()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/usr/local/lib/python3.8/site-packages/sympy/matrices/matrices.py", line 377, in diagonalize
    return _diagonalize(self, reals_only=reals_only, sort=sort,
  File "/usr/local/lib/python3.8/site-packages/sympy/matrices/eigen.py", line 605, in _diagonalize
    raise MatrixError("Matrix is not diagonalizable")
sympy.matrices.common.MatrixError: Matrix is not diagonalizable
</code>```

Matrix $A$ is not diagonalizable for general $h$, though it is trivially diagonalizable when $h=0$. Thus, let us compute a Jordan decomposition $A =  P J P^{-1}$:

```<code>>>> P, J = A.jordan_form()
>>> P
Matrix([[h, 0],
        [0, 1]])
>>> J
Matrix([[1, 1],
        [0, 1]])
</code>```

Body:
Using Python + SymPy:

>>> from sympy import *
>>> h = Symbol("h")
>>> A = Matrix([[1, h],
                [0, 1]])

Let us try to diagonalize the matrix:

>>> A.diagonalize()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/usr/local/lib/python3.8/site-packages/sympy/matrices/matrices.py", line 377, in diagonalize
    return _diagonalize(self, reals_only=reals_only, sort=sort,
  File "/usr/local/lib/python3.8/site-packages/sympy/matrices/eigen.py", line 605, in _diagonalize
    raise MatrixError("Matrix is not diagonalizable")
sympy.matrices.common.MatrixError: Matrix is not diagonalizable

Matrix $A$ is not diagonalizable for general $h$, though it is trivially diagonalizable when $h=0$. Thus, let us compute a Jordan decomposition $A =  P J P^{-1}$:

>>> P, J = A.jordan_form()
>>> P
Matrix([[h, 0],
        [0, 1]])
>>> J
Matrix([[1, 1],
        [0, 1]])


问:
What does it mean to say a functor "factors through" a projection?
There is a question on math.SE defining the notion of a map "factoring through" a set, and this makes sense, although I'd really like a reference for this definition besides wikipedia.

Definition. Let $f: A \to C$ be a map. We say $f$ factors through a set $B$ if there exists $g:A \to B$ and $h: B \to C$ such that $f = h \circ g$.

There's a nice commutative diagram, but I don't know how to easily draw it on this site, since there's no tikz-cd.

But I don't quite understand how this definition applies here:

https://ncatlab.org/nlab/show/dinatural+transformation#eq:domain

The author is saying that the functors $F,G$ may factor through the given projection, but it isn't clear to me which of $h, g$ the projection is analogous to. Could anyone clarify and possibly provide a reference for the definition in a category theory text?

答:
There is actually only one possible way to write a factorization diagram of any of those functors, for if $F:C^{op}\times C\to D$ and the projection is $p:C^{op}\times C\to C$, then it must be that $F$ decomposes as functors $C^{op}\times C\xrightarrow{p} C\to D$, since the overall functor should start at $C^{op}\times C$ and end at $D$, having $p$ in between.

So $p$ plays the role of $h$ and the other map in the factorization that I didn't label plays the role of $g$. In general, for any morphism (including maps of sets) one can talk about factorization through another morphism in these terms.

I know you wanted a different source for the reference, but the Wikipedia also explains in this page that in the situation of your definition, one can say that $f$ factors through any (and all) of $B$, $g$ and $h$.


问:
Function that uniformly converges to 0, but doesn't converges to 0 in norm.
Is there any sequence of function ${f_n}$ that fulfill 3 condition?

Each function ${f_n} \in L^2(0, \infty)$, for every $n$.

${f_n}$ converges uniformly to 0 that is: for every $\epsilon > 0$, there is $N \in \mathbb{N}$ such that $n > N \Rightarrow |f_n(x)| < \epsilon$

${f_n}$ doesn't converges in norm $L^2(0, \infty)$. Norm in $L^2(0, \infty)$ is defined by $$\Vert f \Vert ^2 = \int_{0}^{\infty} |f(x)|^2 dx$$

Basically, the question is: sequence of function $f_n$ that converges uniformly to 0 but doesn't converges in norm to 0.

Thank you.

答: Take$$f_n(x)=\begin{cases}\frac1{\sqrt n}&\text{ if }x\leqslant n\\0&\text{ otherwise}\end{cases}$$and use the fact that $(\forall n\in\Bbb N):\|f\|_2=1$.


问:
Showing existence of an element with order $p$

If a group $G$ has order $p^n$, where $p$ is prime and $n \geq 1$, does there exist some element $a\in G$ s.t. the order of $a$ is $p$?

I happen to know that this is true by Cauchy's theorem, but that theorem has not been presented yet in the book. I only have this so far:

Let $a \in G$, then the order of the cyclic subgroup generated by $a$ divides the order of the group $G$ (by Lagrange's theorem).

But I get stuck because I don't know how to show that the $a$ has order $p$.

答:
To ellaborate on Tobias' comment.
You have that the order of $a$ divides the order of $G$, and the order of $G$ is $p^n$.
This means the order of $a$ can only be $1, p^1, p^2, \cdots, p^n$, as these are the only divisors of $p^n$.

First, you need to pick $a$ such that the order is one of $p^1, \cdots, p^n$ (how?).
Then, you need to pick $k$ such that $a^k$ has order equal to $p$.


答:
Let's say that instead, the probability of a boy winning a prize is $0.9$ and the probability of a girl winning a prize is $0.8.$

Then by your logic, the probability of a random person winning a prize is $0.9^3 \times 0.8^3 = 0.729 + 0.512 = 1.241,$ which is impossible because we can't have probabilities of more than $1$.

So you can't add probabilities like that.

What you can and should do is start by drawing a probability tree diagram for what you want. It should look like this:

The probability of a randomly selected person winning the game is then: $0.5 \times 0.016 + 0.5 \times 0.064 = 0.04,\ $ or $\frac{1}{25}.$

Title: Probability for dummies

Body:
I am the least mathematical person around, so apologies if this question is really dumb, but I'm trying to improve! I've been reading loads of examples everywhere but I'm having a hard time applying the logic/rules of probability to new problems.

Let's say you have to win three out of three rounds of a game in order to win a prize. It is a single player game.

The probability of a boy winning a round is $.25$, and the probability of a girl winning a round is .4. Winning one round doesn't influence the result of the next round. So if I haven't misunderstood, the probability of a girl winning a prize is $.4 \cdot .4 \cdot .4 = 0.064$ and the probability of a boy winning a prize is $.25 \cdot .25 \cdot .25 = 0.016$

Now, this is where I'm stuck. What's the overall probability of a person winning a prize if $50\%$ of the players are girls and $50\%$ of the players are boys? Is it just $.016 + 0.064 = 0.08$? Or should I be dividing by $2$ here somewhere given that it's $50$ percent boys and $50$ percent girls.

Thanks in advance for your help.


问:
Euler's phi function and Multiplicative inverse
We can solve the linear congruence with Euler's phi function:
$$105\hspace{3pt} X ≡ 15\hspace{3pt} mod \hspace{3pt}24 $$
Where $$a = 105, b=15,m=24$$
As
$$\phi(m)=\phi(24)=8\rightarrow (105)^7 ≡ 9\hspace{3pt} mod \hspace{3pt}24 $$
$$\rightarrow X ≡ 15\hspace{3pt}\hspace{3pt}(9)\hspace{3pt} mod \hspace{3pt}\hspace{3pt}\hspace{3pt}24 ≡ 15\hspace{3pt} mod \hspace{3pt}24$$
The solution is: $$ X ≡ 7\pm 8 N $$
Where N is a natural number.$$$$
But when we try to solve this linear congruence with Euler's phi function:
$$192\hspace{3pt} X ≡ 24\hspace{3pt} mod \hspace{3pt}27 $$
Where $$a = 192, b=24,m=27$$
We find that
$$\phi(m)=\phi(27)=18\rightarrow (192)^{17} ≡ 0\hspace{3pt} mod \hspace{3pt}27 $$
That's mean we cannot solve this second linear congruence with Eules's phi function$$$$
I know that the solution for the second linear congruence is:
$$ X ≡ 8\pm 9 N $$
I do not know why i can not solve this second linear congruence with Euler's phi function ? Can any one help me?

答:
First you can divide all by three to make things easier:

$$192x=24\pmod{27}\iff 192x=24+27k\,,\,\,k\in\Bbb Z\stackrel{\div3}\implies 64x=8+9k$$

and since $\;64=1\pmod 9\;$ , we get

$$64x=8\pmod 9\implies x=8\pmod 9$$

and we get the solution.


问:
Carleson embedding theorem
The Carleson embedding theorem gives a criterium by which to decide when for a measure $\mu$ the operator that takes a function on the real line to its harmonic extension (by convolution with Poisson kernel) on the upper half-plane is bounded from $L^2(\mathbb{R})$ to $L^2(\mathbb{R}^2_+,\mu)$.

Why is this theorem so significant? Why is this question interesting? Why is it called "embedding" theorem?

答:
Why is this theorem so significant?

Because it's used to prove other theorems. See Operators, Functions, and Systems: An Easy Reading (Volume 1) by Nikolski, page 130.

Why is this question interesting?

This is subjective. If you do not find a result interesting, then it's not interesting to you.

Why is it called "embedding" theorem?

Because a bounded injective operator between spaces of functions is often called an embedding.

Title: Carleson embedding theorem


答:
The problem is that $(192,27)=3\gt1$.  That's $192$ and $27$ are not coprime.  Thus Euler is not directly applicable.  $192$ doesn't have an inverse $\bmod27$.

In fact you run into the same problem on the first one.  Recall that the theorem says that $a^{\varphi(n)}\equiv1\bmod n\color{blue}{\text{ when }(a,n)=1}$.

One way is to invoke CRT when this happens.

Title: Euler's phi function and Multiplicative inverse

Body:
We can solve the linear congruence with Euler's phi function:
$$105\hspace{3pt} X ≡ 15\hspace{3pt} mod \hspace{3pt}24 $$
Where $$a = 105, b=15,m=24$$
As
$$\phi(m)=\phi(24)=8\rightarrow (105)^7 ≡ 9\hspace{3pt} mod \hspace{3pt}24 $$
$$\rightarrow X ≡ 15\hspace{3pt}\hspace{3pt}(9)\hspace{3pt} mod \hspace{3pt}\hspace{3pt}\hspace{3pt}24 ≡ 15\hspace{3pt} mod \hspace{3pt}24$$
The solution is: $$ X ≡ 7\pm 8 N $$
Where N is a natural number.$$$$
But when we try to solve this linear congruence with Euler's phi function:
$$192\hspace{3pt} X ≡ 24\hspace{3pt} mod \hspace{3pt}27 $$
Where $$a = 192, b=24,m=27$$
We find that
$$\phi(m)=\phi(27)=18\rightarrow (192)^{17} ≡ 0\hspace{3pt} mod \hspace{3pt}27 $$
That's mean we cannot solve this second linear congruence with Eules's phi function$$$$
I know that the solution for the second linear congruence is:
$$ X ≡ 8\pm 9 N $$
I do not know why i can not solve this second linear congruence with Euler's phi function ? Can any one help me?