问:
Low bound for the probabilty that every node in a graph has at least one neighbor in a set of nodes from the graph
I have been given a question and I dont even know where I should begin:

Given a graph with $n$ nodes where every node has at least $\sqrt{n}$ neighbors,

We build the set $S$ in the following order: for $0.3n$ times we pick each time one node randomly and independently and put it in $S$ (repeatition is possible).

Find a low bound for the probability that for every node in the graph there is at least one neighbor in $S$.

Assume that $n$ is a big number.

I have tried all sort of things but I can't seem to get close to the solution, I thought to use indicators to try and sum the number of different nodes in $S$ and by using the Central limit theorem (n is a big number) to somehow get the answer, but when I started defining the indicators it got messy and I could not define them.

P.s . The way $S$ is built reminded me of a Binomial distribution but i failed there too.

I would appreciate if someone could give me the solution.

答:
Let $I_v$ be the indicator of the event that $v$ has no neighbors in $S$. Then
$$E(I_v) \le (1-\sqrt{n}/n)^{3n/10} \le \exp(-3\sqrt{n}/10)\,,$$
since $1-x \le \exp(-x).$
Summing over all vertices in the graph, we find that the probability there is a vertex with no neighbor in $S$ is at most
$$E (\sum_v  I_v) \le n\exp(-3\sqrt{n}/10)\,.$$
Thus  the probability that for every node in the graph there is at least one neighbor in $S$ is at least $1-n\exp(-3\sqrt{n}/10)\,.$


问:
Solve Diophantine Equation
Show that $ x^3 - 6 = 25y^2  + 35y $ doesn't have any non-zero integer solution.

What I have tried

$$ 25y^2 +  35y - x^3 + 6 = 0 $$
solving for y we get
$$ y = \frac{-35 \pm \sqrt { 625 + 100 x^3} } {50} $$
so  I need to show that
$$ 25 + 4x^3= z^2 $$
doesn't have a solution
or if it does I need to show that
the fraction is not equal to an integer
but I can't proceed.

答:
Alternative solution

Suppose $x^3 = 25y^2 + 35y + 6 = (5y+6)(5y+1).$

If $(5y+6)$ and $(5y+1)$ have a common factor, then this common factor must divide $5$.  If the common factor is $> 1$, then it must be $5$.  But this would imply that $5$ is a factor of $x^3$, which would imply that $5 | x^3$ which would imply that $5|(5y+6)(5y+1)$.

This is impossible.  Therefore $(5y+6)$ and $(5y+1)$ are relatively prime.  Therefore $(5y+6)$ and $(5y+1)$ must each be a perfect cube.  This is impossible, because for any positive integer $n$,

$[(n+1)^3 - n^3] = 3n^2 + 3n + 1 > 5.$

Note: It is therefore trivial that for $(5y+6)$ and $(5y+1)$ any perfect cubes, $(a^3)$ and $(b^3),$ where each of $(a^3)$ and $(b^3)$ may be positive or negative, you can't have $|(a^3) - (b^3)| = 5.$

ContentLicense: CC BY-SA 4.0

Title: Solve Diophantine Equation


问:
Is the topology induced by the bases of a finite-dimensional real vector space equal to the topology induced by the norms?
It is well known that all norms on a $n$-dimensional real vector space induce the same topology (e.g. John Lee, Introduction to Smooth Manifolds, 2nd edition, Example 1.24). However, bases induce a "natural" topolgy, too: Let $(v_1,\ldots,v_n)$ be a basis, then we can define a topolgy by requiring that the isomorphism
\begin{align}
\Phi\colon\mathbf{R}^n&\to V\\
(x_1,\ldots,x_n)&\mapsto\sum_{i=1}^nx_iv_i
\end{align}
(and its inverse) are continuous: That is, $U\subset V$ is open if and only if $\Phi^{-1}(U)$ is open. One can show that the topology doesn't depend on the basis by using the fact that all linear maps from $\mathbf{R}^n$ to $\mathbf{R}^n$ are continuous, $L(\mathbf{R}^n,\mathbf{R}^n)\subset C(\mathbf{R}^n,\mathbf{R}^n)$.

My question is: Is the topology induced by the bases of a finite dimensional real vector space equal to (or contained in) the topology induced by the norms?

答:
Short answer: Yes.

Notation: If $X$ is a normed vector space, $x\in X$ and $0< r$, $B(x,r)=\{y\in X:|x-y|<r\}$ is the open ball with center $x$ and radius $r$.

Lemma: Let $V$ and $W$ be vector spaces over a subfield $F$ of $\mathbf{C}$ and let
\begin{equation}
W\ni w\mapsto\|w\|\in[0,\infty)
\end{equation}
be a norm. If $A\colon V\to W$ is a vector space isomorphism,
\begin{equation}
V\ni v\mapsto\|v\|:=\|A(v)\|\in[0,\infty)
\end{equation}
is a norm on $V$ and
\begin{equation}
B(A(v),r)=A(B(v,r)).
\end{equation}

Corollary: Let $V$ be a $n$-dimensional real vector space with a norm $V\ni v\mapsto\|v\|\in[0,\infty)$. If $T$ is the topology induced by the norm and $T'$ is the topolgy induced by the isomorphism $\Phi$ defined in my question, $T=T'$.

Proof:
From the Lemma we know that the norm on $V$ and the isomorphism $\Phi$ induce a norm on $\mathbf{R}^n$. Since all norms on $\mathbf{R}^n$ are equivalent, the topology on $\mathbf{R}^n$ induced by this norm is the standard topology. Thus, if $U\subset V$, the following statements are equivalent:

$U\in T$.

For each $v\in U$, there is $0<r$ such that $B(v,r)\subset U$.

For each $x\in\Phi^{-1}(U)$, there is $0<r$ such that $B(\Phi(x),r)=\Phi(B(x,r))\subset U$.

For each $x\in\Phi^{-1}(U)$, there is $0<r$ such that $B(x,r)\subset\Phi^{-1}(U)$.

$\Phi^{-1}(U)$ is open/contained in the standard topology of $\mathbf{R}^n$.

$U\in T'$.

Q.E.D.


问:
Finite sum involving Stirling number of first kind and Pochhammer symbol
I'm trying to find a closed form of

$$\sum_{m=0}^n\,s(n,m)\,(\alpha)_m\,z^m$$

where $(\alpha)_m$ means pochhammer symbol and $s(n,m)$ are the Stirling numbers of first kind.

I've had a look in related books but I have not luck.

Any help is welcomed.

答:
We can rewrite your sum as
$$
\eqalign{
  & F(z,n,\alpha ) = \sum\limits_{k = 0}^n {\left( { - 1} \right)^{n - k}
 \left[ \matrix{  n \cr  k \cr}  \right]\alpha ^{\,\overline {\,k\,} } z^{\,k} }
  = \sum\limits_{0\, \le \,k} {\left( { - 1} \right)^{n - k}
 \left[ \matrix{  n \cr  k \cr}  \right]z^{\,k} \alpha ^{\,\overline {\,k\,} } }  =   \cr
  &  = \left( { - 1} \right)^n \sum\limits_{0\, \le \,k}
 {\left[ \matrix{  n \cr   k \cr}  \right]z^{\,k} \left( { - \alpha } \right)^{\,\underline {\,k\,} } }
  = \left( { - 1} \right)^n \sum\limits_{0\, \le \,k}
 {k!\left[ \matrix{  n \cr   k \cr}  \right]z^{\,k} \left( \matrix{   - \alpha  \cr  k \cr}  \right)}  \cr}
$$
where

$\alpha ^{\,\overline {\,k\,} }$ represents the Rising factorial;

$ \left[ \matrix{  n \cr  k \cr}  \right]$ represents the unsigned Stirling N. 1st kind.

I can't provide at the moment a closed formula for that, but considering the well known
e.g.f. for the Stirling N. 1st kind
$$
{1 \over {m!}}\left( {\ln \left( {{1 \over {1 - x}}} \right)} \right)^{\,m}
  = \sum\limits_{0\, \le \,k} {\left[ \matrix{  k \cr   m \cr}  \right]\,{{x^{\,k} } \over {k!}}}
$$

we can get this e.g.f.
$$
\eqalign{
  & G(z,y,\alpha ) = \sum\limits_{0\, \le \,n} {F(z,n,\alpha ){{y^{\,n} } \over {n!}}}  =   \cr
  &  = \sum\limits_{0\, \le \,n} {\sum\limits_{0\, \le \,k} {{{k!} \over {n!}}
\left[ \matrix{  n \cr  k \cr}  \right]\left( { - y} \right)^{\,n} z^{\,k} \left( \matrix{   - \alpha  \cr   k \cr}  \right)} }  =   \cr
  &  = \sum\limits_{0\, \le \,k} {\left( {\sum\limits_{0\, \le \,n}
 {\left[ \matrix{  n \cr  k \cr}  \right]{{\left( { - y} \right)^{\,n} } \over {n!}}} } \right)k!z^{\,k}
 \left( \matrix{   - \alpha  \cr   k \cr}  \right)}  =   \cr
  &  = \sum\limits_{0\, \le \,k}
 {\left( \matrix{   - \alpha  \cr  k \cr}  \right)\left( {\ln \left( {{1 \over {1 + y}}} \right)} \right)^{\,k} z^{\,k} }  =   \cr
  &  = \left( {1 + z\ln \left( {{1 \over {1 + y}}} \right)} \right)^{\, - \alpha }
  = {1 \over {\left( {1 - z\ln \left( {1 + y} \right)} \right)^{\,\alpha } }} \cr}
$$

which means
$$
F(z,n,\alpha ) = \left. {{\partial^n  \over {\partial y^n}}
{1 \over {\left( {1 - z\ln \left( {1 + y} \right)} \right)^{\,\alpha } }}\;} \right|_{\,y\, = \,0}
$$


问:
Prove if the set is closed? Bounded? Compact?
Consider the metric space $(\mathbb{Q},d)$ where $\mathbb{Q}$ denotes the rational numbers and $d(x,y)=|x-y|$.  Let $$E:=\{x \in\mathbb{Q}:x>0, 2<x^2<3\}$$

Is $E$ closed and bounded in $\mathbb{Q}?$  Is it compact?  Justify your answers.

答:
Hint

About boundness when do you call a set unbounded? What is your intuition about this set? Can a $x\in E$ be too much large?

Regarding compactness, a characterization of compact metric spaces is:

$(X,d)$ is compact iff every sequence has a convergent subsequence

Does it happen in the set $E$? Can you find a sequence of elements in $E$ which does not converge in $E$?

ContentLicense: CC BY-SA 3.0


问:
Kernel of a linear functional
I have $L^2((0,1))$ space and functional
$$
F(f) = \int_0^1 f(x)(1-x) dx.
$$
How do I find the kernel?

From definition we want to find $f$ such that $\int_0^1 f(x)(1-x) dx = 0$. I suspect we want the function $f$ to satisfy $f(x) = x f(x)$ a. e. Is this correct? How do I proceed from here?

答:
Try integration by parts. Let h'(x) = g(x) and $g'(x)=f(x)$:

$$\int f(x)(1-x)dx = g(x)(1-x) - \int - g(x) dx = g(x)(1-x) + h(x)$$

So if $h(1)-h(0)-g(0) = 0$ then $\int_{0}^{1} f(x)(1-x)dx = 0$

So take a function $G(x)$ so that:

$$G(1)-G(0)-G'(0)=0$$

Then G''(x) is in $ker(F)$.

Title: Kernel of a linear functional


问:
Necessary and sufficient conditions for the existence of a derivative at one point
Let $f:[-1, 1] \rightarrow \mathbb{R}$ be differentiable in $[-1, 1] \setminus \{0\}$. Consider the following condition: The left and right limits of $f'(x)$ at $0$ exist and are finite and equal.

a. Show, through examples, that the condition is not sufficient nor necessary for the existence of $f'(0)$.

b. Add the condition.

This exercise reminded me of the definition of a point of discontinuity of the third kind but I don't know how to proceed.

答:
a. Not sufficient:

$$f(x) = \begin{cases} x & x \neq 0, \\ 1 & x = 0. \end{cases}$$

Not necessary:
$$f(x) = \begin{cases} x^2\sin(1/x) & x \neq 0, \\ 0 & x = 0. \end{cases}$$
Note that $f$ is indeed differentiable at $0$, as can be seen by computing $f'(0)$ using the first principles. However, $\lim_{x \to 0^+} f'(x)$ does not exist. (Neither does the other side limit.)

b. Requiring continuity at $0$ gives us differentiability.

The proof is essentially a use of the Mean Value Theorem. For each $x > 0$, there exists $c_x \in (0, x)$ such that
$$\dfrac{f(x) - f(0)}{x} = f'(c_x).$$
Conclude from the above that
$$\lim_{x \to 0^+}\dfrac{f(x) - f(0)}{x} = \lim_{y \to 0^+} f'(y).$$
The right hand side is given to exist. Similarly, you get a similar equality for the left hand limit. Conclude.


问:
What is $\text{inf} \{ \prod_{i=0}^{n}(x-i) ; n \in \mathbb{N_{0}} \}$ of relation $\sqsubseteq$?
On the set of polynomials with real coefficients we define $\sqsubseteq$ as:

$p \sqsubseteq  q \iff p=q\ \vee \exists n \in \mathbb{N_{0}}:(p(n) <q(n)\ \wedge \forall m \in \mathbb{N_{0}}:(m<n \Rightarrow p(m)=q(m)))$

First part of the question was to prove that this is total order, which I had no problem with. Then it also asks what is $\text{inf} \{ \prod_{i=0}^{n}(x-i) ;  n \in \mathbb{N_{0}} \}$?

My initial though was to take $\prod_{i=0}^{\infty}(x-i)$, which would have all natural numbers as its zeroes, but if I remember correctly, a polynomial can't have infinitely many zeroes unless it is constant. Now I have two questions:

Why can't polynomial have infinitely many zeroes, as does $\prod_{i=0}^{\infty}(x-i)$?

If not $\prod_{i=0}^{\infty}(x-i)$, then what is $\text{inf} \{ \prod_{i=0}^{n}(x-i) ;  n \in \mathbb{N_{0}} \}$ equal to?

答:
$\prod_{i=0}^{\infty}(x-i)$
is undefined for
any $ x\not\in \mathbb{N_{0}}$.


问:
Recurrence relation $a_{n+1} = 2 a_n -1$ with $a_0=3$. What is $a_n$ expressed as a function of $n$?
I'm having a hard time finding an explanation how to solve the following problem.

Given the recurrence relation $a_{n+1} = 2 a_n -1$ with $a_0=3$.   What is $a_n$ when expressed as an explicit function
of $n$?

Supposedly the correct answer is $a_n = 2^{n+1}+1$. But what is the step by step approach to arrive at this solution? All of the resources I've found begin with a recurrence relation in the form an = (some expression) but how to handle this type of problem in the form an+1 = (some expression) ?

Is it a matter of moving the variable to the other side and then solving iteratively? Or is there some clever/magic approach we should use here? Super appreciate if anyone can detail the steps to take to arrive at the correct solution.

Thanks in advance!

Body:
Get rid of $-1$ by taking $$a_{n+1}=2a_n-1$$

minus $$a_{n}=2a_{n-1}-1$$

to get $$a_{n+1}-a_n=2a_n-2a_{n-1}$$

or $$a_{n+1}-3a_n+2a_{n-1}=0.$$

The characteristic equation is $r^2-3r+2=(r-1)(r-2)$, so $a_n=C+D2^n$.

You can solve for $C$ and $D$ using $a_0=3$ and $a_1=5$.


问:
Solving $\log\vert z \vert = -2\arg(z)$
I'm struggling to solve the following equation.
$\log\vert z \vert = -2\arg(z)$

I've played around with it. Dividing by $-2$ I get
$\log\vert z \vert ^{-\frac{1}{2}} = \arg(z)$ and then taking the exponential and after that  both sides to the power of two and rearranging.

$\frac{1}{\vert z \vert} = e^{2\arg(z)} \Leftrightarrow \frac{1}{e^{2\arg(z)}} =  \vert z \vert  $

Since $z \in \mathbb{C}$ we can write it as $z= re^{i\theta}$ for some real number $r$ and an angle $\theta$. But $\vert z \vert = r$ and by taking the principal branch of the argument of z we can say (here I'm a little unsure, please correct me if I'm wrong) that $\arg(z) "=" Arg(z) = \theta$. Thus

$$
z= re^{i\theta} = \frac{1}{e^{2\theta}} \cdot e^{i\theta} = e^{-2\theta}e^{i\theta} = e^{\theta(i-1)}
$$

Assuming this is correct so far, now what? My idea is to solve for $\theta$ and
I've tried various things yielding nothing. Taking the logarithm on both sides. setting it equal to its explicit trigonometric expression, its Cartesian form and trying to solve etc. It feels like I'm missing something maybe obvious. Any hints and or corrections would be really appreciated. I also want to apologize for my $\LaTeX$ being somewhat rusty.

答:
There's a small slip in your working:- $ z=e^{\theta(i-2)}$.

That's the answer!

Title: Solving $\log\vert z \vert = -2\arg(z)$


问:
Addition of numbers with same base but different exponents
The problem itself is $(288\sqrt{3})^{1/5}+(288\sqrt{3})^{4/5}$. The answer is supposed to be of the form $a+b\sqrt{c}$, but I have no idea how to simplify it.</p>

答: HINT: Note that $288\sqrt3=2^5\cdot3^{5/2}$, so the first term is simply $2\sqrt3$. Now what’s the second term?</p>


问:
There are $2^{\omega(a)}$ pairs $(b,c)$ such that $a=bc$ and $\gcd(b,c)=1$
It looks as there are $2^{\omega(a)}$ ordered pairs $(b,c)$ such that $a=bc$ and $\gcd(b,c)=1$, where $\omega(a)$ is the number of different prime factors of $a$.

Proof?

It's also true that there are  $n^{\omega(a)}$ n-tipples $(a_1,\dots ,a_n)$ such that $a=a_1\cdots a_n$.

答:
The easiest way to see this is to use kernels. If $$x=\prod_{i=1}^{\omega(x)}{p_i^{k_i}}$$ then the kernel is defined as $$\ker(x)=\prod_{i=1}^{\omega(x)}{p_i}$$ Plainly, $\omega(x)$ is the same for $x$ and $\ker(x)$. By the well known formula for the number of divisors, there are $2^{\omega(x)}$ divisors of $\ker(x)$.

Each divisor $d_i$ of $\ker(x)$ is the first member of an ordered pair $\Bigl (d_i,\ \frac{\ker(x)}{d_i}\Bigr)$ whose members are coprime. In particular, because $\ker(x)$ is square free, it is always the case that $$d_i\ne \frac{\ker(x)}{d_i}$$

Restoring the ignored exponents $k_i$ of the prime factors of $x$ back into the analysis changes nothing with respect to counting the ordered pairs possible.

Applying this to the question posed, let $a=x$, $b=d_i$ with relevant exponents restored, $c=\frac{\ker(x)}{d_i}$ with relevant exponents restored, and the proof is complete. There are $2^{\omega(a)}$ coprime ordered pairs $(b,c)$ of $a$


答:
If $z=re^{i\theta}$, $|z|=r$ and $\arg(z)=\theta\in[0,2\pi)$.
So
$$\log|z|=-2\arg(z)\iff \log r=-2\theta\iff r=e^{-2\theta}.$$
So $$z=e^{-2\theta}e^{i\theta}=e^{\theta(i-2)}.$$

Title: Solving $\log\vert z \vert = -2\arg(z)$

Body:
I'm struggling to solve the following equation.
$\log\vert z \vert = -2\arg(z)$

I've played around with it. Dividing by $-2$ I get
$\log\vert z \vert ^{-\frac{1}{2}} = \arg(z)$ and then taking the exponential and after that  both sides to the power of two and rearranging.

$\frac{1}{\vert z \vert} = e^{2\arg(z)} \Leftrightarrow \frac{1}{e^{2\arg(z)}} =  \vert z \vert  $

Since $z \in \mathbb{C}$ we can write it as $z= re^{i\theta}$ for some real number $r$ and an angle $\theta$. But $\vert z \vert = r$ and by taking the principal branch of the argument of z we can say (here I'm a little unsure, please correct me if I'm wrong) that $\arg(z) "=" Arg(z) = \theta$. Thus

$$
z= re^{i\theta} = \frac{1}{e^{2\theta}} \cdot e^{i\theta} = e^{-2\theta}e^{i\theta} = e^{\theta(i-1)}
$$

Assuming this is correct so far, now what? My idea is to solve for $\theta$ and
I've tried various things yielding nothing. Taking the logarithm on both sides. setting it equal to its explicit trigonometric expression, its Cartesian form and trying to solve etc. It feels like I'm missing something maybe obvious. Any hints and or corrections would be really appreciated. I also want to apologize for my $\LaTeX$ being somewhat rusty.


问:
$\mathcal A$ is a Banach algebra. Can there be two different $*$ operators which both make $\mathcal A$ a $C^*$-algebra?
I really have no idea where to start. The only thing I know is that there can not be different $C^*$-norms (whether complete or not) on a $C^*$-algebra, but I find that I barely know nothing about algebraic isomorphisms between $C^*$-algebras.

If such isomorphism (which is not a $*$-isomorphism) exists, it must map some self-adjoint element to a non self-adjoint one. I can neither find an example nor show it is impossible.

答:
The answer pointed out by @Mark presumes that you want to keep the norm, that is, there is no adjoint operation
"$^{\bigstar}$" on $\mathcal A$, other than the defaul adjoint, such that
$(\mathcal A, ^{\bigstar}, \|\cdot\|)$
is a
C*-algebra, where $\|\cdot\|$ is the default norm.

However,  it is possible to find a different adjoint operation "$^{\bigstar}$", and a different norm $|||\cdot|||$,
such that
$(\mathcal A, ^{\bigstar}, |||\cdot|||)$
is a C*-algebra.

All you need to do is choose an automorphism
$$
  \varphi :\mathcal A \to \mathcal A,
  $$
which preserves everything but the star and norm,  and define a new star and norm by
$$
  a^\bigstar := \phi^{-1}\big (\phi(a)^*\big ), \quad \text{and} \quad |||a||| := \|\phi(a)\|.
  $$
One such automorphism may be taken to be
$$
  \phi(a) = uau^{-1},
  $$
where $u$ is a non-unitary, invertible element, such that $u^*u$ is non-central.

One more point: on a commutative C*-algebra one cannot find another adjoint operation, even if one is willing to consider a
change of norm.  The reason is that self-adjoint elements may be characterized as those with real spectrum and,
moreover, the space of self-adjoint elements determines the adjoint operation.


问:
Calculate the gradient and Hessian of $x_0^T(X\operatorname{Diag}(w)X^T)^{-1}x_0$ w.r.t. vector $w$ in matrix calculus?
I'm trying to calculate the Hessian matrix of the formula
$$
f(w) = x_0^T(XWX^T)^{-1}x_0
$$
where $w=(w_1,\cdots ,w_n)^T$, $W=\operatorname{Diag}(w)$ is a diagonal matrix that has $w$ as diagonal elements, $x_0$ is a column vector, $X$ is a matrix.

To calculate the first-order derivative, do I need to first transform $W$ into $$\sum_i w^T e_i E_i,$$
where $e_i$ is a column vector whose $i$-th element is $1$ and others $0$, and $E_i$ is a matrix whose $(i,i)$ element is $1$ and others $0$? Any reference is appreciated.

答:
$\def\o{\operatorname}\def\t{\o{Tr}}\def\v{\o{vec}}\def\d{\o{diag}}\def\D{\o{Diag}}\def\p#1#2{\frac{\partial #1}{\partial #2}}$For
typing convenience, define the symmetric matrices
$$\eqalign{
A = X\,\D(w)\,X^T \quad\qquad B = A^{-1}x_0x_0^TA^{-1} \\
}$$
and use a colon as a product notation for the trace, i.e.
$$\eqalign{A:B = \t(A^TB) = B:A}$$
Write the function in terms of the new variables.
Then calculate the differential and gradient.
$$\eqalign{
f &= x_0x_0^T:A^{-1} \\
df &= x_0x_0^T:dA^{-1} \\
  &= -x_0x_0^T:A^{-1}dA\,A^{-1} \\
  &= -B:dA \\
  &= -B:X\,\D(dw)\,X^T \\
  &= -X^TBX:\D(dw) \\
  &= -\d\left(X^TBX\right):dw \\
  &= -\d\left(X^TA^{-1}x_0x_0^TA^{-1}X\right):dw \\
\p{f}{w} &= -\d\left(X^TA^{-1}x_0x_0^TA^{-1}X\right)
 \;\doteq\; g\quad\big({\rm gradient}\big) \\
}$$
where the $\d(M)$ function takes the main diagonal from its matrix argument and returns it as a column vector, while the $\D(v)$ function takes its vector argument and creates a diagonal matrix.

NB: The properties of the trace allow the terms in a colon product to be rearranged in many equivalent ways, e.g.
$$\eqalign{
A:B &= B:A = B^T:A^T \\
CA:B &= C:BA^T = A:C^TB \\
}$$

A comment asked for the Hessian.

If you define
$$\eqalign{
y &= X^TA^{-1}x_0 \\
dy &= -X^TA^{-1}X\,dW\,X^TA^{-1}x_0 \\
Y &= \D(y) \\
}$$
then you can write the gradient as
$$\eqalign{
g &= -\d(yy^T) = -Yy \\
}$$
The Hessian is the gradient of this, so
$$\eqalign{
dg &= -\d(dy\,y^T + y\,dy^T) \\
 &= -2Y\,dy \\
 &= +2Y\Big(X^TA^{-1}X\,dW\,X^TA^{-1}x_0\Big) \\
 &= 2YX^TA^{-1}X\,dW\,y \\
 &= 2YX^TA^{-1}XY\,dw \\
\p{g}{w} &= 2 YX^TA^{-1}XY \;\doteq\; H \qquad\big({\rm Hessian}\big) \\
}$$
$(Y,dW)$ are diagonal matrices corresponding to the $(y,dw)$ vectors, therefore
$$dW\,y = Y\,dw$$
The Hessian is obviously symmetric, and since $(X,Y)$ appear an even number of times, its positivity is determined by the $A^{-1}$ term, which in turn is determined by $\D(w)$, which in turn is determined by the signs on the individual $(w_k)$ elements.


问:
What does it take to get a perfect score on the Putnam?
There are many guides on how to prepare for the Putnam, and so many people participate in the Putnam. However scoring perfect scores or anything reasonably close to a perfect score on the Putnam is so rare. It seems to me that the present guides make the basic assumption that answering everything on the Putnam is impossible, thus people who follow those guides would end up scoring 'normal' scores, just like everyone else. So, what specific steps should one take in order to get a perfect score on the Putnam? Is it important to master all of undergraduate mathematics and some graduate mathematics? Or is it really just a matter of practicing endless practice problems, just like all the other participants who don't get anything close to a perfect score?

答: There is no useful advice, because the grading rubric on the Putnam makes it difficult to gain complete marks for an essentially correct solution.   It is not all that unrealistic for some of the top competitors to have solved the important parts of every question, but to gain maximum points they need also to have checked every boundary case or small $n$, and not made some minor errors.  You would need a combination of extreme preparation, tremendous speed, checking and re-checking of details, and still depend on luck to deliver up an exam "easy" enough (in the year to year variation) to be able to really solve everything.  The last is probably the limiting factor, as there are enough multiple IMO gold winners taking the test now, every year, and an effectively unlimited amount of preparation material.</p>


问:
Is it possible that some Integrals do not have solutions? What about this one?
I cannot figure out how to solve this integral with the commonly known methods of integration. The different online integral calculators that I have tried fail to solve it.

$ \int  \sin(a x) ((b-x)^2-c^2)^{-3/2} dx $

where a, b and c are constants.

Is there a solution? Thanks.

答:
The problems with your integral are the unknown parameters $a, b, c$.
Assuming they are all real, it's basically impossible to solve the general integral due to their unknown nature indeed.

Even the very trivial case in which $a = b = c = 1$ brings lots of problems.
There are special cases, for example:

Special case 1: when $a = b = 1$, $c = 0$

The solution reads
$$\frac{(x-1)^2 \sin (1) (-\text{Ci}(1-x))+(x-1)^2 \cos (1) \text{Si}(1-x)-\sin (x)-x \cos (x)+\cos (x)}{2 (x-1) \sqrt{(x-1)^2}}$$

Where special functions Integral Sine and Integral Cosine pop out.

Special case2: $a = 1$, $b = c = 0$

The solutions reads

$$-\frac{x^2 \text{Si}(x)+\sin (x)+x \cos (x)}{2 x \sqrt{x^2}}$$

It's always hard when there are too many unknown parameters in the integrand function.

In any case, the number of integrals without a solution is huge! You know, to differentiate is easy, but to integrate is art.


问:
How do I explain to students that $0 \bmod n$ equals $0$?
I am teaching a beginning programming class in Visual Basic (for non-CS majors). I told my students that the mod operator basically gives the remainder of the division. So, when seeing $0 \bmod 10$, some students (apparently) reasoned that, "$10$ goes into $0$ zero times and there are $10$ leftover."

What would be the best way to explain this to (non-math and non-CS major) students?

I would rather counter the "$10$ goes into $0$ zero times with $10$ leftover" reasoning than change my definition, if possible.  I checked the documentation for the language, and it gave the same "remainder" definition that I gave.

答:
Think of it this way.

x mod n where n > 0 and x > 0.

For example x = 49 and n = 10.
As we no there are 9 remaining.

However, how does x have anything remaining if there was nothing there to start with.

I.E x = 0.

0 mod n == 0  as there was nothing to divide into to begin with so it can't have a remainder.


问:
t-derivative of $f\left(\frac{x-g(t)}{h(t)}\right)$
I'm certain I'm just overthinking this, I've been staring at this for a while! My question is, if a function was written in the form $f\left(\frac{x-g(t)}{h(t)}\right)$, what would the t-derivative be? I know that if the function was written as the form $a(x, g(t), h(t))$, it'd be done via the chain rule and would be
$$\frac{\partial a}{\partial t} = \frac{\partial a}{\partial g}\frac{dg}{dt} + \frac{\partial a}{\partial h}\frac{dh}{dt}$$ but in my eyes now, $f\left(\frac{x-g(t)}{h(t)}\right)$ isn't dependent on $h(t)$ but rather $h^{-1}(t)$. Would this affect it? Or have I just been staring at this for far too long?

答:
Check this link for "Faa' di Bruno" and "Leibniz" formulae:
https://en.wikipedia.org/wiki/Differentiation_rules#Derivatives_to_nth_order

Using the "Faa' di Bruno" and "Leibniz" formulae, you shall first assume $s(t)=\frac{x-g(t)}{h(t)}$. Now use "Faa' di Bruno" formula to evaluate $f(s(t))$. Now for evaluating this derivative using "Faa' di Bruno" formula you also need the n-th derivative to $s(t)$. For finding this, assume $r_1(t)=x-g(t)$ and $r_2(t)=\frac{1}{h(t)}$ and use "Leibniz" rule to find the n-th derivative to $s(t)=r_1(t)r_2(t)$ and then you're done!


问:
Power Series with Random Coefficients
First of all, I don't know how such series are called, but I assume there exists some theoretical work behind that kind of power series. So for example assume we have the random sequence $(a_n)_{n\in\mathbb{N}}$ with $a_n\in\{-1,1\}$ and $P(a_n=-1)=P(a_n=1)=\frac{1}{2}$, e.g. $(1,-1,-1,1,1,\dots)$, and define the power series $$f(x)=\sum_{n=0}^\infty\frac{a_n}{n!}x^n.$$
As example, the series could look like $$f(x)=1-x+\frac{x^2}{2}-\frac{x^3}{6}-\frac{x^4}{24}+\dotsm$$ or
$$f(x)=1+x+\frac{x^2}{2}-\frac{x^3}{6}+\frac{x^4}{24}-\dotsm.$$
Clearly the resulting series will lie (in absolute terms) between $e^x$ and $-e^{-x}$, but can we write down further properties, e.g. if we averge over all possible function $$\lim_{N\to\infty}\frac{f^{(1)}(x)+f^{(2)}(x)+\dotsm+f^{(N)}(x)}{N}$$what would be the result, my intuition says $0$, but I'm definitly not sure. If anyone knows some theory behind these series or how they're called, it would be highly appreciated.

答:
You'd have to be clear about what such a series means, convergence and all that. But you do know the finite sum $Y = \sum_{n = 0}^N a_n z^n/n!$ has mean zero and, assuming $a_0, a_1, \dots$ are all independent, has variance
\begin{equation} \text{var}\,Y = \sum_{n = 0}^N \frac{z^{2n}}{n! n!}
\end{equation}
which tends to $I_0(2z)$ as $N \rightarrow \infty$. Same result, for example, as when each $a_0, a_1, \dots$ are normally distributed with zero mean and unit variance.

If you want to dig in deeper, you might take advantage of the fact that each $a_n$ has characteristic function $\cos t$, so the finite sum will
have characteristic function $\cos(tz^0/0!)\cdots\cos(tz^N/N!)$.

Title: Power Series with Random Coefficients


问:
Degree of a map $f: S^1 \to S^1$.
I’m reading the following definition.

Definition B11.1. Given a map $f: S^1 \rightarrow S^1$, we define its degree $\operatorname{deg} f \in \mathbb{Z}$ as follows. Choosing a basepoint $x_{0} \in \mathbb{S}^{1},$ we get
$$
f_{*}: \mathbb{Z} \cong \pi_{1}(\mathbb{S}^{1}, x_{0}) \mapsto \pi_{1}(\mathbb{S}^{1}, f(x_{0})) \cong \mathbb{Z}
$$
Any such homomorphism $\mathbb{Z} \rightarrow \mathbb{Z}$ is given as multiplication by some $n \in \mathbb{Z}$, meaning the map $k \mapsto n k$. We then set $\operatorname{deg} f:=n$

While in our class we’ve covered why $\mathbb{Z} \cong \pi_{1}(\mathbb{S}^{1}, x_{0})$, I’m not sure I get how that translates into equivalency between the maps $\pi_{1}(\mathbb{S}^{1}, x_{0}) \to \pi_{1}(\mathbb{S}^{1}, f(x_{0}))$ and the maps $S^1 \to S^1$. Say we embed $S^1$ into $\mathbb{R}^2$, and $f$ rotates $x_0 = (1,0)$ by an angle of $\pi/3$ around the origin. What would $n$ be in this case? And given $k \mapsto nk$, what does it mean in terms of the homotopy classes of loops in $S^1$? If it helps please use concrete numbers for $n$ and $k$.

Edit: It’s been pointed out to me that orientation matters. Indeed this comes right after the definition in my materials. For now, I’m just assuming we map the loop going counter-clockwise around $S^1$ once to $1$.

答:
It seems as though the degree is well-defined only up to a sign. The reason I say this is because while it is true that $\pi_1(\Bbb S^1, f(x_0)) \cong \Bbb Z$, there are two isomorphisms, depending on which generator you pick.

In your example, you technically haven't given a map $f: \Bbb S^1 \to \Bbb S^1$, you've only told what $f$ does to $(1, 0)$. I assume that you mean that $f$ rotates $\Bbb S^1$ by $\pi/3$. In this case, the degree could either be $+1$ or $-1$.

To get a geometric interpretation: Recall what the isomorphism $G = \pi_1(\Bbb S^1, x_0) \cong \Bbb Z$ explicitly is. There are exactly two generators of $G$. One is (the equivalence class of) the loop that goes around exactly once and the other is the reverse of that. We get the isomorphism by mapping one of them to $1$. This means that the degree of the map is simply measuring how many times the generating loop gets twisted, under $f$. The only problem is that you have a choice in fixing the isomorphism, which means that the orientation matters.

Title: Degree of a map $f: S^1 \to S^1$.


问:
doubt about the calculation of the rest for a taylor's series
I'm trying to calculate some numbers with the help of series.
My problem is the calculation of the rest of the series.

For example I have to calculate $log 2$.

I can consider the function $$log (\frac{1+x}{1-x})$$ in which I can put $x= \frac{1}{3}$.

The series of Taylor for this function is:
$$log (\frac{1+x}{1-x}) = 2*\sum_{k=0}^ \infty \frac{x^{2k+1}}{2k+1}$$
$$log (\frac{1+x}{1-x}) = 2*\sum_{k=0}^ n \frac{x^{2k+1}}{2k+1}+ 2*\sum_{k=n+1}^ \infty \frac{x^{2k+1}}{2k+1} = 2*\sum_{k=0}^ n \frac{x^{2k+1}}{2k+1}+ R_n$$

The passage that I can't understand it this:
$$R_n = 2*\sum_{k=n+1}^ \infty \frac{x^{2k+1}}{2k+1} < \frac{2 |x|^{2n+3}}{2n+3}\sum_{h=0}^ \infty x^{2k}$$ and then it becomes $\frac{2 |x|^{2n+3}}{2n+3} \frac{1}{1-x^2}$

NOTE
$$R_n = 2*\sum_{k=n+1}^ \infty \frac{x^{2k+1}}{2k+1} = 2*\sum_{h=0}^ \infty \frac{x^{2h+3}}{2h+3} =2 x^3 \sum_{h=0}^ \infty \frac{(x^2)^h}{2h+3}$$

答:
For $k \ge n+1$, $\vert x \vert^{2k+1} = \vert x \vert^{2n+3} x^{2(k-n-1)}$ and $\frac{1}{2k+1} \le \frac{1}{2n+3}$. Hence

$$2\sum_{k=n+1}^ \infty \frac{x^{2k+1}}{2k+1} \le \frac{2 |x|^{2n+3}}{2n+3}\sum_{h=0}^ \infty x^{2h}.$$

Finally $\sum_{h=0}^ \infty x^{2h}$ is the sum of the terms of a geometric sequence and therefore

$$\sum_{h=0}^ \infty x^{2k} = \frac{1}{1-x^2}$$ which provides the desired result.


问:
Show that an Ehresmann connection on a principal G bundle is equivalent to a Lie Algebra Valued one form.
Let $E$ be a smooth principal $G$-bundle on M.

The vertical bundle $V$ is defined as $V=\ker(d\pi:TE\to \pi^*TM)$. An Ehresmann connection on $E$ is a smooth subbundle $H$ of $TE$ (also called the horizontal bundle), such that $TE=H\oplus V$.

Now let $H$ be invariant with respect to the $G$ action on $E$, so $H_{eg}=d(R_g)_e(H_e)$ for all $e\in E$ and all $f \in G$, where $d(R_g)_e$ is the differential of the right action of $g$ at $e$.

The Ehresmann connection, $H$ should be equal to a 1-form $\omega$ on $E$ with values in the Lie algebra $\mathfrak{g}$ of $G$. Can anyone provide me any insight as to why this is true and how one constructs this 1-form from the Ehresmann connection? I have poor knowledge of differential geometry.

答:
Not quite right. The horizontal subspace, $H_e$, will be given by $\ker\omega(e)$.

For the direction you didn't ask for, the transformation properties of $\omega$ under $R_g$ (which you didn't state) will give you the invariance of $H$. For the direction you asked for, note that for any $\xi\in\mathfrak g$ you get a vector field tangent to $V$ by taking $\xi(e)=\frac{d}{dt}\big|_{t=0}R_{\exp t\xi}(e)$.  Now define $\omega$ by specifying its values on $T_e E = H_e \oplus V_e$ by $\omega(e)(y) = 0$ when $y\in H_e$ and $\omega(e)(y) = \xi\in\mathfrak g$ when $y=\xi(e)$.


问:
If $A\preceq B$ and $C \preceq D$ then $C^A \preceq D^B$
I was hoping I could get a little supervision on "prettying-up" this proof. I think the logic is correct, but the presentation is probably cumbersome...and I have a clarification question at the end.

The claim to prove:

If $A\preceq B$ and $C \preceq D$ then $C^A \preceq D^B$

I use the following n0tation:

Let $f$ represent the injective function from $A$ to $B$ $\quad f: A \xrightarrow{1-1}B$

Let $g$ represent the injective function from $C$ to $D$ $\quad g:C \xrightarrow{1-1} D$

$h \in C^A \iff h: A \to C$

Construct the following function $\Phi: C^A \to D^B$ with a mapping rule that satisfies the following formula:

For non-empty sets...

$\forall a \in A \Big (g \circ h(a) =\Phi(h)\circ f(a)\Big) \ \land \ \forall x \in B\setminus f[A]\Big(\Phi(h)(x)=d_0\Big)$ where $d_0$ is an element within the range of $g$. (i.e. $d_0 \in g[C]$)

I am uncertain how to turn the above FOL statement into a piecewise function presentation...my best guess is:

$$\Phi(h)(b)=\begin{cases}
g\circ h(a),&\text{if }b=f(a)\text{ for some }a\in A\\
d_0,&\text{if } b \in B\setminus f[A]
\end{cases}$$

Injectivity: Let $\Phi(h_1) = \Phi(h_2)$. Remembering that $\Phi(h)$ is a function in $D^B$:

$\Phi(h_1) = \Phi(h_2)\rightarrow \forall b \in B [ \Phi (h_1)(b)=\Phi (h_2)(b) ]$

Because of this, note that $f[A] \subseteq B$, which implies that $\forall a \in A [\Phi(h_1)f(a)=\Phi(h_2)f(a)]$

By definition of our function $\Phi$, $\quad g \circ h(a) =\Phi(h)\circ f(a)$. Thus:

$\forall a \in A [\Phi(h_1)f(a)=\Phi(h_2)f(a)] \rightarrow \forall a \in A [g \circ h_1(a)=g \circ h_2(a)]$

$g$ is an injective function and therefore $g(c_1)=g(c_2) \rightarrow c_1=c_2$. We then see that, $\forall a \in A [h_1(a)=h_2(a)]$. But that is the definition of function equality, so $h_1=h_2$ $\square$

Extraneous cases: $\color{red}{\text{ this reasoning is incorrect }}$

If $A$ or $B = \emptyset$, then the injective function that satisfies $A \preceq B$ is the empty function (same idea for $C$ or $D$).

W.L.G. suppose $A$ is the empty set. Then $C^A = \emptyset$. Regardless of what $D^B$ equals, there exists an injective function that maps $\emptyset \to D^B$...namely, $\emptyset$. The argument can be similarly repeated for any of the other sets equaling the empty set.

Clarification Side Note: I was a little surprised when I didn't require the injectivity of $f$ to demonstrate the  injectivity of $\Phi$. After some thinking, I believe the injectivity of $f$ is implicitly required in the statement $\forall a \in A \Big (g \circ h(a) =\Phi(h)\circ f(a)\Big)$ in order for the domain of $\Phi$ to equal $C^A$. If $f$ is not injective, there could, depending on $C$'s cardinality, be functions in $C^A$ that do do not satisfy the relevant criteria to be admitted into $\Phi$.

For this demonstration, suppose $A \prec C$ and the injective function's name is $h_\alpha$. Let $f$ be non-injective. Then $h_\alpha$ cannot be admitted into $\Phi$'s domain because, due to $g$'s injectivity, there will exist an $a \in A$ such that $g \circ h(a) \neq \Phi(h)\circ f(a)$. To be slightly more formal, I think it is the case that:

$$f \text{ not injective, } h_\alpha \text{ is injective, and } A \prec C \rightarrow f[A] \prec h_\alpha[A]$$

If I could receive some further input on this final point, it would be greatly appreciated!

答:
Assume for now that $A,B,C$, and $D$ are non-empty. Let $B_0=f[A]$ and $D_0=g[C]$, and fix $d_0\in D_0$. For each $b\in B_0$ denote by $f^{-1}(b)$ the unique $a\in A$ such that $b=f(a)$. For $h\in C^A$ define a function $\Phi(h):B\to D$ as follows:

$$\Phi(h)(b)=\begin{cases}
(g\circ h\circ f^{-1})(b),&\text{if }b\in B_0\\
d_0,&\text{if }b\in B\setminus B_0\,.
\end{cases}$$

Clearly $\Phi:C^A\to D^B$. To show that $\Phi$ is injective, suppose that $h_1$ and $h_2$ are distinct elements of $C^A$. Then there is an $a\in A$ such that $h_1(a)\ne h_2(a)$. Let $b=f(a)$. Then

$$\Phi(h_i)(b)=(g\circ h_i\circ f^{-1})(b)=(g\circ h_i)(a)=g\big(h_i(a)\big)$$

for $i=1,2$. Now $h_1(a)\ne h_2(a)$, and $g$ is injective, so

$$\Phi(h_1)(b)=g\big(h_1(a)\big)\ne g\big(h_2(a)\big)=\Phi(h_2)(b)\,,$$

and therefore $\Phi(h_1)\ne\Phi(h_2)$, i.e., $\Phi$ is injective.

Now we consider the edge cases in which one or more of $A,B,C$, and $D$ is empty. Recall that $X^\varnothing=\{\varnothing\}$ for all sets $X$, and $\varnothing^X=\varnothing$ for all non-empty sets $X$.

If $A=\varnothing$, then $C^A=\{\varnothing\}$, which clearly injects into any non-empty set. However, it is possible that $B\ne\varnothing=D$, in which case $D^B=\varnothing$, and the result fails.

If $B=\varnothing$, then $A=\varnothing$, $C^A=\{\varnothing\}=D^B$, and the result holds.

If $C=\varnothing\ne A$, then $C^A=\varnothing$, and the result holds. If $C=\varnothing=A$, then $C^A=\{\varnothing\}$, and the result holds iff $B=\varnothing$.

If $D=\varnothing$, then $C=\varnothing$. If $A\ne\varnothing$, then $B\ne\varnothing$, $C^A=D^B=\varnothing$, and the result holds. If $A=\varnothing$ and $B\ne\varnothing$, however, the result fails.

To see why you need the injectivity of $f$, suppose that there are distinct $a_1,a_2\in A$ and $b\in B$ such that $f(a_1)=f(a_2)=b$. Let $h$ be any function from $A$ to $C$ such that $h(a_1)\ne h(a_2)$. Then $\Phi(h)(b)$ is not well-defined: is it $g\big(h(a_1)\big)$, or is it $g\big(h(a_2)\big)$? These are different, since $g$ is injective.


问:
any clue how to solve this lim
Having trouble starting to solve this limit:
$$
\lim_{n\to \infty} \left(1+\frac{n+1}{n^2+1}\right)^{3n+1}
$$

I understand it's related to $e$ some how, but having hard time getting there.

Title: any clue how to solve this lim

Body:
You just need to recall that when $u_n \to +\infty$
$$
\lim\left(1+\frac{1}{u_n}\right)^{u_n} = e.
$$

In this case,

$$
\lim \left(1+\frac{n+1}{n^2+1}\right)^{3n+1} = \lim \left(\underbrace{\left(1+\frac{1}{\frac{n^2+1}{n+1}}\right)^{\frac{n^2+1}{n+1}}}_{\to e}\right)^{\frac{(3n+1)(n+1)}{n^2+1}}=e^{\lim \frac{(3n+1)(n+1)}{n^2+1}} = e^3
$$


答:
REcall that if $a_n\to0$ then $(1+a_n)^{1/a_n}\to e$.

So, let's call $a_n:=\frac{n+1}{n^2+1}\to0$ and $b_n:=3n+1$. So your limit is $$L=\lim(1+a_n)^{b_n}=\lim \left[(1+a_n)^{1/a_n}\right]^{a_n b_n}.$$
So If you can calculate $l:=\lim a_nb_n$, then $L=e^l$.

In your case, you only need to calculate $\lim \frac{(n+1)(3n+1)}{n^2+1}$.

Title: any clue how to solve this lim

Body:
Having trouble starting to solve this limit:
$$
\lim_{n\to \infty} \left(1+\frac{n+1}{n^2+1}\right)^{3n+1}
$$

I understand it's related to $e$ some how, but having hard time getting there.


问:
help needed to find the 5th term on the number series
Can someone help me calculate the 5th term of these two series of numbers? They are somewhat difficult because they have no common difference or ration, which means they are neither arithmetic nor geometric.
7,1,2,3,___ the 5th term seems to be 5, but I don't know how to get it.
6,7,7,12___ The 5th term is 7, also I don't know how to get it.

Many thanks in advance.

答:
The online encyclopedia of integer sequences (oeis.org) is a great resource. Unfortunately, your cases are weird.

Searches for $7,1,2,3$ and $6,7,7,12$ return hits appearing unsatisfyingly late in their respective sequences. Even if you knew the answer ahead of time, they wouldn't fit any well known sequences.

For better results, we can loosen our affinity for the question's accuracy: For example, the Fibonacci sequence starts $1,1,2,3,5$. Pi times zeta($2$) equals $\frac{\pi^3}{6} = 5.1\textbf{677127}\dots$.


问:
How can we tell the cross point of two half-open lines given by starting points and angles?
Consider having two lines given by their start points ($A$ and $B$) and their angle with respect to the $x$-axis.

These are half-open lines, and the intersection may occur only after the starting points. For example, they do not intersect in the given image if $a>b$ or $a$ is negative.

How can we tell if they cross from their angles?

What is the fastest way to find the intersection?

I have two options: First, using an imaginary far endpoint, calculate the intersection of line segments. Second, treat them as two vectors.

But I think both are overkill, as a simpler solution might be available using the angles (instead of the endpoints).

答:
Let Equation of line from A is:

$y=a_1 x+b_1$

And that of B is:

$y=a_2 x +b_2$

Where $a_1=tan (a)$ and $a_2=tan (b)$ . These lines intersect if the angle between them is less than $180^o$, that means $a\neq b$ or they are not parallel. This angle can be found from following formula:

$$tan(\theta)=\frac {a_2-a_1}{1+a_1a_2}$$

For second question you have to solve following system of equation which gives the ordinates of point of intersection:

$\begin{cases}y=a_1x+b_1 \\y=a_2x+b_2\end{cases}$


问:
What does $O(d^2)$ mean in dimension of vector $2d+{d \choose 2}\approx O(d^2)$?
I understand that it means that the dimension is roughly of order $d^2$. Is that the correct interpretation?

Thanks

答:
This is big-O notation. It means that there is $K$ such that $2d + \binom{d}{2} \le K d^2$ for all sufficiently large $d$.

Since $\binom{d}{2} = \frac{1}{2}d(d-1) = \frac{1}{2}d^2-\frac{1}{2}d$, in fact any $K > \frac{1}{2}$ will witness this fact.


答:
Cf. big O notation; it means there is a positive number $M$ and a number $D$ such that,

for all $d\ge D$, $$\left\vert 2d+\binom d2\right\vert\le Md^2.$$

Title: What does $O(d^2)$ mean in dimension of vector $2d+{d \choose 2}\approx O(d^2)$?

Body:
I understand that it means that the dimension is roughly of order $d^2$. Is that the correct interpretation?

Thanks


问:
math distribution of averages and distribution of standard deviation for comparison of bytes
Let me change my question:

I have a sequence of 512 numbers (these 512 numbers can have a value between 0 and 255) (in computer mode these numbers can be between 00000000 and 11111111 (bits)). For these 512 numbers I calculate the average value, standard deviation value and kurtosis value. Now I want to calculate the values for the distribution of averages, distribution of standard deviations and distribution of Byte values. Finaly, I want to compare the distributions from other sequesces of 512 numbers to find out the similarity (as percentage) that two sequences have. I dont understand how can I calculate these distributions.

for example Distribution of Averages is : the probability that an average chosen from all the averages of a memory block is of value B in the range of 0-255.

Dxb=Pr((B+1))>Xj>=B   ,Xj=Average.

答: For the new question, it appears that you have no problem calculating the mean, standard deviation, and kurtosis of one series.  Then if you have a bunch of series and want to calculated the distribution of means you do it the same way.  You have data that comes as $(i,mean_i,sd_i,kurt_i)$ and ignore where it came from.  So you can calculate the mean and standard deviation of all the means.  As you have not defined "the similarity as percentage" I don't know what to do with that.</p>


问:
Proof of $\lim_{s\to 1^+}\frac{\sum_p p^{-s}}{\ln(s-1)}=-1$
I am looking for a proof of $$\lim_{s\to 1^+}\frac{\sum_p p^{-s}}{\ln(s-1)}=-1.$$
If the proof is too long, a direct reference is fine. Here the sum $\sum_p$ denotes the sum over all prime numbers.

EDIT: Perhaps this is one of the examples Mathematica does not help. I put in $s=1.0001$ and calculate

Sum[(Prime[n])^(-1.0001),{n,1,10000}]/Log(0.0001)

I get $-0.294043$. I even use smaller $s$, and larger upper bound for $n$, but the answer is far from $-1$. If you can get it to close to $-1$, please let me know how you do it.

答:
Since
$$
\lim\limits_{s\to 1^+}(s-1)\zeta(s)=1\tag{1}
$$
it is enough to show that
$$
\lim\limits_{s\to 1^+}\frac{\sum\limits_{p\in P} p^{-s}}{\log\zeta(s)}=1\tag{2}
$$
Using Euler poduct formula we get
$$
\log\zeta(s)
=\log\prod\limits_{p\in P}\frac{1}{1-p^{-s}}
=\sum\limits_{p\in P}\log(1-p^{-s})
=\sum\limits_{p\in P}\sum\limits_{n=1}^\infty\frac{p^{-sn}}{n}\\
=\sum\limits_{p\in P}p^{-s}+\sum\limits_{p\in P}\sum\limits_{n=2}^\infty\frac{p^{-sn}}{n}\tag{3}
$$
We have the following bound
$$
\begin{align}
\left|\sum\limits_{p\in P}\sum\limits_{n=2}^\infty\frac{p^{-sn}}{n}\right|
&\leq\sum\limits_{p\in P}\sum\limits_{n=2}^\infty\frac{1}{n}\left|\frac{1}{p^{sn}}\right|\\
&=\sum\limits_{n=2}^\infty\sum\limits_{p\in P}\frac{1}{n}\frac{1}{|p^{sn}|}\\
&=\sum\limits_{n=2}^\infty\frac{1}{n}\sum\limits_{p\in P}\frac{1}{p^{\Re(s)n}|p^{\Im(s)n}|}\\
&=\sum\limits_{n=2}^\infty\frac{1}{n}\sum\limits_{p\in P}\frac{1}{p^{\Re(s)n}}\\
&\leq\sum\limits_{n=2}^\infty\frac{1}{n}\sum\limits_{p\in P}\frac{1}{p^{n}}\\
&\leq\sum\limits_{n=2}^\infty\frac{1}{n}\int_1^\infty\frac{1}{t^n}dt\\
&=\sum\limits_{n=2}^\infty\frac{1}{n}\frac{1}{n-1}\\
&=\sum\limits_{n=2}^\infty\left(\frac{1}{n-1}-\frac{1}{n}\right)\\
&=1\\
\end{align}\tag{4}
$$
From $(1)$ we know that $\lim\limits_{s\to 1^+}\log\zeta(s)=\infty$, then from bound $(4)$ and equality $(3)$ it follows that $(2)$ holds. And now we are done!</p>


问:
Upper bound for - $\sum_{i=1}^n i {2n \choose n+i} $.
I am tryin to calculate an upper bound for the following sum:
$\sum_{i=1}^n i {2n \choose n+i} $.
I know that the limit should be $\Theta (4^n\sqrt n)$ but I am not sure how to proceed. I already tried to bound by the biggest element, but the bound was too high.

答:
We have that

$$S_n = \sum_{q=1}^n q {2n\choose n+q}
= \sum_{q=1}^n q [w^{n-q}] \frac{1}{(1-w)^{n+q+1}}
= [w^n] \frac{1}{(1-w)^{n+1}}
\sum_{q=1}^n q \frac{w^q}{(1-w)^q}.$$

Now here the coefficient extractor enforces the range and we get

$$[w^n] \frac{1}{(1-w)^{n+1}}
\sum_{q\ge 1} q \frac{w^q}{(1-w)^q}
= [w^n] \frac{1}{(1-w)^{n+1}}
\frac{w/(1-w)}{(1-w/(1-w))^2}
\\ = [w^{n-1}] \frac{1}{(1-w)^n} \frac{1}{(1-2w)^2}
\\ = \mathrm{Res}_{w=0}
\frac{1}{w^n} \frac{1}{(1-w)^n} \frac{1}{(1-2w)^2}.$$

We now apply the change of variables rule 1.8 (5) from Egorychev's
Integral representation and the Computation of Combinatorial sums
(page 16) with $A(w) = \frac{w}{(1-2w)^2}$ and $f(w) = \frac{1}{1-w}.$ We
get $h(w) = w (1-w)$ and find

$$\mathrm{Res}_{z=0} \frac{1}{z^{n+1}}
\left.\left[ \frac{A(w)}{f(w) h'(w)} \right]\right|_{w=g(z).}$$

with $g$ the inverse of $h.$ This becomes

$$\mathrm{Res}_{z=0} \frac{1}{z^{n+1}}
\left.\left[ \frac{w/(1-2w)^2}{(1-2w)/(1-w)}
\right]\right|_{w=g(z)}
\\ = \mathrm{Res}_{z=0} \frac{1}{z^{n+1}}
\left.\left[ \frac{w(1-w)}{(1-2w)^3} \right]\right|_{w=g(z)}
= \mathrm{Res}_{z=0} \frac{1}{z^{n}}
\left.\left[ \frac{1}{(1-2w)^3} \right]\right|_{w=g(z).}$$

Observe that $(1-2w)^2 = 1 - 4w + 4w^2 = 1-4w(1-w) = 1-4z$
so this is

$$\mathrm{Res}_{z=0} \frac{1}{z^{n}}
\frac{1}{(1-4z)^{3/2}} =
4^{n-1} {n-1+1/2\choose n-1}
= 4^{n-1} {n-1/2\choose n-1}.$$

Expanding the binomial coefficient we find

$$4^{n-1} \frac{1}{(n-1)!} \prod_{q=0}^{n-2} (n-1/2-q)
= 2^{n-1} \frac{1}{(n-1)!} \prod_{q=0}^{n-2} (2n-1-2q)
\\ = 2^{n-1} \frac{1}{(n-1)!} \frac{(2n-1)!}{2^{n-1} (n-1)!}
= \frac{1}{2n} n^2 {2n\choose n}
= \frac{1}{2} n {2n\choose n}.$$

Now for the central binomial
coefficient
we have ${2n\choose n} \sim \frac{4^n}{\sqrt{\pi n}}$ so this yields

$$\bbox[5px,border:2px solid #00A000]{
S_n = \frac{1}{2} n {2n\choose n}
\sim \frac{4^n}{2\sqrt{\pi}} \sqrt{n}.}$$

If we seek $\Theta(4^n\sqrt{n})$ from first principles we use from the
same reference the upper bound  ${2n\choose n} \le 4^n/\sqrt{3n+1}$ so
we get

$$S_n \le \frac{1}{2} n \frac{4^n}{\sqrt{3n+1}}
= \frac{1}{2\sqrt{3}} \sqrt{n} \frac{4^n}{\sqrt{1+1/3/n}}
\lt \frac{4^n}{2\sqrt{3}} \sqrt{n}.$$

The lower bound is done similarly and we have the claim.


问:
When $A$ is not a closed set, there exists an element $c\in X\setminus A$ and a sequence ${x_n}$ in A for all $n\ge1$ where $x_n$ converges to $c$
I am told to start with the negation of Open set.  Let $V= X\setminus A$.  I have that $V$ is not Open if for all $r>0$, There exists $a \in V$ such that the open ball $B(a,r)$ is not a subset of $V$.

Is that the correct negation of open set?

答:
No, you must switch $\forall r>0$ and $\exists a\in V$.

$V$ is an open set iff $V$ is an open neighbourhood of any of its points, or equivalently for all $a\in V$, there exists $r>0$ such that the open ball $B(a,r)$ is included in $V$ (if we are in a metric space).

The negation of this is that there exists $a\in V$ such that for all $r>0$, the open ball $B(a,r)$ is not contained in $V$.

So if $A$ is not a closed set, then $X\backslash A$ is not open so there exists $c\in X\backslash A$ such that for all $n\in\mathbb N^*$, the open ball $B(a,\frac1n)$ is not contained in $X\backslash A$. Let $n\in\mathbb N^*$. Since $B(a,\frac1n)$ is not contained in $X\backslash A$, there exists $x_n\in B(a,\frac1n)$ such that $x_n\notin X\backslash A$, or equivalently $d(x_n,a)<\frac1n$ and $x_n\in A$. Then $(x_n)_{n\in\mathbb N^*}$ is a sequence in $A$ which converges to $a$.


问:
Integral involving normal densities
I am trying to solve the integral

$$I(y)=\int_{\mathbb R}f(x,y)g(x)\,dx,$$

where $f(x,y)$ is the bivariate normal density with known mean $(\mu_1,\mu_2)$ and covariance matrix $\Sigma$ , and $g(x)$ is a normal density with mean $\mu$ and variance $\sigma^2$. I am stuck on this and would appreciate some help (I guess there is some sort of trick to do this without too much pain).

In case someone is interested in the motivation: this integral appears in a measurement error model where one observes $(x,y)$ which is a noisy observation of the quantities of interest $(\mu_1,\mu_2)$.

答:
A common approach is to express the integrand in the form $\exp(ax^2+bx+c)$ and then
complete the square in the argument of the exponential. Then, what remains can be
expressed in the form of an integral of a normal density over $\mathbb R$ which has value $1$, and of course anything involving $y$ and various constants is what is left.  In short,
$I(y)$ should work out to be of the form $\alpha\exp(-\beta (y-\mu_Y)^2)$.  For example,
$$\begin{align}
\exp(-(x^2 + y^2 -xy)) &= \exp(-(x^2 - 2x(y/2) + y^2/4 + 3y^2/4))\\
&= \exp(-3y^2/4)\exp(-(x-y/2)^2)
\end{align}$$  where $\exp(-3y^2/4)$ and sundry constants can be
pulled outside the integral, leaving the integral of a normal density
with mean $-y/2$
and variance $\ldots$, which integrates to $1$. So the whole thing simplifies to
an exercise in high-school algebra.

Since the OP requested "a more elegant trick", here goes.
$f_{X,Y}(x,y) = f_{X\mid Y}(x\mid y)f_Y(y)$ where $f_{X\mid Y}(x\mid y)$
and $f_Y(y)$ are
$N(\mu_1+\rho\frac{\sigma_1}{\sigma_2}(y-\mu_2),(1-\rho^2)\sigma_1^2)$ and $N(\mu_2,\sigma_2^2)$ random variables where $\rho$ is
the correlation coefficient of $X$ and $Y$.

Fix $y$ to be some constant, and consider a normal random variable
$\hat{X}$ with mean $\mu_1+\rho\frac{\sigma_1}{\sigma_2}(y-\mu_2)$
and variance $(1-\rho^2)\sigma_1^2$.  Let $Z$ be
a $N(\mu,\sigma^2)$ random variable that is independent of $\hat{X}$
and note that $f_Z(x) = g(x)$.
Then $W = \hat{X}-Z$ is also a normal random variable with mean
$\mu_W = \mu_1+\rho\frac{\sigma_1}{\sigma_2}(y-\mu_2) - \mu$ and
variance $\sigma_W^2 = (1-\rho^2)\sigma_1^2 + \sigma^2$ so that
$$f_W(0)
= \frac{1}{\sigma_W\sqrt{2\pi}}\exp\left(-\frac{\mu_W^2}{2\sigma_W^2}\right).$$
But,
$$\begin{align}
f_W(0) &= \int_{-\infty}^\infty f_{\hat{X}}(x)f_Z(x)\,\mathrm dx\\
&= \int_{-\infty}^\infty \frac{f_{X,Y}(x,y)}{f_Y(y)} g(x)\,\mathrm dx\\
\Rightarrow \qquad f_W(0)f_Y(y)&= \int_{-\infty}^\infty f_{X,Y}(x,y) g(x)\,\mathrm dx\\
\Rightarrow \qquad I(y) &= f_W(0)f_Y(y)\\
&= \frac{1}{\sigma_W\sqrt{2\pi}}\exp\left(-\frac{\mu_W^2}{2\sigma_W^2}\right)
\frac{1}{\sigma_2\sqrt{2\pi}}\exp\left(-\frac{(y-\mu_2)^2}{2\sigma_2^2}\right).
\end{align}$$
Now, plug-and-chug for $\mu_W$ and $\sigma_W$ to get the final result.

Is that any easier or any more elegant?

Title: Integral involving normal densities


问:
When I not seen a formula of prostapheresis in an identity
We suppose that I must to proof this identity

$$\sin(220°)+\cos(10°)=\cos(70°)$$

Easily if I put $\cos(10°)$ to RHS of the identity, I can apply the formula of prostapheresis $\cos(\alpha)-\cos(\beta)$ and I solved all.

But if we not will put the $\cos(10°)$ to the RHS, the identity is it possible to solve with a calculator or is there another alternative?

Body:
We can simplify as follows.

$\sin 220^{\circ} = \sin (180^{\circ}+40^{\circ}) = -\sin 40^{\circ}$

$\cos 10^{\circ}=\sin (90^{\circ}-10^{\circ})=\sin 80^{\circ}$

Then RHS becomes
$$\sin 80^{\circ}-\sin 40^{\circ}=\sin (60^{\circ}+20^{\circ})- \sin(60^{\circ}-20^{\circ}) $$
$$=2\cos 60^{\circ} \sin 20^{\circ}$$
$$=\sin 20^{\circ}=\cos 70^{\circ}$$


问:
Basis of a linear transformation
If $V$ and $W$ are vectorial spaces of finite dimension, we denote $\mathscr{L}(V,W)$ like the set of all linear maps of $V$  in $W$. Let a linear map $L\in \mathscr{L}(\mathbb{R}^{2},\mathbb{R}^{2})$ defined like $L(x,y)=(-x+y,2x+y)$ for each $(x,y)\in \mathbb{R}^{2}$. Find a basis $S=\{u_{1},u_{2}\}$  of $\mathbb{R}^{2}$ such that
$$[L]_{S} := M_{L}^{S \rightarrow S}= \begin{pmatrix} 0 & -1 \\ -3 & 0 \end{pmatrix}$$
Where $M_{L}^{S \rightarrow T}$ denote the matricial representation of the linear map $L\in \mathscr{L}(V,W)$ respective to the basis $S$ of $V$ and $T$ of $W$.

Let $u_{1}=(v,w)$,$u_{2}=(a,b)\in \mathbb{R}^{2}$. We have that the columns in the matricial representation are $L(u_{1})$ and $L(u_{2})$, so

$$L_{u_{1}}=(-v+w,2v+w)= (0,-3)^{T}$$
and
$$L_{u_{2}}=(-a+b,2a+b)=(-1,0)^{T}$$
Then, $v=w=-1$,$a=\frac{1}{3}$ and $b=-\frac{2}{3}$. Am I right?

答:
In your attempt you are forgetting that the columns of $[L]_S=M^{S\to S}_L$ are not the images of $u_1$ and $u_2$ but their coordinates with respect to $S$. Hence, the conditions on the basis $S=\{u_1, u_2\}$ given by $[L]_S$ should have been
\begin{align}
L(u_1) &= \phantom{-}0 u_1 - 3 u_2, \\
L(u_2) &= -1 u_1 + 0 u_2.
\end{align}
Setting $u_1=(v,w)^T$ and $u_2=(a,b)^T$ as you did, this translates to
\begin{align}
(-v+w,2v+w)^T &= (-3a, -3b)^T, \\
(-a+b,2a+b)^T &= (-v, -w)^T.
\end{align}
This is a homogeneous linear system of four equations and four variables:
\begin{align}
-v+w+3a\phantom{{}+3b} &= 0,\\
2v+w\phantom{{}+3a}+3b &= 0,\\
v\phantom{{}+w}-\phantom{3}a+\phantom{3}b &= 0,\\
w+2a+\phantom{3}b &= 0.
\end{align}
Row reduction yields that this is equivalent to
\begin{align}
v&=a-b,\\
w&=-2a-b.
\end{align}
In addition, $u_1$ and $u_2$ need to be linearly independent, which is equivalent to $vb-wa\neq 0$ and hence to $ab\neq 0$ (by substituting $v$ and $w$).

Picking $a=b=1$ we get $v=0$, $w=-3$ and indeed $u_1=(0,-3)^T$ and $u_2=(1,1)^T$ are linearly independent and yield
\begin{align}
L(u_1) &= L((\phantom{-}0,-3)^T) = (-3,-3)^T = -3 u_2, \\
L(u_2) &= L((\phantom{-}1,\phantom{-}1)^T) = (\phantom{-}0, \phantom{-}3)^T = -1 u_1.
\end{align}

Note that any other choice of non-zero $a,b$ also yields a valid solution.

Title: Basis of a linear transformation


问:
On the Ancients' Conception of Limit and "Meeting at Infinity"
I would like gain a better understanding of the ancient Greek notion of limit.

In particular, I am asking for some good references (books especially) that explain well, for example, why the ancients thought that an asymptotic curve meets its asymptote "at infinity.'' I presume that this would have also included numerical sequences as well. I would furthrmore like to discover any connections between this and the idea that parallel lines ``meet at infinity"---which I understand may have motivated Euclid's 5th postulate.

If anyone can offer any good insight to the ancient notion of limit---that will be appreciated; but I am especially looking for good references that expound well on topics related to the ancient notion of limit and the their idea of two things ``meeting at infinity.'' (no wikipedia please)

Thank you.

答:
So according to the modern notation and concepts these ideas of 'meeting at infinity' are meant to be taken as conceptual gimmicks and rules of thumb which are not precisely true, but rather explain in laymen's terms a more sophisticated structure or process which preserves some of these notions as limits are taken, and which are shown through an epsilon delta arguments as per the usual route.

Regarding the notions of the ancient Greeks as they thought them is more useful from an historical perspective. The ancient Greeks generally had a notion and definition of limits which gave rise to many paradoxes which we find easy to resolve. Consider for example Zeno's paradoxes, particularly the paradox of Achilles and the Tortoise. You will have to look at another answer to understand the historical perspective in depth. I'm not a math historian.

It may therefore be more interesting to turn your attention toward modern rigorous mathematical structures which formalize certain notions of behavior at infinity, meeting at infinity, or the notion of the infinitesimal.

Concerning parallel lines in the plane meeting at infinity you may wish to consider the Riemann Sphere representation of the complex plane where this notion can be made precise in a formal way.

Concerning an asymptotic curve meeting it's asymptote at infinity there are several different extensions of the standard real numbers where something like this can be made formal.

One of the most extensive attempts to formalize these notions is called Non-Standard Analysis. Non-Standard Analysis attempts to formalize the earlier intuitive proofs and concepts introduced by Gottfried Wilhelm Leibniz by extending the set of reals with an additional set of hyperreals representing infinite and infinitesimal (infinitely small but non-zero) quantities and defining a set of principles which describe their interactions.


问:
How to convert parametric form of tangent line to Cartesian equivalent
I have calculated the tangent line of the following curve $c(t)=(2cost−cos2t,2sint−sin2t$) @ $t=\pi/4$ to be

$l(t)=(\sqrt{2}, \sqrt{2}-1)+ (t- \frac{\pi}{4})(2-\sqrt{2}, \sqrt{2})$

How do I convert this to $y-y_{0}=f'(x_{0})(x-x_{0})$?

答:
From $$(\sqrt{2}, \sqrt{2}-1)+ \left(t- \frac{\pi}{4}\right)(2-\sqrt{2}, \sqrt{2})$$
we get
$$\begin{cases}
x=\left(2-\sqrt{2}\right) \left(t-\frac{\pi }{4}\right)+\sqrt{2}\\
y=\sqrt{2} \left(t-\frac{\pi }{4}\right)+\sqrt{2}-1\\
\end{cases}
$$
from the second equation we have

$t-\frac{\pi}{4}=\frac{y}{\sqrt{2}}+\frac{1}{\sqrt{2}}-1$

Plug in the first, expand and simplify
$$\sqrt{2} x+\left(\sqrt{2}-2\right) y=6-3 \sqrt{2}$$

in another form
$$y+3=\left(\sqrt{2}+1\right) x$$
edit

look at the image below. The curve and the tangent at $t=\pi/4$

$$...$$


问:
Taking the limit with 2 more variables
$\lim_{x\to \infty} (1+a/x)^{bx}$

I came across this question and this is my approach:

$\lim_{x\to \infty}(e^{bx\ln(1+a/x)})$ and I stucked here.

I used calculators to compute the answer and it is $e^{ab}$. Any idea how I can get to the answer? Thank you!

答:
Hint.

In the limit, $a$ and $b$ are not "variables".

Assuming that $a\ne 0$, the limit can be written as
$$
\lim_{x\to\infty}\left((1+\dfrac{1}{\frac{x}a})^{\frac{x}{a}}\right)^{ab}
=\left(\lim_{x\to\infty}(1+\dfrac{1}{\frac{x}a})^{\frac{x}{a}}\right)^{ab}
$$

Also, depending on the sign of $a$, you will need one of the following limits:
$$
\lim_{y\to\infty}\left(1+\frac1y\right)^y=e,\quad \lim_{y\to-\infty}\left(1+\frac1y\right)^y=e
$$

Title: Taking the limit with 2 more variables


问:
Do the BAC-CAB identity for triple vector product have some intepretation?
As in the title, I was wondering if the formula: $$a\times (b\times c)=b(a\cdot c)-c(a \cdot b)$$ for $\mathbb R ^3$ cross product has some geometrical interpretation. I've recently seen a proof (from Vector Analysis - J.W. Gibbs) that's not at all difficult to understand, however I would hardly remember the steps of the proof (and I keep forgetting the correct order of A,B,C's), since it appears to me just as algebraic manipulation. So, why is this true? Or is it just an accident?</p>

答:
No, it's not an accident. The cross product is orthogonal to each factor, so the vector has to be orthogonal to $b\times c$, hence in the plane spanned by $b$ and $c$. But it also has to be orthogonal to $a$. So, writing
$$a\times(b\times c) = xb + yc$$
and dotting with $a$, you get $x(b\cdot a) + y(c\cdot a)=0$. So the answer must be some scalar multiple of the correct formula. Now you only have to check that that scalar is $1$ by substituting $a=b$ and $a=c$. Better yet, let $a$ be a unit vector in the plane spanned by $b$ and $c$ that is orthogonal to $b$.</p>


问:
Solve $\lim_{x\to -\infty }\frac{-1+\sqrt{1+4x^2}}{2x}$
$\lim_{x\to -\infty }\frac{-1+\sqrt{1+4x^2}}{2x} = \frac12\lim_{x\to -\infty }\frac{-1+\sqrt{1+4x^2}}{x} = \frac12\left(\lim_{x\to -\infty}\frac{-1}{x}+\lim_{x\to -\infty}\frac{\sqrt{1+4x^2}}{x}\right)=\frac12\left(0+\lim_{x\to -\infty}\frac{\sqrt{1+4x^2}}{x}\right) = \frac12\lim_{x\to -\infty}\sqrt{\frac{1+4x^2}{x^2}} = \frac12\lim_{x\to -\infty}\sqrt{\frac1{x^2}+4}=\frac22=1$

But the graph of the function is clearly approaching $-1$ when $x$ is large and negative. Where did I make a mistake?

答:
You can not put $x$ under the square since it is negative. You can put $-x$:

$$\lim_{x\to -\infty}\frac{\sqrt{1+4x^2}}{x} = -\lim_{x\to -\infty}\sqrt{\frac{1+4x^2}{(-x)^2}}$$


答:
Theorem 1. Let $a,b\in\mathbb{R}$ and $a\neq0$. Then
$$\lim_{x\to\infty}\left(1+\frac{a}{x}\right)^{xb} = e^{ab}.$$
Proof.
We can extract $b$ from the exponent.
$$\left(\lim_{x\to\infty}\left(1+\frac{a}{x}\right)^{x}\right)^b.$$
We may define
$$L = \lim_{x\to\infty}\left(1+\frac{a}{x}\right)^x$$
so our final result is equal to $L^b$. Now we will define a new limit variable
$y\cdot a=x$. Since $a\in\mathbb{R}$ and $a\neq0$ we are alloud to write only the $y\to\infty$ in the limit, because $y\cdot a\to\infty \Rightarrow y\to\infty$.
$$L = \lim_{y\to\infty}\left(1+\frac{a}{ay}\right)^{ay}.$$
Lemma 2.
The number $e$ is defined as
$$e = \lim_{y\to\infty}\left(1+\frac{1}{y}\right)^{y}.$$
In equation $(4)$ we can cancel the $a$'s and extract the exponent like we did in equation $(2)$. Now we have
$$L=e^a$$
and including $b$ that will give us
$$L^b = e^{ab}.$$ QED

Title: Taking the limit with 2 more variables

Body:
$\lim_{x\to \infty} (1+a/x)^{bx}$

I came across this question and this is my approach:

$\lim_{x\to \infty}(e^{bx\ln(1+a/x)})$ and I stucked here.

I used calculators to compute the answer and it is $e^{ab}$. Any idea how I can get to the answer? Thank you!


答:
Since the exponential is a continuous function, you have that:
$$\lim_{x \to \infty} e^{f(x)} = e^{\lim_{x \to \infty} f(x)}$$
From there, you should be able to use the Hôpital's rule to evaluate
$$\lim_{x \to \infty} bx \ln(1 + \frac{a}{x})$$ by rewriting
$$\lim_{x \to \infty} bx \ln(1 + \frac{a}{x}) = \lim_{x \to \infty} \frac{\ln(1+\frac{a}{x})}{\frac{1}{bx}}$$ which is of the form $\frac{0}{0}$.

Title: Taking the limit with 2 more variables


问:
If $y'' \leq y$ and $y'(0) = 0$, is it true that $y = 0$ on $[0,\infty)$?

Assume $y(t)$ is a function such that $y,y'$ and $y''$ are continuous on $[0,\infty)$, $y(0) = 0$ and $y(t) \geq 0$. Show that:

If $y' \leq y$, then $y = 0$ on $[0,\infty)$;

If $y' \leq y^2$, then $y = 0$ on $[0,\infty)$;

If $y'' \leq y$ and $y'(0) = 0$, is it true that $y = 0$ on $[0,\infty)$?

This is a classic problem: If $y' \leq y$, then multiply the inequality by $e^{-t}$ gives
$$
(e^{-t}y)' \leq 0
$$
this is a function which is greater or equal to $0$ and $e^{-t}y(0) = 0$, as the derivative tells this forces $y = 0$.

My intuitive says to multiply both sides by $e^{\int y}$ so to get
$$
(e^{\int y}y)' \leq 0
$$
and apply the same argument as above, however I'm not sure if this is valid: there might be something I didn't catch so that this argument might fail.

and it seems like 3 is false, but I haven't found an counterexample yet.

Edit: bountied because the need for verification of 2 and an answer for 3.

答:
You might want to try $(e^{-\int y}y)'=e^{-\int y}(y'-y^2)\le 0$. Or use $(y^{-1})'=-y^{-2}y'$ for some interval $[t_1,t_2]$ where $y$ is strictly positive.

Add $y'$ and multiply with $e^t$, then this integrates to
$$
e^ty'(t)\le e^ty(t),
$$
since $y(0)=y'(0)=0$. One further integration leads to $e^{-t}y(t)\le 0$. Or refer to case 1.


问:
Evaluating $\sum_{n=1}^{\infty} \int_{1}^{\infty} \frac {\cos(2 \pi n t)}{t^z} dt$
I need help evaluating any of:

$$\sum_{n=1}^{\infty} \int_{1}^{\infty} \frac {\cos(2 \pi n t)}{t^z} dt$$

$$\sum_{n=1}^{\infty} \int_{1}^{\infty} \frac {\cos(2 \pi (2n - 1) t)}{t^z} dt$$

Where $z \in \Bbb{C}$

Also I would like to ask if this is true:

$$\sum_{n=1}^{\infty} \int_{1}^{\infty} \frac {\cos(2 \pi n t)}{t^z} dt = \int_{1}^{\infty} \frac {\sum_{n=1}^{\infty} \cos(2 \pi n t)}{t^z} dt$$

If yes then how to in clever way calculate:
$$\sum_{n=1}^{\infty} \cos(2 \pi n t)$$

答:
In a very general manner, the sum and integration operations can be switch iff they both converge in both cases.

For the evaluation of the integral I would do the following:
$$
\sum_{n=1}^\infty \int_0^\infty \frac{\cos(2\pi nt)}{t^z}dt=\sum_{n=1}^\infty \int_0^\infty \frac{\frac{e^{i2\pi nt}-e^{-i2\pi nt}}{2}}{t^z}dt\\
\frac{1}{2}\sum_{n=1}^\infty \int_0^\infty \frac{e^{i2\pi nt}-e^{-i2\pi nt}}{t^z}dt
$$
Now try interchanging the sum and integral, I think this would be a good way to approach the problem


Title: When I not seen a formula of prostapheresis in an identity

Body:
We suppose that I must to proof this identity

$$\sin(220°)+\cos(10°)=\cos(70°)$$

Easily if I put $\cos(10°)$ to RHS of the identity, I can apply the formula of prostapheresis $\cos(\alpha)-\cos(\beta)$ and I solved all.

But if we not will put the $\cos(10°)$ to the RHS, the identity is it possible to solve with a calculator or is there another alternative?

Body:
You can also have:

$\cos(10)=\sin(90-10)=\sin( 80)$

$\sin (220)+\sin(80)=2 \sin (\frac{220+80}2) \cos(\frac{220-80}2)=2 \sin(150)\cos(70)=\cos(70)$

beacause

$\sin(150)=\sin(180-30)=\sin(30)=\frac 12$


问:
Solving System of Differential equations
The general solution to differential equation
$$x'=Ax$$ where A is a square matrix is given by solving for the eigenvalues and then eigen vectors of matrix $A$. However, is there a general method if I have $$x'=Ax+b$$ where $b$ is a vector of length same a $x$?

The straight method to attack it would be to integrate to obtain:
$$x-A\frac{x^2}{2}-b=0$$
and solve:
$$x^2-2A^{-1}x+2A^{-1}b=0$$
$$x=\frac{2A^{-1}\pm\sqrt{(2A^{-1})^2-8A^{-1}}}{2}$$
Is there any steop ahead of this?
Thank you.

答:
If there exists a vector $v$ such that $Av=b$ then consider the variable $y=x-v$.  One finds $y'=Ay$.

If there is no such $v$ then $A$ is not invertible. If $A$ is still diagonalizable, however, then the tricky part of $b$ is the bit lying in the kernel of the matrix. Thus write $b=b_0+c$ where $Ab_0=0$ but there is some $Av=c$ solution. Try computing $y=x-v+tb_0$.

That's the easiest cases to deal with.

By the way your attempt to integrate is doomed to failure, because these are vectors not numbers, and you've integrated one thing with respect to $t$ andtheother to $x$!


问:
How do you find the area of a Venn diagram?
If we have a Venn diagram (only two circles)

And the larger circles have a radius of 2...

If the middle circle has a radius of 1, how can we find the area of the intersection part?

I understand that side is the middle of the larger circles, but I'm still confused on what to do to find the area of the middle intersection part.

Can anyone help? Thanks.

答:
Note the centre of each circle is on the other circle. The area of the intersection is the area of the circular sector $AIB$ minus the area of triangle $AIB$.

Elementary trigonometry shows that angle $\widehat{AIH} $ is equal to $\frac{\pi}3$,  so the length $AH$ is $2\sin\frac\pi 3=\sqrt 3$ and the area of triangle $AIB$ is twice the area of triangle $AIH$, i.e. also $\color{red}{\sqrt{3}}$.

On the other hand, as angle $\widehat{AIB}=\frac{2\pi} 3$ by symmetry, the area of the circular sector $AIB$ is equal to $\frac12\cdot 4\cdot\frac{2\pi}3=\color{red}{\frac{4\pi}3}$.

Ultimately, the area of the intersection is twice the difference, i.e.
$$\frac{8\pi}3-2\sqrt 3.$$


问:
set function $G＝f-g$, then $f≦P$ and $g≦N$
I want to prove the following statement.

Let $G$ be a set function, and let its Jordan decomposition $G＝P-N$.If $G$ can be written like $G＝f-g$ using finite measure $f,g$,
then $f≦P$ and $g≦N$.

I assumed that $P＜f$ and $N＜g$,but I couldn't find contradiction.I may use finiteness of measure,$f,g$.Thank you for your help.Hints or references are also appreciated.

答:
This is clearly false. Let $\mu$ a signed measure on $(X, \mathcal{M})$, let $\mu = \alpha - \beta$ be its Jordan decomposition, suppose that both $\alpha, \beta$ are finite. Let $x \in X$ and suppose $\{x\} \in \mathcal{M}$. Let $\lambda$ be measure on $(X,\mathcal M)$ given by the Dirac mass at $x$, i.e. $\lambda(E) = 1_{E}(x)$. Then $\mu = (\alpha + \lambda) - (\beta + \lambda)$, and clearly it is not the case that $\alpha + \lambda \leq \alpha.$

Note that really the opposite statement holds, as you can read about on the pertinent Wikipedia article. Namely if $\mu = \alpha - \beta$ is the Jordan decomposition and $\mu = \gamma - \lambda$ then $\alpha \leq \gamma, \beta \leq \lambda$ where $\gamma, \lambda$ are positive measures.


问:
Conditional expectation of a bounded almost sure random variable
Let $Z$ a random variable which is $\mathscr{p}$-measurable  and bounded almost sure, ie, there is a positive number $M$ such that $|Z| \leq M $ a.s.Then show thatfor the conditional expectation $$Y = \mathbb{E}[X|\mathscr{p}]$$ we have :
$$\mathbb{E}[YZ] = \mathbb{E}[XZ]$$

and more general if $Z$ is $\mathscr{p}$-measurable and $$\mathbb{E}[|ZX|]< \infty$$ then the equality to be proven is still true.

答: For $Z$ that is an indicator this is the definition of conditional expectation. For a general bounded $Z$ you can approximate it uniformly by a linear combination of indicators. Any non-negative $Z$ is a monotone increasing limit  of bounded variables. Finally for a general  $Z$ (with $|ZX|$ integrable) express it as a difference of two non-negative   variables (with the same property.)  It might help to first reduce to the case where $X \ge 0$.


Title: Solve $\lim_{x\to -\infty }\frac{-1+\sqrt{1+4x^2}}{2x}$

Body:
$\lim_{x\to -\infty }\frac{-1+\sqrt{1+4x^2}}{2x} = \frac12\lim_{x\to -\infty }\frac{-1+\sqrt{1+4x^2}}{x} = \frac12\left(\lim_{x\to -\infty}\frac{-1}{x}+\lim_{x\to -\infty}\frac{\sqrt{1+4x^2}}{x}\right)=\frac12\left(0+\lim_{x\to -\infty}\frac{\sqrt{1+4x^2}}{x}\right) = \frac12\lim_{x\to -\infty}\sqrt{\frac{1+4x^2}{x^2}} = \frac12\lim_{x\to -\infty}\sqrt{\frac1{x^2}+4}=\frac22=1$

But the graph of the function is clearly approaching $-1$ when $x$ is large and negative. Where did I make a mistake?

Body:
Since $x = sign(x)\sqrt{x^{2}}$ we have:

$$\lim_{x\to -\infty}\frac{\sqrt{1+4x^2}}{x} = -\lim_{x\to -\infty}\sqrt{\frac{1+4x^2}{x^2}}$$

Since $$\lim_{x\to -\infty} \frac{1}{x^{2}} = 0$$

and $$\lim_{x\to -\infty} \frac{4x^{2}}{x^{2}} = 4$$

then $$0+4 = \lim_{x\to -\infty} \frac{4x^{2}}{x^{2}} + \frac{1}{x^{2}} = \lim_{x\to -\infty} \frac{4x^{2}+1}{x^{2}}$$

Since $f(x)=\sqrt{x}$ is continuous in a neighborhood of 4,

$$\lim_{x\to -\infty}\sqrt{\frac{1+4x^2}{x^2}} = \sqrt{\lim_{x\to -\infty} \frac{4x^{2}+1}{x^{2}}} = \sqrt{4} = 2.$$

So $$\lim_{x\to -\infty}\frac{\sqrt{1+4x^2}}{x} = -2$$

$$\frac{1}{2}(-2) = -1$$

Just like it shows on that graph.


问:
Unifying the connections between the trigonometric and hyperbolic functions
There are many, many connections between the trigonometric and hyperbolic functions, some of which are listed here. It is probably too optimistic to expect that a single insight could explain all of these connections, but is there a holistic way of seeing the parallels between $\sin$ and $\sinh$, $\cos$ and $\cosh$? Can all of these seemingly disparate connections be shown to be essentially the same, or at least very similar?

Geometric connections

Sine and cosine parameterise the unit circle $x^2+y^2=1$, just as hyperbolic sine and cosine parameterise the 'unit hyperbola' $x^2-y^2=1$. Both circles and hyperbolas are conic sections.

The sector of the circle connecting the points $(0,0)$, $(1,0)$, and $(\cos t,\sin t)$ has an area of $t/2$. The region of the hyperbola connecting the points $(0,0)$, $(1,0)$, and $(\cosh t,\sinh t)$ has an area of $t/2$. This can even be used to define the hyperbolic functions geometrically, and many authors do the same with the trigonometric functions.

Sine and hyperbolic sine are odd, whereas cosine and hyperbolic cosine are even. But sine and cosine are periodic functions, unlike the hyperbolic counterparts.

The analogue of the identity $\cos^2x+\sin^2x \equiv 1$ is $\cosh^2x-\sinh^2x \equiv 1$. The compound angle formulae are almost identical to their hyperbolic counterparts, save for a pesky minus sign:
\begin{align}
\sin(x+y) &= \sin(x)\cos(y)+\cos(x)\sin(y) \\
\sinh(x+y) &= \sinh(x)\cosh(y) + \cosh(x)\sinh(y) \\[4pt]
\cos(x+y) &= \cos(x)\cos(y) \color{red}{-} \sin(x)\sin(y) \\
\cosh(x+y) &= \cosh(x)\cosh(y) \color{blue}{+} \sinh(x)\sinh(y) \, .
\end{align}

In general, given a trigonometric function, it is possible to write down the corresponding hyperbolic identity using Osborn's rule: replace every occurrence of $\cos$ with $\cosh$; replace every occurrence of $\sin$ by $\sinh$; but negate the product of two $\sinh$ terms.

Analytic connections

$\sin$ is the unique solution to the initial value problem
\begin{align}
f''(x) &= \color{red}{-}f(x) \\
f'(0) &= 1 \\
f(0) &= 0 \, ,
\end{align}
and the corresponding initial value problem for $\sinh$ is the same, except $f''(x) = \color{blue}{+}f(x)$.

Likewise, the initial value problem for $\cos$ is
\begin{align}
f''(x) &= \color{red}{-}f(x) \\
f'(0) &= 0 \\
f(0) &= 1 \, ,
\end{align}
and again we see a mysterious sign change for $\cosh$: $f''(x) = \color{blue}{+}f(x)$.

It follows that the higher-order derivatives of $\sin$ and $\sinh$ form periodic sequences.

If we solve the initial value problems shown above, we obtain the exponential forms of all $4$ functions:
\begin{align}
\sin x &= \frac{e^{\color{green}{i}x}-e^{-\color{green}{i}x}}{2\color{green}{i}} \quad{} \cos x = \frac{e^{\color{\green}{i}x}+ e^{-\color{green}{i}x}}{2} \\[3pt]
\sinh x &= \frac{e^{x}-e^{-x}}{2} \quad{} \cosh x = \frac{e^x + e^{-x}}{2} \, .
\end{align}

All $4$ functions are analytic, and their Taylor series bear a striking resemblance to each other:
\begin{align}
\sin x &= x \color{red}{-} \frac{x^3}{3!} + \frac{x^5}{5!} \color{red}{-} \frac{x^7}{7!} + \ldots \\[4pt]
\sinh x &= x \color{blue}{+} \frac{x^3}{3!} + \frac{x^5}{5!} \color{blue}{+} \frac{x^7}{7!} + \ldots \\[4pt]
\cos x &= 1 \color{red}{-} \frac{x^2}{2!} + \frac{x^4}{4!} \color{red}{-} \frac{x^6}{6!} + \ldots \\[4pt]
\cosh x &= 1 \color{blue}{+} \frac{x^2}{2!} + \frac{x^4}{4!} \color{blue}{+} \frac{x^6}{6!} + \ldots
\end{align}

And Euler's formula
$$
e^{ix} = \cos x + i \sin x
$$
is replaced by the underwhelming
$$
e^x = \cosh x + \sinh x \, .
$$

答:
The connection between the trigonometric and hyperbolic functions becomes more intimate when one introduces  split-complex numbers: the numbers of the form $a+bj$, where $j^2=1$ and $j\ne \pm 1$.

Using Taylor series you can easily find that
$$
e^{jx} = \cosh x + j\sinh x , \quad
\sinh x = \frac{e^{jx}-e^{-jx}}{2j}, \quad \cosh x = \frac{e^{jx} + e^{-jx}}{2}.
$$

One can understand $\cos x$ and $\sin x$ as the real and imaginary parts of the complex exponential $e^{ix}$. Likewise, $\cosh x$ and $\sinh x$ are the real and 'imaginary' parts of the split-complex exponential $e^{jx}$.

The analogy can be summarized as follows:
$$
\begin{align}
& &&\textbf{Complex numbers} && \textbf{Split-complex numbers} \\
&\text{Imaginary unit:} &&  i^2 = -1, &&    j^2 = 1, \\
&\text{Numbers:}  && z = a+bi,  &&    z = a+bj,  \\
&\text{Modulus:}  && |z| = \sqrt{a^2+b^2}, &&  |z| = \sqrt{a^2-b^2} \; (\text{for } a^2 \ge b^2), \\
&\text{Euler's formula:}  &&  e^{ix} = \cos x + i \sin x, &&   e^{jx} = \cosh x + j \sinh x, \\
&\text{Pythagorean theorem:} &&  |e^{ix}| = \cos^2 x + \sin^2 x =1,  &&   |e^{jx}| = \cosh^2 x - \sinh^2 x = 1,  \\
&\text{Geometric meaning:} &&  \text{Unit circle,}  &&   \text{Unit hyperbola (right branch)}.  \\
\end{align}
$$

The connection is in the correspondence:
$$
\begin{pmatrix} \text{complex} \; i \\ \sin, \; \cos  \end{pmatrix} \leftrightarrow \begin{pmatrix} \text{split-complex} \; j \\ \sinh, \; \cosh  \end{pmatrix}.
$$

One can use imagination to say that $\sin$, $\cos$ and complex numbers come from the elliptic world, while $\sinh$, $\cosh$ and split-complex numbers are their counterparts from the hyperbolic world.


问:
Can we define "Algebra of fractions" like ring/field of fractions?
According to the answers to this question, if $R$ is a ring with no zero divisors, we can define a ring of fractions of $R$ if $R$ satisfies the Ore condition and that ring will be a skew-field. What would happen if $R$ is an algebra over a field is? Say $A$ is an division algebra over a field that is not necessarily associative and may have zero divisors. Is there a condition like the Ore condition that guarantees that an algebra of fractions can be defined?

Can we not just adjust the Ore condition to make it suitable for algebras? I'm not sure if associativity and having no zero-divisors affect the fact that we can extend a ring to a ring of fractions.

答:
The (right or left) Ore conditions are already defined for any multiplicative subset in any ring.

Using them you can define a ring of fractions by considering the subset of regular elements, of the conditions are satisfied.

Allowing zero divisors into the set does not make any sense, because if you try to make an inverse for them, the resulting ring will collapse to zero.


问:
tensor product and direct product of algebra presentations
Let $R$ be a commutative unital ring and $R\langle x_i\mid f_j\rangle$ denote a unital $R$-algebra presentation.

Q1: What is the presentation of $R\langle x_i\mid f_k\rangle\otimes R\langle y_j\mid g_l\rangle$? It must be $R\langle x_i,y_j\mid f_k,g_l,h_m\rangle$ for some $h_m$.

Q2: What is the presentation of $R\langle x_i\mid f_k\rangle\oplus R\langle y_j\mid g_l\rangle$?

答:
I misunderstood the problem. Martin pointed out that it is for $R$-algebras, not $R$-modules.

So, this is what I think is correct:
$$R\langle x_i \mid f_i\rangle \otimes R\langle y_i \mid g_i\rangle = R\langle x_i, y_j \mid f_i, g_i, x_iy_j - y_jx_i\rangle.$$
The direct sum (based on the coproduct of semigroups) is
$$R\langle x_i \mid f_i \rangle \oplus R\langle y_i \mid g_i\rangle = R\langle x_i, y_i \mid f_i, g_i \rangle.
$$
The direct product (based on the product of semigroups) is
$$
R\langle x_i \mid f_i \rangle \times R\langle y_i \mid g_i\rangle = R\langle x_i, y_i, e_1, e_2 \mid f_ie_1, g_ie_2, x_iy_j - y_jx_i, e_1 + e_2 - 1, e_1x_i - x_i, x_i - x_ie_1, e_2y_i - y_i, y_i - y_ie_2, e_1^2 - e_1, e_2^2 - e_2 \rangle.
$$
Intuitively, $e_1$ represents $(1, 0)$ and $e_2$ represents $(0, 1)$ in the direct product, while elements originally from $X$ and $Y$ become $(x, 0)$ and $(0, y)$. The unit in the product is $(1, 1) = e_1 + e_2$.


答:
Firstly we need to realise that $\sin 220^\circ=\cos130^\circ$, as Blue has noted in comments. Now, using the angle addition formulae
$$\begin{align}\cos 130^\circ&=\cos(120^\circ+10^\circ)=\cos120^\circ \cos 10^\circ-\sin120^\circ\sin10^\circ\\
&=-0.5\cos10^\circ-\frac{\sqrt{3}}{2}\sin10^\circ
\end{align}$$
Hence,
$$\begin{align}
\sin220^\circ+\cos10^\circ&=\cos10^\circ-0.5\cos10^\circ-\frac{\sqrt{3}}{2}\sin10^\circ\\
&=0.5\cos10^\circ-\frac{\sqrt{3}}{2}\sin10^\circ
\end{align}$$
Now we can easily recognise this, again using the angle addition formulae, as equal to
$$\cos 60^\circ\cos10^\circ-\sin60^\circ\sin10^\circ$$
which is in fact
$$\cos(60^\circ+10^\circ)$$
and so we have:
$$\sin220^\circ+\cos10^\circ=\cos70^\circ$$
as required.


问:
Construct a set of intersections using a Lebesgue measurable function
Let $G_\delta$ denote the collection of all intersections of countably many sets in $G$.

Show that if $f$ is Lebesgue measurable, then there is a $G_\delta$ set $G$ with $m(G)=0$ so that $f \chi_{G^c}$ is Borel measurable.

Hint: Write $f$ as a linear combination of positive functions. If $f\geq0$, let $\{ r_n : n \geq 1\}$ be dense in $[0, \infty)$ and consider $f^{-1}([0,r_n))$. Remove a set of measure $0$ from each to get an $F_\sigma$ set, then find. a $G_{\delta}$ set G with $m(G)=0$ containing all of them.

I'm confused about the linear combination part of the hint. I know that every simple function can be written as a linear combination of characteristic functions. But not sure why a Lebesgue measurable function can also do so. From my understanding, we know that for each $n\in \mathbb{N}$, the set $f^{-1}([0, r_n ))$ is Lebesgue measurable. So we're taking the union of each $F_\sigma$ to get the $G_\delta$ set?

Thank you in advance

答:
Note that Lebesgue measure is just the completion of the associated Borel measure. We have the following lemma.

Lemma: Suppose $(X, \mathcal M, \mu)$ is a measure space and $(X, \overline{\mathcal M}, \overline \mu)$ is it completion, then $f: X \to Y$ for some measurable space $(Y, \mathcal N)$ is $\overline{\mathcal M}$ measurable iff $f = g$ $\overline{\mu}$ a.e. for some $\mathcal M$ measurable function $g$. See here for a proof in the special case we care about here.

Thus if $f$ is Lebesgue measurable, $f = g$ a.e. where $g$ is Borel measurable. Then let $E$ a Lebesgue null set s.t. $f=g$ on $E^C$. Then by outer regularity of Lebesgue measure, $m(E) = \inf \{m(U) : U \text{ open}, E \subseteq U\}$, so let $U_n$ a sequence of open sets containing $E$ s.t. $m(U_n) \to m(E) = 0$. Let $G = \bigcap_n U_n$, then $G$ is $G_\delta$ set with $0$ measure containing $E$. Thus $f =g $ on $G^C$ also, and so $f \circ \chi_{G^C} = g \circ \chi_{G^C}$, but $\chi_{G^C}$ is Borel measurable and $G$ is Borel, so $g \circ \chi_{G^C}$ is Borel. Thus $f \circ \chi_{G^C}$ is also. Thus $G$ is the set we were looking for.


Body:
hint

$$\sqrt{1+4x^2}=\sqrt{x^2(4+\frac{1}{x^2})}$$

$$=|x|\sqrt{4+\frac{1}{x^2}}$$

$$=\color{red}{-}x\sqrt{4+\frac{1}{x^2}}$$

because, near $ -\infty $,  $|x|=-x$.


问:
Gauss formula to calculate integral
Using Gauss formula calculate:$\int_S x^3dydz+y^3dxdz+z^2dxdy$ where $S$ is down part of $z=x^2+y^2$ cut out with plane $z=2x$. Using divergence theorem it comes to find:$\iiint_D (3x^2+3y^2+2z)dxdydz$, where $D$ is area bounded with(after cylindrical coordinates) $-\frac{\pi}{2}\le \phi\le \frac{\pi}{2}, 0\le r\le 2\cos\phi, 2r\cos\phi\le z\le r^2$. What is left is to plug in formula. Correct?

答:
I will go ahead and show all three and how we choose right orientation for different integrals so they are in sync with each another. Hope this helps for future questions too. I think you already did the first two integrals.

a) By applying divergence theorem which you already did,

Divergence = $3(x^2+y^2)+2z = 3r^2 + 2z$ in cylindrical coordinates,

$\displaystyle \int_{-\pi/2}^{\pi/2} \int_0^{2\cos\theta} \int_{r^2}^{2r\cos\theta} (3r^2 + 2z) \ r \ dz \ dr \ d\theta = \frac{11\pi}{3}$

This is flux for positively oriented surface (outward normal vector).

b) Flux through paraboloid surface (surface integral, without applying divergence theorem), which again you mentioned you knew how to find.

Surface is $z - x^2 - y^2 = 0$, taking derivative, normal vector $(2x, 2y, -1)$ or $(-2x, -2y, 1)$. The question does not state but since here we assume outward normal vector as in (a), that would be pointing generally downward for paraboloid surface $z = x^2+y^2$. Hence we choose $(2x, 2y, -1)$ - note the sign of $z$ component.

$\vec{F} = (x^3,y^3,z^2),$ so $\vec{F} \cdot \vec{n} = 2(x^4+y^2) - (x^2+y^2)^2$ (as on the paraboloid surface $z = x^2 + y^2$)

Simplifying, $\vec{F} \cdot \vec{n} = (x^2-y^2)^2$. Now the projection of paraboloid surface in $XY$ plane is the intersection of the plane and the paraboloid which is $x^2+y^2 = 2x$. In cylindrical coordinates the surface integral becomes,

$\displaystyle \int_{-\pi/2}^{\pi/2} \int_0^{2\cos\theta} r^4 \cos^2(2\theta) r dr d\theta = \frac{13\pi}{6}$.

c) Flux through the planar surface on top

Plane is $z = 2x$ and hence normal vector is $(2,0,-1)$ or $(-2, 0, 1)$. As we are finding flux for outward normal, in this case, the normal vector will be pointing upward in positive $z$ direction. So we choose $(-2, 0, 1)$.

Again $\vec{F} \cdot \vec{n} = (x^3,y^3,z^2) \cdot (-2, 0, 1) = 4x^2 - 2x^3$ (as on the plane, $z = 2x$). In cylindrical coordinates, the integral becomes,

$\displaystyle \int_{-\pi/2}^{\pi/2} \int_{0}^{2\cos\theta} (4r^2 \cos^2\theta - 2r^3 \cos^3 \theta) \ r \ dr \ d\theta = \frac{3\pi}{2}$

Now for this question, if you are not mandated to use Divergence theorem, you can go with $(b)$ as it is more straightforward. But if you are asked to use Divergence theorem, you find $(a)$ and $(c)$ and then $(b) = (a) - (c)$.

Title: Gauss formula to calculate integral


问:
Question about meaning of derivative of vector-valued function
So I was wondering about the interpretation of the derivative of a vector-valued function in the sense of how it is analogous to a function with one variable input and one output. With the derivative in single variable calculus, it is interpreted as the slope of the line that is tangent to the curve.

With a vector valued function, when we take the derivative, it is equivalent to taking the derivative of each of the components of the function's output. So are we saying that instead of a representing a line that is tangent to the curve, it is representing a vector that is tangent to the curve?

But what is confusing me about the analogy is this. In the single variable situation, we can travel along the tangent line that is given by the derivative. What is the meaning of "travelling along the tangent line" when we have a tangent vector? Do we travel along the tangent vector?

答:
If $f: \mathbb R \to \mathbb R^n$ is differentiable, then $\{ f(t) \mid t \in \mathbb R
\}$ is a curve in $\mathbb R^n$. The tangent line to this curve at a point $f(t_0)$  consists of all points in $\mathbb R^n$ of the form $f(t_0) + f'(t_0)(t - t_0)$ where $t \in \mathbb R$. You can visualize that tangent line.


答:
To solve
$$
x'=Ax+b
$$
use
$$
x=e^{At}x_0-A^{-1}b
$$
where $e^{At}x_0$ is the solution to $x'=Ax$.

Often times it is useful to look at the $1$-dimensional analog:
$$
x'=ax+b
$$
where the solution is
$$
x=x_0e^{at}-b/a
$$
I just put that solution into matrix form and checked it out to make sure it worked. Low tech math.

Title: Solving System of Differential equations

Body:
The general solution to differential equation
$$x'=Ax$$ where A is a square matrix is given by solving for the eigenvalues and then eigen vectors of matrix $A$. However, is there a general method if I have $$x'=Ax+b$$ where $b$ is a vector of length same a $x$?

The straight method to attack it would be to integrate to obtain:
$$x-A\frac{x^2}{2}-b=0$$
and solve:
$$x^2-2A^{-1}x+2A^{-1}b=0$$
$$x=\frac{2A^{-1}\pm\sqrt{(2A^{-1})^2-8A^{-1}}}{2}$$
Is there any steop ahead of this?
Thank you.


问:
Re{...} function specification
I am doing a proof that uses the $Re\{\}$ function (the function that outputs the Real part of a complex number $a + bi$). Along the way I realized that I am not sure how the function behaves when it comes to multiplied inputs. Specifically, is

$Re\{\phi_1 • \phi_2\}$ where $\phi_1, \phi_2$ are complex numbers equivalent to $Re\{\phi_1\} • Re\{\phi_2\}?$

答: Since $$(a+bi)(c+di) = (ac-bd)+(ad+bc)i$$ we have that $$\operatorname{Re}(zw) = \operatorname{Re}(z)\operatorname{Re}(w) - \operatorname{Im}(z)\operatorname{Im}(w).$$

Title: Re{...} function specification


答:
No, it's not. This is how complex multiplication works:
$$
(a+b\mathrm i)\cdot(c+d\mathrm i) = (ac-bd)+(ad+bc)\mathrm i.
$$
Hence, when $z,w$ are complex numbers, we have
$$
\operatorname{Re}(z\cdot w) = \operatorname{Re}(z)\cdot\operatorname{Re}(w)-\operatorname{Im}(z)\cdot\operatorname{Im}(w).
$$

Title: Re{...} function specification

Body:
I am doing a proof that uses the $Re\{\}$ function (the function that outputs the Real part of a complex number $a + bi$). Along the way I realized that I am not sure how the function behaves when it comes to multiplied inputs. Specifically, is

$Re\{\phi_1 • \phi_2\}$ where $\phi_1, \phi_2$ are complex numbers equivalent to $Re\{\phi_1\} • Re\{\phi_2\}?$


答:
$x \ne \sqrt {x^2}$.

$|x| =\sqrt{x^2}$ so $x = \begin{cases}\sqrt{x^2}& x\ge 0\\-\sqrt{x^2}& x < 0\end{cases}$.

And as we are taking $x \to -\infty$ we know $x < 0$.

So

$\lim_{x\to -\infty }\frac{-1+\sqrt{1+4x^2}}{2x} = \frac12\lim_{x\to -\infty }\frac{-1+\sqrt{1+4x^2}}{x} = \frac12\left(\lim_{x\to -\infty}\frac{-1}{x}+\lim_{x\to -\infty}\frac{\sqrt{1+4x^2}}{x}\right)=\frac12\left(0+\lim_{x\to -\infty}\frac{\sqrt{1+4x^2}}{x}\right) = \color{red}{\frac12\left(0+\lim_{x\to -\infty}\frac{\sqrt{1+4x^2}}{-\sqrt{x^2}}\right)}=\frac12\lim_{x\to -\infty}\color{red}-\sqrt{\frac{1+4x^2}{x^2}} = \frac12\lim_{x\to -\infty}\color{red}-\sqrt{\frac1{x^2}+4}=\color{red}{\frac 12(\color{green}- 2)}=\color{red}-\frac22=\color{red}-1$


问:
Does the linear recurrence theory hold for inequalities?
Let $\alpha > 0$ and $a_n$ be a sequence of positive real numbers such that $a_{n+2} \leq a_{n+1} - \alpha a_n$ for all $n\geq 0$. Is it true that $a_n \leq A_1x_1^n + A_2x_2^n$ for some constants $A_1$ and $A_2$, where $x_1$ and $x_2$ are the roots of $t^2 - t + \alpha = 0$?

If $\leq$ was instead $=$, then this is well known from the theory of linear recurrences (and follows by induction). However, here assuming $a_n \leq A_1x_1^n + A_2x_2^n$, we cannot say $a_{n+1} -\alpha a_n \leq A_1x_1^{n+1} + A_2x_2^{n+1} - \alpha(A_1x_1^{n} + A_2x_2^{n})$ when $\alpha > 0$ because of the minus sign. Is this fixable at all?

Any help appreciated!

答:
Well, with the hypotheses you presented you can show that the sequence converges to $0$. If this counts as 'fixable' then yes it is fixable. If you want to get precise bounds compared to the solution to linear equations then that is more difficult but still possible.

{$a_{n}$} is positive monotone non-increasing and bounded below by 0, hence converges to $L\geq 0$. If $L > 0$ you can use epsilon delta arguments with the limsup and liminf to show:

$$L < L -\alpha L$$.

Which will lead to a contradiction, forcing $L=0$.


问:
u-substitution of definite integral
$$\int_1^3 \frac{(\ln(v+1))^2}{(v+1)}dv$$

Here is what I have done:

$u =  v+1$

$\frac{du}{dx}=1$

$du = dv$

$$\int_2^4 \frac{(\ln(u))^2}{(u)}du$$

Then

$z=\ln(u)$

$\frac{dz}{du}=\frac{1}{u}$

$dz = \frac{du}{u}$

$$\int_2^4 \frac{(z)^2}{(u)^2}dz$$

And here I was told I was wrong. As there should only be $z^2$ left in the equation but I cant tell what is the problem. Any hint? Thanks.

答: $$\int \frac{(\log u)^2}{u}du = \int (\log u)^2 \color{red}{\frac{du}u} = \int z^2 \color{red}{dz}.$$

Title: u-substitution of definite integral


答:
Hint

With the substitution $$z=\ln(u)$$
and

$$dz=\frac{du}{u}$$

the integral $$\int_2^4\frac{(\ln(u))^2}{u}du$$

becomes

$$\int_{\ln(2)}^{\ln(4)}z^2dz$$

Title: u-substitution of definite integral

Body:
$$\int_1^3 \frac{(\ln(v+1))^2}{(v+1)}dv$$

Here is what I have done:

$u =  v+1$

$\frac{du}{dx}=1$

$du = dv$

$$\int_2^4 \frac{(\ln(u))^2}{(u)}du$$

Then

$z=\ln(u)$

$\frac{dz}{du}=\frac{1}{u}$

$dz = \frac{du}{u}$

$$\int_2^4 \frac{(z)^2}{(u)^2}dz$$

And here I was told I was wrong. As there should only be $z^2$ left in the equation but I cant tell what is the problem. Any hint? Thanks.


问:
Operations on smooth functions
I am wondering what operations preserve smoothness of functions. For example, sum of two smooth functions, product of two smooth functions and composition of two smooth functions is also smooth. Is it the case for integrals of smooth functions?

Also, do there exist functions that have well-defined 2nd derivative but not 1st derivative? I would like to delve further into the topic, links to references and sources would be appreciated.

Thanks!

答:
If we talk about "simple" smooth functions, i.e $f:\mathbb R \rightarrow \mathbb R$, then the set of all smooth function form a vector space over the field of real numbers.

So any property of a vector space can be apllied to smooth functions.

As for your specific question, a smooth function on a closed interval $[a,b]$ is continous there, therefore it is integrable. And we know that integrating a function makes it "nicer", or "smoother" so of course the integral operator also preserve the smoothness of a function.

Furthemore, if the function is $C^{\infty}$ (which is usually what we mean when we say smooth), then the derivative operator also preserve the smoothness.

As for your second question - No. A function can't have a second derivative if it does not have a first derivative, because the second derivative of $f$ is simply the derivative of $f'$, but since $f'$ is not defined, then it is not differentiable.

Title: Operations on smooth functions


问:
Denurable - onto function $\mathbb N \to A$ then $A$ is denurable
I know that if there is a bijective function from $\mathbb N$ to $A$ then $A$ has the same cardinality of $\mathbb N$, therefore, $A$ is denurable. However, I was reading one answer that $A$ is denurable if there is a onto function from $\mathbb N$ to $A$. How can I prove this?

答:
Hint:

If $A$ is finite, then it is denumerable.

If $A$ is infinite, then there is an injection $\mathbb N \hookrightarrow A$, which coupled with the assumed surjection $\mathbb N \to A$ shows that a bijection exists. Do you see how to use, say, the Cantor-Bernstein-Schroder theorem to show this?

I hope this helps ^_^


答:
Q1. I assume by $A \otimes_R B$ you mean the $R$-algebra which has as underlying $R$-module the tensor product of the underlying $R$-modules, and whose algebra is structure is given by $1 := 1 \otimes 1$ and $(a \otimes b) (a' \otimes b'):=aa' \otimes bb'$. This $R$-algebra has the following universal property: Homomorphisms $A \otimes_R B \to C$ correspond 1:1 to pairs of homomorphisms $f : A \to C$ and $g : B \to C$ which commute in the following sense: For $a \in A$ and $b \in B$ we have $f(a) g(b)=g(b) f(a)$. Therefore, it might be called the "commutative coproduct". It follows immediately that

$R\langle X,F \rangle \otimes_R R\langle Y,G \rangle = R \langle X \cup Y : F \cup G \cup \{xy-xy : x \in X, y \in Y\}\rangle$

Q2. Do you mean the coproduct in the category of $R$-algebras? In that case, we have $R \langle X,F \rangle \oplus R \langle Y,G \rangle = R\langle X \cup Y ,F \cup G \rangle$. Or do you mean the direct product (which is, unfortunately, often denoted by $\oplus$, see also here)? In that case, we have $R \langle X,F \rangle \times R \langle Y,G \rangle = R \langle X,Y,e : e^2=e, Fe, Ge^{\perp}, x=ex=xe, y=e^{\perp} y=e^{\perp} y\rangle$. By $Fe$ I mean that every occurence of $1$ in a relation has to be replaced by $e$. The reason is the following (co)universal property of the direct product of two $R$-algebras: A homomorphism $A \times B \to C$ corresponds to an idempotent $e \in C$ (the image of $(1,0)$) and two homomorphisms of algebras $A \to Ce$ and $B \to Ce^{\perp}$. Here, $e^{\perp}:=1-e$.

Title: tensor product and direct product of algebra presentations

Body:
Let $R$ be a commutative unital ring and $R\langle x_i\mid f_j\rangle$ denote a unital $R$-algebra presentation.

Q1: What is the presentation of $R\langle x_i\mid f_k\rangle\otimes R\langle y_j\mid g_l\rangle$? It must be $R\langle x_i,y_j\mid f_k,g_l,h_m\rangle$ for some $h_m$.

Q2: What is the presentation of $R\langle x_i\mid f_k\rangle\oplus R\langle y_j\mid g_l\rangle$?


问:
Let $z_1$, $z_2$ and $z_3$ be complex vertices of an equilateral triangle. Show $z_1^2 + z_2^2 + z_3^2 = z_1 z_2 + z_2 z_3 + z_3 z_1$.
Prompt:

Let $z_1$, $z_2$ and $z_3$ represent vertices of an equilateral triangle in the complex plane. Show $z_1^2 + z_2^2 + z_3^2 = z_1 z_2 + z_2 z_3 + z_3 z_1$.

Question:

I hope the following question isn't too soft. I know a solution to the above question by taking the sides of the triangle and equating them by rotations in the complex plane of $e^{i\pi/3}$.

What, if anything, is the significance of the result other than showing some clever algebraic manipulation? Does it tell us something unique about equilateral triangles defined by vertices in the complex plane?

Is there any intuition to be gained here?

答:
The significance of the expression
$$z_1^2 + z_2^2 + z_3^2 = z_1 z_2 + z_2 z_3 + z_3 z_1\tag1$$
is that it is the most concise and easiest to apply in analyzing an equilateral triangle in the complex plane.

There may be some suspicion on the statement. It is tempting, for example, to point out that the expression
$$|z_1-z_2| = |z_2-z_3| = |z_3-z_1|\tag2 $$
is simpler. However, it is only so in appearance, not in complex algebraic operation. It is actually the following complex conjugate expression in disguise
$$(z_1-z_2)(\bar z_1-\bar z_2 )= (z_2-z_3)(\bar z_2-\bar z_3 ) =( z_3-z_1)(\bar z_3-\bar z_1 )\tag3$$
which is more involved than (1). Besides, the expression (2) is in fact a system of two equations, which is again more involved than the single equation (1). In practical application, it is more straightforward to check the condition (1) in determining whether $z_1$, $z_2$ and $z_3$ form an equilateral triangle than, say, checking the condition (2).