问:
Prove $\mathbf b\times(\mathbf aM)-\mathbf a\times(\mathbf bM)=(\mathbf a\times\mathbf b)M-(\mathrm{tr}M) (\mathbf a\times\mathbf b)$
I want to prove the following identity:
For $\mathbf a,\mathbf b\in \mathbb R^3, M\in M_3(\mathbb R)$, we have
$$\mathbf b\times(\mathbf aM)-\mathbf a\times(\mathbf bM)=(\mathbf a\times\mathbf b)M-(\mathrm{tr} M )(\mathbf a\times\mathbf b)$$

One direct way to prove it is to write out the components of $\mathbf a,\mathbf b$ and $M$, then compare the result of $LHS$ and $RHS$ by brute force, but this way seems not so beautiful. Is there any nice (elegant) way to prove this identity? I guess probably there's a geometric interpretation of this identity.

答:
The proposed identity may be written as
$$(\epsilon_{ijk}\delta_{ml} -\epsilon_{imk}\delta_{lj} -\epsilon_{lmj}\delta_{ki} +\epsilon_{imj}\delta_{lk})a_mb_jM_{lk}=0$$
where the expression in brackets is almost one of the identities listed in equations (5)-(9) in "The Isotropic Invariants of Fifth-rank Cartesian Tensors", Int. J. Quantum Chem. vol V 381-386 (1971).

Is the author of the question certain the first term on the right hand side of their identity is not $M \cdot ({\bf a}\times{\bf b})$, rather than $({\bf a}\times{\bf b})\cdot M$? (Assuming $M$ is not symmetric.) If it were, then the third term in the brackets would be $\epsilon_{kmj}\delta_{li}$, which yields a more symmetric result, and the expression in the brackets does vanish!


问:
Find flow of the vector field $\overrightarrow{\operatorname{rot}F}$
We've given 5 points in $\mathbb{R}^3$: $A=(1,0,0)$, $B=(0,1,0)$, $C=(0,0,1)$, $D=(1,1,0)$, $E= (1,1,1)$. We have a surface $S$ given by triangles $ADE, DBE, BCE, CAE$. We have a vector field: $$\overrightarrow{F}= \begin{bmatrix} xz\\-yz\\\frac{xyz}{\sqrt{x^2+y^2+z^2}}\end{bmatrix}.$$ My task is to compute flow of the vector field $\overrightarrow{\operatorname{rot}F}$ from negative side (seen from point $(\frac{1}{2}, \frac{1}{2}, \frac{1}{2})$) to the positive side.
Can somebody show me the solution. My attempt was the following: I found $$\overrightarrow{\operatorname{rot}F} = \begin{bmatrix} \frac{xz(x^2 + z^2)}{(x^2+y^2+z^2)^{\frac{3}{2}}} + y\\x -\frac{yz(y^2 + z^2)}{(x^2+y^2+z^2)^{\frac{3}{2}}} \\0\end{bmatrix}.$$ And my problem is that in every method used to compute the flow of some vector field, we need a parametric representation- here whether parametric representation of $S $ or parametric representation of are bounded by $S$ and triangle $ABC$. I don't know how to find neither of them. Can somebody show me how to solve my exercise?</p>

答: Do not try to compute the flux directly. Use Stokes's Theorem. When you piece the four triangles together, what is the boundary curve?

ContentLicense: CC BY-SA 3.0


问:
Reducible/Irreducible 2x2 matrices
as far as I understand, one considers that an irreducible matrix only contains trivial invariant subspaces (the whole space, the null vector and its eigenvectors). So, are there 2x2 reducible matrices?, they only contain trivial invariant spaces. Looking for examples of reducible/irreducible matrices, I only find at most 3x3 matrices.

Thanks in advance. Sorry if I am being imprecise.

答:
How about
$A=\begin{pmatrix}
  1 & 0\\
  1 & 1
\end{pmatrix}$?

$A^n=\begin{pmatrix}
  1 & 0\\
  n & 1
\end{pmatrix}$.

ContentLicense: CC BY-SA 4.0

Title: Reducible/Irreducible 2x2 matrices


问:
Can $f(\Theta)$ be open in $\mathbb{R}^k$, where $\Theta\subset\mathbb{R}^{k-1}$?
Can $f(\Theta)$, the image of $f$ defined as $\{f(\theta),\theta\in\Theta\}$ be open in $\mathbb{R}^k$, where $\Theta\subset\mathbb{R}^{k-1}$? $f$ is a function from $\Theta$ to $\mathbb{R}^k$.

I don't think so, otherwise $f(\Theta)$ would contain a open ball in $\mathbb{R}^k$, which is not possible. Is this correct? If not, can you please give me an example?

答: We don't have to get as fancy as a Peano curve. The cardinality of $\mathbb R$ equals the cardinality of  $\mathbb R^k.$ That alone says the answer is yes.


问:
Why is the set of elements in a valuation ring $R$ with positive valuation the unique maximal ideal of $R$?
Definition. Let $K$ be a field and let $G$ be a totally ordered abelian group. A valuation of $K$ with values in $G$ is a map $v: K-\{0\} \rightarrow G$ such that for all $x, y \in K, x, y \neq 0,$ we have
$$
\begin{array}{l}
\text { (1) } v(x y)=v(x)+v(y) \\
\text { (2) } v(x+y) \geqslant \min (v(x), v(y)) .
\end{array}
$$

If $v$ is a valuation,show that :

$R=\{x \in K \mid v(x) \geqslant 0\} \cup\{0\}$ is a subring of $K$.

$m=\{x \in K \mid v(x)>0\} \cup \{0\}$ is unique maximal ideal of $R$.

I can solve 1), but how can we show $m$ is an ideal of $R$ and also the unique maximal ideal of $R$?

答:
That $m$ is an ideal follows readily from the definition of the valuation since:

if $x$, $y$ are in $m$ then $v(x+y)\geq\min(v(x),v(y))>0$ so that $x+y\in m$;

if $x\in m$ and $r\in R$ then $v(xr)=v(x)+v(r)>0$ so that $xr\in m$.

To show that $m$ is maximal, observe that if $0\neq r\in R$ but $r\notin m$ then $v(r)=0$. Thus $v(r^{-1})=-v(r)=0$ and $r^{-1}\in R$, i.e. $r$ is invertible.

Then you can use the following well known fact: If $R$ is a ring and $m\subset R$ an ideal such that all elements in $R$ not in $m$ are invertible, then $m$ is a maximal ideal.


问:
How to prove that the average velocity is $0.5\alpha$
Question: Suppose $g$ is a constant. If $d(t) = 0.5gt^2$ represents the distance of a particle falling from rest at time $t$ ($t = 0$). Let the velocity after some time $T$ be given by $\alpha$. Prove the average velocity is $0.5\alpha$.

I am not exactly sure how to prove this, but here is my attempt.

If $d(t) = 0.5gt^2$, then the derivative of the function would be $v(t) = gt$. I am assuming this exercise is asking us to do this using the average function value, from $0$ to $\alpha$, in order to find the average velocity, so: $$v_{avg}(t) = \frac{1}{b - a}\int_{a}^{b} v(t) \ dt = \frac{1}{\alpha}\int_0^\alpha gt \ dt = \frac{1}{\alpha}\left[\frac{1}{2}gt^2\right]_0^\alpha = 0.5g\alpha$$

This is what I got, and it is not the answer I am supposed to get. I would appreciate the assistance. Where did I go wrong here?

答:
The velocity $v$ at time $t$ is given by $d'(t)=v(t)=gt$.

At $t=T$ we get that $v(T)=gT=\alpha\implies g=\frac{\alpha}{T}$.

Therefore, $d(t)=\frac{\alpha}{2T}t^2$. To get the average speed we use $$v_\text{avg}=\frac{\frac{\alpha T}{2} - 0}{T-0}=\frac{\alpha}{2}$$


问:
I don't understand equivalence classes with relations
I am not quite understanding equivalence classes. For example I have this problem:

Let $A$ be the set of integers and
$\quad a\;R\;b\quad$ if and only if $\quad |a| = |b|$.

I have proved that this is an equivalence relation, (its reflexive, symmetric and transitive), but how do I show the equivalence classes?

答:
Its best to begin with the following definition.

The equivalence class associated with $a \in A$ with respect to $\sim$ will be denoted $[a],$ and has defining property: $b \in [a]$ iff $b \sim a$.

We can then define the term 'equivalence class' as follows.

An equivalence class of $A$ with respect to $\sim$ is subset $B \subseteq A$ such that $B=[a]$ for some $a \in A$.

We can now discover the equivalence classes iteratively. Choose $a_0 \in A$ and work out what $[a_0]$ is. Then choose $a_1 \in A$ such that $a_1 \notin [a_0]$ and work out what $[a_1]$ is. Then choose $a_2 \in A$ such that $a_2 \notin [a_1] \cup [a_0]$, and work out what $[a_2]$ is. etc.

For example, in the case of $A =$ the integers, we might proceed as follows.

Choose $a_0 = 0$.

Then $b \in [a_0]$ is equivalent to $b \sim a_0$, which is equivalent to $|b| = |a_0|$, which is equivalent to $|b|=0$, which is equivalent to $b=0$. So $b \in a_0$ iff $b=0$, or in other words $a_0 = \{0\}$.

Next choose $a_1 = 1$. Notice $1 \notin [a_0]$.

Then $b \in [a_1]$ is equivalent to $b \sim a_1$, which is equivalent to $|b| = |a_1|$, which is equivalent to $|b|=1$, which is equivalent to $b=-1$ or $b=1$. So $b \in a_1$ iff $b=-1$ or $b=1$, or in other words $a_1 = \{-1,1\}$.

Next choose $a_2 = 2$. Notice $2 \notin [a_1] \cup [a_0]$.

Wash, rinse and repeat.

At some point, you'll intuit that the equivalence classes of $A$ with respect to $\sim$ are precisely the $B \subseteq A$ such that there exists natural $n$ such that $B = \{-n,n\}$. Note this is only true if we regard zero as a natural number.


问:
What operation is $\circ$?
Given the following exercise:

"Consider the operation $\circ$ defined on $G_4 =\{ 0, 1, 2, 3\}$ by Table 3. Check whether it satisfies each of the three axioms from the definition of a group, and decide whether it is a group."

Table 3:

The properties of a group:

$1$. Asssoiativity: for any $a, b, c \in G_4$, we have $a\circ (b\circ c)= a\circ (b\circ c)$

$2$. Identity: for any $a \in G_4$, we have $a\circ e = e\circ a=a$

$3$. Inverse: for every $a \in (G_4,\circ ) $, theres is an inverse $a^{-1}$, such that $a\circ a^{-1}=a^{-1}\circ a=e$

Previosly I have concluded whether the sets $(G_4,+)$ and $(G_4,*)$ are groups by doing the following:

But I dont know what the $\circ$ operation for Table 3 is doing. Since we are working with numbers and not functions I am stuck trying to figure this out. I know that to find the inverses I can check if there's a value in every row that's equal to the identity value $e$, but I cannot find the identity value because I'm stuck wondering what the operation is doing. How do I read from Table 3 if it is a group or not?

答: It is not a group, since $2\circ 2=3=2\circ 3$; the Latin square property is not satisfied.

Title: What operation is $\circ$?


问:
Trouble understanding intersection of subspaces
I am having trouble understanding the intersection of subspaces.

My book says $U\cap W = \{\mathbf{v}:\mathbf{v}\in U$ and $\mathbf{v}\in W\}$. But it doesn't have good examples.

Let $U$ be the subspace generated by $(1,2)$ and $W$ the subspace generated by $(2,3)$. What does "generated" mean in this context? Does it mean $U = u_1+2u_2$ and $W=2w_1+3w_3$? I am very confused by the terminology in my book.

Lastly, is $U\cap W = \{\mathbf{0}\}$?

答:
Based on your last question, are you assuming that U and V are 1-D vector spaces here? If so, if $U = span(\{ (1,2 )\} )$ and $V = span( \{ (2,3) \} )$, then $U \cap V = \{ (0,0) \}$, which qualifies as a vector space.

A more visual example: think of two distinct planes in $\mathbb R^3$ intersecting the origin. These correspond to two 2-D vector spaces, and their intersection is a line, which is a 1-D vector space.


问:
Writing a projection as a matrix
Here is a simple question which I can't figure out.

I know that a projection is a linear mapping, so it has a matrix representation. What I am interested is finding the matrix which represents:

$$\pi_d : \mathbb{R}^{d+1} \rightarrow \mathbb{R}^d$$

which maps $\mathbb{R}^{d+1}$ onto the hyperplane $H=\{x \in \mathbb{R}^d : \langle x, \mathbf{1} \rangle = 1\}$, where $\mathbf{1}$ is the all-ones vector.

I can easily compute the position of a given vector projection; for example, considering $\pi_1$:

$$\begin{bmatrix} 1 \\ 0 \end{bmatrix} \mapsto \begin{bmatrix} 1/2 \\ -1/2 \end{bmatrix}$$

but I am unsure on how to get this explicitly as a $1$-dimensional vector.

If you could provide a hint rather than a full answer, I would be appreciative!

答:
An orthogonal projection in $R^d$ is a type of linear transformation from $R^d$ into $R^d$. So its image is a subspace of $R^d$. Your $H$ is not a subspace of $R^d$, so there is no such thing as an orthogonal projection onto $H$.

You can fix that by moving $H$ to the origin; take $H$ to be the set of vectors orthogonal to ${\bf 1}$.

Now the hint. You can find the projection ${\bf p}$ of any vector ${\bf x}$ onto the span of ${\bf 1}$ by using the dot product as described in calculus books; it is the component of ${\bf x}$ in that direction. Once you have that, the difference ${\bf x-p}$ is what you are looking for. The matrix that you work out will be square of course, not $d$ by $d+1$.

Title: Writing a projection as a matrix


问:
Finding standard deviation from unknown distribution percentage
"Runzheimer International publishes data on overseas business travel costs. They report that the

average per diem total for a business traveller in Paris, France, is $349. Suppose the shape of the

distribution of the per diem costs of a business traveller to Paris is unknown, but that 51% of the

per diem figures are between $320 and $378. What is the value of the standard deviation?"
Runzheimer International publishes data on overseas business travel costs. They report that the average per diem total for a business traveller in Paris, France, is $349. Suppose the shape of the distribution of the per diem costs of a business traveller to Paris is unknown, but that 51% of the per diem figures are between $320 and $378. What is the value of the standard deviation?

mean=349

1-1/k^2*100=0.51

k=1.00

320-419/SD=1

SD=-99 is wrong?

What did I do wrong/supposed to do?

答: Let $X$ represent the per diem cost of a randomly selected business traveler. From Chebychev's inequality (taking $k=10/7$) we have $$P\Bigg(\Bigg|\frac{X-349}{{\sigma}}\Bigg|\geq \frac{10}{7}\Bigg)\leq \bigg(\frac{7}{10}\bigg)^2 $$ Notice $$\Bigg|\frac{X-349}{{\sigma}}\Bigg|\geq \frac{10}{7} \iff X\in\bigg(-\infty,349-\frac{10 \sigma}{7}\bigg]\cup\bigg[349+\frac{10 \sigma}{7},\infty\bigg)$$ We need to find $\sigma$ so that $$349-\frac{10 \sigma}{7}=320$$ $$349+\frac{10 \sigma}{7}=378$$ Solving these for $\sigma$ gives $\sigma =\frac{203}{10}$


问:
Books for Geometry processing
Please suggest some basic books on geometry processing.

I want to learn this subject for learning algorithms in 3d mesh generation and graphics.

Please suggest me subjects or areas of mathematics i have learn in order to be understanding 3d mesh generation. I am doing self study and i am very new to this topic

Please suggest me online links to any videos or resources, if available.

答:
I hope these links helpful for you. ;)

Book

Dynamic Geometry Processing

Software

Introduction Software

Videos

some Videos

Title: Books for Geometry processing


答:
As a comment explains, the table itself is the defining mechanism for the operation. You are familiar with definitions for $+$, and then understand how to make a table from that. Likely, that understanding is misguiding you here. Here, the defintion of $\circ$ just is that table. Want to know what $a\circ b$ is? To find out you are supposed to look up where row $a$ and column $b$ meet in the table.

This operation does not make a group. $\circ$ing by $3$ is not a one-to-one map. So $3$ would not have an inverse.

Title: What operation is $\circ$?

Body:
Given the following exercise:

"Consider the operation $\circ$ defined on $G_4 =\{ 0, 1, 2, 3\}$ by Table 3. Check whether it satisfies each of the three axioms from the definition of a group, and decide whether it is a group."

Table 3:

The properties of a group:

$1$. Asssoiativity: for any $a, b, c \in G_4$, we have $a\circ (b\circ c)= a\circ (b\circ c)$

$2$. Identity: for any $a \in G_4$, we have $a\circ e = e\circ a=a$

$3$. Inverse: for every $a \in (G_4,\circ ) $, theres is an inverse $a^{-1}$, such that $a\circ a^{-1}=a^{-1}\circ a=e$

Previosly I have concluded whether the sets $(G_4,+)$ and $(G_4,*)$ are groups by doing the following:

But I dont know what the $\circ$ operation for Table 3 is doing. Since we are working with numbers and not functions I am stuck trying to figure this out. I know that to find the inverses I can check if there's a value in every row that's equal to the identity value $e$, but I cannot find the identity value because I'm stuck wondering what the operation is doing. How do I read from Table 3 if it is a group or not?


问:
If $f:\mathbb R \to \mathbb R$ then for any $A\subset \mathbb R$, $\mathbb R-f^{-1}(A)=f^{-1}(\mathbb R-A)$
Let $f:\mathbb R\to \mathbb R$ and $A\subset \mathbb R$. Show that $f^{-1}(\mathbb R-A)=\mathbb R-f^{-1}(A)$

I never got this question when I was reading Set theory & functions. But while in Continuity there is a Theorem in my book which used this.

答: The elements of $f^{-1}(\mathbb R-A)$ and the elements of $\mathbb R-f^{-1}(A)$ can both be described simply as the elements $x\in\mathbb R$ that satisfy $f(x)\notin A$. And sets with the same elements are equal.


问:
Proving $f(x) = x^2 \sin(1/x)$, $f(0)=0$ is differentiable at $0$, with derivative $f'(0)= 0$ at zero
I need a solution for this question. I've been trying out this question for days and I haven't been able to find out its solution yet. And some explanation would help too.

Show that the function f defined by:
  $$f(x):= \begin{cases}
  x^2\sin(1/x) &:\text{if $x \ne 0$} \\
  0 &:\text{if $x=0$}
\end{cases}$$
  is differentiable at $x=0$, and that $f'(0)=0$.

Body:
Hints:

$$\bullet\;\;\text{If $\,f,g\,$ are two functions defined in some punctured neighborhood I$_0$ of $\;x_0\;$ and s.t.:}$$

$$\begin{align*}(1)&\;\;\lim_{x\to x_0}f(x)=0\\{}\\
(2)&\,\exists\,M\in\Bbb R\;\;s.t.\;\;|g(x)|\le M\;\;\forall\,x\in\text{I}_0\end{align*}$$

$\;\;\;\;\;\;\;\;\;\;\;$then $\;\displaystyle{\lim_{x\to x_0}f(x)g(x)=0}$

$$\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\bullet\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\lim_{x\to x_0}\frac{f(x)-f(0)}{x}=\lim_{x\to x_0}\;x\sin\frac1x\;\;\ldots\ldots$$


问:
How did Euler obtain these two formulae? Are they correct?
I am reading this book on trigonometric series, "Тригонометрические ряды от Эйлера до Лебега" (Trigonometric series from Euler to Lebesgue) , it is in Russian, and my Russian is abysmal. But there is a very interesting formula, it is:

$$\dfrac{1-r\cos(x)}{1-2r\cos(x)+r^2}=1+r\cos(x)+r^2\cos(2x)+r^{3}\cos(3x)...$$

$$\dfrac{r\sin(x)}{1-2r\cos(x)+r^2}=r\sin(x)+r^2\sin(2x)+r^{3}\sin(3x)...$$

As $r=1$ or $r=-1$

We have: $$-\dfrac{1}{2}=\cos(x)+2\cos(x)+3\cos(3x)+...$$

$$\dfrac{1}{2}=\cos(x)-2\cos(x)+3\cos(3x)+...$$

He integrated the second series and obtain the trigonometric series:

$$\dfrac{x}{2}=\sin(x)-\dfrac{1}{2}\sin(2x)+\dfrac{1}{3}\sin(3x)-...$$

The first and second series are divergent, so I think the result doesn't make sense, although the formal method of deriving it may seem correct.

The third series is correct on the interval $[\pi, -\pi]$, because the series expansion converges to that function (Correct me if I am wrong here, I haven't studied Fourier series).

I wish to ask how did he obtain the above formula? Are these formula incorrect because the issue of convergence? Also, which papers contain these results?

答:
In Euler's time, the concept of convergence had not been properly defined. Much of his work lacks the rigor we are now used to.

At that time, even renowned mathematicians used to do things like manipulate divergent series freely and, at the end of their calculations, examine the result. If it made sense, they save it, if not, they discard it.

They developed an intuition about what could and could not be done, but they did not have the theory to back up many of the algebraic manipulations they did.

The formulas can be proved using $\displaystyle \cos(x) =\frac{e^{i x}+e^{-ix}}{2}$ and $\displaystyle\sin(x) =\frac{e^{i x} - e^{-ix}}{2i}$ to turn the sums into geometric series.


问:
The Exponential Group in a Banach Algebra: a proof of Rudin.
Define $G$  to be the group of invertible elements in a Banach algebra $A$ and $G_1$ the component of $G$ that contains $e\in G.$

Now, $\exp x=\sum_{n=0}^\infty \frac{x^n}{n!}$ and if $\Gamma$ is the subgroup generated by $\exp (A),$ a straightforward argument shows that $\Gamma$ is open: if we define $y=\sum_{n=0}^\infty \frac{-1}{n}(1-x)^n$ then the series converges absolutely whenever $\|1-x\|<1$ and a (at least for me) somewhat messy calculation shows that $\exp y=x$ so $\exp (A)$ contains a neighborhood of $e$, and since $\Gamma$ is a group, it follows that $\Gamma$ is open.

Unfortunately, Rudin does not do it this way. He cites the following theorem:

Theorem Suppose $A$ is a Banach algebra, $x\in A$, and the spectrum $\sigma(x)$ of $x$ does not separate $0$ from $\infty$. Then

(a) $x$ has roots of all orders in $A$,

(b) $x$ has a logarithm in $A$, and

(c) if $\varepsilon>0$, there is a polynomial $P$ such that $\|x^{-1}-P(x)\|<\varepsilon$.

to claim that $\exp(A)$ has nonempty interior in $G$ and uses this fact to prove that $\Gamma$ is open. I tried to do this as follows: take $y\in\exp(A)$ and note that as soon as $\|x-y\|<\|y^{-1}\|^{-1}$ then $\|e-y^{-1}x\|<1.$ It follows that $\sigma(y^{-1}x)\subseteq \{|\lambda-1|<1|\}$ and now the theorem applies to show that $y^{-1}x=\exp z$ for some $z\in A$ and so $x=y\exp z.$ Of course, unless $A$ is commutative, we can't conclude.

Since Rudin claims that $\exp(A)$ is open, as a passing remark, I guess it's easy, but I have not been able to prove it. I would like a hint to get me going in the right direction.

答: I don't think Rudin claims that $\exp(A)$ is open, but rather that it has a nonempty interior.  Moreover the proof I believe he has in mind is not much different from what you have above:  in a neighborhood of the identity all elements have spectrum close to 1, which therefore does not separate 0 from $\infty$.  By Theorem 10.30.(b) these elements have a logarithm, which is to say that they lie in $\exp(A)$.


问:
Help understanding proof of "$\sum |a_n|$ converges $\Rightarrow$ $\sum a_n$ converges"
I'm self-studying Real Analysis. I can't understand a proof of this commonly known theorem:

Theorem. Every absolutely convergent series is convergent.

Proof. If $\sum |a_n|$ converges, given an arbitrary $\varepsilon>0$, there exists $n_0 \in \mathbb{N}$ such that $n>n_0 \Rightarrow |a_{n+1}| + ... + |a_{n+p}|<\varepsilon$, whichever $p \in \mathbb{N}$. So $|a_{n+1} + ... + a_{n+p}| \leq |a_{n+1}| + ... + |a_{n+p}|<\varepsilon$. Therefore, $\sum a_n$ converges by virtue of the Cauchy Criterion for series.

The passage that I don't understand is
$$n>n_0 \Rightarrow |a_{n+1}| + ... + |a_{n+p}|<\varepsilon.$$

Why is that the case? It doesn't seem obvious... and it doesn't seem to follow from any previous discussion. I don't know what I'm missing here.

答:
The series $\sum_{n=1}^{\infty}a_{n}$ is defined to converge if, for all $\epsilon>0$, there exists an $N\in\mathbb{N}$ such that $\left|\sum_{n=p}^{q}a_{n}\right|<\epsilon$ for all $N\leq p<q$. If, in particular, $\sum_{n=1}^{\infty}|a_{n}|$ converges, then given $\epsilon>0$, there exists (by definition) $N\in\mathbb{N}$ such that
\begin{equation*} \left|\sum_{n=p}^{q}|a_{n}|\right|=|a_{p}|+\ldots+|a_{q}|<\epsilon \end{equation*}
if $N\leq p<q$ but, of course,
\begin{equation*} \left|\sum_{n=p}^{q}a_{n}\right|\leq |a_{p}|+\ldots+|a_{q}|<\epsilon \end{equation*}
so $\sum_{n=1}^{\infty}a_{n}$ is convergent (by definition) as well.


问:
Best books to self study number theory and especially diophantine equations?
I'm an undergraduate math student. I've already passed linear algebra and abstract algebra and recently passed a course in elementary number theory and I really enjoyed it. (Our source was Daniel Flath's "Introduction to Number Theory" and Kenneth Rosen's "Elementary Number Theory".)

Now, I want to self-study more books this semester and I'm searching for good books with exercises. My main interest is in solving diophantine equations and their applications.

Thanks for your suggestions and sorry for my bad English. :)

答:
If you are interested in arithmetic geometry and the theory of elliptic curves (Diophantine equations with geometry involved) then I can recommend The Arithmetic of Elliptic Curves, written by Joseph H. Silverman. It may or may not be a little difficult to read, but it is a rewarding journey.

About the Book

The first two chapters is preliminaries from algebraic geometry. If you don't know this theory from before, then you may spend a little time on these two chapters and you may also want to learn from another source.

The next two chapters is about the understanding of the geometry of elliptic curves and the group structure one can give on the points of an elliptic curve, this gives rise to a Projective Group Scheme.

The next five chapters spends time to look at elliptic curves defined over different fields and arithmetic objects.

Some things you will see are (among other things):

How elliptic curves over the complex numbers relates to elliptic integrals and elliptic functions.

The Weil Conjectures for elliptic curves. You define a generating function which encodes the number of points of the elliptic curves defined over finite fields. This object is called a $\zeta$-function. The Weil Conjecture is then a list of statements regarding this newly defined object.

See the Mordell-Weil Theorem for elliptic curves defined over the rational numbers. This means that you can find a finite list of rational points of the elliptic curve, from which you can generate all other points. This is an important result in the foundations of the Birch and Swinnerton-Dyer conjecture. This conjecture gives a fascinating relationship between arithmetic, geometry and analysis, I think.

How Elliptic Curves relates to problems in Diophantine Approximations

I think the book is great, but it can be a little bit difficult to read - perhaps.

$\text{ }$

If you want something slightly easier, then you maybe want to have a look at Rational Points on Elliptic Curves, which is by Silverman and John T. Tate.

Bonus Section

The Birch and Swinnerton-Dyer Conjecture is a great motivation for studying elliptic curves and their arithmetic properties. For that reason, I also want to add an extra book which I think is a really good introduction to that conjecture. The book is called  Elliptic Tales: Curves, Counting, and Number Theory and is written by Avner Ash and Robert Gross. It begins at a very elementary level and successively builds up to the conjecture. It is entertaining and describes most of the concepts in a non-technical language.


问:
Calculate the intersection point of $m$ with $AC$.
Let $ABC$ be the triangle with $A=(0,0)$, $B=(16,0)$ and $C=(4,16)$.

Let $B'$ be the midpoint of the side $\overline{AC}$, let $C'$ be the midpoint of the side $\overline{AB}$ and let $I$ be the point $(2\sqrt{17}-2, 9-\sqrt{17})$.

Calculate the line that goes through $B$ and $I$.

The angle bisector $m$ at $B$ of the triangle $ABC$ divides the opposite $\overline{AC}$ proportional to the adjacent sides. Calculate the intersection point of $m$ with $AC$. Show that this intersection point is on the line $BI$.

Show that $I$ has the same distance to $AB$ as to $AC$.

Conclude that $I$ is the center of the inscribed circle of the triangle $ABC$and calculate the line that goes through $I$ and the contact point of the inscribed circle and the side $\overline{BC}$.

$$$$

For question 1 we have:

\begin{equation*}y-0=\frac{9-\sqrt{17}-0}{2\sqrt{17}-2-16}(x-16)  \Rightarrow y=-\frac{1}{2}(x-16) \end{equation*}

For question 2 we have:

We have to calculate the side lengths $|AB|$ and $|BC|$, right?

We have that $|AB|=16$ and $|BC|=\sqrt{12^2+16^2}=20$. Therefore $m$ divides the side $\overline{AC}$ in relation $\frac{20}{16}=\frac{5}{4}$, right? But how can we calculate the coordinates of that point?

For question 3 we have:

Here we just use the formula for distances and calculate both and see that these are equal, right?

答:
What you need to do for part (2) is to compute the intersection point of the line you got from part (1), with the line that passes through $A$ and $C$, which you have not yet calculated.  This is simply $$y = \frac{16}{4} x = 4x,$$ consequently the intersection point, which we will call $P$, satisfies the system $$\begin{align} y &= -\frac{x}{2} + 8 \\ y &= 4x. \end{align}$$  Then compute the distances $|PA|$ and $|PC|$, and show that because $$\frac{|PA|}{|AB|} = \frac{|PC|}{|CB|},$$ point $P$ obeys the proportionality criterion, thus $P$ lies on the angle bisector of $\angle B$.  Conclude that, since this line is also the line passing through $BI$ from part (1), that $I$ lies on the angle bisector of $\angle B$.

For part (3), simply perform the distance calculation.  The distance of $I$ to $AB$ is easy.  However, the distance of $I$ to $AC$ requires us to find the line perpendicular to $AC$ passing through $I$, computing its intersection point $Q$ with $AC$, then computing the distance $|QI|$.

For part (4), observe from parts (2) and (3) that if $I$ lies on the angle bisector of $\angle B$, then the distance of $I$ to $BC$ equals the distance of $I$ to $AB$; therefore, $I$ is equidistant to all three sides of $\triangle ABC$, hence $I$ is the incenter.


问:
And integer proof question involving fractions
Sorry to ask two questions but I would ask: What conditions will these two fractions$$\frac {p^2+t^2} {p-t} $$
and $$\frac {pt(p+t)} {p-t}$$
Both be integers. I found that not all the time the fraction can both be function. Obviously $t=p-1$ will work, but what if $t=p-n$?
Note that both $p $ and $t$ are positive integers on both fractions

What have I tried:
Assuming $t=p-n$ getting $\frac {2p^2-2pn+n^2} n$ and $\frac {a(a-n)(2a-n)} n$. Or assuming $t=pk$ and getting $\frac {p(k^2+1)} {1-k}$ and $\frac {p^2k(1+k)} {1-k}$. How can I go further than that?

答:
I called them $x,y$ and the ratio $z.$ There are two infinite families that make up all coprime integer solutions to
$$  x^2 + y^2 + yz - zx = 0 $$
The first family is
$$  x=u^2 + uv \, , \; \;  y = -u^2 + uv \, , \; \; z = u^2 + v^2 $$
For this family $$ \frac{xy(x+y)}{x-y}  $$
is always an integer. There is, in particular, no need to investigate common factors anywhere.

The second family is
$$  x=2u^2 + 3uv \, , \; \;  y = u^2 + uv \, , \; \; z = 5u^2 +4uv v^2 $$
Let me look at that again, I don't appear to have written it out

For this family
$$ \frac{xy(x+y)}{x-y} = \frac{u^2 (u+v)(2u+3v)(3u+4v)}{u+2v} $$
and this may not be an integer. It ought to be possible, and quick, to check the individual conditions, (pretending the the denominator is prime) and working out the possible patterns for $u,v.$ Note that we can take coprime $u,v$ and just multiply both by $u+2v,$  since the new denominator is $(u+2v)^2$ but the numerator has been multiplied by $(u+2v)^5$

BACKGROUND:

Solving a homogeneous quadratic equation in three variables over the integers.

Help solving $ax^2+by^2+cz^2+dxy+exz+fzy=0$ where $(x_0,y_0,z_0)$ is a known integral solution

Describe the rational points on $3x^2 + y^2 = 4$

Question about non-degenerate polynomials, and a proof

Restrictions on the [hypothetical] solutions of a ternary quadratic form

Ternary Quadratic Forms  MY PROOF!!! isotropic ternary!!!

$(a^2+b^2-ab)(a^2+b^2-2ab)$ is a perfect square eisenstein

Solutions to a quadratic diophantine equation $17x^2 + 29y^2 = 41z^2$

Identifying relation between numbers based on equation relating them


问:
Boundedness in $W^{2,p}(\Omega)$ for all $1\leq p<\infty$
Assume $\Omega$ is nice and $\{u_n\}$ is a bounded sequence in $W^{2,p}(\Omega)$ for all $p\in [1,\infty)$, could we find a subsequence that converges strongly $u_{n_k}\to u$ in $\mathrm{C^2}(\Omega)$?

答: No, not even if you have boundedness in $C^2(\overline \Omega)$. Since this space is infinite dimensional, closed balls are not compact.


问:
Prove n is always a multiple of 4 for this sum to be 0.
Each of the numbers $x_1,x_2,x_3,...,x_n$ equals 1 or -1, and
$$x_1x_2x_3x_4 + x_2x_3x_4x_5 + ... + x_nx_1x_2x_3 = 0.$$
Prove that n is divisible by 4.

I attempted to solve this problem first by letting $x_1x_2x_3...x_n$ = $c$ where $c=-1$ or $c=1$ and rewriting but it didn't provide much use. It's pretty clear that n must be even, so from there we must prove that if n is even with only 1 factor of 2 this fails. Another idea I had was that I noticed we can write the term i+1 in terms of term i, by multiplying by $x_{i+5}/{xi}$. In the case of n=6 this provides some nice cancellations, but overall, nothing that I think helps me solve the problem. Any insights would be appreciated.

答: Hint:  prove that changing the sign of one variable always changes the sign of the sum by a multiple of $4$.  You can do that just by looking at the possible values of the three variables before and after.  Then if all the variables are $1$ the sum is $n$.  You need this to be a multiple of $4$ to have a hope to get to $0$


问:
Formulas to find the end point of an arc given the start point and midpoint on the unit sphere?
So the radius and the center of the arc are $1$ and $(0, 0, 0)$ respectively because the points and the rac is on the unit sphere $S^2$. Given the start and the midpoint of the arc, is there any formula to deduce the endpoint? I know that normalizing the midpoint on the straight line between the start and end returns the midpoint of the arc. Thanks in advance.

答:
Let $O$ be the center of the sphere, $S$ and $E$ be the endpoints of the arc, and $M$ be the midpoint. These all lie within the plane though $O, S, M$, so this becomes a two-dimensional problem in a circle instead of a three dimensional problem in a sphere.

Let $N$ be the midpoint of the chord $\overline{SE}$, which is also the intersection of $\overline{SE}$ and $\overline{OM}$. Because $N$ is the midpoint, the chord and radius intersect in a right angle. Let $\theta$ be the angle $\angle SOM$ and $h = ON$, the distance from the center to $N$. Then $h = \cos \theta$.

If we make $O$ the origin, and consider all the points to be vectors
$$\mathbf s = S\\\mathbf m = M\\\mathbf n = N\\\mathbf e = E$$
Then $$\mathbf n = \frac{\mathbf s + \mathbf e}2\\\mathbf n = (\cos\theta)\mathbf m\\\cos\theta = \mathbf s \cdot \mathbf m$$
so
$$\mathbf e = 2(\mathbf s\cdot \mathbf m)\mathbf m - \mathbf s$$
Where $\mathbf s\cdot\mathbf m = s_1m_1 + s_2m_2 + s_3m_3$ is the dot product.


问:
I don't understand equivalence classes with relations
I am not quite understanding equivalence classes. For example I have this problem:

Let $A$ be the set of integers and
$\quad a\;R\;b\quad$ if and only if $\quad |a| = |b|$.

I have proved that this is an equivalence relation, (its reflexive, symmetric and transitive), but how do I show the equivalence classes?

答:
The quotient set of integers modulo your relation $R$ can be identified with the set of natural numbers $\geq 0$, via the following bijection: set
$$f:\frac{\mathbb{Z}}{R}\longrightarrow\mathbb{N}$$
sending the class $[a]$ to the natural number $|a|$, hence you get:

$\bullet$ a well-defined map: if $a$ and $a'$ both represent the class $[a]$, then they differ by the signum, hence their absolute value $|a|=|a'|$ is the same.

$\bullet$ suppose $|a|=|a'|$ for a couple $a,a'$ of integers. Then they differ by the signum, hence they represent the same equivalence class in the quotient $\frac{\mathbb{Z}}{R}$, and this shows that $f$ is injective

$\bullet$ given a natural number $a\geq 0$, then $f([a])=a$, meaning that $f$ is also surjective.


问:
Evaluating $\lim_{\beta\to 0^-} \left(-\ln|\beta| + e^{\beta}\right) + \lim_{\beta\to 0^+} \left(\ln|\beta|\right)$?
Here is my problem:
$$\begin{align}
&\lim_{\beta\to 0^-} \left(-\ln|\beta| + e^{\beta}\right) + \lim_{\beta\to 0^+} \left(\ln|\beta|\right) \\
=& \lim_{\beta\to 0} \left(\ln|\beta| - \ln|\beta| + e^{\beta}\right) \\
=& \lim_{\beta\to 0} e^{\beta} = 1
\end{align}$$

First of all sorry i can't give the whole context since will consume a lot of time to type it by hands. So, in short i'm evaluating the integral using cauchy principal value and came up with this.

Is it legal to do that limit? Since i have an absolute value and if i split the first limit, the first term will be the same as the second limit and for convenience i change the limit goes to 0 (both sides)? Please help me to understand about limit. Or maybe if it's illegal you can tell me why. Thanks in advance!

答:
Unfortunately, this is not legal. Why? Well - when you say something like
$$
\lim_{\beta\to 0}\ln|\beta|
$$
we are talking about the number that this limit converges to. If it doesn't converge, it's not well defined. The property of limits that
$$
\lim_{x\to a}f(x)+g(x) = \lim_{x\to a}f(x) + \lim_{x\to a}g(x)
$$
(which is essentially what you used on the first line, or I guess, the reverse of this) only holds if the individual limits exist (in your case, they definitely don't).


问:
Why does this partial fraction decomposition work, even with division by $0$?
Here is how partial fraction decomposition works. First, take a fraction like $\frac{1}{n(n+1)}$. You can express this fraction as the sum $\frac{A}n + \frac{B}{n+1}$ for some constants $A$ and $B$. Thus, you have the equation $\frac{1}{n(n+1)} = \frac{A}n + \frac{B}{n+1}$. Now, multiply both sides by $n(n+1)$ to get $1=A(n+1)+B(n)$. However, note that the expression $\frac{x}{x} = 1$ ONLY when $x \ne 0$ because division by $0$ is forbidden. Thus, $1=A(n+1)+B(n)$ only if $n \ne 0,-1$ because these are the values that cause division by $0$. However, many people plug in $n=0,-1$ to solve for $A,B$ even though that is an illegal move, and it works.

Plugging in $n=0$, we get $1=A$. Plugging in $n=-1$, we get $1 =-B$, thus $B=-1$. So, we have $\frac{1}{n(n+1)} = \frac{A}n + \frac{B}{n+1} = \frac{1}n - \frac{1}{n+1}$. This is true.

However, I don't understand why it works, even though you used an illegal move?

答:
This is a good question, and it's quite subtle and almost no one mentions it; the reason is continuity. Let $P:\Bbb{R}\to\Bbb{R}$ be a polynomial function, $\{a_1,\dots, a_k\}$ all of its roots, and let $f:\Bbb{R}\to\Bbb{R}$ be a continuous function. Suppose you know that for all $x\in\Bbb{R}\setminus\{a_1,\dots, a_k\}$
\begin{align}
\frac{1}{P(x)}&=\frac{f(x)}{P(x)}
\end{align}
Then, for these values of $x$, it follows that $1=f(x)$. Now since $f$ is a continuous function, the same equality holds even if $x\in\{a_1,\dots, a_k\}$. Therefore, for every $x\in\Bbb{R}$ without exception, we have the equality $1=f(x)$.

In your case, we're applying this reasoning to $P(n)= n(n+1)$ and $f(n)=A(n+1)+Bn$.

More generally, we can make the following observation:

Let $P:\Bbb{R}\to\Bbb{R}$ be a polynomial function, $f,g:\Bbb{R}\to\Bbb{R}$ be continuous functions, and suppose that for every $x\in\Bbb{R}$ which is not a zero of $P$, we have
\begin{align}
\frac{g(x)}{P(x)}&= \frac{f(x)}{P(x)}
\end{align}
Then, for every $x\in\Bbb{R}$ (without exception) we have $f(x)=g(x)$.


问:
Proving that a surface is a plane
Hey everyone I'm loosing my mind on this one..

Let's say we have a a parametric surface:

$A(u,v) = (u^2+v^2,u^2-v^2,uv)$

and we want to determine whether it is a plane or not. From what I understand if we compute the partial derivatives $\frac{\partial }{\partial u}$ and $\frac{\partial }{\partial v}$ in this case:

$\vec{n_1}=\frac{\partial }{\partial u} = (2u,2u,v)$ and $\vec{n_2} = \frac{\partial }{\partial v} = (2v,-2v,u)$

these will be two vectors defining the normal plane of the the surface at a point A(u,v). Their cross product is the normal vector at that point, in this case:

$\vec{n} = (2u,2u,v) \times (2v,-2v,u) = (2u^2+2v^2,2v^2-2u^2,-8uv)$

Well, if the surface is a plane shouldn't the normal vector have the same gradient at every point on the surface? Which means there should be a vector $\vec{w} = \left \langle x,y,z \right \rangle$ such that the normal vector $\vec{n} $ could be expressed as a scalar of $\vec{w}$ :

$\vec{n} = (a,b,c) \cdot \vec{w}$

Am I missing something here? Is there such a way to express $\vec{n}$?

答:
Hint: With some algebraic manipulation you can get that the cartesian equation of the surface is

$$x^2-y^2-4z^2=0$$

This is a cone.

Title: Proving that a surface is a plane


问:
Ratio of PDF to complementary CDF
Let $f(x)$ be a probability density function, and $F(x)$ be the cumulative distribution function of $f(x)$.

$$F(x) = \int_{-\infty}^{x}f(u)du$$

Then intuitively, what does the following ratio mean?  I'm trying to understand the "common sense" logic behind this ratio, if there is any..

$$ r(x) = \frac{f(x)}{1-F(x)}$$

Thanks!

答:
The function
$$
\lambda:x\mapsto \frac{f(x)}{1-F(x)}
$$
is called the hazard function (at least in survival analysis) and it measures the instantaneous rate of change in the sense that it can be described as
$$
\lambda(x)=\lim_{\Delta x\downarrow 0}\frac{P(X\in [x,x+\Delta x]\mid X>x)}{\Delta x}
$$
where $X$ is a random variable with distribution function $F$.

In survival analysis $X$ often measures the time from some origin (e.g. diagnosis of cancer) until an event happens (e.g. death) and in this case $\lambda(x)$ describes the instantaneous risk of dying shortly after time $x$ given that you're alive at time $x$.

Title: Ratio of PDF to complementary CDF


答: $A(n+1)+Bn$ is a polynomial in $n$, as is $1$. You want values of $A$ and $B$ that make these two polynomials equal for all real numbers except $0$ and $-1$. But two polynomials that agree at infinitely many real numbers are necessarily identically equal and will therefore agree at all real numbers. In particular, $A(n+1)+Bn=1$ will hold at $n=0$ and $n=-1$ if it holds at all other integers, so we can use $n=0$ and $n=-1$ as shortcuts to finding the correct values of $A$ and $B$.

Title: Why does this partial fraction decomposition work, even with division by $0$?

Body:
Here is how partial fraction decomposition works. First, take a fraction like $\frac{1}{n(n+1)}$. You can express this fraction as the sum $\frac{A}n + \frac{B}{n+1}$ for some constants $A$ and $B$. Thus, you have the equation $\frac{1}{n(n+1)} = \frac{A}n + \frac{B}{n+1}$. Now, multiply both sides by $n(n+1)$ to get $1=A(n+1)+B(n)$. However, note that the expression $\frac{x}{x} = 1$ ONLY when $x \ne 0$ because division by $0$ is forbidden. Thus, $1=A(n+1)+B(n)$ only if $n \ne 0,-1$ because these are the values that cause division by $0$. However, many people plug in $n=0,-1$ to solve for $A,B$ even though that is an illegal move, and it works.

Plugging in $n=0$, we get $1=A$. Plugging in $n=-1$, we get $1 =-B$, thus $B=-1$. So, we have $\frac{1}{n(n+1)} = \frac{A}n + \frac{B}{n+1} = \frac{1}n - \frac{1}{n+1}$. This is true.

However, I don't understand why it works, even though you used an illegal move?


答:
Let's think of these expressions as functions on subsets of the real numbers. i.e. $\frac{1}{x(x+1)} = \frac{A}{x} + \frac{B}{x+1}$ for $x \neq -1, 0$. Then $1=A(x+1)+B(x)$ for $x \neq -1, 0$.

You're correct to note that we can't plug in $-1$ and $0$ straight away because the functions aren't defined there. However, since the functions are defined on a set with $-1$ and $0$ as limit points we can take the limit as $x$ tends to the forbidden points. Since the functions $1$ and $A(x+1)+B(x)$ are equal everywhere else they have the same limit.

Since the functions are continuous the limit is exactly the same as plugging in the forbidden value.


问:
Proving inequalities with analysis
Suppose that for all $0<c \in \mathbb{R}$ that $a < bc + b$ where $a, b \in \mathbb{R}$. Prove that $b \geq a$ and $b \geq 0$.

My attempt:

We start by considering $a < bc + b$. We have that

$$a < bc + b \implies bc + b  - a > 0$$

Using the archimedian property of real numbers, it follows

$$bc + b - a > \frac{1}{n}$$

$$nbc + nb  - na > 1$$

Now, by the archimedian property involving integers, there is an $m \in \mathbb{Z}$ such that we have

$$m \leq nbc + nb < m + 1$$

This is as far as I have gotten with this proof, and I'm not sure if I'm even right thus far. I am new to this topic and I am not sure how to manipulate the archimedian property/ies to complete the proof. Any assistance is much appreciated.

答:
$a<bc+b$

We know that this must be true for any real number $c$ greater than, but not equal, to 0. Let's first assume $b>0$, then we can establish an upper bound for $a$ by letting $c$ approach zero.

$a<\lim_{c\to 0}(bc+b)\  \to \ (a<b)$

Since $c$ cannot equal zero, than $a=b$ also satisfies $a<bc+b$.

Therefore:

$b>0\Rightarrow a\leq b$

Let's consider when $b<0$.

$a<\lim_{c\to\infty}(bc+b)\ \to \ (a<-\infty)$

This is impossible.

$b\nless 0$

Lets consider when $b=0$

$a<bc+b$

$a<0c+0$

$a<0$

This is a perfectly valid possibility.

Therefore:

$b\geq 0$

$b \geq a$

Title: Proving inequalities with analysis


问:
Grid Problem - How to get from Point A to B while avoiding segment CD
Very curious on how to solve this problem. Let's say I'm given a grid problem but we want to avoid segment BC and get to point D from point A with the condition that you can only go right and up. How would we go about solving this?
Hand drawn pic of a graph with points</p>

答:
Hints:

How many total ways from $A$ to $D$?

How many ways from $A$ to $D$ using $BC$?

How many ways from $A$ to $B$?

How many ways from $C$ to $D$?

How many ways from $A$ to $D$ not using $BC$?


问:
Proving $\prod_{n=1}^{\infty}\left(1+\frac{1}{n^{3}}\right)=\frac{1}{\pi}\cosh\frac{\pi\sqrt{3}}{2}$
First I rewrite $\prod_{n=1}^{\infty}\left(1+\frac{1}{n^{3}}\right)$ as $\prod_{n=1}^{\infty}\left(\frac{1+n^{3}}{n^{3}}\right)$, then by factor out polynomial I get $\prod_{n=1}^{\infty}\left(\frac{(1+n)(n^{2}-n+1)}{n^{3}}\right)$ which is a problem because I can't factor any further which makes me stuck on this step I would hope for any help.

答:
If you're familiar with the gamma function, then this (or this... etc) is the way to go.

Otherwise, write $$1+\frac{1}{n^3}=\left(1+\frac1n\right)\left(1-\frac{1}{2n}\right)^2\left(1+\frac{3}{(2n-1)^2}\right)$$ and see that $$\prod_{n=1}^N\left(1+\frac1n\right)\left(1-\frac{1}{2n}\right)^2=(N+1)\left(\frac{(2N-1)!!}{(2N)!!}\right)^2=(N+1)\left(\frac{(2N)!}{2^{2N}N!^2}\right)^2$$ tends to $1/\pi$ as $N\to\infty$ (by Wallis product, or Stirling's formula), and $$\prod_{n=1}^\infty\left(1+\frac{3}{(2n-1)^2}\right)=\cosh\frac{\pi\sqrt3}{2}$$ is a special case of the infinite product for the [hyperbolic] cosine.


答:
[Note] : I shall use the following identites :

Identity $(1)$ : $\quad\displaystyle\left|\Gamma\left(\frac{1}{2}+\beta i\right)\right|=\frac{\pi}{\cosh(\pi\beta)}$

Identity $(2)$ :$\quad\displaystyle\lim_{n\to\infty}\frac{\Gamma(n+\alpha)}{\Gamma(n+\beta)}\approx n^{\alpha-\beta}$

Identity $(3)$ : $\quad\displaystyle\prod_{k=1}^{n}(x+k)=\frac{\Gamma(n+x+1)}{\Gamma(x+1)}$

We proceed as follow :
\begin{align*}
\prod_{n=1}^{\infty}\left(1+\frac{1}{n^{3}}\right)&=\prod_{n=1}^{\infty}\left(\frac{n^{3}+1}{n^{3}}\right)\\ \\
&=\prod_{n=1}^{\infty}\left[\frac{(1+n)(n^{2}-n+1)}{n^{3}}\right] \\ \\
&=\lim_{\gamma\to\infty}\prod_{n=1}^{\gamma}\left[\frac{(n+1)\left(n+\frac{-1+i\sqrt{3}}{2}\right)\left(n+\frac{-1-i\sqrt{3}}{2}\right)}{n^{3}}\right] \\ \\
&=\lim_{\gamma\to\infty}\frac{\Gamma(\gamma+2)\cdot\displaystyle\frac{\Gamma\left(\gamma+\displaystyle\frac{1+i\sqrt{3}}{2}\right)}{\Gamma\left(\frac{1+i\sqrt{3}}{2}\right)}\displaystyle\cdot\frac{\Gamma\left(\gamma+\frac{1-i\sqrt{3}}{2}\right)}{\Gamma\left(\frac{1-i\sqrt{3}}{2}\right)}}{\Gamma(\gamma+1)\Gamma(\gamma+1)\Gamma(\gamma+1)} \\ \\
&=\frac{1}{\displaystyle\underbrace{\Gamma\left(\frac{1+i\sqrt{3}}{2}\right)\Gamma\left(\frac{1-i\sqrt{3}}{2}\right)}_{\text{Reflection}}} \\ \\
&=\frac{1}{\displaystyle\frac{\pi}{\cosh\left(\frac{\pi\sqrt{3}}{2}\right)}} \\ \\
&=\frac{\cosh\left(\frac{\pi\sqrt{3}}{2}\right)}{\pi}
\end{align*}

Title: Proving $\prod_{n=1}^{\infty}\left(1+\frac{1}{n^{3}}\right)=\frac{1}{\pi}\cosh\frac{\pi\sqrt{3}}{2}$

Body: First I rewrite $\prod_{n=1}^{\infty}\left(1+\frac{1}{n^{3}}\right)$ as $\prod_{n=1}^{\infty}\left(\frac{1+n^{3}}{n^{3}}\right)$, then by factor out polynomial I get $\prod_{n=1}^{\infty}\left(\frac{(1+n)(n^{2}-n+1)}{n^{3}}\right)$ which is a problem because I can't factor any further which makes me stuck on this step I would hope for any help.


答:
Here provides another method:
Note that
$$a_n=\frac{|a_n|+a_n}{2}-\frac{|a_n|-a_n}{2},n\in \mathbb N.$$
Obviously, we have
$$0\leq\frac{|a_n|+a_n}{2}\leq|a_n|\ \ \text{and}\ \ 0\leq\frac{|a_n|-a_n}{2}\leq|a_n|,$$
by comparison test of positive series，both of the series
$$\sum_{n=1}^{\infty}\frac{|a_n|+a_n}{2}\ \ \text{and}\ \ \sum_{n=1}^{\infty}\frac{|a_n|-a_n}{2}$$
converge.
So the series $$\sum_{n=1}^{\infty}a_n$$
converges.

Title: Help understanding proof of "$\sum |a_n|$ converges $\Rightarrow$ $\sum a_n$ converges"

Body:
I'm self-studying Real Analysis. I can't understand a proof of this commonly known theorem:

Theorem. Every absolutely convergent series is convergent.

Proof. If $\sum |a_n|$ converges, given an arbitrary $\varepsilon>0$, there exists $n_0 \in \mathbb{N}$ such that $n>n_0 \Rightarrow |a_{n+1}| + ... + |a_{n+p}|<\varepsilon$, whichever $p \in \mathbb{N}$. So $|a_{n+1} + ... + a_{n+p}| \leq |a_{n+1}| + ... + |a_{n+p}|<\varepsilon$. Therefore, $\sum a_n$ converges by virtue of the Cauchy Criterion for series.

The passage that I don't understand is
$$n>n_0 \Rightarrow |a_{n+1}| + ... + |a_{n+p}|<\varepsilon.$$

Why is that the case? It doesn't seem obvious... and it doesn't seem to follow from any previous discussion. I don't know what I'm missing here.


问:
Lindeberg-Feller CLT for weighted sumation.
Problem : Let $\{a_n\}_{n\ge 1}$ be an increasing positive sequence of integers $s.t.$ the following condition holds $$\frac{a_n^2}{a_1^2+\cdots+a_n^2} \rightarrow 0.$$
Assume $\{X_n\}_{n \ge 1}$ are $iid$ sequence of random variables with mean $0$ and variance $1$. TO SHOW
$$\frac{a_1X_1 + a_2X_2 + \cdots + a_nX_n}{\sqrt{a_1^2+\cdots+a_n^2}} \rightarrow N(0,1)$$

Direction: Well I feel we need to use the Linderberg's condition for CLT. By the definition of triangular arrows define $X_{n,k} = \frac{a_kX_k}{\sqrt{a_1^2+\cdots+a_n^2}}$, where $k \le n$. The Lindeberg condtion states $\sum_{k=1}^n\mathbf{E}[X_{n,k}^2\mathbf{1}_{X_{n,k} > \delta}] \rightarrow 0$ as $n \rightarrow \infty$ for any arbit $\delta$ implies the result we set out to show.

Thus trying to show the condition holds:
$$\mathbf{E}[X_{n,k}^2\mathbf{1}_{X_{n,k} > \delta}]$$
$$= \frac{a^2_k}{a_1^2+\cdots+a_n^2} \mathbf{E} \left[X_1^2\mathbf{1}_{|X_1| > \delta\frac{\sqrt{a_1^2+\cdots+a_n^2}}{a_k}}\right] $$
$$\leq \frac{a^2_n}{a_1^2+\cdots+a_n^2}\mathbf{E} \left[X_1^2\mathbf{1}_{|X_1| > \delta\frac{\sqrt{a_1^2+\cdots+a_n^2}}{a_n}}\right]$$
But now summing over $n$ terms give me
$$\frac{na^2_n}{a_1^2+\cdots+a_n^2}\mathbf{E} \left[X_1^2\mathbf{1}_{|X_1| > \delta\frac{\sqrt{a_1^2+\cdots+a_n^2}}{a_n}}\right]$$
I can conclude $\mathbf{E} \left[X_1^2\mathbf{1}_{|X_1| > \delta\frac{\sqrt{a_1^2+\cdots+a_n^2}}{a_n}}\right]$ converges to $0$ because of finite variance and the condtion stated. However there remains a $\frac{na^2_n}{a_1^2+\cdots+a_n^2}$ term which I am pretty sure I cant show it goes to $0$.

Any help is appreciated

Edit : I was wondering if it could be shown that $\frac{na^2_n}{a_1^2+\cdots+a_n^2}$ this could be shown to be a constant as $\frac{na^2_n}{a_1^2+\cdots+a_n^2} \rightarrow 0$ for increasing sequences of $(a_n)$

答:
Assume $ \{X_n,n\ge 1\} $ are $ iid $ sequence of random variables with $ \mathsf{E}[X_n]=0 $ and $ \mathsf{var}[X_n]=1 $.
Let $ \{a_n, n\ge1\} $ be a sequence of real number and $ s_n^2=\sum_{i=1}^{n}a^2_i $, $ S_n=\sum_{i=1}^{n}a_iX_i $.

1. If
\begin{equation*}
 \lim_{n\to\infty}s_n^2=\lim_{n\to\infty}\sum_{i=1}^{n}a^2_i<+\infty, \tag{1}
\end{equation*}
then
\begin{equation*}
 \lim_{n\to\infty}S_n=\lim_{n\to\infty}\sum_{i=1}^{n}a_iX_i=S, \quad \text{a.s.}
\end{equation*}
where $ S $ is a nondegenerated random variable, its distribution depends on the $ X_1 $'s. This is a consequence of Kolmogorov theorem(For series of independent r.v. terms with zero mean, the convergence of series of variance implies the convergence a.s.).

2. If
\begin{gather*}
 \lim_{n\to\infty}s_n^2=\lim_{n\to\infty}\sum_{i=1}^{n}a^2_i=+\infty, \tag{2}\\
 \lim_{n\to\infty}\frac{a_n^2}{s_n^2}=0.     \tag{3}
\end{gather*}

Notice that, for $ 1\le k\le n $,
\begin{gather*}
 b_n\triangleq\max_{1\le j\le n}|a_j|\le b_k+\max_{k\le j\le n}\Big(\frac{|a_j|}{s_j}\Big) s_n,\\
 \frac{b_n}{s_n}\le \frac{b_k}{s_n}+\max_{k\le j\le n}\Big(\frac{|a_j|}{s_j}\Big),\\
  \limsup_{n\to\infty}\frac{b_n}{s_n}\le \max_{j\ge k}\frac{|a_j|}{s_j},  \\
\limsup_{n\to\infty}\frac{b_n^2}{s_n^2}\le \limsup_{j\to\infty}\frac{a_j^2}{s_j^2}=0.\tag{4}\\
  \mathsf{E}[a_k^2X^2_k1_{|X_k|>\delta s_n/|a_k|}]\le
  a_k^2\mathsf{E}[X^2_k1_{|X_k|>\delta s_n/b_n}]\\
  \sum_{k=1}^{n}\mathsf{E}[a_k^2X^2_k1_{|X_k|>\delta s_n/|a_k|}]
  \le \sum_{k=1}^{n}a_k^2\mathsf{E}[X^2_11_{|X_1|>\delta s_n/b_n}] \tag{5}
\end{gather*}
hence from (4),
\begin{equation*}
 \lim_{n\to\infty}\frac{\max\limits_{1\le i\le n}a_i^2}{s_n^2}=
   \lim_{n\to\infty}\frac{b_n^2}{s_n^2}=0, \tag{6}
\end{equation*}
and the Lindeberg's condition holds  from (5)(6), i.e.,
\begin{equation*}
 \lim_{n\to\infty}\frac{1}{s_n^2}\sum_{k=1}^{n}\mathsf{E}[a_k^2X^2_k1_{|X_k|>\delta s_n/a_k}]=0,\quad \forall\delta>0. \tag{7}
\end{equation*}

3  If  $ \{a_n, n\ge1\} $ is a sequence of increasing positive real numbers and (3) holds also, then
\begin{equation*}
 s_n^2\ge a_1^2\Big[\frac{a_n^2}{s_n^2}\Big]^{-1}\to+\infty,\quad \text{ as } n\to+\infty,
\end{equation*}
(6) and Lindeberg's condition (7) holds too.


问:
Show that the sequence $x_n$ converges.
$$x_n = 1 + \frac{2}{4} + \frac{3}{4^2} + \dotsb + \frac{n}{4^{n-1}}$$

So I need to see if the sequence is monotone and bounded. By doing $x_{n+1} - x_n$ I get $\frac{n+1}{4^n} > 0$ so its increasing. How do I show that it is bounded? Also I have a hint to use the $2^n > n$ inequality.

Title: Show that the sequence $x_n$ converges.

Body:
[Note] :
$$
1 + \frac{2}{4} + \frac{3}{4^2} + \frac{4}{4^{3}}+\cdots + \frac{n}{4^{n-1}}
$$
$$
\leq
$$
$$
1+\frac{2}{4}+\frac{3}{4}+\frac{4}{4}+\cdots+\frac{n}{4(n-1)}
$$


答: Use $$n\leq 2^n\leq 4^n$$ which implies $$\frac{n}{4^n}\leq \frac{2^n}{4^n}=\frac{2^n}{(2^{n})^2}=\frac{1}{2^n}\leq 1$$ and use the comparison test, as $$\sum_{k=0}^{\infty}\frac{1}{2^k}=2 $$ is a convergent geometric series. You might need to make some slight edits, as this isn't your exact series, but the same principle will work.

Title: Show that the sequence $x_n$ converges.

Body:
$$x_n = 1 + \frac{2}{4} + \frac{3}{4^2} + \dotsb + \frac{n}{4^{n-1}}$$

So I need to see if the sequence is monotone and bounded. By doing $x_{n+1} - x_n$ I get $\frac{n+1}{4^n} > 0$ so its increasing. How do I show that it is bounded? Also I have a hint to use the $2^n > n$ inequality.


答:
Hint When $n$ is large enough we have
$$
\frac{n}{4^{n-1}} <\frac{1}{2^n}
$$

Title: Show that the sequence $x_n$ converges.


问:
How many comparisons does the insertion sort use to sort the lists in question
I have two lists to sort using insertion sort:

How many comparisons does the insertion sort use to sort the list
$n, n − 1, . . . , 2, 1$?

How many comparisons does the insertion sort use to sort the list $1, 2, . . . , n$?

I know that the answer for each respectively is:

$1+1+...+1=\sum\limits_{i=1}^{n-1}1=n-1$

$\frac{(n^2-n)}{2}$

I don't understand why the first does not have $O(n^2)$ complexity instead of $O(n)$? I guess we need the same amount of comparison for both. I am confused about what the series of numbers in both questions mean?

Insertion Sort Algorithm:

```<code>    void insertionSort(int arr[], int n)
{
    int i, key, j;
    for (i = 1; i < n; i++)
    {
        key = arr[i];
        j = i - 1;

        /* Move elements of arr[0..i-1], that are
        greater than key, to one position ahead
        of their current position */
        while (j >= 0 && arr[j] > key)
        {
            arr[j + 1] = arr[j];
            j = j - 1;
        }
        arr[j + 1] = key;
    }
}
</code>```

答:
It looks like you have a mistake in (1). You wrote:
$$1 + 1 +  \cdots + 1 = \sum_{i=1}^{n-1} i = (n-1).$$

That sum should have been:
$$\sum_{i=1}^{n-1} 1 = n-1.$$

Now in terms of the comparisons, those get made when percolating the element $$A[i]$$ forward. If the elements are already in ascending order, then $$A[i] \geq A[i-1]$$ and no swap is made. So the inner loop does not execute.


问:
Prove that matrix is non-negative
Problem:

Given $A_{1}, A_{2}, ..., A_{n}$ - finite sets and $a_{ij} = |A_{i}\cap A_{j}|$ - number of elements in intersection of sets. Prove, that matrix  $(a_{ij})_{i=1,2,..,n}^{j=1,2,.., n}$ is non-negative.

I've cleared out that this matrix is symmetric and the largest elements are on its diagonal. I need to prove that $x^{T}Ax \ge 0$ or

$$\sum_{i=1}^n\sum_{j=1}^na_{ij}x_{i}x_{j} \ge 0$$

or its eigenvalues are non-negative.

I don't know, what need I do next? Any hint will be useful!

答:
A fun problem. I think that the following approach is natural (at least it is the first that occurred to me, YMMV).

Let us consider the union
$$
A=\bigcup_{i=1}^nA_i.
$$
I will work in the space $F_A$ of real valued functions from $A$ to $\mathbb{R}$.
If you list the elements of $A$ like
$$
A=\{a_1,a_2,\ldots,a_m\},
$$
you can identify the space $F_A$ with vectors of $\mathbb{R}^m$. The identification is natural. We identify the function $f:A\to\mathbb{R}$ with the vector
$$
\vec{f}=(f(a_1),f(a_2),\ldots, f(a_m))\in\mathbb{R}^m.
$$
Let us denote by $\chi_i$ the characteristic function of the subset $A_i$, that is the function in $F_A$ defined by $\chi_i(a)=1$, if $a\in A_i$, and $\chi_i(a)=0$, if $a\notin A_i$.

Let $x_i,i=1,2,\ldots,n$ be arbitrary real numbers as in your question. Consider the function
$$
f=\sum_{i=1}^n x_i\chi_i\in F_A.
$$
Let us look at the vector $\vec{f}\in\mathbb{R}^m$. I denote by $\langle\ ,\ \rangle$ the usual inner product of the space $\mathbb{R}^m$. We have trivially
$$
\langle \vec{f},\vec{f}\rangle=\Vert\vec{f}\Vert^2\ge0.
$$
On the other hand we have
$$
\vec{f}=\sum_{i=1}^n x_i\vec{\chi_i}.
$$
Bilinearity of the inner product gives us thus that
$$
\langle\vec{f},\vec{f}\rangle=\sum_{i=1}^n\sum_{j=1}^n x_i x_j\langle\vec{\chi_i},\vec{\chi_j}\rangle.
$$
But, more or less obviously (think about the inner product of two distinct vectors both with components zero or one only), we also have
$$
\langle\vec{\chi_i},\vec{\chi_j}\rangle=|A_i\cap A_j|=a_{ij}.
$$
The claim about non-negativity follows from this.

Title: Prove that matrix is non-negative


问:
$f:\mathbb{N}\to\mathbb{Z}$ which is surjective but not injective
I'm looking for a really simple example of $f:\mathbb{N}\to\mathbb{Z}$ which is surjective but not injective. I thought about:
$$
f(x)=\begin{cases}
0 & x=0\\
\frac{x+1}{2} & x\text{ is odd}\\
-\frac{x}{2} & x\text{ is even}\wedge x\neq0
\end{cases}
$$
Is there a better and more simple  example I can use?

答:
$$
f(x)=\begin{cases}
0 & x=0\\
\frac{x+1}{2} & x\text{ is odd}\\
-\frac{x}{2} & x\text{ is even}\wedge x\neq0
\end{cases}
\implies
f(n)=(-1)^{n+1}\bigg\lfloor\frac{n+1}{2} \bigg\rfloor
$$

$$
f(0)=(-1)^1\bigg\lfloor\frac{1}{2}\bigg\rfloor=0\quad
f(1)=(-1)^2\bigg\lfloor\frac{2}{2}\bigg\rfloor=1\quad
f(2)=(-1)^3\bigg\lfloor\frac{3}{2}\bigg\rfloor=-1\\
f(3)=(-1)^4\bigg\lfloor\frac{4}{2}\bigg\rfloor=2\quad
f(4)=(-1)^5\bigg\lfloor\frac{5}{2}\bigg\rfloor=-2\quad
f(5)=(-1)^6\bigg\lfloor\frac{6}{2}\bigg\rfloor = 3$$

Of course, the following alone also meets the criteria: $$f(x)=(-1)^x$$

as does the floor function alone unless you talk about infinities.


问:
Proving Integration by Parts for Brownian Motion and function of bounded variation
Looking for someone to help me finish my proof for a version of Integration by Parts. I am trying to prove that if $f$ is a continuous function of Bounded Variation, then $$\int_0^t f(s)dBs = f(t)B_t - \int_0^t B_sdf(s)$$
By definition of the Ito Integral we know that for any partition $\Pi = \{s_0,\ldots s_n\}$ where $x_0=0,x_n=t$ that
$$\lim_{|\Pi|\to0} \sum_{i=1}^n f(s_{i-1}) (B_{s_i}-B_{s_{i-1}})=\int_{0}^tf(s)dBs$$
Where the limit converges in $L^2(P)$. Now Brownian Motion has continuous path almost surely. For any $\omega$ so that $s\to B_s(\omega)$ is continuous, then $B_s(\omega)$ is integrable with respect to $f$ since $f$ is continuous and of bounded variation. Thus $\int_0^t B_s(\omega)df(s)$ exists almost surely. In particular
$$\lim_{|\Pi|\to 0}\sum_{i=1}^n B_{s_i}(\omega)(f(s_i)-f(s_{i-1}))=\int_0^t B_s(\omega)df(s) $$
What I need help is showing that the convergence is also in $L^2(P)$.

If I can show that the convergence is also in $L^2(P)$, then
\begin{align*}\int_0^t f(s)dBs + \int_0^t B_sdf(s) &= \lim_{|\Pi|\to 0}\sum_{i=1}^n f(s_{i-1}) (B_{s_i}-B_{s_{i-1}})+\sum_{i=1}^n B_{s_i}(f(s_i)-f(s_{i-1})) \newline
&= \lim_{|\Pi|\to 0}\sum_{i=1}^n f(s_{i-1}) B_{s_i}-f(s_{i-1})B_{s_{i-1}}+f(s_i)B_{s_i}- f(s_{i-1})B_{s_i}\newline
&= \lim_{|\Pi|\to 0}\sum_{i=1}^n f(s_i) B_{s_i}-f(s_{i-1}) B_{s_{i-1}}\\
&= f(t)B_t - f(0)B_0= f(t)B_t
\end{align*}
where this is the $L^2(P)$ limit.

答:
Consider the a sequence of partitions $\{\Pi_n\}$ on $[0,t]$ where $\Pi_n\subset \Pi_{n+1}$. Here $|\Pi_n| = \max |s_i-s_{i-1}|$ where $\Pi_n = \{s_0, s_1, \ldots s_{m_n}\}$, $0=s_0 < s_1 < \ldots < s_{l_n}=t$ and we assume that $|\Pi_n|\to 0$. Put
$$g_n(\omega)=\sum_{i=1}^{l_n}B_{s_{i}}(\omega)(f(s_i)-f(s_{i-1})) $$
Since Brownian motion is continuous almost surely, then if $t\to B_t(\omega)$ is continuous we see that $g_n(\omega)\to \int_0^t B_s(\omega)df(s)$ since every continuous function is Riemann-Stieltjes integrable with respect to any function of bounded variation.

We now show that $\{g_n\}$ is Cauchy in $L^2(P)$. Assume $n < m$ and without loss of generality, we may assume that for any $s_{i-1},s_i\in \Pi_n$ we have at least one $\tau\in (s_{i-1},s_i) \cap \Pi_m$. We can denote all of the $\tau$ in the following manner $s_{i-1} = \tau_{i,0}< \tau_{i,1} < \ldots < \tau_{i,j_i} = s_{i}$. Then we can write $g_m$ in the following manner
$$g_m(\omega)= \sum_{i=1}^{l_n} \sum_{k=1}^{j_i} B_{\tau_{i,k}}(\omega)(f(\tau_{k,i})-f(\tau_{i,k-1}))$$
We can also rewrite $g_n$ in the following manner
$$g_n(\omega)= \sum_{i=1}^{l_n} \sum_{k=1}^{j_i} B_{s_{i}}(\omega)(f(\tau_{k,i})-f(\tau_{i,k-1})) $$
By our assumption on $\tau$, we know that $i_j \geq 2$ for all $j$ so that $i_j-1\geq 1$ as well. With this in mind, we compute that
$$\begin{align*}\mathbb{E}[(g_n-g_m)^2] &= \mathbb{E}\Big[\Big(\sum_{i=1}^{l_n}\sum_{k=1}^{j_i -1} (B_{s_i}- B_{\tau_{i,k}})(f(\tau_{i,k})-f(\tau_{i,k-1}))\Big)^2\Big]  \\
&= \sum_{r=1}^{l_n} \sum_{u=1}^{l_n} \sum_{w=1}^{j_r-1}\sum^{j_s-1}_{v=1} \mathbb{E}[(B_{s_r}-B_{\tau_{r,w}})(B_{s_u}-B_{\tau_{u,v}})](f(\tau_{r,w})-f(\tau_{r,w-1}))(f(\tau_{u,v})-f(\tau_{u,v-1}))
\end{align*}$$
Now if $u<r$, then we have that $\tau_{u,v}< s_u < \tau_{r,w}<s_r$. Since Brownian Motion has independent increments, then $B_{s_r}-B_{\tau_{r,w}}$ and $B_{s_u}-B_{\tau_{u,v}}$ are independent and both have $0$ mean. Thus $$ \mathbb{E}[(B_{s_r}-B_{\tau_{r,w}})(B_{s_u}-B_{\tau_{u,v}})]=\mathbb{E}[B_{s_r}-B_{\tau_{r,w}}]\mathbb{E}[B_{s_u}-B_{\tau_{u,v}}]=0$$
The same also holds if $r < u$. Hence we can write the previous statement as
$$\begin{align*} \mathbb{E}[(g_n-g_m)^2] &= \sum_{i=1}^{l_n}\sum_{k=1}^{j_i -1} \mathbb{E}[(B_{s_i}- B_{\tau_{i,k}})^2](f(\tau_{i,k})-f(\tau_{i,k-1}))^2\\
&=  \sum_{i=1}^{l_n}\sum_{k=1}^{j_i -1} (s_i-\tau_{i,k})(f(\tau_{i,k})-f(\tau_{i,k-1}))^2\\
&\leq |\Pi_n|\sum_{i=1}^{l_n}\sum_{k=1}^{j_i -1} (f(\tau_{i,k})-f(\tau_{i,k-1}))^2\\
&\leq |\Pi_n|\Big(\sum_{i=1}^{l_n}\sum_{k=1}^{j_i} |f(\tau_{i,k})-f(\tau_{i,k-1})|\Big)^2\\
&\leq |\Pi_n|(V_0^t(f))^2\end{align*}$$
Here $V_0^t(f)$ is the total variation of $f$, since $f$ is assumed to be of bounded variation then $V_0^t(f)<\infty$. Since $|\Pi_n|\to 0$, then we can now observe that $\{g_n\}$ is a Cauchy sequence in $L^2(P)$. Since $L^2(P)$ is a Banach space, then $g_n \to g$ in $L^2(P)$. But we already know that $g_n\to \int_0^t B_sdf(s)$ $a.e$, since we have convergence in both topologies, then the limits must coincide. Hence $g_n \to \int_0^t B_sdf(s)$ in $L^2(P)$.


问:
Probability of hanging n keys on n hooks
Question:

a) One evening, a bemused lodge-porter tried to hang $n$ keys on their $n$ hooks, but only managed to hang them independently and at random. There was no limit to the number of keys which could be hung on any hook. Otherwise, find an expression for the probability that at least one key was hung on its own hook.

b) The following morning, the porter was rebuked by the Bursar, so that in the evening she was careful to hang only one key on each hook. But she still only managed to hang them independently and at random. Find an expression for the probability that no key was then hung on its
own hook.

Find the limits of both expressions as $n$ tends to infinity

I know how to find the number of ways that the hooks can be placed using combinatorics methods, but I don't know how to find the probabilities of these events. Does the probability use combinatorics? Is there a simple formula I don't know? Any help appreciated.

答:
Suppose we have $k$ hooks and $n$ keys. You may think of $k = n$, but for the sake of clarity, I'd like to (for the moment) use different variables. I'd like to think of the hooks as letters and the keys as positions in a string. For each key, we may select a hook in $k$ ways. There are $n$ such keys, so there are $k^{n}$ ways of placing the keys on the hooks.

For (a), take the count you got and divide by $k^{n}$ (in this case, $k = n$, so divide by $n^n$). Here, the probability is a relative frequency. In particular, we are selecting a configuration of keys on hooks uniformly at random (that is, we are selecting the configuration uniformly at random).

For (b), our sample space is now permutations. We have $n!$ permutations. How many ways are there of placing $n$ keys on $n$ hooks, so that no key is on its own hook?

 The number here is the derangement number. Have you had a chance to work out the inclusion-exclusion expression for derangements?


答:
Let $\mathcal{A} = \cup_{i=1}^n A_i$ and $\chi_i : \mathcal{A} \to \mathbb{N}$ be the indicator function for $A_i, i = 1,\ldots n$.

For any $n$ real numbers $x_1, \ldots, x_n$, not all zero. Consider the function $f: \mathcal{A} \to \mathbb{R}$ defined by:

$$f(a) = \sum_{i=1}^{n} x_i \chi_i(a)$$

We have:

$$
  \sum_{i=1}^{n}\sum_{j=1}^{n} a_{ij} x_i x_j
= \sum_{i=1}^{n}\sum_{j=1}^{n} x_i x_j |A_i \cap A_j|
= \sum_{i=1}^{n}\sum_{j=1}^{n} x_i x_j \left(\sum_{a\in A} \chi_i(a) \chi_j(a)\right)\\
= \sum_{a\in \mathcal{A}} \left(\sum_{i=1}^n x_i \chi_i(a)\right)\left(\sum_{j=1}^n x_j \chi_j(a)\right) = \sum_{a\in \mathcal{A}} f(a)^2
\ge 0
$$

Title: Prove that matrix is non-negative

Body:
Problem:

Given $A_{1}, A_{2}, ..., A_{n}$ - finite sets and $a_{ij} = |A_{i}\cap A_{j}|$ - number of elements in intersection of sets. Prove, that matrix  $(a_{ij})_{i=1,2,..,n}^{j=1,2,.., n}$ is non-negative.

I've cleared out that this matrix is symmetric and the largest elements are on its diagonal. I need to prove that $x^{T}Ax \ge 0$ or

$$\sum_{i=1}^n\sum_{j=1}^na_{ij}x_{i}x_{j} \ge 0$$

or its eigenvalues are non-negative.

I don't know, what need I do next? Any hint will be useful!


问:
How may random unique combinations a 4x4 matrix can make
Let's say we have a matrix (grid) of 4x4 that has values from 0 to 15 in it.

```<code>    0  1  2  3
a = 4  5  6  7
    8  9  10 11
    12 13 14 15
</code>```

If this array is randomized or shuffled how many unique combination can it make?

For example:

By unique I mean let's say 10 in the matrix a above is at 3x3 and in an attempt of a shuffle of the whole matrix, only 10 position shuffled with let's say 15 (10 is at the place 15 is and vice versa) so the new position fo 10 is now 4x4 and the matrix will look like this:

```<code>    0  1  2  3
b = 4  5  6  7
    8  9  15 11
    12 13 14 10
</code>```

Now, b is a unique combination of a 4x4 matrix a. So, like that how many unique combinations can array a can have?

Body: I'd say $16!=16\times15\times\cdots\times2\times1$.


问:
Proof if $f$ is continuous then the preimage is closed.
I want to show that if $f: \mathbb{R}\to\mathbb{R}$ is continuous then $f^{-1}(l) = \{x\in\mathbb{R}|f(x) = l\}$ is closed.

I am not sure of my proof as I feel like I am missing a step:

I decided to use sequential continuity and consider a convergent sequence $(x_n)\subset f^{-1}(l)$.

By sequential continuity if $x_n \to L \implies f(x_n) \to f(L) = L'$

I want to say that I am now done - but I think I need to prove that $L \in f^{-1}(l)$, or $L' \in Imf$.

答:
[Note] : As Robert mentioned, it is easier to show that continuity implies inverse of an open set is open which I will use as a Lemma.

Lemma $(1)$ : Consider $(X,d_{X})$ and $(Y,d_{Y})$ metric spaces and let $f : X\longrightarrow Y$. We claim that $f$ is continuous in $X$ if and only if for every open set $G\subseteq Y$, $f^{-1}(G)$ is open in $X$.

Proof :

$(\implies)$ :

Let $G$ be open in $Y$, we have to show that $f^{-1}(G)$ is open in $X$. Let $x_{0}\in f^{-1}(G)\implies f(x)\in G$ and since $G$ is open, then $f(x_{0})$ is an interior point of $G$ that is to say $\exists\varepsilon>0$ such that $N_{\varepsilon}(f(x_{0}))\subseteq G$. By continuity of $f$ at $x_{0}$, $\exists\delta>0$ such that $\forall x\in N_{\delta}(x_{0})\implies f(x)\in N_{\varepsilon}(f(x_{0}))$. We now show that $x\in f^{-1}(G)$, notice that :
\begin{align*}
\text{If}\;x\in N_{\delta}(x_{0})&\implies f(x)\in N_{\varepsilon}(f(x_{0}))\subseteq G \\
&\implies x\in f^{-1}(G),\;\forall x\in N_{\delta}(x_{0})\\
&\implies N_{\delta}(x_{0})\subseteq f^{-1}(G)
\end{align*}
Hence, $x_{0}$ is an interior point of $f^{-1}(G)$ implying that $f^{-1}(G)$ is open in $X$.

$(\impliedby)$ :

We have that $\forall G\subseteq Y$ open, we have that $f^{-1}(G)$ is open in $X$. Let $x_{0}\in X$, we want to show that $f$ is continuous at $x_{0}$. Let $N_{\varepsilon}(f(x_{0}))$ be a neighborhood in $Y$ with $\varepsilon>0$, then since $N_{\varepsilon}(f(x_{0}))$ is open in $Y$, then $f^{-1}(N_{\varepsilon}(f(x_{0}))$ is open in $X$. Now $x_{0}\in f^{-1}(N_{\varepsilon}(f(x_{0}))$ because $f(x_{0})\in N_{\varepsilon}(f(x_{0}))$. Therefore, $x_{0}$ is an interior point of $f^{-1}(N_{\varepsilon}(f(x_{0}))$ which means there exist $\delta>0$ such that $N_{\delta}(x_{0})\subseteq f^{-1}(N_{\varepsilon}(f(x_{0}))$. Therefore, $\forall x\in N_{\delta}(x_{0})$ we have that $x\in f^{-1}(N_{\varepsilon}(f(x_{0}))\implies f(x)\in N_{\varepsilon}(f(x_{0}))$. Hence, $f$ is continuous at $x_{0}$.

Corollary : Consider $(X,d_{X})$ and $(Y,d_{Y})$ metric spaces and let $f : X\longrightarrow Y$. We claim that for every open set $G\subseteq Y$, we have that $f^{-1}(G)$ is open in $X$ if and only if for every closed set $F\subseteq Y$, we have $f^{-1}(F)$ is closed in $X$

Proof : The proof is trivial since it follows from the fact that $f^{-1}(E^{c})=(f^{-1}(E))^{c}$. Thus, if $f : X\longrightarrow Y$ is continuous and $F\subseteq Y$ is closed, then $F^{c}$ is open which implies $f^{-1}(F^{c})$ is open which means $(f^{-1}(F))^{c}$ is open by the $\text{(Lemma)}$ and thus $f^{-1}(F)$ is closed. The reverse implication also follows from that basis.


问:
Rigorous proof for Theorem 7.7 in Loring Tu
Theorem 7.7 is stated and proved as follows in Loring Tu's An Introduction to manifolds:

However, it seems that the deduction which involves Figure 7.4 in the proof is not so rigorous, so I want to know if there's a deduction involving only pure logic, rather than graphs.

答:
The product topology on $S\times S $ is definitionally that generated by the basis of products $A\times B$ of open subsets $A,B$ of $S$. Thus every open set $J$ in $S\times S$ can be written as a union $J= \bigcup_{n=1}^\infty A_i \times B_i$, where $A_i, B_i$ are open in $S$.

Setting $J:=S\times S \setminus R$, we obtain the result.


问:
Find subdifferential of function $f: \mathbb{R} \rightarrow \mathbb{R}, f(x) = \max(1, \exp(x^2))$
By subdifferential I mean the set of all subderivatives, where the subderivative at point $x$ is a real number $c$ such that for all $y \in \mathbb{R}$ it is the case that:
$$f(y) - f(x) \geq c(y-x)$$

It seems to me that for $x < 0$ it is just $0$ and for $x > 0$ it is $2x\exp(x^2)$. So I am interested in subdifferential at point $x = 0$.

So I have to find $c$ such that for all $y$
$$f(y) - f(x) \geq c(y -x) \\
f(y) - 1 \geq cy \\$$

$f(y)$ has to be greater or equal to $1$.

Is there other subdifferential at point $x = 0$ then $c = 0$?

答:
In general at a breakpoint the subgradient is the convex hull of the left and right derivative. In this case you have $f(y) - 1 \geq cy$ for all $y \in \mathbb{R}$, so
\begin{align}
0 \geq cy \qquad \forall y \leq 0 \\
\exp(y^2) - 1 \geq cy  \qquad \forall y \geq 0
\end{align}
The first inequality gives $c \geq 0$. The second inequality gives $c \leq [\exp(y^2) - \exp(0)] / y$, which if you take the limit of $y \downarrow 0$ gives $c \leq 0$, so $c=0$.


问:
Conjugate function of negative logarithm
I'm trying to understand how conjugate function is built for example like

$f(x) = -\log x$ where $\operatorname{dom} f = \mathrm{R_{++}}$

One of the lectures which I'm checking

By definition, $f^{*}(y) = \sup_x(yx + \log x)$

Questions:

what does it mean that the difference is bounded?

why the difference is bounded if and only if $y < 0$?

why is it reaching its maximum at $x = \frac{-1}{y}$?

答:
You have to understand that when it says $log(x)$ it really means $ln(x)$

Here's my work below

$$f(x)=-ln(x)$$

$$f'(x)=-1/x$$

Because we know that $f$ is differentiable and this point of maximum gap on $f(x)$ occurs at $f'(x)=y$, therefore the maximum is at

$-1/x=y$ or $x=-1/y$

So now, to find the $f^*(y)$, we know our slope is $y$, and our x-point is $-1/y$ which makes our y-point $f(-1/y)=-ln(-1/y)$ which also equals $ln(-y)$

Now if I put this in point-slope formula

$$y-(ln(-y))=y(x-(-1/y))$$

Then with $x=0$

$$y-(ln(-y))=y(0-(-1/y))$$

$$y-(ln(-y))=1$$

$$y=ln(-y)+1$$

And therefore my $f^*(y)$ will equal the negative of my $y$ result and I get $-ln(-y)-1$


问:
Finding a recursive for virus outbreak
Suppose a contagious person infects 3 new people in one day. After a day, these newly infected people become contagious themselves. Every contagious person goes into quarantine after one day, meaning they do not infect any more people. To summarize:

a person becomes infected and is not contagious

after 1 day they become contagious and infect 3 people

1 day after that they get put in quarantine

I want to find a recursive formula that gives me the number of people, that are not in quarantine (Infected or Contagious)

My approach was the following:

I have found an recursive Formula:

$$\overline{Q}_n =3* \overline{Q}_{n-1}$$
under the conditions
$$ n>1, \overline{Q}_{0}=1, \overline{Q}_{1}=4$$

Is there a way to make the recursive formula so, such that only one initial value is required? Thanks in advance

答:
You have two sequences involved:

$I_n =$ number of infected but not contagious people on day $n$.

$C_n =$ number of contagious people on day $n$.

There is no particular reason to worry about the number of quarantined people for this problem. The givens are:

$I_0 = 1$

$C_0 = 0$

$C_n = I_{n-1}$

$I_n = 3C_n$, $n > 0$

The $n > 0$ clause is necessary because the given data for day $0$ does not match the information that $I_n = 3C_n$, since $1 \ne 3\times 0$. I.e., the initial infection arrived in a way other than in accordance to the rules that hold going forward.

And it is this exception that you are running into in your sequence. $\overline Q_n = \frac 43 I_n$, but only for $n > 0$. That additional $\frac 13$ contagious person required by the later rules just isn't there on day $0$.

Title: Finding a recursive for virus outbreak


问:
Some have more than one stationary distributions?
Given a Markov chain transition matrix $P$, I want to find its stationary distribution. According to the definition, if $\pi$ satisfies the equation $\pi = \pi P$, then $\pi$ is called a stationary distribution.

The above definition is what my professor said in the class. When I learned linear algebra, I was told that stationary distribution is the eigenvector of the matrix $P$ (row sum =1) associated with the eigenvalue 1. I verify this is correct in R.

However, the eigenvector method got problem when I face the 3 by 3 identity matrix. This is the example of "have more than one stationary distributions". I use the eigenvector method in R, and find $(1,0,0), (0,1,0), (0,0,1)$ are all eigenvectors corresponding to eigenvalue 1. However, in fact, the true answer is any row vector sum to 1 is the stationary distribution. However, the eigenvector method doesn't show it. Why?

答:
"$(1,0,0),(0,1,0),(0,0,1)$ are all eigenvectors corresponding to eigenvalue 1"

What you have found is not all eigenvectors, but a basis for the vector space of the eigenvectors.

If you want all eigenvectors, you want the space spanned by these:

$$\{t(1,0,0)+u(0,1,0)+v(0,0,1):t,u,v\in\mathbf{R}\}$$

that is,

$$\{(t,u,v):t,u,v\in\mathbf{R}\}$$

that is,

$$\mathbf{R}^3$$

Notice that for any $\mathbf{x}\in\mathbf{R}^3$ you have $\mathbf{x}.I=1\mathbf{x}$, so these all are, in fact, (left) eigenvectors with eigenvalue 1.

Since you only want distributions, you actually have

$$\{(t,u,v):t+u+v=1\}$$

That's infinitely many stationary distributions. Indeed, if I have probability $t$ of being in state 1, and probability $u$ of being in state 2, then after 1 step, I still (since I never change state) have probability $t$ of being in state 1, and probability $u$ of being in state 2, so all of these are stationary distributions - applying a step of the Markov process doesn't change the probability distribution of the state I might be in.


Body:
Use $$\frac{e^{-\gamma z}}{\Gamma(1+z)}=\prod_{n=1}^{\infty} (1+\frac{z}{n})e^{-z/n}~~~~(1)$$
Next, $$\left(1+\frac{1}{n^3}\right)=\left(1+\frac{1}{n}\right) \left(1+\frac{w}{n}\right) \left(1+\frac{w^2}{n}\right),$$
where $w$ is cube root of unity $w^3=1, w^2+w+1=0, w^2=w^*$
$$P=\prod_{1}^{\infty} \left(1+\frac{1}{n^3}\right)$$
From (1), we get
$$ e^{-(1+w+w^2)\gamma}[\Gamma(2)\Gamma(1+w)\Gamma(1+w^2)]^{-1}= [\Gamma(-w) \Gamma(-w^2)]^{-1}=\prod_{n=1}^{\infty} \left(1+\frac{1}{n^3}\right) e^{(1+w+w^2)/n}$$
So $$P=\frac{1}{|\Gamma(-w)|^2}=$$
$$P=\frac{1}{\Gamma[(1-i\sqrt{3})/2]~ \Gamma[(1+i\sqrt{3})/2]}$$
Use  $\Gamma(1-z)\Gamma(z)=\frac{\pi}{\sin \pi z}$ to evaluate $\Gamma(1/2-it) \Gamma(1/2+it)=\frac{\pi}{\cosh \pi t}$. Then the required $$P=\Gamma(1/2-it) \Gamma(1/2+it), t=\pi\sqrt{3}/2$$
Hence $$P=\frac{1}{\pi} \cosh(\sqrt{3}\pi/2).$$.


问:
Equivalence relation existence
Suppose an equivalence relation is defined as: for $a, b \in \mathbb{R}$, $a R b \iff a - b = n/m$, where we have $n, 0 \neq m \in \mathbb{Z}$. For any natural number $N$ and any $a\in \mathbb{R}$, show that there is a $b\in \mathbb{R}$ so that $b R a$ and $b< \frac{1}{N}$ and $b \geq 0$.

I seem to struggle with "existence" proofs and I'm new to proofs using analysis. For the second part, I think I'm supposed to use the archimedian property/ies but I'm not very sure where $0$ and $\frac{1}{N}$ come from. I'd appreciate any hints or assistance!

答: Intuitively, you are being asked to show that if you are given any real $a$ you can find a rational $\frac nm$ such that $0 \le a-\frac nm \lt \frac 1N$ or in other words you can find a rational as close as needed to any real.  Note that if $m$ is huge the spacing between  the options is very small.  In fact, if $m \gt N \ldots$

Title: Equivalence relation existence


问:
In $\mathcal{P}(\mathbb{Z})$, $A \equiv B \iff \exists_{r \in \mathbb{Z}} ( \forall_{x \in A} (x + r \in B) ∧ \forall_{y \in B} (y - r \in A))$...
We define an equivalence relation $≡$ in the set $\mathcal{P}(\mathbb{Z})$ as:

$$A \equiv B \iff \exists_{r \in \mathbb{Z}} ( \forall_{x \in A} (x + r \in B) ∧ \forall_{y \in B} (y - r \in A))$$

Determine $|\mathcal{P}(\mathbb{Z})/_≡|$ and $∣[\mathbb{N}]_≡∣ $.

I know that for $r = 0$: $A = B$, for $r = 1$: $A$ and $B$ lay next to each other, have $1$ common element and both cover a part of $X$ axis that is $1$ wide. Similar facts are true for $r = 2$, $r =3$...

That would mean that: $|\mathcal{P}(\mathbb{Z})/≡| = |\mathbb{N}|$.

However, I don't know what to do with $∣[\mathbb{N}]_≡∣ $.

答:
HINT: It is not necessarily true that $A$ and $A+1$ are one unit wide and lie next to each other: $A$ could be all of $\Bbb Z$, in which case $A+1=A$. (And in fact this is the crucial observation that lets you show that $\left|[\Bbb N]_\equiv\right|=|\Bbb N|$.)

Nor do they necessarily have one element in common. In that example they are identical and have infinitely many elements in common, and if $A=\{0\}$, $A+1=\{1\}$, and they have to element in common. Finally, no subset of $\Bbb Z$ covers a one-unit segment of the $x$-axis.

In fact the quotient has cardinalities larger than $|\Bbb N|$. I’ll point you towards one way to answer that question.

Let $\Sigma$ be the set of strictly increasing sequences in $\Bbb N$ whose first term is $1$. If $\sigma=\langle n_k:k\in\Bbb N\rangle\in\Sigma$, let $$A_\sigma=\left\{\sum_{i=0}^kn_k:k\in\Bbb N\right\}\,.$$ Show that if $\sigma,\tau\in\Sigma$, and $\sigma\ne\tau$, then $A_\sigma\not\equiv A_\tau$. Conclude that

$$\left|\wp(\Bbb N)/\!\sim\right|=\left|\wp(\Bbb Z)/\!\sim\right|\ge|\Sigma|\,.$$

Then show that

$$|\Sigma|=|\wp(\Bbb N)|=|\Bbb R|$$

and deduce that $\left|\wp(\Bbb N)/\!\sim\right|=\left|\wp(\Bbb Z)/\!\sim\right|=|\Bbb R|$.


问:
Upper bound on the difference between two elements of an eigenvector
Let $W$ be the non-negative, symmetric adjacency/affinity matrix for some connected graph. If $W_{ij}$ is large, then vertex $i$ and vertex $j$ have a heavily weighted edge between them. If $W_{ij} = 0$, then no edge connects vertex $i$ to vertex $j$.

Now $L = \mathrm{diag}(W\mathbf{1})-W$ is the (unnormalized) graph Laplacian. Let $v$ be the Fiedler vector of $L$, that is, a unit eigenvector corresponding to the second smallest eigenvalue of $L$. As $W_{ij}$ increases, all else equal, $|v_i - v_j|$ tends to decrease---at least this is the idea behind spectral clustering. What is an upper bound on $|v_i - v_j|$, given quantities that don't require computing $v$, like $W_{ij}$ and $\|W\|$?

Any suggestions or thoughts would be greatly appreciated.

答:
Let me at least point out a trivial bound.

If $L$ is a Laplacian, then $x^TLx=\sum_{i,j}L_{i,j}(x_i-x_j)^2$. Let
$\lambda$ be the eigenvalue for the Fiedler vector. Then
$$\lambda = x^TLx = \sum_{i,j} w_{i,j}(x_i-x_j)^2,$$ whence
$|x_i-x_j| \le (\lambda/w_{i,j})^{1/2}$.


问:
Boundedness of inverse in Evans Chapter 6 (why is the sequence bounded in Sobolev spaces?)
Evans says this can be estimate by energy estimates,but this is true if α＜β,what about α≥β,could the estimate be valid again (screenshot transcribed below)

THEOREM 6 (Boundedness of the inverse). If $\lambda \notin \Sigma$, there exists a constant $C$ such that

$$
\|u\|_{L^{2}(U)} \leq C\|f\|_{L^{2}(U)} \tag{29}
$$
whenever $f \in L^{2}(U)$ and $u \in H_{0}^{1}(U)$ is the unique weak solution of
$$
\left\{\begin{array}{cl}
L u=\lambda u+f & \text { in } U \\
u=0 & \text { on } \partial U
\end{array}\right.
$$
The constant $C$ depends only on $\lambda, U$ and the coefficients of $L .$
This constant will blow up if $\lambda$ approaches an eigenvalue.

Proof. If not, there would exist sequences $\left\{f_{k}\right\}_{k=1}^{\infty} \subset L^{2}(U)$ and $\left\{u_{k}\right\}_{k=1}^{\infty} \subset$ $H_{0}^{1}(U)$ such that
$$
\left\{\begin{aligned}
L u_{k} &=\lambda u_{k}+f_{k} & & \text { in } U \\
u_{k} &=0 & & \text { on } \partial U
\end{aligned}\right.
$$
in the weak sense, but
$$
\left\|u_{k}\right\|_{L^{2}(U)}>k\left\|f_{k}\right\|_{L^{2}(U)} \quad(k=1, \ldots)
$$
As we may with no loss suppose $\left\|u_{k}\right\|_{L^{2}(U)}=1,$ we see $f_{k} \rightarrow 0$ in $L^{2}(U)$. According to the usual energy estimates the sequence $\left\{u_{k}\right\}_{k=1}^{\infty}$ is bounded in $H_{0}^{1}(U) .$ Thus...

The energy estimate is below (screenshot):

THEOREM 2 (Energy estimates). There exist constants $\alpha, \beta>0$ and $\gamma \geq 0$ such that

$$
|B[u, v]| \leq \alpha\|u\|_{H_{0}^{1}(U)}\|v\|_{H_{0}^{1}(U)}\tag{i}
$$
and

$$
\beta\|u\|_{H_{0}^{1}(U)}^{2} \leq B[u, u]+\gamma\|u\|_{L^{2}(U)}^{2}\tag{ii}
$$

Could someone help me, thanks a lot.

答:
You don't need to use (i), (ii) is enough and gives
\begin{align}
 \|u\|_{H^1_0(U)}^2
&\le \frac1\beta B[u,u] + \frac\gamma\beta \|u\|_{L^2(U)}^2
\\&= \frac1\beta (u,f)_{L^2(U)} + \frac\gamma\beta\|u\|^2_{L^2(U)}
\\&\le \frac1\beta\|u\|_{L^2(U)}\|f\|_{L^2(U)}+\frac\gamma\beta\|u\|_{L^2(U)}^2
\\&=\frac1\beta\|f\|_{L^2(U)}+\frac\gamma\beta
\end{align}
where we used the definition of a weak solution $B[u,v]=(f,v)_{L^2(U)}$ valid for all $v\in H^1_0(U)$ (in particular valid for $v=u$).


问:
Help confirming my proof and the answer key of $(\exists!x)P(x)$ as equivelant to the following?
In a transition to advanced mathematics, fourth edition, chapter 1.3 #11. It want to prove

$(\exists!x)P(x)$ is equivelant to $(\exists x)A(x) \land (\forall y)(\forall z)\left[A(y)\land A(z)\!\!\implies \!\! y=z\right]$

$(\exists!x)P(x)$ is equivelant to $(\exists x)\left[A(x)\!\!\implies \!\! x=y\right]$

Attempt at Solving Problem #1

$(\exists!x)P(x)$ is true in Universe U

iff There exists a unique x where P(x) is true in Universe U

iff There exists some x where A(x) is true and two elements of property A are the same element

iff $(\exists x)A(x) \land (\forall y)(\forall z)\left[A(y)\land A(z)\!\!\implies \!\! y=z\right]$

Answer Key To Problem #1

I couldn't fully undestand the answer key

$(\exists !)A(x)$ is true in Universe U

Then $(\exists ! x) A(x)$ implies that A(x) is true for unique x in U

iff $(\exists ! x) A(x)$ is not true for every x in U

iff there exists some $x$, $y$ and $z$ such that $(\exists x)A(x) \land (\forall y)(\forall z)\left[A(y)\land A(z)\right]$ for $y=z$

iff $(\exists ! x) A(x)$ is equivelant to $(\exists!x)P(x)$ is equivelant to $(\exists x)A(x) \land (\forall y)(\forall z)\left[A(y)\land A(z)\!\!\implies \!\! y=z\right]$

Questions

In the answer key I don't understand step 4. Shouldn't some $y$ and $z$ give $(\exists x)A(x) \land (\exists y)(\exists z)\left[A(y)\land A(z)\right]$ for $y=z$? Is my answer correct?

Since this post is long I will save the second part of #11 for another post.

答:
In the answer key I don't understand step 4. Shouldn't some y and z give (∃x)A(x)∧(∃y)(∃z)[A(y)∧A(z)] for y=z?

For sure, that answer key is rather misleading.

You do not want to say that there exists some $y$ and $z$ where if each satisfies $A$ then they are then same. The claim needs to be that when any $y$ and $z$ both satisfy $A$ then they must be the same.

$$\Big(\forall y~\forall z~\big((A(y)\land A(z))\to (y=z)\big)\Big)$$

Alternatively, we could say that when two variables are distinct, then at least one will not satisfy $A$: $$\Big(\forall y~\forall z~\big((y\neq z)\to(\lnot A(y)\lor\lnot A(z))\big)\Big)$$

Is my answer correct?

Yes, you have got it.

The full claim of uniqueness is that there is something that satisfies $A$ and when any $y$ and $z$ both satisfy $A$ then they must be the same.

$$\Big(\exists x~A(x)\Big)~\land~\Big(\forall y~\forall z~\big((A(y)\land A(z))\to (y=z)\big)\Big)$$

Of course, you may compact this to claim: There is something which satisfies $A$ and anything which satisfies $A$ is that thing.  $$\exists x~\big(A(x)\land\forall y~(A(y)\to x=y)\big)$$

Or "There is something which satisfies $A$, and anything else does not."$$\exists x~\big(A(x)\land\forall y~(x\neq y\to\lnot A(y)\big)$$


问:
Leibniz Integration Rule
Can anything be said of the following expression? $$\frac{\partial{}}{\partial{\pi}}\int_{-\infty}^{g(\pi)}v(s_1,s_2)f(s_1|s_2)ds_1$$

$f(s_1|s_2)$ is a conditional pdf, however I'm not sure that's relevant to answer the question I'm pondering. In the course of a project I stumbled upon this expression and wonder if anything can be simplified. I've been looking for ways to apply Leibniz's Integration rule however I do not think I can because the lower bound is infinite. Any ideas?

答:
$$\frac{\partial}{\partial\pi}\int_{-\infty}^{g(\pi)}v(s_1,s_2)f(s_1|s_2)ds_1 = \frac{\partial}{\partial\pi}\int_{-\infty}^0v(s_1,s_2)f(s_1|s_2)ds_1 + \frac{\partial}{\partial\pi}\int_0^{g(\pi)}v(s_1,s_2)f(s_1|s_2)ds_1\\
= 0 + \frac{\partial}{\partial\pi}\int_0^{g(\pi)}v(s_1,s_2)f(s_1|s_2)ds_1$$

The integral from $-\infty$ to $0$ does not change with $\pi$, so it is constant, and its derivative is $0$.

I.e., assuming that the integral from $-\infty$ exists, you can consider $-\infty$ as a constant.

Title: Leibniz Integration Rule


问:
A naive inquiry of Godel's incompleteness--or why does mathematics need proofs of unprovability?
My question stems from reading Swetz, 1994 (mostly excerpts from the journal Mathematics Teacher) and Berlinski, 2005 (a popular book on 10 most important mathematical breakthroughs in history).

1) I'm having difficulty understanding why Godel's theorem (if I understand what I have read) requires both formulas provable and unprovable. My naive concept of mathematics is a system containing only "positive" theorems and laws and I imagine the same for science. That the scientific method culls anything proven wrong, so as not to carry failures as baggage together with its laws. Where failure, mistakes, and problems are separated as historical study or challenge. Why do unprovable statements need be admitted to a proper system?

In anticipation of my enlightenment, I have a few follow-up Qs,

2) Its my understanding the Hilbert project meant to take the point of view that mathematics and formalism needed to take a meta-point of view for the purpose of separating formulas (as symbols) from the discussion (in natural language) about mathematics. If these unprovable formulas are necessary to prove other provable proofs, then why not segregate them to yet another meta-level?

3) What I gather about Godel's argument from these sources includes a point about mathematics' use of symbols (representing numbers, variables for formulas, variables for sets and sets of sets, etc.). Then is it his argument to suggest that where a proof can be admitted to the system which can be neither proved or disproved, the substitution of variables will propagate this error?

Answers need not be technical, only sufficient to sort my "Godel baggage".

Chris

PS. I found this January post very useful for its references and look forward to finding a volume for my naive appetite for the history of formalism:Understanding Gödel's Incompleteness Theorem---

答:
It is important to see the theorems in the context of Hilbert's attempt to formalize mathematics. Formalization here means reducing the whole Mathematics-making art into something as simple, in principle, as manipulating strings according to some fixed rules. In Hilbert's times there wasn't computers yet, but I'm sure he had in mind a goal that can be stated as "allowing computers to do math without thinking".

This formalization has two strengths. One is obvious - you can find theorems by letting your computer run, in an ordered way much different from current-day's guessing and searching in the dark game. The second one was no less important to Hilbert - if you can show all mathematics can be derived from extremely simple, "obviously true" rules, then our faith in the correctness of mathematics will be high. Remember, this was around the time of the "great crisis in the foundations of mathematics", raised by Russel's paradox (which requires only the extremely plausible hypothesis that you can define sets freely using mathematical predicates) and other paradoxes.

So Hilbert was looking out for a simple axiomatic system from which all mathematics could be derived. But what is "all" mathematics? It is not only what can be proven in some sort of system, but everything that is true about the objects we are interested in. One such object are the natural numbers. Now consider Goldbach's conjecture - that every even number greater than 4 can be written as the sum of two odd primes. This is surely either true or false (or is it...? But we adopt the viewpoint that believes this). The only question remaining is - can Hilbert's wonderful system prove it using no thinking at all, only manipulation of symbols? If it can't prove nor disprove it, surely it does not cover "all" mathematics.

What Godel showed is exactly that - that no matter what proof system Hilbert will give, it will have some flow: Either it won't be consistent (can prove and disprove the same thing, and from it usually follows you can prove everything), or it won't be complete (there will be some claim you won't be able to prove or disprove), or it won't be effective (in the sense you won't be able to check if a proof is legal or not - there will be no simple mechanical rules for doing it) or it won't talk about the nautral numbers (with the operations of addition and multiplication; Presburger arithmetic eludes Godel's theorems because it leaves the operation of multiplication out of the game).

What makes Godel's proof all the more interesting is that it uses the strengths of the system against it. On the other hand, the system can talk about natural numbers, and we can encode very intricate claims using natural numbers; on the other hand, the systems' rules are very simple, so they can be encoded themselves with natural numbers; and hence the system can indirectly prove things about itself, and this leads to some sort of "explosion". This is a very powerful idea, which has also given birth to the theory of Computability.


问:
a recurrence equation interpolating linear and exponential
$f(n+1) = f(n) + f(n)^{a}$ where $a \in (0,1)$ and $n \ge 1$ with $f(1) = m$.

If $a=0$, we see $f(n) = m + n - 1$ and if $a=1$, we see $f(n) = 2^{n-1}m$. So the recursion seems to interpolate between linear and exponential forms.

Is there a closed form or tight approximation or tight asymptotics for $f(n)$ in terms of $n$, $a$ and $m$?

Update after leshik's answer

Can the same asymptotics hold if $a \ge 1$? What is the asymptotics/tight approximation/exact expression if $a \ge 1$?

答:
We "guess" the asymptotic by making use of Stolz-Cesaro theorem. The idea is to show that the sequence behaves like $n^{\alpha}$ for an appropriate choice of $\alpha.$ By Stolz- Cesaro,
$$A=\lim_{n\to\infty}\frac{x_{n+1}}{(n+1)^{\alpha}}=\lim_{n\to\infty}\frac{x_{n+1}-x_n}{(n+1)^{\alpha}-n^{\alpha}}=\lim_{n\to\infty}\frac{x_n^a}{\alpha n^{\alpha-1}}=\frac{1}{\alpha}\lim_{n\to\infty}\left(\frac{x_n}{ n^{{\alpha-1}/a}}\right)^a.$$

Now, we want to have $\frac{\alpha-1}{a}=\alpha$ or $\alpha=\frac{1}{1-a}.$
The equation for our limit can be rewritten as $A=\frac{1}{\alpha}A^a$ which gives
$A=(1-a)^{\frac{1}{1-a}}.$


问:
How do you propagate uncertainty for the equation $\sqrt{\left( \frac{2 \sin{\theta}}{\sqrt{3}}+\frac{1}{2} \right)^{2}+\frac{3}{4}}$
I am calculating  $n(\theta) = \sqrt{\left( \frac{2 \sin{\theta}}{\sqrt{3}}+\frac{1}{2} \right)^{2}+\frac{3}{4}}$. But I have a problem as $n = 59.96 \pm 0.01$.

I have tried using a first order Taylor approximation (based on this post):
$$\Delta n \approx \left.\frac{dn}{d\theta}\right|_{\theta=59.96} \Delta\theta \approx (-0.24125)(0.01) \approx -0.002$$
Is this a valid method to obtain new uncertainty values? If someone knows a better way please let me know.

答:
Using Taylor is for sure the good idea but I have the feeling that you are working with degrees instead of radians.

So by Taylor,around $\theta=a$, the error is
$$\Delta=\frac{3 \cos (a)+4 \sqrt{3} \sin (a) \cos (a)}{3 \sqrt{4 \sin ^2(a)+2 \sqrt{3} \sin
   (a)+3}}\Delta \theta$$ and we have
$$a=59.96 \times \frac \pi{180}\qquad \text{and} \qquad \Delta \theta=0.01\times \frac \pi{180}$$ This makes
$$\Delta=0.5005708179\times0.01\times \frac \pi{180}\sim 0.000087$$

Let us check computing in degrees

for $\theta=59.96 {}^{\circ}$, the expression is $1.731701542$

for $\theta=59.95 {}^{\circ}$, the expression is $1.731614164$

for $\theta=59.97 {}^{\circ}$, the expression is $1.731788896$

which confirm the given result.

Please, make me a favor : in mathematics, forget degrees for ever and always use radians.


问:
Is it true that $P(A|B)+P(A|\overline B) = 1$?

tl;dr

$P(A|B) + P(A|\overline B)=1$. My question is, is this true?

More detail

The book I'm reading (Statistics for Business and Economics, Paul Newbold et al) has this example (paraphrased a little).

10% athletes have used performance enhancing drugs. A test is available that correctly identifies an athlete's drug usage 90% of the time. If an athlete is a drug user, the probability is 0.9 that the athlete is correctly identified by the test as a drug user. Similarly is not a drug user, the prob is 0.9 that the athlete is correctly identified as not using drugs.

So this is fine so far. From this I'd get the following:

Let D = athlete is a drug user, + = test calls positive, - = test calls negative

$ \begin{align}
   P(D) &= 0.10  \implies P(\overline D)\\
   P(+|D) &= 0.90 \\
   P(-|\overline D) &= 0.90
  \end{align}
$

Where the example looses me is that it says that the following can also be defined from the above information...

$ \begin{align}
   P(+|\overline D) &= 0.10 \\
   P(-|D) &= 0.10
  \end{align}
$

And this is where I'm struggling... it seems that they can only be getting this by assuming $P(+|D)+P(+|\overline D)=1$, or more generally $P(A|B) + P(A|\overline B)=1$. My question is, is this true?

Here's what I've tried so far to see if it is true...

$
\begin{align}
P(A|B)+P(A|\overline B) &= \frac{P(A \cap B)}{P(B)} + \frac{P(A \cap \overline B)}{P(\overline B)}\\
&= \frac{P(A \cap B)P(\overline B) + P(A \cap \overline B)P(B)}{P(B)P(\overline B)} \\
&= \frac{P(A \cap B)(1-P(B)) + P(A \cap \overline B)P(B)}{P(B)P(\overline B)}\\
&= \frac{P(A \cap B) - P(A \cap B)P(B) + P(A \cap \overline B)P(B)}{P(B)P(\overline B)}\\
&= \frac{P(A \cap B)}{P(B)(1 - P(B))} + \frac{P(A \cap \overline B) - P(A \cap B)}{P(\overline B)} \\
&= \frac{P(A \cap B)}{P(B)(1 - P(B))} + \frac{P(A \cap \overline B) - (1 - P(A \cap \overline B))}{P(\overline B)} \\
&= \frac{P(A \cap B)}{P(B)(1 - P(B))} + \frac{2\cdot P(A \cap \overline B) - 1}{P(\overline B)} \\
&= \frac{P(A \cap B)}{P^2(B)} + \frac{2\cdot P(A \cap \overline B) - 1}{P(\overline B)} \\
&= ?????
\end{align}
$

And then stuck... can't see how this is true. And looking at a Venn diagram representation it didn't enlighten me... any thoughts on how this question's answer in the book is making the assertions it is? Many thanks.

答:
First, I want to introduce the proof of Law of total probability:

\begin{align}
P(A \mid B) P(B) + P(A \mid \bar B) P(\bar B)
&= \frac{P(A \cap B)}{P(B)} P(B) + \frac{P(A \cap \bar B)}{P(\bar B)} P(\bar B) \\
&= P(A \cap B) + P(A \cap \bar B) \\
&= P[(A \cap B) \cup (A \cap \bar B)] + P[(A \cap B) \cap (A \cap \bar B)] \\
&= P(A) + P(\varnothing) \\
&= P(A) + 0 \\
&= P(A)
\end{align}

Let $P(B) = p$, $P(\bar B) = 1 - p$, $P(A) = q$, $P(A \mid B) = x$, $P(A \mid \bar B) = y$, we can form the following linear equation:

\begin{align}
\left\{
\begin{array}{l}
p x + (1-p) y = q \\
x + y = 1
\end{array}
\right.
\end{align}

We can solve it:

\begin{align}
x &= \frac{\begin{vmatrix}
q & 1 - p \\
1 & 1
\end{vmatrix}}
{\begin{vmatrix}
p & 1 - p \\
1 & 1
\end{vmatrix}}
= \frac{p + q - 1}{2p - 1}
\\
y &= \frac{\begin{vmatrix}
p & q \\
1 & 1
\end{vmatrix}}
{\begin{vmatrix}
p & 1 - p \\
1 & 1
\end{vmatrix}}
= \frac{p - q}{2p - 1}
\end{align}

In conclusion, $P(A \mid B) + P(A \mid \bar B) = 1$ does NOT hold unless the equation above holds.