问:
If $f$ is a bilinear function then prove that $Df(a,b).(x,y)=f(a,y)+f(x,b)$
Let $f:\mathbb{R}^n \times \mathbb{R}^m \rightarrow \mathbb{R}^p$ be a bilinear function.

Prove that

$$Df(a,b).(x,y)=f(a,y)+f(x,b)$$

Here $Df(a,b)$ is the Jacobian.

My Problem

I'm not able to understand what does the dot(.) right after $Df(a,b)$ means? Is it a typo? OR does it mean something?

What I think

I think the $(x,y)$ after the dot is input to the Jacobian but
the Jacobian $(Df(a,b))$ will be A $p \times 2$ matrix and RHS is $p \times 1$. Also I'm not entirely sure that the Jacobian is $p \times 2$ matrix, if Jacobian is some combination of $f$ then that would make a lot of sense because then it will take $(x, y)$ as input and output a $p \times 1$ vector(matrix).

答:
Here $\Bbb R^n\times\Bbb R^m$ is identified with $\Bbb R^{n+m}$, and the Jacobian is a $p\times(n+m)$ matrix.

Its input $(x,y)$ is considered as a column vector $\in\Bbb R^{n+m}$ with the first $n$ coordinates coming from $x$ and the last $m$ coordinates coming from $y$.

Dot means matrix multiplication, yielding both sides of the equation to be proved an element of $\Bbb R^p$.


问:
if vectors are linearly independent then their images are linearly independent?
Let $V$ and $W$ be finite-dimensional vector spaces and suppose  $l\in L(V, W)$. If $v_{1},...,v_{n} \in V$ are linearly independent then can we say that $l(v_{1}),....,l(v_{n}) \in W$ are linearly independent?
What i've tried if $\dim(V)>\dim(W)$ then it's not true and if $\dim(V)\leq \dim(W)$ then linear map $l$ must be an injection.
EDIT::: Note that $v_{1},...,v_{n} \in V$ need not be a basis in that case my assumption fails.

答:
If $l$ is an injection, the image of a linearly independent set is linearly independent. To see this let $T: V \rightarrow W$ be a linear injection and assume that $S = \{v_1, \cdots, v_n \} \subseteq V$ is linearly independent.

Suppose

$$c_1 T(v_1) + \cdots + c_n T(v_n) = 0$$

Since $T$ is linear, we have

$$T(c_1v_1 + \cdots + c_n v_n ) = 0$$

Since $T$ is injective, we have

$$c_1 v_1 + \cdots + c_n v_n = 0, $$

which gives $c_1 = \cdots = c_n = 0$ since $S$ was assumed linearly independent.

ContentLicense: CC BY-SA 4.0


问:
Fredholm operators -- index
Atkinson's theorem states:

$T ∈ L(H)$ is a Fredholm operator if and only if T is invertible modulo compact perturbation, i.e. $TS = I + C_{1}$ and $ST = I + C_{2}$ for some bounded operator S and compact operators $C_{1}$ and $C_{2}$.

$L(H)$= espace of linear and bounded operators.

Let $A$ and $B$ two fredholm operators. How to prove that $A\oplus B\ $  is a fredholm operator using the the Atkinson's theorem?

答:
$\newcommand{\dim}{\text{dim}}\newcommand{\codim}{\text{codim}}\newcommand{\Ind}{\text{Ind}}\newcommand{\Ker}{\text{Ker}}\newcommand{\Ran}{\text{Ran}}$Notice that $\Ker(A\oplus B) =  \Ker(A)\oplus \Ker(B)$, hence a finite dimensional space,  and
$\Ran(A\oplus B) =  \Ran(A)\oplus \Ran(B)$,
hence a finite co-dimensional space.  Therefore $A\oplus B$ is Fredholm and
$$
  \Ind(A\oplus B) = \dim(\Ker(A\oplus B)) - \codim(\Ran(A\oplus B)) = $$ $$ =
  \dim(\Ker(A)) +   \dim(\Ker(B)) - \codim(\Ran(A)) - \codim(\Ran(B)) = $$ $$ =
  \Ind(A) +   \Ind(B).
  $$

Now, since $\Ker(A^*)^\perp = \Ran(A)$, we have
$$
  \Ind(A) =   \dim(\Ker(A)) - \dim(\Ker(A^*)),
  $$
from where it follows that   $\Ind(A^*) =  -\Ind(A)$.

Title: Fredholm operators -- index


问:
I cannot solve these separable differential equations
There is no explanation in the book on how to solve these and I can't find any help online. Step-wise calculators also don't make sense.

(1) $y'=(y^2-1)x, \; y(0)=0$,

(2) $xy'=y^2-2y, \; y(1)=1 \; x\geq0$.

I'll post the question and write my attempts underneath straight after

答:
For the first one, note that you can re-write

\begin{align}
y' &= (y+1)(y-1)x \\
\implies \displaystyle \int \frac{dy}{(y+1)(y-1)} &= \int x dx
\end{align}
Whence
\begin{align}
\frac{1}{2}\displaystyle \int \frac{1}{y-1} - \frac{1}{y+1}&= \int x dx
\end{align}
Thus
$$\ln\frac{y-1}{y+1} = x^2+C$$

Then high school algebra can assist in re-arranging to make $y$ the subject.

For the second DE this can be written on the right as $y(y-2)$ and upon finding partial fractions
\begin{align}
\displaystyle \int \frac{dy}{y(y-2)} &= \int \frac{1}{x}dx \\
\implies \frac{1}{2}\displaystyle \int \frac{1}{y-2}-\frac{1}{y} &= \int \frac{1}{x}dx \\
\implies \ln(y-2)-\ln y &= \ln x + C \\
\implies \ln \frac {y-2}{y} &= \ln x+C\\
\implies \ln \left( 1-\frac{2}{y} \right) &= \ln x +C
\end{align}

and the result follows.


答:
Concerning the dot: When a linear map $A$ is applied to a vector $x$ one usually writes $Ax$. But in some cases, like the one in your problem there are various nested maps at stake. Then one often writes a bottom dot between the linear map  and the vector  to which the map is applied. The map here is $Df(a,b):\>{\mathbb R}^{n+m}\to{\mathbb R}^p$, and the vector is $(x,y)\in{\mathbb R}^{n+m}$. Write $(a,b)=:c$, $\>(x,y)=:z$.

In order to find $Df(c)$ we compute
$$\eqalign{f\bigl(c+z\bigr)-f(c)&=f(a+x,b+y)-f(a,b)\cr
&= f(a,y)+f(x,b)+f(x,y)\ ,\cr}$$
by bilinearity of $f$. The map $L:\>{\mathbb R}^{n+m}\to{\mathbb R}^p$ defined by
$$Lz:=f(a,y)+f(x,b)\qquad\bigl(z=(x,y)\in{\mathbb R}^{n+m}\bigr)$$
is linear (note that $x$ and $y$ depend linearly on $z$), and we have
$${f\bigl(c+z\bigr)-f(c)-Lz\over
|z|}={f(x,y)\over|z|}\ .\tag{1}$$
Now there is a constant $C$ such that $$|f(x,y)|\leq C|x|\,|y|\leq C|z|^2\ ,$$ and this implies $\lim_{z\to0}{f(x,y)\over|z|}=0$.
From $(1)$ we then can read off that $Df(c)=L$, according to the general definition of the derivative in a multidimensional situation.

Title: If $f$ is a bilinear function then prove that $Df(a,b).(x,y)=f(a,y)+f(x,b)$

Body:
Let $f:\mathbb{R}^n \times \mathbb{R}^m \rightarrow \mathbb{R}^p$ be a bilinear function.

Prove that

$$Df(a,b).(x,y)=f(a,y)+f(x,b)$$

Here $Df(a,b)$ is the Jacobian.

My Problem

I'm not able to understand what does the dot(.) right after $Df(a,b)$ means? Is it a typo? OR does it mean something?

What I think

I think the $(x,y)$ after the dot is input to the Jacobian but
the Jacobian $(Df(a,b))$ will be A $p \times 2$ matrix and RHS is $p \times 1$. Also I'm not entirely sure that the Jacobian is $p \times 2$ matrix, if Jacobian is some combination of $f$ then that would make a lot of sense because then it will take $(x, y)$ as input and output a $p \times 1$ vector(matrix).


问:
Prove Inequalities $\frac{x(2+x)}{1+x} \leq f(x) \leq \frac{x(2-x)}{1-x}$ Using Calculus
I am often asked to prove inequalities with natural logarithms. An example question I am stuck on is this:

Let $f(x) = \ln(\frac{1+x}{1-x})$ for $-1 < x < 1$. Prove that

$$\frac{x(2+x)}{1+x} \leq f(x) \leq \frac{x(2-x)}{1-x}$$ with equality iff $x=0$.

My general approach has been to try to find a function which upper and lower bounds the function $1/x$ such that the integral of the function will result in the upper and lower bound. I know that the definition of $\ln(x) = \int_{1}^{x} \frac{1}{t}dt$.

Example I want to prove (1)$f_1(x) < \ln(x) <f_2(x)$ . If I can find (2)$F_1(x) < 1/x < F_2(x)$, then by taking the integral I can get the original inequality. Problem is I find it difficult to come up with the (1) relationship.  Espically when it gets complicated.

Are there better ways / tricks that can help me solve these proofs more easily? What would be your approach to this problem?

答:
I don't know if you like this method which doesn't need derivatives. Suppose $x\ge 0$. For $t\in[0,x]$,$1-t\ge 1-x,1+t\ge 1$ and hence
$$ \ln(\frac{1+x}{1-x})=\int_0^x\bigg(\frac{1}{1+t}+\frac{1}{1-t}\bigg)\le\int_0^x\bigg(1+\frac{1}{1-x}\bigg)dt=\frac{x(2-x)}{1-x}. \tag1$$
Similarly for $t\in[0,x]$,$1-t\ge 1$
$$ \ln(\frac{1+x}{1-x})=\int_0^x\bigg(\frac{1}{1+t}+\frac{1}{1-t}\bigg)\ge\int_0^x\bigg(\frac{1}{1+x}+1\bigg)dt=\frac{x(2+x)}{1+x}. \tag2$$
Combining (1) and (2) gives the desirable inequality. If $x<0$, you can do the same.


问:
Finding the structure of a group without using sylows theorem.

If$ |G|=pq $ and $p $ doesnt divide $(q-1)$ and $p <q$ then $G$ is cyclic.

My proof stands as this .I have used the fact that

$(i)$ If $|G|=pq$ then I showed that there will only be one element of order $p$ and one element of order $q$.

My approach in proving this part has been to show that if there are two elements of order $p$ say $x_1$ and $x_2$ , then say $H_1$ is a group of order $p$ generated by $x_1$ and $H_2$ is a group of order $p$ generated by $x_2$ .We assume that the intersection is {e} if not then we can get $H_1$=$H_2$(by property of subgroup).

Proving that the elements are distinct .

We assume that the elements $(x_1)^{i}.(x_2)^j$ are not distinct then $(x_1)^{i}.(x_2)^j =(x_1)^{i'}(x_2)^{j'}$.From here we can arrive at a contradiction as $H_1 \cap H_2 =e$.So if there are $p^2$ elements then we can arrive at a contradiction as $p$ and $q$ are both primes.

Similar results will hold in the case of $q$.

$(ii)$ Now  there is only one subgroup of order $p$ and one subgroup of order $q$ so they are both normal

$(iii)$ let $H$ and $K$ be two subgroups of order $p$ and order $q$.Then we know that $H \cap K={e}$.$H$ and $K$ are both normal .Then I showed that $x^{-1}y^{-1}xy \in H \cap K$ and $xy=yx$. So the order of the element $xy$ is $pq$.Where am I going wrong in my proof and since I have not used the fact that $p $ doesnot $q-1$.

答:
As Gerry Meyerson pointed out, it does not hold in general as it is stated. You need to infer that $p \lt q$. Choose $Q$ a subgroup of $G$ of order $q$ (you can use Cauchy's Theorem for its existence!). Then $|G:Q|=p$ is the smallest prime dividing $|G|$ (ah yes here we are using $p \lt q$), hence $Q \lhd G$ (I hope you know this theorem ... see here for example).

Now $P$ acts on $Q$ by conjugation, but since $p \nmid q-1$ and Aut$(Q) \cong C_{q-1}$ (here we use that $q$ is prime), the action must be trivial: $P$ centralizes $Q$ and the other way around. Since $G=PQ$, $G$ must be abelian and the Chinese Remainder Theorem does the rest: $G \cong C_p \times C_q \cong C_{pq}$.

Bonus remark if $|G|=n$ and gcd$(\varphi(n),n)=1$, then $G$ is cyclic (as a matter of fact the only group of order $n$).


答:
for (1)
First separating the variables and integrating both sides with respect to y and x respectively
$\int \frac{1}{y^2-1}dy=\int xdx$.

The RHS is trivial but the LHS needs partial fractions, so

$\frac{-1}{2}ln\mid y+1\mid+\frac{1}{2}ln\mid y-1\mid=\frac{x^2}{2}+C$.
Using log identity and getting rid of the ln function gives

$\frac{\mid y-1\mid}{\mid y+1\mid}=Ce^{x^2}$, because the RHS is positive and using the initial conditions we get

$\frac{y-1}{y+1}=Ce^{x^2} => C=-1$, algebraic manipulation eventually gives the solution:

$ y=\frac{1-e^{x^2}}{1+e^{x^2}}$.

Title: I cannot solve these separable differential equations

Body:
There is no explanation in the book on how to solve these and I can't find any help online. Step-wise calculators also don't make sense.

(1) $y'=(y^2-1)x, \; y(0)=0$,

(2) $xy'=y^2-2y, \; y(1)=1 \; x\geq0$.

I'll post the question and write my attempts underneath straight after


问:
Evaluate $\lim_{x \rightarrow \infty}\left( \sqrt[3]{x^3+1} -\sqrt[4]{x^4+1}\right)$
Could you help to evaluate $$\lim_{x \rightarrow \infty}\left( \sqrt[3]{x^3+1} -\sqrt[4]{x^4+1}\right)?$$
I tried all methods I know of to no avail

Body:
\begin{align*}
\lim_{x \to \infty}{\sqrt[3]{x^3 + 1} - \sqrt[4]{x^4 + 1}} &= \lim_{x \to \infty}{\left(\sqrt[3]{x^3 + 1}-x + x- \sqrt[4]{x^4 + 1}\right)}\\
&=\lim_{x \to \infty}{\frac{x^3+1-x^3}{\sqrt[3]{x^3+1}^2+x\sqrt[3]{x^3+1}+x^2}-\frac{x^4-x^4-1}{x^3+x^2\sqrt[4]{x^4+1}+x\sqrt{x^4+1}+\sqrt[4]{x^4+1}^3}}\\ &=\lim_{x \to \infty}{\frac{1}{\sqrt[3]{x^3+1}^2+x\sqrt[3]{x^3+1}+x^2}-\frac{1}{x^3+x^2\sqrt[4]{x^4+1}+x\sqrt{x^4+1}+\sqrt[4]{x^4+1}^3}}\\ &=0-0\\ &=0
\end{align*}


答:
An argument close to the OP's idea could proceed as follows. Fill in the details.

If $x$ and $y$ are elements of order $q$, then among the products $x^iy^j$, $0\le i,j<q$, there must be repetitions. This is because $q^2>|G|$. Show that this implies that the subgroup $H$ of order $q$ is unique. Let's fix a generator $x$ of $H$.

If $z\notin H$ has order $pq$ then $G$ is cyclic. Therefore the remaining possibility is that all such elements $z$ have order $p$.

Because $H$ is a unique subgroup of its order, $H\unlhd G$. Why does it follow that $zxz^{-1}=x^i$ for some $i, 1\le i<q$?

Why do we have $z^pxz^{-p}=x$?

On the other hand we also have $z^pxz^{-p}=x^{i^p}$, why? Why does this imply the congruence $$i^p\equiv1\pmod q?$$

It follows that the coset of $i$ in the multiplicative group $\Bbb{Z}_q^*$ has either order $1$ or order $p$. Why?

If the order of the coset of $i$ is equal to one, then $zx$ has order $pq$. Why?

If the order of the coset of $i$ is equal to $p$, why does it follow that $p\mid q-1$?

Title: Finding the structure of a group without using sylows theorem.

Body:
If$ |G|=pq $ and $p $ doesnt divide $(q-1)$ and $p <q$ then $G$ is cyclic.

My proof stands as this .I have used the fact that

$(i)$ If $|G|=pq$ then I showed that there will only be one element of order $p$ and one element of order $q$.

My approach in proving this part has been to show that if there are two elements of order $p$ say $x_1$ and $x_2$ , then say $H_1$ is a group of order $p$ generated by $x_1$ and $H_2$ is a group of order $p$ generated by $x_2$ .We assume that the intersection is {e} if not then we can get $H_1$=$H_2$(by property of subgroup).

Proving that the elements are distinct .

We assume that the elements $(x_1)^{i}.(x_2)^j$ are not distinct then $(x_1)^{i}.(x_2)^j =(x_1)^{i'}(x_2)^{j'}$.From here we can arrive at a contradiction as $H_1 \cap H_2 =e$.So if there are $p^2$ elements then we can arrive at a contradiction as $p$ and $q$ are both primes.

Similar results will hold in the case of $q$.

$(ii)$ Now  there is only one subgroup of order $p$ and one subgroup of order $q$ so they are both normal

$(iii)$ let $H$ and $K$ be two subgroups of order $p$ and order $q$.Then we know that $H \cap K={e}$.$H$ and $K$ are both normal .Then I showed that $x^{-1}y^{-1}xy \in H \cap K$ and $xy=yx$. So the order of the element $xy$ is $pq$.Where am I going wrong in my proof and since I have not used the fact that $p $ doesnot $q-1$.


问:
Solving limit using both a substitution and L'Hôpital's Rule
While evaluating this limit to find a, b, and c; $$\lim_{x \to 0}\frac{axe^x-b\ln(1+x)+cxe^{-x}}{x^2\sin x}=2$$

I first substituted $\frac{\ln(1+x)}{x}=1$ in the numerator, and then used L'Hôpital's Rule twice to get the expression $\frac{ae^x + ce^{-x}}{\cos x -x\sin x +\cos x}$. I also derived that a+c=b, and that a=c, from the numerators of the intervening functions.

But the values I got for $a,b$ and $c$ weren't correct. Am I not supposed to use the limit substitution and LHR simultaneously? If so, why?

答:
Expand numerator and denominator with MacLaurin. We get
$$\frac{x (a-b+c)+x^2 \left(a+\frac{b}{2}-c\right)+\frac{1}{6} x^3 (3 a-2 b+3 c)+O\left(x^4\right)}{x^3+O\left(x^4\right)}$$
As we know that the limit is $2$ the following must be true
$$
\begin{cases}
a-b+c=0\\
a+\frac{b}{2}-c=0\\
\frac{1}{6} (3 a-2 b+3 c)=2
\end{cases}
$$
Solve the system to get $a = 3, b = 12, c = 9$.


问:
Show that $\lim\limits_{n \to +\infty}(\sin(\frac{1}{n^2})+\sin(\frac{2}{n^2})+\cdots+\sin(\frac{n}{n^2})) = \frac{1}{2}$
Show that the sequence defined as
$$x_n = \sin\left(\frac{1}{n^2}\right)+\sin\left(\frac{2}{n^2}\right)+\cdots+\sin\left(\frac{n}{n^2}\right)$$
converges to $\frac{1}{2}$.

My attempt was to evaluate this limit by using squeeze theorem. I managed to show that $x_n < \frac{n+1}{2n}$ by using $\sin(x) < x$, but I haven't been able to find a sequence smaller than $x_n$ that also converges to $\frac{1}{2}$. I tried showing by induction that $x_n > \frac{1}{2}-\frac{1}{n}$, but I got nowhere with that.

Any help would be appreciated.

答:
Start with $x-x^3/6<\sin(x)<x$; there are many nice proofs here.

Then we have
$$
\sum_{k=1}^{n}(k/n^2)-(k/n^2)^3/6 <\sum_{k=1}^{n}\sin(k/n^2) < \sum_{k=1}^{n}k/n^2
$$Using the fact that $\sum_{k=1}^{n}k^p = \frac{n^{p+1}}{p+1}+\text{ lower order terms}$, we have
$$
\frac{n^2+\cdots}{2 n^2}- \frac{n^4+\cdots}{24 n^6} <\sum_{k=1}^{n}\sin(k/n^2)  < \frac{n^2+\cdots}{2 n^2},
$$and the result follows by squeezing. We can actually do a bit better with the sums using closed-form identites:
$$
\frac{n^2+n}{2 n^2}- \frac{n^4+2n^3+n^2}{24 n^6} <\sum_{k=1}^{n}\sin(k/n^2)  < \frac{n^2+n}{2 n^2}
$$


答:
If you do not know Taylor series, use the equivalent
$$\sin(x)\sim x \qquad \text{when} \qquad x \text{ is small}$$

So
$$x_n \sim \frac {1} {n^2}+\frac {2} {n^2}+\cdots+\frac {n-1} {n^2}+\frac {n} {n^2}=\frac {1+2+\cdots+(n-1)+n} {n^2}$$

I am sure that you can take it from here.

Title: Show that $\lim\limits_{n \to +\infty}(\sin(\frac{1}{n^2})+\sin(\frac{2}{n^2})+\cdots+\sin(\frac{n}{n^2})) = \frac{1}{2}$

Body:
Show that the sequence defined as
$$x_n = \sin\left(\frac{1}{n^2}\right)+\sin\left(\frac{2}{n^2}\right)+\cdots+\sin\left(\frac{n}{n^2}\right)$$
converges to $\frac{1}{2}$.

My attempt was to evaluate this limit by using squeeze theorem. I managed to show that $x_n < \frac{n+1}{2n}$ by using $\sin(x) < x$, but I haven't been able to find a sequence smaller than $x_n$ that also converges to $\frac{1}{2}$. I tried showing by induction that $x_n > \frac{1}{2}-\frac{1}{n}$, but I got nowhere with that.

Any help would be appreciated.


问:
IMO 2019 Problem N5 Solution 2
IMO 2019 Problem N5 Solution 2.

https://www.imo-official.org/problems/IMO2019SL.pdf

Let $a$ be a positive integer. We say that a positive integer $b$ is $a$-good if $\binom{an}{b}-1$ is divisible by $an+1$ for all positive integers $n$ with $an\geqslant b$. Suppose $b$ is a positive integer such that $b$ is $a$-good, but $b+2$ is not $a$-good. Prove that $b+1$ is prime.

Solution 2. $\quad$ We show only half of the claim of the previous solution: we show that if $b$ is $a$-good, then $p\mid a$ for all primes $p\leqslant b$. We do this with Lucas' theorem.
$\quad$ Suppose that we have $p\leqslant b$ with $p\not\mid a$. Then consider the expansion of $b$ in base $p$; there will be some digit (not the final digit) which is nonzero, because $p\leqslant b$. Suppose it is the $p^t$ digit for $t\geqslant 1$.
$\quad$ Now, as $n$ varies over the integers, $an+1$ runs over all residue classes modulo $p^{t+1}$; in particular, there is a choice of $n$ (with $an\gt b$) such that the $p^0$ digit of $an$ is $p-1$ (so $p\mid an+1$) and the $p^t$ digit of $an$ is $0$. Consequently, $p \mid an+1$ but $p\mid \binom{an}{b}$ (by Lucas’ theorem) so $p\not\mid\binom{an}{b}-1$. Thus $b$ is not $a$-good.
$\quad$ Now we show directly that if $b$ is $a$-good but $b+2$ fails to be so, then there must be a prime dividing $an+1$ for some $n$, which also divides $(b+1)(b+2)$. Indeed, the ratio between $\binom{an}{b+2}$ and $\binom{an}{b}$ is $(b+1)(b+2)/(an-b)(an-b-1)$. We know that there must be a choice of $an+1$ such that the former binomial coefficient is $1$ modulo $an+1$ but the latter is not, which means that the given ratio must not be $1\ \text{mod}\ an+1$. If $b+1$ and $b+2$ are both coprime to $an+1$ then the ratio is $1$, so that must not be the case. In particular, as any prime less than $b$ divides $a$, it must be the case that either $b+1$ or $b+2$ is prime.
$\quad$ However, we can observe that $b$ must be even by insisting that $an+1$ is prime (which is possible by Dirichlet’s theorem) and hence $\binom{an}{b}\equiv (-1)^b=1$. Thus $b+2$ cannot be prime, so $b+1$ must be prime.

I struggle with "$an+1$ is prime (which is possible by Dirichlet’s theorem)". For Dirichlet's theorem to be applicable here, there has to be an arithmetic progression of numbers $an+1$ for which $\binom{an}{b}\equiv1$ and $\binom{an}{b+2}\not\equiv1$ modulo $an+1$. Because $b$ is $a$-good and $b+2$ is not, we are guaranteed that at least one such $n$ exists. But I don't see how a single $n$ leads to a whole arithmetic progression of such numbers.

Any ideas? Thanks.

答:
Now, as $n$ varies over the integers, $an+1$ runs over all residue classes modulo $p^{t+1}$; in particular, there is a choice of $n$ (with $an > b$) such that the $p^0$ digit of $an$ is $p-1$ (so $p | an+1$) and the $p^t$ digit of $an$ is $0$.

The possible values of $n$ here are in an arithmetic progression. Indeed every $n$ denotes a class here.

class 0: $1 \pmod{p^{t+1}}$

class 1: $a+1 \pmod{p^{t+1}}$

class 2: $2a+1 \pmod{p^{t+1}}$

class 3: $3a+1 \pmod{p^{t+1}}$

...

class S: $(????0????????(p-1))\pmod{p^{t+1}}$ where the $p^t$ digit of $an$ is 0

...

The class S above satisfies the hypothesis of Dirichlet's theorem: it is an arithmetic progression with step $p^t$ and first term coprime with $p^t$.

Then we can choose $aS+1$ as our class. Every term of it - including the (infinitely many) prime terms of it - satisfies the construction.

Title: IMO 2019 Problem N5 Solution 2


问:
A limit of integrals
Let $f:[0,T]\to\mathbb{R}$ be a Lebesgue integrable function. For each $h>0$ we define the piecewise function $f_h$ by
$$f_h(t)=f(h\left[\frac{t}{h}\right])\quad\mbox{for}\quad t\in[0,T].$$

Can we affirm that
$$\lim_{h\to 0}\int_0^Tf_h(t)dt=\int_0^Tf(t)dt ?$$

*N.B :*$\left[ x\right]$  is the entire part of the real $x$.

答:
Let $T=1$ and let $f:[0,1]\to\mathbb R$ be the function which takes the value $0$ on rational numbers and $1$ on the irrational ones.

If $n\in\mathbb N$, then $f_{1/n}$ is the zero function. It follows easily from that that $\liminf\limits_{h\to0}\int_0^1f_h=0$ and therefore your equality cannot be true.

ContentLicense: CC BY-SA 3.0

Title: A limit of integrals


问:
A circle with diameter $AC$ is intersected by a secant at points $B,D$. The secant and the diameter intersect at point $P$ outside the circle.

A circle with diameter $AC$ is intersected by a secant at points $B,D$. The secant and the diameter intersect at point $P$ outside the circle. Perpendiculars $AE$ and $CF$ are drawn at the secant. If $EB = 2$ and $BD = 6$, find $DF$.

What I Tried: Here is a picture :-

Unfortunately I have no idea to start with this. A first idea is to find similar triangles, which I find a lot, but I cannot find a way on how to use the lengths of $EB$ and $BD$.

Can anyone help me? Thank You.

答:
$\angle ADE = \angle DCF\Rightarrow \triangle AED \sim \triangle DFC $
$$\Rightarrow \dfrac{AE}{DF} = \dfrac{AD}{DC} $$

$\angle ABE = \angle ACD\Rightarrow\triangle AEB \sim \triangle ADC $
$$\Rightarrow \dfrac{AE}{EB} = \dfrac{AD}{DC} $$

$$\therefore DF=EB =2$$


答:
Denote the center of the circle as $O$. Draw a perpendicular line from $O$ to the secant, and denote the point of intersection $Q$.

We can prove that $BQ = QD$. Moreover $EA // QO // FC$.

Since $AO = OC$ as radii, $EQ = QF$ by intercept theorem.

Now $FD = QF - QD = EQ - BQ = EB = 2$.

Body:
A circle with diameter $AC$ is intersected by a secant at points $B,D$. The secant and the diameter intersect at point $P$ outside the circle. Perpendiculars $AE$ and $CF$ are drawn at the secant. If $EB = 2$ and $BD = 6$, find $DF$.

What I Tried: Here is a picture :-

Unfortunately I have no idea to start with this. A first idea is to find similar triangles, which I find a lot, but I cannot find a way on how to use the lengths of $EB$ and $BD$.

Can anyone help me? Thank You.


答:
For a given $x_n$, the largest argument inside of $\sin$ is $1/n$. You can bound $\sin$ on $[0,1/n]$ by below with the linear function that passes by $(0,0)$ and $(1/n,\sin(1/n)$. You can therefor bound $x_n$ by below with

$$y_n=(n\sin(1/n))(\sum_{k=1}^n \frac{k}{n^2})=\frac{\sin(1/n)(n+1)}{2}. $$

As $n \to \infty$, $y_n \to 1/2$.

Hope this helps!


问:
Determine the set of positions of $ M$
let M be the affix of $Z$ Determine the set of positions of $ M$ when : $$ Z=2\sin(\theta)e^{i({\theta-\frac{\pi}{2}})}  , \ \ \ with  \ \ \ \ \theta \in [0,\pi] $$
I need indication to finding a solution to the exercise

答:
$$2 e^{i \left(t-\frac{\pi }{2}\right)} \sin (t)=2 e^{i \left(t-\frac{\pi }{2}\right)} \left(\frac{1}{2} i e^{-i t}-\frac{1}{2} i e^{i t}\right)=1-e^{2 i t},\;t\in[0,\pi]$$
Represents a circle $radius=1$ and center $(0,1)$

Title: Determine the set of positions of $ M$


问:
$f$ meromorphic prove that $f$ is not constant
I'm struggling with the following problem. Any help would be a really welcomed.

Let $f$ a meromorphic function on $\mathbb C$ s.t $g(w)= f(\frac{1}{w})$ is meromorphic for $w=0$. We say that $f$ is meromorphic over $\hat{ \mathbb C}$.

I  proved that $f$ can be extended to an analytic (i.e continuous) function $\hat{f}:\hat{ \mathbb C}\rightarrow \hat{ \mathbb C}$ but I have some trouble proving:

$\hat{f}$ is onto if $f$ is not a constant.

If $\hat{f}$ is bijective, then $f$ is a Mobius function.

For 1) I tried using the compactness of $\hat{ \mathbb C}$ and the open mapping theorem but I failed.

For 2) I know that $\hat{f}$ is bijective i.e a holomorphic automorphism of $\hat{ \mathbb C}$ and that if $f$ is Mobius then $f(z)= \frac{ax+b}{cz+d},\forall a,b,c,d\in \mathbb C$ s.t $ac-bd \neq 0$ , but I don't know how I can proceed.

Thanks in advance for your help.

答:
If $f$ is not constant, it is open. Since its domain ($\hat{\mathbb{C}}$) is compact, $f$ is also closed, so $f(\hat{\mathbb{C}})$ is clopen and not empty, a contradiction since $\hat{\mathbb{C}}$ is connected: a connected set $A$ cannot have a non empty clopen subset $B$, since otherwise $A=B\sqcup A\backslash B$ with both $B, A\backslash B$ open set.

Since $f$ is meromorphic at $\infty$, it must be a rational function (every non rational function has an essential singularity at $\infty$, can you prove it?). Thus $f=\frac{P(z)}{Q(z)}$. Let us suppose that $P,Q$ are coprime. Since it is bijective, we can have only one point sent to $\infty$, i.e. one and only one pole, which implies $\text{deg}(Q)\le1$. Similarly, $\#f^{-1}(0)=1$, so $\text{deg}(P)\le 1$. We can now write $f(z)=\frac{az+b}{cz+d}$, and it is straightforward to prove that since $f$ is bijective $ad-bc\neq 0$, which gives us the claim.


问:
Derivative of Linear Map
I'm reading Allan Pollack's Differential Topology and got stuck on this argument: In the second paragraph of page 9, section 1.2 he said

"Note that if $f:U\to \mathbf{R^m}$ is itself a linear map $L$, then $df_x=L$ for all $x\in U$. In particular, the derivative of the inclusion map of $U$ into $\mathbf{R^n}$ at any point $x\in U$ is the identity transformation of $\mathbf{R^n}$."

I don't quite understand his first sentence. How could $f=L$ and $L$ linear imply $df_x=L$? A counter example is $f(x)=x$ linear, then $f'(x)=1 \neq L$.

Thanks a lot for everyone's help!

答:
Formally, the situation is like this: there is a function $\phi:{\cal M}\to{\cal N}$ (between manifolds), and we want something to describe infinitesimal change. We associate to each $x\in{\cal M}$ a linear map, the differential, which is a transformation of tangent spaces $df_x:T_x{\cal M}\to T_{f(x)}{\cal N}$.

$\hskip 0.6in$

Image source: Wikipedia.

In Euclidean space, points themselves may be viewed as vectors and hence added or subtracted, and so the manifold and all of its tangent spaces may be identified together. That is, every tangent vector exists as a point in the original space (codomain). If $f:{\bf R}^n\to{\bf R}^m$ is differentiable, then the differential is the "directional derivative" as a linear function of the "direction." Explicitly, the matrix of this linear map $df_x$ is given by the Jacobian. The write-up says that the Jacobian matrix of the linear map $L:x\mapsto Ax$ is just $A$, hence $dL_x=L$ is an equality of maps for all $x$.

In the one-dimensional case, $f:{\bf R}\to{\bf R}$, the matrix of a linear map is $1\times1$, so essentially just a scalar value. The scalar is in fact $f'(x)$, so the differential is $df_x:v\mapsto f'(x)v$. In particular, $f(x)=x$ implies $f'(x)=1$ so $df_x:v\mapsto1v$ is the identity map, i.e. the same as $f$.

Title: Derivative of Linear Map


问:
Dijkstra Algorithm recovery from long dead ends behaviour
I would like to reference this question: Is Dijkstra's algorithm possible with graphs containing dead ends?

It is somewhat similar however i need to know the Dijkstra behaviour if the dead end is many nodes on from a single edge i.e see the image link below

If the travel of minimal distance calculations continues down the direction as indicated by the arrow until it reaches the dead end, will a simple version of Dijkstra be able to recover and at what point if any will it not be able to recover, to back track through many edges to begin the correct route again?

I hope this makes sense and someone can help me understand this. Thank you. If you need any more information please ask.

Dijkstra Example illustration image

答:
Dijkstra's Algorithm is basically Dynamic Programming. The process reduces the dimension by one each step. After updating the current shortest distances, the next closed node is the one with the shortest distance. It will end up with finding the shortest distances from the origin to all other nodes. It doesn't have any issue such as dead end. If the graph has n nodes, it has to go through n steps (including the initialization step).

Of course, in your example, in case you only care about the shortest distance between the source and node $G$, you might end up with reaching $G$ before reaching the end of the long path. However, this doesn't mean Dijkstra's Algorithm fails. You need to work through a couple of examples. Once you fully understand it, you will get it.


答: If questioner or others have not encountered Taylor series yet, it is notable that an elementary trig identity (for the sin of a sum) applied to the $k$th term in the series can be shown to be $k$ times the first term in the series, thus it can be deduced that the limit of the series is equal to the limit of $$\frac{n(n+1)}{2}\cdot \sin(1/n^2)$$ Therefore, the limit becomes $$\frac{1}{2}\cdot \lim (u\cdot \sin(1/u)) = \frac{1}{2}$$


答: More generally, if $f(0)=0$ and $f$ is twice-differentiable at $0$,$$f(\tfrac{k}{n^2})\in\tfrac{k}{n^2}f^\prime(0)+O\left(\tfrac{k^2}{n^4}\right)\implies\lim_{n\to\infty}\sum_{k=1}^nf(\tfrac{k}{n^2})=\lim_{n\to\infty}\sum_{k=1}^n\tfrac{k}{n^2}f^\prime(0)=\tfrac12f^\prime(0),$$because $\sum_{k=1}^n\frac{k^2}{n^4}\sim\tfrac{1}{3n}\in o(1)$.


问:
Find PACF of sum on a die.
Problem:

A die was thrown three times. The results of the throws are $X_1,X_2,X_3$. Let $A = X_1 + X_2$, $B = X_2 + X_3$ and $C = X_1 + X_2 + X_3$.

Find $p \Bbb Corr(A,C; B)$.

Note, $p \Bbb Corr(X, Y ;Z)$ means a partial correlation between $X$ and $Y$, "cleared" of the $Z$ effect.

Is it possible to use some formula? Using the formula for Pearson partial correlation I've obtained $p \Bbb Corr(A,C; B) = \sqrt{\frac{2}{3}}$ but is it right?

答:
I believe $\sqrt{\frac{2}{3}} \approx 0.8165$ is correct

As a hand-waving check, it seems fitting $A$ against $B$ gives a coefficient of $\frac12$ to $B$, while fitting $C$ against $B$ gives a coefficient of $1$ to $B$.

So you are looking for the correlation between $A-\frac12B$ and $C-B$, i.e. between $X_1+\frac12X_2-\frac12X_3$ and $X_1$.

We do not need to know the variance for a single die is $\sigma^2 =\frac{35}{12}$, as the variance of the first term is $\sigma^2+\frac1{2^2}\sigma^2+\frac1{2^2}\sigma^2=\frac32 \sigma^2$ while the variance of the second is $\sigma^2$ and the covariance between these is the varaince of $X_1$ so is also $\sigma^2$, making the correlation $\dfrac{\sigma^2}{\sqrt{\frac32 \sigma^2\, \sigma^2}} = \sqrt{\dfrac{2}{3}}$

Title: Find PACF of sum on a die.


问:
How to solve a quadratic equation with floor function
I want to solve $\lfloor{x^2-5x\rfloor} = -6$. So far, I wrote

$$x^2-5x \ge -6\implies x^2-5x+6 \ge 0$$
and get $(x-2)(x-3) \ge 0$. Therefore, $x$ must satisfy
$$x \le 2 \text{ or } x\ge 3.$$
However, I do not know what to do next. I also tried writing $x = k + \alpha$ for some integer $k=\lfloor x \rfloor$ and $\alpha \in [0,1)$. Then, plug $k+\alpha$ into the equation $x^2-5x$ but could not get something meaningful.

How can I solve this equation?

答: By the definition of the floor function, if $\lfloor x^2-5x\rfloor=-6$, then it is necessarily the case that $-6\leq x^2-5x<-5$. You can, of course, consider this inequality as two separate inequalities, namely $-6\leq x^2-5x$ (which you've already correctly solved), and $x^2-5x<-5$. The set of values of $x$ which solve your original inequality may then be obtained by taking the intersection of the solution sets to the two aforementioned inequalities. Do you reckon you can complete the question using this fact?


问:
I think that's true but I don't know how I can prove it logically.
Firstly I got it in my exam so I didn't know how I can prove that this sum cannot equal 2 or more.

By: $0\lt a \lt 1$

$\sum_{k=0}^n\ (a^k) \lt 2$

答:
You can't because it's not true. Take $a=\frac{2}{3}$, and $n=2$.
\begin{align*}
\sum_{k=0}^na_k&=\sum_{k=0}^2 \left(\frac{2}{3}\right)^k\\
&=1+\frac{2}{3}+\frac{4}{9}\\
&=\frac{19}{9}\\
&>2
\end{align*}

You tagged this with Cauchy Sequences; what you could do is notice that your sum is the $n^{th}$ partial sum for the sequence $a^k$. Since $0<a<1$, this sequence of partial sums is convergent, and therefore Cauchy. This implies the differences between terms gets small, so you could get $\left | \sum_{k=0}^na^k - \sum_{k=0}^ma^k\right|<2 $ for $n,m>N$ for sufficiently large $N$.


问:
Riemann integral with intervals?
Let $f(x) = \begin{cases} 3 && 0 \leq x \leq 1 \\ 0 && 1 \leq x  \leq 2 \end{cases}$

Compute $\displaystyle \ \ \int_0^2 f(x)dx\,\,\,$.

You can use the definition of Riemann integral to compute this where there is an $inf$ and a $sup$.

I was looking at this question and was wondering how would you prove this. I did not learn this but I want to know how this proof would look like. It seems really good to know.

From reading an article I came up with some of this which might help:

$P_n$ $=$ {$0$ , $1$-$({1/n})$, $1$+$({1/n})$, ...} where $n$ $=$ $2$, $3$, ...

I am just thinking what the proof should look like for fun. Can someone please show me? I would like to see the proof and then learn it.

This seems like a good place to learn it.

答:
The Riemann integral is defined in the following way:

$$\int_{a}^{b}f(x)\text{d}x = \lim_{n\rightarrow\infty}\sum_{i=1}^{n}f(x_{i}^{*})\Delta x_{i}.$$

In the Riemann sum on the right-hand side $f(x_{i}^{*})$ is a sample point in the interval $(x_{i-1},x_{i})$ and $\Delta x_{i}$ is the width of that interval (which must go to zero as $n\rightarrow \infty$ for every $i$).

I'm going to partition the interval into $2^m$ equal sub-intervals and I will choose my sample points in the middle of each interval. Then:

$$\int_{0}^{2}f(x)\text{d}x = \lim_{m\rightarrow\infty} \sum_{i=1}^{2^{m}} f(x_{i}^{*})\frac{2}{2^{m}}$$
$$ = \lim_{m\rightarrow \infty}\left(\sum^{2^{m-1}}_{i=1}3\frac{2}{2^{m}} + \sum_{i=2^{m-1}+1}^{2^{m}}0\frac{2}{2^{m}} \right)$$
$$ = \lim_{m\rightarrow\infty} \left( 3 \times 2^{m-1} \times \frac{2}{2^{m}}\right) = 3$$
which is exactly what we would expect since the area under that graph is just the area of a rectangle with width $1$ and height $3$.

Title: Riemann integral with intervals?


问:
Convolution of a function and its Fourier transform
I would like to analytically evaluate the following
$$
\left(g*\hat g\right)(x)
$$
where I have defined $g(k)=e^{\frac i 3 (2\pi k)^3}$ and hence its Fourier transform is
$$
\mathcal F[g(k)](x)=\hat g(x)=\text{Ai}(x).
$$
Are there any tricks that I could use to evaluate this for $x\in\mathbb R$?

Note that
$$\left(g*\hat g\right)(x)=\mathcal F[\hat g(k) g(k)](x).$$

答:
The answer is of the form

$$c_0\text{Ai}(x)+c_1\text{Ai}'(x)$$

The coefficients can be found by evaluating the integrals at x=0.


问:
Free group of rank-$2$ has no non-trivial abelian normal subgroup
I am doing this problem and write a solution. Can someone help me by giving some feedback? Thanks in advance.

Let $p:X\to \Bbb S^1\lor\Bbb S^1$ be a covering with $X$
path-connected and $\pi_1(X)$ is non-trivial abelian. Then, $p:X\to \Bbb S^1\lor\Bbb S^1$ is irregular.

$X$ is a graph, as it is a covering of a graph.  The fundamental group of $X$ is a free group with some number of generators and this number is the same as the number of edges outside a maximal tree of $X$.

Also, $\pi_1(X)$ is non-trivial abelian implies there is exactly one edge outside a maximal tree of $X$, i.e. $\pi_1(X)=\Bbb Z$.

So, $X$ has the following form: Consider a circle and choose
$n$-points on this circle, and add exactly two infinite-tree at
each point. For $n=1,2,3$ some coverings are given.

Now, in each of these images consider two vertices, bold black, and bold green. There is no deck-transformation that interchange bold black and bold green vertices: after removing the green one there is a component with loop, on the other hand removing the black one each component is contractible.

Edit After Lee Mosher's Answer

答:
The examples that you have worked out exhibit a pattern, of which you have stated some special cases. Here's the general pattern:

Proposition: If $X$ has an abelian fundamental group $\pi_1(X)$, then $X$ is homeomorphic to the quotient of a circle $C$ and a finite collection of trees $T_1,...,T_k$, by choosing one vertex $x_i \in T_i$ on each tree and a collection of distinct vertices $y_i \in C$, and identifying $x_i \sim y_i$ for each $i=1,...,k$.

And although in your application the graph $X$ is a connected covering space of $Y$, in fact this proposition holds in general for any connected graph $X$. The proof is not hard, and uses only facts about connected graphs and their fundamental groups.

Furthermore, once you've proved that proposition, then you can easily prove the additional fact that any graph automorphism of $X$ takes $C$ to $C$ (Hint: For each open edge $e \subset X$, $X-e$ is connected if and only if $e \subset C$). Thus, knowing in your special case that the set of trees $T_i$ is not trivial, you can conclude that the group of graph automorphisms of $X$ does not act transitively on the vertex set of $X$. It follows that the deck transformation group of the covering map $X \mapsto Y$ does not act transitively on the vertex set of $X$.


问:
Number of ways to express $n$ as a sum of squares of integers
Let $n$ be a positive integer.

The question is, how many ways are there, to write $n$ as
$\sum_i a_i^2$, where $a_i$ are some positive integers.

The lenght of $a_i$ is irrelevant so that one could formulate this abstractly as follows.
Let $$P_n := \{ a = (a_i)_{i\in \mathbb{N}} \in \mathbb{N}^{(\mathbb{N})} | \sum_{i} a_i^2 = n \text{ and } a_j = 0 \implies a_k = 0 \ \forall \ k \geq j \}.$$ Then, what is the size of $P_n$?

(I write $\mathbb{N}^{(\mathbb{N})}$ for sequences with values in $\mathbb{N}$ that are eventually $0$).

So far, I have a recursive formula:

$P_n = \coprod_{x \leq \sqrt{n}} P_{n - x^2}$

With this, one could compute the exact value for a given number, but the number of terms gets large quickly. (I wrote a basic program but the results neither enable me to find a formula, nor to reach numbers greater than ~60).

If a closed formula for the size of $P_n$ seems to be difficult, I would also be interested in upper bounding it, such that the upper bound is not to far off.

I couldn't find anything in literature (e.g. by using Google) but maybe I am missing the correct words, so I would also appreciate any hints on that direction.

答:
If $p_n$ is the cardinality of $P_n$ then $\log_2 p_n$ is approximately $\frac n2$.  I wrote a little python script to compute the values and graph them.

```<code>from math import log2
from matplotlib import pyplot as plt

P = [1, 1]
for n in range(2,1001):
    a = 0
    k = 1
    while k*k <= n:
        a += P[n-k*k]
        k += 1
    P.append(a)
y = [log2(x) for x in P[1:]]
x = list(range(1,1001))

plt.plot(x,y)
plt.show()
</code>```

This produced

This looks like a good starting point to me.

Body:
If $p_n$ is the cardinality of $P_n$ then $\log_2 p_n$ is approximately $\frac n2$.  I wrote a little python script to compute the values and graph them.

from math import log2
from matplotlib import pyplot as plt

P = [1, 1]
for n in range(2,1001):
    a = 0
    k = 1
    while k*k <= n:
        a += P[n-k*k]
        k += 1
    P.append(a)
y = [log2(x) for x in P[1:]]
x = list(range(1,1001))

plt.plot(x,y)
plt.show()

This produced

This looks like a good starting point to me.


问:
On Suprema and limits
Let be $f:I\to\mathbb{R}$ a continuous function, $[x,x_0]\subseteq I$ and $M:=\sup\{f(y)\mid y\in[x,x_0]\}$. Show that the supremum attains the value of $f(x_0)$ when $x\to x_0$.

My original idea was to argue with the rules of limits:

By definition of the supremum we know that for each $\epsilon>0$ there exists a $y\in [x,x_0]$ such that $f(y)<M<f(y)+\epsilon$. Then $x\to x_0$ leads to
$$\implies y\to x_0\implies\lim\limits_{y\to x_0}f(y)=f(x_0)\\\implies f(x_0)<M<f(x_0)+\epsilon \implies M=f(x_0),$$

as we can make $\epsilon>0$ arbitrarily small.

However, when I submitted this answer my tutor put a comment saying that this conclusion is not completely wrong (and maybe intuitively right) but not quite rigorous. Further he asked which specific rule would allow me to do that and finally deducted points.

Second attempt with $\epsilon$-$\delta$-criterion:

Again by definition of the supremum we know that for each $\epsilon>0$ there exists a $y\in [x,x_0]$ such that $f(y)<M<f(y)+\epsilon$. So if I now consider $x\to x_0$ then for an arbitrarily small $\epsilon>0$ there exists a $\delta>0$ such that for all $x$ with $|x-x_0|<\delta\implies |f(x)-f(x_0)|<\epsilon$. As $y\in [x,x_0]$ it follows $f(x_0)-\epsilon<f(y)<M<f(y)<f(x_0)+\epsilon$. Hence $M=f(x_0)$.

I don't understand what the problem was with my first attempt? Is my second one right?

EDIT:

Beside my both approaches I am wondering if I am able to simply argue $\lim\limits_{x\to x_0}M=\sup\{f(y)\mid y\in[x,x_0]\}=\sup\{f(y)\mid y\in[x_0,x_0]\}=f(x_0)$ because if I shrink the interval $[x,x_0]$ then $x_0$ is the only value that's left. However in this approach continuity doesn't play a role which seems strange!?

答:
The problem in your first attempt is that, $y$ depends on $\varepsilon .$ Roughly speaking, you can regard this dependence as a one-to-one correspondence(N.B. this saying may not be logically correct because some $y$ may corresponds to many different $\varepsilon .$ But what I want to say is that, when you change $y$, $\varepsilon$ may be changed too.) So, when you send $y$ to $x_0$, $\varepsilon$ may change too. And it's quite hard to manage two changes simultaneously. And actually the real problem is that, you need to fix $\varepsilon$ first when $y\to x_0$, but this can not be done as what I just illustrated.

While in your second attempt, you send $\varepsilon$ to $0$, which forces $f(y)=f(x_0)$ by $$ f(x_0)-\epsilon<f(y)<M<f(y)+\varepsilon<f(x_0)+2\epsilon .$$

The reason why you can let $\varepsilon \to 0^+$ is that $x\to x_0,$ which promises that $|x-x_0|<\delta$ for any $\delta >0.$

You don't need $f(y)$ is fixed when $\varepsilon \to 0^+$.

So this is just the essential difference between your first attempt and the second one.

Title: On Suprema and limits


问:
Monte carlo methods and the curse of dimensionality for solving PDEs
Do Monte Carlo methods suffer from the curse of dimensionality when solving PDEs? I was under the impression that they do not, since the variance scales like $\sigma /\sqrt{N}$, and thus it only depends on the number of samples and not on the number of dimensions.

However, I've read in several references (for example in this one), that direct Monte Carlo methods for simulating the Fokker-Planck equation also suffer from the curse of dimensionality, hence my question. I'm not sure I understand where this dependence is coming from, are there any good references about this topic? I would like to know how this type of Monte Carlo algorithms to solve Fokker-Planck equations scale with the dimensions.

Also, is there any Monte Carlo method to solve this type of equations that doesn't suffer from the curse of dimensionality?

Thank you very much.

答: If your discrete simulation of the system in some region of $\mathbb R^d$ uses a rectangular grid with $N$ points in each coordinate direction, that's $N^d$ points, and thus $N^d$ values that must be all be updated in each iteration.  So right there you have a factor in the complexity that is exponential in $d$.  That's the curse of dimensionality.  It has little to do with the details of the particular method you are using.


问:
Find an appropriate transformation
Consider

$$-\frac{d}{dx}(p(x)y’)+q(x)y=f(x),0\le x \le 1,y’(0)=a, y’(1)=b$$
I have to transform this boundary value problem into

$$-\frac{d}{dx}(p(x)z’)+q(x)z=F(x),0\le x \le 1,z’(0)=0, z’(1)=0$$

I have spent hours on hours on this problem and still could find the right transformation.

It would be great if someone could help me with this
Thanks in advance

答:
If we set

$$
z(x) := y(x) + \underbrace{\left[ (a-b) \frac{x^2}{2} -ax\right]}_{h(x)},
$$

then

$z'(0)=y'(0)-h'(0)=a-a = 0$;

$z'(1) = y'(1)-h'(1)=b -b = 0$.

As regards the equation:

$$
\begin{split}
-(pz')'+qz &= -(p(y+h)')' +q(y+h) \\
&=-\left(py'+ph'\right)' + qy + qh\\
&= \underbrace{-(py')'+qy}_{=f(x)} - (ph')'+qh\\
&= f(x) + q(x)h(x) -p'(x)h'(x)-p(x)h''(x)\\
&= f(x) + q(x)\left[ (a-b) \frac{x^2}{2} -ax\right] -\left[(a-b)x-a\right]p'(x)-(a-b)p(x)
\end{split}
$$
So,

$$
F(x) = f(x) + q(x)\left[ (a-b) \frac{x^2}{2} -ax\right] -\left((a-b)x-a\right)p'(x)-(a-b)p(x)
$$

Example:

As an example, if $p \equiv 1$ and $q \equiv 0$, then the initial problem

$$
\begin{cases}
y''(x) = f(x) \\
y'(0) = a\\
y'(1) = b\\
\end{cases}
$$

is transformed into

$$
\begin{cases}
z''(x) = f(x) -(a-b)\\
z'(0) = 1\\
z'(1) = 1\\
\end{cases}
$$

Title: Find an appropriate transformation


问:
Is my limit proof ok? , with $f(x)^2$
Thank you for your time!

I would like to check if my proof is ok here:

Prove: Let $\lim_{x\to\infty} f(x)= 4$. Prove that for every $M>0, x>M$, $15.5 < [f(x)]^2 < 16.5$.

So, I said that: $-\sqrt{15.5} < f(x) < \sqrt{16.5}$

$ |f(x)-L| < \epsilon$,

I chose $\epsilon = 0.05$, and than $-0.05 < f(x)-4 < 0.05$

From here, $3.95 < f(x) < 4.05$ and $15.6025 < [f(x)]^2 < 16.4025$

So $15.5 < [f(x)]^2 < 16.5$

Is that legit?

Thanks a lot!

答:
I suspect the question should read, prove that if $\lim_{x\to\infty} f(x) = 4$ then there exists a number $M$ such that for all $x > M$,
$$15.5 < f(x)^2 < 16.5.$$

With this understanding, by definition of the concept of limit, for every $\varepsilon > 0$ there exists $M$ (dependent on $\varepsilon$) such that when $x > M$, $|f(x) - 4| < \varepsilon$.  That means,
$$ 4-\varepsilon < f(x) < 4 + \varepsilon.$$
Now you can square the inequality provided all terms are positive;  that will be the case if $\varepsilon$ is small enough, and so,
$$16 - 8 \varepsilon + \varepsilon^2 < f(x)^2 < 16 + 8 \varepsilon + \varepsilon^2.$$
Now just choose $\varepsilon$ so small that the left extreme is greater than $15.5$ and the right extreme is less than $16.5$.  For example $\varepsilon = 0.01$.  Then with the corresponding $M$ the required statement is true.

Of course you can get many different values for $M$ but your are only asked to show one exists.  Once you have on, any larger $M$ would also work.  Some smaller $M$ might also work depending on how finely you chose the first one.

Title: Is my limit proof ok? , with $f(x)^2$


答:
The solutions of the given equation are the solutions of the following system of inequalities
$$\begin{cases}
x^2-5 x<-5\\
x^2-5 x\geq -6\\
\end{cases}
$$
That is
$$\frac{1}{2} \left(5-\sqrt{5}\right)<x\leq 2\lor 3\leq x<\frac{1}{2} \left(\sqrt{5}+5\right)$$

Title: How to solve a quadratic equation with floor function

Body:
I want to solve $\lfloor{x^2-5x\rfloor} = -6$. So far, I wrote

$$x^2-5x \ge -6\implies x^2-5x+6 \ge 0$$
and get $(x-2)(x-3) \ge 0$. Therefore, $x$ must satisfy
$$x \le 2 \text{ or } x\ge 3.$$
However, I do not know what to do next. I also tried writing $x = k + \alpha$ for some integer $k=\lfloor x \rfloor$ and $\alpha \in [0,1)$. Then, plug $k+\alpha$ into the equation $x^2-5x$ but could not get something meaningful.

How can I solve this equation?


问:
Proving that Geometric Brownian Motion is positive
An Ito process is defined as:

$$X_t = X_0 + \int_{h=0}^{h=t}\left(a(X_h,h)\right)dh + \int_{h=0}^{h=t}\left(b(X_h,h)\right)dW_h$$

Where $a(X_t, t)$ and $b(X_t, t)$ are square integrable processes.

If one chooses $a(X_t,t)=\mu X_t$ and $b(X_t,t)=\sigma X_t$, with $\mu$ and $\sigma$ being constant, we obtain the well known Geometric Brownian Motion (where $\mu$ can be regarded as the "drift" and $\sigma$ as the "volatility"):

$$X_t = X_0 + \int_{h=0}^{h=t}\left(\mu X_h\right)dh + \int_{h=0}^{h=t}\left(\sigma X_h\right)dW_h$$

The above is of course still an Ito process. Ito's lemma states that for any (well-behaved), twice-differentiable function $F()$ of $X_t$ and $t$, the function $F(X_t, t)$ follows the process:

$$F(X_t,t)= F(X_0,t_0) + \int_{h=0}^{h=t}\left(\frac{\partial F}{\partial t}+\frac{\partial F}{\partial X}a(X_h,h)_{=\mu X_h}+\frac{1}{2}\frac{\partial^2 F}{\partial X^2}b(X_h,h)^2_{=\sigma^2 X_h^2}\right)dh + \int_{h=0}^{h=t}\left(\frac{\partial F}{\partial X}b(X_h,h)_{=\sigma X_h}\right)dW_h$$

Using the Ito's lemma above, we can easily find the solution to the GBM equation for $X_t$ by taking $F(X_t,t)=ln(X_t)$ to obtain:

$$X_t=X_0e^{\left(\mu t -0.5 \sigma^2 t + \sigma W_t\right)}$$

Question: I just got asked how come we can take the log of $X_t$, without knowing whether the equation $X_t = X_0 + \int_{h=0}^{h=t}\left(\mu X_h\right)dh + \int_{h=0}^{h=t}\left(\sigma X_h\right)dW_h$ potentially produces negative values for $X_h$ along some specific paths.

Looking at the solution $X_t=X_0e^{\left(\mu t -0.5 \sigma^2 t + \sigma W_t\right)}$, it is obvious that for any initial positive $X_0$, $X_t$ cannot be negative. However, using this argument to justify taking the log is a bit sketchy because we are using the solution to validate the method we use to derive that solution.

Is there a way to show that the GBM SDE for $X_t$ produces strictly positive values if the initial value $X_0>0$ for all $X_h: 0\leq  h \leq t < \infty$?

I suppose we could start by writing out the integrals by definition:

$$X_t = X_0 + \int_{h=0}^{h=t}\left(\mu X_h\right)dh + \int_{h=0}^{h=t}\left(\sigma X_h\right)dW_h= \\= \mu \lim_{n\to\infty}\sum_{h=t_0}^{h=t_{n-1}}X_h((h+1)-h)+\sigma \lim_{n\to\infty}\sum_{h=t_0}^{h=t_{n-1}}X_{h}(W_{h+1}-W_{h})=\\=\mu \lim_{n\to\infty} \left(x_0(t_{1}-t_{0})+...+x_{t_{n-1}}(t_n-t_{n-1}) \right) +\sigma \lim_{n\to\infty}\left(x_0(W_{t_1})+...+x_{t_{n-1}}(W_{t_n}-W_{t_{n-1}})\right)$$

Not sure where to go from here and whether the above is a good idea?

答:
We don't need to justify taking the log beforehand because we get an explicit solution, i.e. "using the solution to validate the method we use to derive that solution" is perfectly fine.  A good way to think about it is that applying Ito's lemma to $F(t,X_t) = \ln(X_t)$ doesn't give us the solution, it just gives us something we think is likely to be a solution.  When we use that to obtain $X_t = X_0 e^{\sigma W_t + (\mu - \frac 12 \sigma^2)t}$, we haven't shown that $X_t$ is actually the solution yet.  Once we apply Ito to $X_t$ we can verify that it does indeed satisfy $dX_t = \mu X_t dt + \sigma X_t dW_t$ and now can say that $X_t$ is a solution.

Note that if $X_0 < 0$ then we couldn't actually take $\ln(X_t)$, but the "candidate solution" we get from doing the same derivation is $X_t = X_0 e^{\sigma W_t + (\mu - \frac 12 \sigma^2)t}$ and we can use Ito's lemma to show that this is still a solution, even though the method we used to obtain it wasn't fully justified.


问:
Definition of a centralizer of a weight
The following definition is from Takesaki's book, Vol2, Chapter 8, Definition 2.1.

Let $\varphi$ be a faithful weight in a von Neumann algebra $M$. Set $M_{\varphi}=\{x\in M:\sigma_{\varphi}^t(x)=x,t\in \Bbb R\}$. We call $M_{\varphi}$ the centralizer of $\varphi$.

What is the explicit expression of $\sigma_{\varphi}^t(x)$ in the above definition?

答:
Here
$$
\sigma_\varphi^t(x)=\Delta_\varphi^{it} x \Delta_\varphi^{-it},
$$
where $\Delta_\varphi$ is the modular operator of the GNS representation of $\varphi$.

I have to say that if you are reading Tomita Theory and you have to ask what the modular group is, you are reading way ahead of where you should be reading.

Title: Definition of a centralizer of a weight


问:
Help approximating the leading eigenvalue of this Jacobian
I have a system of 4 ordinary differential equations with positive parameters, which has two stable fixed points. The Jacobian matrix evaluated at the first equilibrium leads easily to finding the conditions for linear stability. These conditions are necessary but are not sufficient for the second equilibrium to be stable. At the second equilibrium, the determinant of the Jacobian is:

\begin{equation*}
  \small
  \boldsymbol{J}  =
  \begin{pmatrix}
    % pds of Capital
    0 & 0 & 0 & 0 \\
    % pds of Inventory
    0 &
    - q - \frac{ b }{ 4 }&
    - \frac{ b }{ 4 }      &
    0 \\
    % pds of Demand
    0 & 0 & - c & - c \big( \frac{k b}{a + \frac{b}{2}} \big)^2 \\
    % pds of Price
    0 &
    - d \big( \frac{a + \frac{b}{2}}{\kappa b} \big)^2 &
    d \big( \frac{a + \frac{b}{2}}{\kappa b} \big)^2 &
    0\\
  \end{pmatrix}
\end{equation*}

which has characteristic polynomial $\rho(\lambda)$:

\begin{equation}
\rho(\lambda) = -\lambda \Big[ -\lambda^3 - (a + b + c) \lambda^2 - c(a+b+d) \lambda - cd(a+b) \Big]
\end{equation}

Using the Routh-Hurwitz criterion, I know the the 3rd order polynomial inside the bracket has negative roots always (all parameters are positive), which means the remaining eigenvalue must control the second equilibrium's stability. However, the remaining eigenvalue is zero, so the linearisation fails.

I have been attempting to approximate the leading eigenvalue by taking the Taylor series approximation around the point $a \approx 0$ but I haven't been having much luck. That is, making $\lambda = \lambda_0 + \lambda_{1} a + \lambda_{2} a^2 + ...$ in $\rho(\lambda)$ above, and computing $\lambda_0$, $\lambda_1$, $\lambda_2$ etc. at $a = 0$, just returns each coefficient equal to zero, so it seems that the Taylor series approximation is also failing in higher powers.

Does anyone know how I can assess the stability of the second equilirium? Thank you

答:
The fact that one row (or one column) of the matrix is $0$ implies that it is singular,
and thus has an eigenvalue of $0$.  That means that you can't decide the stability by just looking at this Jacobian.  On the other hand, you do know in what direction to look for  potential trouble: the first coordinate direction.  Seeing what happens (in the nonlinear system) to points near the equilibrium in that direction may help decide.


问:
Two commuting weights in a von Neumann algebra
Let $\varphi$ be a faithful weight on $M$. $\psi$ is a weight commuting with $\varphi$.

In Takesaki's book, Vol 2, Chapter VIII , Proposition 3.15. The author mentions that $\psi \varphi=\varphi \psi$ is equivalent to $\psi=\psi\circ \sigma_t^{\varphi}$.

How to derive the above conclusion?

答: The relation $\psi=\psi\circ\sigma_t^\varphi$ is how Takesaki defines "$\psi$ commuting with $\varphi$".


问:
How to prove $a \ln(b) \sqrt{b} > b \ln(a) \sqrt{a}$ for $a,b > 0$ and $a>b$
Visually, it is easy to see that the inequality holds for $a \neq b$ and $a,b > 0$. Do note that I'm only considering $a,b \in \mathbb{N}$ so I may miss something if they were real numbers.

I did think earlier about the inequality $\ln(x) < \sqrt{x}$ but I don't know how to prove $\ln(x)\sqrt{x} < x$, which I think would help.

答:
Hint:

The inequality holds iff
$$
\frac {\ln(b)} {\sqrt{b}} > \frac {\ln(a)} {\sqrt{a}}.
$$
Let $f(t)=\frac{\ln(t)}{\sqrt{t}}$. This function is slowly decreasing after a certain $x$, so for a large $b$ we get the result (you can compute $f'$ to see this). Otherwise, if $b$ is small enough, let's say $b=1$, we have $a>1$ and $a<1$ if the inequality still holds, which is absurd.


问:
Solve in real numbers the system of equations $x^2+2xy=5$, $y^2-3xy=-2$
Solve in real numbers the system of equations $x^2+2xy=5$, $y^2-3xy=-2$

I couldn't do this question and hence I looked at the solution which goes as follows:

if $x=0$ then $x^2+2xy=0\ne5$ hence $x\ne0$

I state that $y=lx$. Hence the system becomes:

$x^2(1+2l)=5$, $x^2(l^2-3l)=-2$ which becomes $5l^2-11l+2=0$, hence $l=2$ or $l=\frac{1}{5}$.

Hence the possible solutions are:

$(1, 2), (-1,-2), (\frac{5}{\sqrt{7}}, \frac{1}{\sqrt{7}}), (-\frac{5}{\sqrt{7}}, -\frac{1}{\sqrt{7}})$.

My question is why was I supposed to think of substituting $y=lx$? Why was this supposed to be intuitive and is there a more intuitive approach?

Body:
$x^2+2xy=5$, $y^2-3xy=-2$

Multiply the first by $2$, the second by $5$ and add
$$2 x^2-11 x y+5 y^2=0$$
Divide all terms by $y^2$ and set $\frac{x}{y}=z$
$$2z^2-11z+5=0\to z=\frac{1}{2};\;z=5$$
So we have $\frac{x}{y}=\frac{1}{2}\to y=2x$ and $\frac{x}{y}=5\to x = 5y$

Plug $y=2x$ into the first equation
$$x^2+2x(2x)=5\to x=\pm 1;\; (1,2);\;(-1,-2)$$
and then $x=5y$
$$25y^2+10y^2=5\to y=\pm\frac{1}{\sqrt 7};\; \left(\frac{5}{\sqrt 7},\frac{1}{\sqrt 7}\right);\;\left(-\frac{5}{\sqrt 7},-\frac{1}{\sqrt 7}\right)$$


问:
Basic understanding of sampling from a continuous distribution.
For continuous distribution (on R) the probability of a single point is $0$.
So I'm not sure what does it mean to sample $M$ elements from a continuous distribution.

Let say there is a continuous distribution D and there is a number z and a function f  such that:
$f(x)=1$ for $x < z$ except of a finite number of cases.

$f(x)=0$ for $x \ge z$ except of a finite number of cases.

$D(x < z) > 0  , D(x \ge z) > 0$

So if I have a random sample  $(X_1, \dots ,X_m)$ from $D$, And assume $X_1>z$ , can I conclude that $f(X_1)=0$ ?

And assume $f(X_2)=1$ , can I conclude that $X_2 < z$ ?

答:
Sampling $m$ elements simply means obtaining $m$ values from $X_i$, call them $x_i$, according to the distribution $D$, exactly as for the discrete case. If you prefer, think about it as choosing a number which has the correct probability of being in any interval.

If your observation $x_i$ of $X_i$ satisfies $x_i>z$, then with probability one it satisfies the result you give. Strictly the statement is $P(f(X_i)=0|X_i>z) =1$. That is, with probability 1 your conclusion holds. Similarly for $x_2$.

The key points are

The fact that something happens with probability 0/1 doesn't mean it's impossible/certain.

You make statements about specific observed values of random variables by referring to particular observations, and statements about probabilistic deductions you can make given random variables by using conditional probabilities.


问:
Combinatorics with repetition - is my textbook right?
I am suspicious that my textbook is incorrect.

Here's the question:

In how many ways can we distribute 20 candies among 6 children so that the youngest gets at most 2 candies?

My solution:

There are 20 candies (identical) and 6 children.

This is a combinations with repetition question so it can be represented by $x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 20$

And we can consider the cases where $x_1 = 0, 1, \text{or } 2$ where $x_1$ is the youngest child.

$x_1 = 0$: No candies for the youngest one. Then $r=20$ and $n = 5$ and so there are $\binom{20+5-1}{20}$ combinations.

$x_1 = 1$: We give one candy to the youngest one. Then $r=19$ and $n = 5$ and so there are $\binom{19+5-1}{19}$ combinations.

$x_1 = 2$: We give two candies to the youngest one. Then $r=18$ and $n = 5$ and so there are $\binom{18+5-1}{18}$ combinations.

The number of ways we can distribute the candies is $\binom{24}{20} + \binom{23}{19} + \binom{22}{18} = 26,796$.

Textbook solution:

Who is right?

答:
You are correct.  The flaw in the textbook's solution is that if the youngest child gets $0$, $1$, or $2$ candies, only the number of candies the other children may receive can vary.  In the textbook's solution, $y_1$ should be $0$ in each case.

That said, there is another way to do this problem.

If we ignore the restriction that the youngest child can receive at most two candies, the number of ways to distribute the indistinguishable candies to the children is the number of solutions of the equation
$$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 20 \tag{1}$$
in the nonnegative integers.  A particular solution corresponds to the placement of $6 - 1 = 5$ addition signs in a row of $20$ ones.  The number of such solutions is
$$\binom{20 + 6 - 1}{6 - 1} = \binom{25}{5} = \binom{25}{20}$$
since we must choose which five of the $25$ positions required for twenty ones and five addition signs will be filled with addition signs or, equivalently, which $20$ of those $25$ positions will be filled with ones.

From these, we must subtract those cases in which the youngest child receives more than two candies.  Suppose the youngest child does receive at least three candies.  Let $x_1' = x_1 - 3$.  Then $x_1$ is a nonnegative integer.  Substituting $x_1' + 3$ for $x_1$ in equation 1 yields
\begin{align*}
x_1' + 3 + x_2 + x_3 + x_4 + x_5 + x_6 & = 20\\
x_1' + x_2 + x_3 + x_4 + x_5 + x_6 & = 17 \tag{2}
\end{align*}
Equation 2 is an equation in the nonnegative integers with
$$\binom{17 + 6 - 1}{6 - 1} = \binom{22}{5} = \binom{22}{17}$$
solutions.

Hence, the number of ways the candies can be distributed to the six children so that the youngest child receives at most two candies is
$$\binom{25}{5} - \binom{22}{5} = 26,796$$
as you found.


问:
Continuous fractions theorem
$$k_n h_{n-1} - k_{n-1} h_n = (-1)^n$$

Can you guys explain to me about the continued fractions and about this particular theorem?  What does this mean?

(I have some general info about this, but it is kind of vague for me)

答:
Take a continued fraction (infinite or finite). Truncate the calculations after step $n-1$ to arrive at a rational number $\frac{k_{n-1}}{h_{n-1}}$ and truncate it after step $n$ to arrive at the rational number $\frac{k_n}{h_n}$. These rational numbers will satisfy $k_n h_{n-1} - k_{n-1} h_n = (-1)^n$ which is also equivalent to $\frac{k_n}{h_n} - \frac{k_{n-1}}{h_{n-1}}=\frac{(-1)^n}{h_n h_{n-1}}$.

This is important for a few reasons. Firstly, your equation is a linear diophantine equation. If I wanted to solve $83x-36y=1$, then we can begin by looking at the continued fraction expansion of $\frac{83}{36}=[2;3,3,1,2]$ and finding some partial convergents. We see that $[2;3,3,1]=\frac{30}{13}$ so we try $83(13)-36(30)=-1$ which actually wasn't what we wanted to solve initially....so the trick is to actually use $\frac{83}{36}=[2;3,3,1,1,1]$ and try to use the partial convergent of $[2;3,3,1,1]=\frac{53}{23}$. This leads to $83(23)-36(53)=1$ which works. So solving $ax-by=\pm1$ is solvable by looking at the continued fraction expansion of $\frac{a}{b}$.

The second equation shows the difference between successive partial convergents becomes increasingly small (as you move farther into the continued fraction expansion, the $q_n$ and $q_{n-1}$ terms will get larger). This is used to prove that infinite continued fractions actually converge to their limit and that for an irrational real number $x=[a_0;a_1,a_2,a_3,a_4,...,a_n,...]$, you'll always have $\left|x-\frac{k_n}{h_n}\right| < \frac{1}{h_n h_{n-1}}$. This is useful for finding close rational approximations to irrational numbers. For example, $\pi=[3;7,15,1,292,1,1,1,2,...]$. Stopping at $\pi \approx [3;7,15,1]=\frac{335}{113}$, you'll see that this approximation gives an error of $\left|\pi-\frac{335}{113}\right| < \frac{1}{113 \cdot 33102}$. The $33102$ came from the next stopping convergent of $[3;7,15,1,292]$. So if you find an irrational number with a big number in the continued fraction expansion, you can find a very good rational approximation using this method.

Title: Continuous fractions theorem


问:
Dini derivatives of Dirichlet function
I'm hoping someone can verify my answers. I have that for $x\in\mathbb{Q}$ we have $$D^+f(x)=\limsup_{h\to 0^+}\frac{f(x+h)-f(x)}{h}=\limsup_{h\to 0^+}\frac{f(x+h)-1}{h}$$ and since $\mathbb{Q}$ is dense in $\mathbb{R}$, $(x+h)$ is rational as $h\to 0^+$ and we have $$D^+f(x)=\limsup_{h\to 0^+}\frac{1-1}{h}=0$$  Similarly for $D^-f(x)$ we have $$D^-f(x)=\limsup_{h\to 0^+}\frac{f(x)-f(x+h)}{h}=\limsup_{h\to 0^+}\frac{1-1}{h}=0$$And for $D_+f(x)$ $$D_+f(x)=\liminf_{h\to 0^+}\frac{f(x+h)-f(x)}{h}=\liminf_{h\to 0^+}\frac{0-1}{h}=-\infty$$ Finally for $D_-f(x)$ we have $$D_-f(x)=\liminf_{h\to 0^+}\frac{f(x)-f(x+h)}{h}=\liminf_{h\to 0^+}\frac{1-0}{h}=\infty$$ Now for $x\in\mathbb{R}\setminus\mathbb{Q}$ we have that $f(x)=0$ and we have $$D^+f(x)=\limsup_{h\to 0^+}\frac{1-0}{h}=\infty$$ $$D^-f(x)=\limsup_{h\to 0^+}\frac{0-1}{h}=-\infty$$ $$D_+f(x)=\liminf_{h\to 0^+}\frac{0-0}{h}=0$$ $$D_-f(x)=\liminf_{h\to 0^+}\frac{0-0}{h}=0$$ I am having trouble convincing myself of the effect (and proper explanation) of $\liminf$ vs $\limsup$ and their impact on the derivates. Is this correct? Could someone help me better conceptually understand what is going on?

Thanks

答:
In order for $f$ to be differentiable at any $x\in\mathbb{R}$ we must have that $D^+=D^-=D_{-}=D_{+}$. Let's calculate our derivates. We must calculate separately for the cases when $x\in\mathbb{Q}$ and $x\in\mathbb{R}\setminus\mathbb{Q}$. Note that since $\mathbb{Q}$ is dense in $\mathbb{R}$ we can get arbitrarily close to zero and find a rational number.  Then for $x\in\mathbb{Q}$ we have $$D^+f(x)=\limsup_{h\to 0^+}\frac{f(x+h)-f(x)}{h}=\limsup_{h\to 0^+}\frac{f(x+h)-1}{h}$$ Now, note that $$\forall\varepsilon>0, \{f(x+h):0<h<\varepsilon\}=\{0,1\}$$ depending on whether or not $(x+h)$ is rational or not (the same can be said for $(x-h)$ in subsequent cases). and then $$\left\{\frac{f(x+h)-f(x)}{h}:0<h<\varepsilon\right\}=\{0,\frac{-1}{h}\}$$ and we have $$D^+f(x)=\limsup_{h\to 0^+}\{0,\frac{-1}{h}\}=0$$  Similarly for $D^-f(x)$ we have $$D^-f(x)=\limsup_{h\to 0^+}\frac{f(x)-f(x-h)}{h}=\limsup_{h\to 0^+}\{\frac{1}{h},0\}=\infty$$And for $D_+f(x)$ $$D_+f(x)=\liminf_{h\to 0^+}\frac{f(x+h)-f(x)}{h}=\liminf_{h\to 0^+}\{0,\frac{-1}{h}\}=-\infty$$ Finally for $D_-f(x)$ we have $$D_-f(x)=\liminf_{h\to 0^+}\frac{f(x)-f(x-h)}{h}=\liminf_{h\to 0^+}\{\frac{1}{h},0\}=0$$

Now for $x\in\mathbb{R}\setminus\mathbb{Q}$ we have that $f(x)=0$ and since $\mathbb{R}\setminus\mathbb{Q}$ is also dense in $\mathbb{R}$, following the same lines of reasoning from above we have $$D^+f(x)=\limsup_{h\to 0^+}\{\frac{1}{h},0\}=\infty$$ $$D^-f(x)=\limsup_{h\to 0^+}\{\frac{-1}{h},0\}=0$$ $$D_+f(x)=\liminf_{h\to 0^+}\{\frac{1}{h},0\}=0$$ $$D_-f(x)=\liminf_{h\to 0^+}\{\frac{-1}{h},0\}=-\infty$$ Since our derivates are not equal we have that $f$ is not differentiable for any $x$.

Title: Dini derivatives of Dirichlet function


问:
The variance of Brownian motion
Currently I'm learning about Brownian motion. In the lecture slides the following definition is given.

Definition: A Wiener process $W_{t}, t \geq 0,$ is a process with
$W_{0}=0$ and with increments $W_{t}-W_{s}$ that are Gaussian random
variables with mean $E\left\{W_{t}-W_{s}\right\}=0$ and variance
$\operatorname{Var}\left\{W_{t}-W_{s}\right\}=t-s .$ Non overlapping
increments $W_{t}-W_{s}$ and $W_{v}-W_{u}$ for any $0 \leq s<t \leq
> u<v$ are independent of each other.

Also the following properties are given:

\begin{array}{l} W_{t+1}-W_{t}=\Delta W_{t}, E\left\{\Delta
> W_{t}\right\}=0, E\left\{\Delta W_{t}^{2}\right\}=\Delta t \\
W_{t+1}=W_{+}+\sqrt{\Delta t} \cdot N(0,1), \quad W_{n}=0 \end{array}

Since the variance of a increment is t-s or Δt. I would actually think that $W_(t+1) = W_t +N(0,Δt)$. The expectation of t-s or Δt is 0. Is this the same as $W_(t+1) + \sqrt(Δt)N(0,1)$? If so, how is this converted?

答:
By definition, $W_t$ has Normally distributed independent increments with Variance proportional to the increment size, that is to say that $W(t-s)=W_t-W_s\sim N(0,t-s)$ for: $0<s<t$.

For any random variable, it is true that $Var(aX)=a^2Var(X)$

For $W_t$, we can use this property to say that $W(t-s)$ equals in distribution to $(\sqrt{t-s})W(1)$, because:

$Var((\sqrt{t-s})W(1))=(t-s)Var(W(1))=t-s$

We know that $W(1)\sim N(0,1)$, and from the above, we know that $Var((\sqrt{t-s})W(1))=t-s$, so we can conclude that $\sqrt{t-s})W(1)\sim N(0,t-s)$ and so has the same distribution as $W(t-s)$.

Title: The variance of Brownian motion


问:
Is it true that the dimension of an eigenspace of a square matrix is at most the multiplicity of the corresponding eigenvalue?
By "the multiplicity of the corresponding eigenvalue", I mean the multiplicity of the eigenvalue as a root of the characteristic polynomial $p(x)=\det(xI-A)$, i. e. the maximum integer $k$ such that $(x-\lambda)^k$ divides $p(x)$.

I believe that this is true, but I failed to find a proof.

Thanks a lot.

答:
Yes, this is true.

Hint:

To see this, take a basis of $E_\lambda(A)$ (the eigenspace of the eigenvalue $\lambda$ of the matrix $A$) and expand it in a basis of the vector space $E=\Bbb R^n$. Using the matrix of $A$ in this basis and the fact that the characteristic polynomial is the same as in the previous basis, you find that $p(x)=(\lambda I_n -x)^{\dim{E_\lambda(A)}}Q(x)$ where $Q(x)$ is a polynomial. So, by definition of the multiplicity of a root, you get $\dim{E_\lambda(A)} \le$ the multiplicity of $\lambda$ as a root of $p$.


答:
The notation on the second section of quotation and in your block of questions is not quite consistent, but the underlying answer is yes.

For a standard Wiener process, as stated in the first section of quotation, the change in a period of time $\Delta t$ is a random variable normally distributed with mean $0$ and variance $\Delta t$, i.e. standard deviation $\sqrt{\Delta t}$,

so is equal to $\sqrt{\Delta t}$ times a random variable normally distributed with mean $0$ and variance $1$

Title: The variance of Brownian motion

Body:
Currently I'm learning about Brownian motion. In the lecture slides the following definition is given.

Definition: A Wiener process $W_{t}, t \geq 0,$ is a process with
$W_{0}=0$ and with increments $W_{t}-W_{s}$ that are Gaussian random
variables with mean $E\left\{W_{t}-W_{s}\right\}=0$ and variance
$\operatorname{Var}\left\{W_{t}-W_{s}\right\}=t-s .$ Non overlapping
increments $W_{t}-W_{s}$ and $W_{v}-W_{u}$ for any $0 \leq s<t \leq
> u<v$ are independent of each other.

Also the following properties are given:

\begin{array}{l} W_{t+1}-W_{t}=\Delta W_{t}, E\left\{\Delta
> W_{t}\right\}=0, E\left\{\Delta W_{t}^{2}\right\}=\Delta t \\
W_{t+1}=W_{+}+\sqrt{\Delta t} \cdot N(0,1), \quad W_{n}=0 \end{array}

Since the variance of a increment is t-s or Δt. I would actually think that $W_(t+1) = W_t +N(0,Δt)$. The expectation of t-s or Δt is 0. Is this the same as $W_(t+1) + \sqrt(Δt)N(0,1)$? If so, how is this converted?


问:
General formula for derivative of multiplication
I'm struggling with numerical differentiation, Newton forward interpolation formula

To calculate my derivative i need to take derivative from the newton-forward interpolation polynom

So i need to take the derivatives of the next n polynomials in series:

$(t)' = 1$

$((t)(t-1))' = 2t-1$

$((t)(t-1)(t-2))' = 3t^{2}-6t+2$

$((t)(t-1)(t-2)(t-3))' = 4t^{3}-18t^{2}+22t-6$

$...$

$((t)(t-1)(t-2)...(t-n+1))' =  ?$

I cant figure out the general formula for it, is there any at all?

Thank you!

答:
Use logarithmic derivative:
$$
\frac{df}{dt} = f \frac{df}{f dt} = f \frac{d\log f}{dt}.
$$
Now
$$
\left(
\prod_{i=0}^{n-1} (t-i)
\right)' =
\left(\prod_{i=0}^{n-1} (t-i)\right)
\sum_{j=0}^{n-1} \frac{d\log (t-j)}{dt} =
\left(\prod_{i=0}^{n-1} (t-i)\right)
\sum_{j=0}^{n-1} \frac{1}{t-j}.
$$

The expression might be simplified further if one puts the product into the sum:
$$
\left(
\prod_{i=0}^{n-1} (t-i)
\right)' =
\sum_{j=0}^{n-1} \frac{\prod_{i=0}^{n-1} (t-i)}{t-j} =
\sum_{j=0}^{n-1} \prod_{i=0, i\neq j}^{n-1} (t-i)
$$


问:
What is the order of $\bar{2}$ in the multiplicative group $\mathbb{Z}_{221}^\times$?
What is the order of $\bar{2}$ in the multiplicative group $\mathbb{Z}_{221}^\times$?

I keep computing $2^0, 2^1,...$ until we get $1\ (\operatorname{mod} 221)$, but this will take forever! Is there an easier way to do this? A hint that was given is that $221= 13 \times 17$.

答:
The order of $2$ mod $13$ is $\color{blue}{12}$ because $2^{12}\equiv1\bmod13$ by Fermat's little theorem,

but $2^6=64\equiv-1\not\equiv1$ and $2^4=16\equiv3\not\equiv1\bmod13$.

The order of $2 $ mod $17$ is $\color{blue}8$ because $2^4=16\equiv-1\bmod17$, so $2^8\equiv(-1)^2\equiv1\bmod17$.

The order of $2$ mod $221$ is the least common multiple of $\color{blue}{12}$ and $\color{blue}8$.

See this Wikipedia article about the Carmichael function.


问:
Follow up question about translation of a limit expression
This is a follow up to this.

Given $$0 \leq \lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n} < 1$$

I think this is equivalent as saying

$$ \exists N: \forall n: n > N \implies \frac{a_{n+1}}{a_n} < 1$$

Apparently, this is wrong but I fail to see why. Can someone explain to me why? Thank you!

答:
I'm posting the proof of the claim in the original question in my own words:

claim: If $$ \forall n \in \mathbb{N}: a_n > 0$$ and $$ \lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n} < 1$$

then

(i) $\{ a_n \}_n $ converges

(ii) $\lim a_n = 0$

proof:

$$ 0 \leq \lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n} < 1 \implies \exists 0< s <1: \lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n} = s$$

$$ \iff \forall \varepsilon > 0 \exists N: n > N \implies \frac{a_{n+1}}{a_n} \in [s-\varepsilon, s+\varepsilon]$$

Choose $\varepsilon$ s.t. $s+\varepsilon < 1$ and $s-\varepsilon > 0$.
Then $$ 0 < s-\varepsilon \leq \frac{a_{n+1}}{a_n} \leq s+\varepsilon < 1$$

$$ \iff (s-\varepsilon) a_n \leq a_{n+1} \leq (s+\varepsilon) a_n$$

and setting $\delta := s+\varepsilon$:

$$ \implies a_{n+k} \leq \delta a_{n+k-1} \leq \delta^2 a_{n+k-2} \leq \dots (\forall n > N)$$

$$ \implies a_{n+k} \leq \delta^{k} a_n (\forall n > N)$$

Then $0 \leq a_{n+k} \leq a_n \delta^k$ and $a_n \lim_k \delta^k = 0$ implies

$$\lim_k a_{n+k} = 0$$

$$ \implies \lim_n a_n = 0$$


问:
How to evaluate incomplete elliptic integrals?
From what I learned here, for evaluating complete elliptic integrals, using the binomial theorem is quite effective as it puts the elliptic integral in the form of.
$$\sum_{n=1}^{\infty} some-constants\int_0^{\dfrac{\pi}{2}} \sin^{2n} \phi d\phi $$ which turns out to be easy to evaluate. Yet when it turns in to the form $$\sum_{n=1}^{\infty} some-constants\int_0^{\theta} \sin^{2n} \phi d\phi $$ which is quite common among elliptic integrals, I start to not know what to do. I resolved to go on with conducting a Taylor expansion where the derivatives(as I do not know how to evaluate the derivatives) I computed using Matlab. Yet what is the best way or the standard way(if there is such a way) to conduct/approximate these kinds of incomplete elliptic integrals?
As a reference, I want am trying to compute
$$ F(\dfrac{\theta + \gamma}{2}|\dfrac{2a}{a-1})$$</p>

答:
The modern way to compute incomplete elliptic integrals is via the Carlson symmetric form:
$$F(\varphi,m)=\sin\varphi R_F(\cos^2\varphi,1-m\sin^2\varphi,1)$$
$$R_F(x,y,z)=\frac12\int_0^\infty\frac1{\sqrt{(t+x)(t+y)(t+z)}}\,dt$$
$R_F$ itself can be computed efficiently by a so-called duplication theorem, as detailed in the "Numerical evaluation" section of the linked Wikipedia article. The process is reminiscent of the AGM formula for computing complete elliptic integrals.


问:
Show that $\lim\limits_{n \to +\infty}(\sin(\frac{1}{n^2})+\sin(\frac{2}{n^2})+\cdots+\sin(\frac{n}{n^2})) = \frac{1}{2}$
Show that the sequence defined as
$$x_n = \sin\left(\frac{1}{n^2}\right)+\sin\left(\frac{2}{n^2}\right)+\cdots+\sin\left(\frac{n}{n^2}\right)$$
converges to $\frac{1}{2}$.

My attempt was to evaluate this limit by using squeeze theorem. I managed to show that $x_n < \frac{n+1}{2n}$ by using $\sin(x) < x$, but I haven't been able to find a sequence smaller than $x_n$ that also converges to $\frac{1}{2}$. I tried showing by induction that $x_n > \frac{1}{2}-\frac{1}{n}$, but I got nowhere with that.

Any help would be appreciated.

答:
Because $\sin x$ is convex on $x \in (0, \pi)$ then the sum of $\sin (kx)$ inside that interval will be not less than $n$ times the linear average on the secant $(\sin x+sin (nx))/2$ and not greater than $n$ times the linear average on the tangent $n \sin((n+1)/2 x)$, so in this case

$$
\eqalign{
  & n\left( {{{\sin \left( {1/n^{\,2} } \right) + \sin \left( {n/n^{\,2} } \right)} \over 2}} \right) \le x_{\,n}  =   \cr
  &  = \sin \left( {{1 \over {n^{\,2} }}} \right) + \sin \left( {{2 \over {n^{\,2} }}} \right) +  \cdots  + \sin \left( {{n \over {n^{\,2} }}} \right)
 \le n\sin \left( {{{\left( {n + 1} \right)/2} \over {n^{\,2} }}} \right) \cr}
$$


问:
Infinitesimals | limits question
I have a question that I am not sure if I did well. can you give me your way to prove it?

Here is the question:

Suppose $\displaystyle{\lim_{n \to \infty}}{(a_n-b_n)}=l$, with $l\in\mathbb{R}$, and $\displaystyle{\lim_{n \to \infty}}{(a_n)}=\infty$ . Does $b_n$ converge?

答:
We have
$$
b_n = (b_n - a_n) + (a_n) \to -l + \infty = \infty.
$$
So, no.

Title: Infinitesimals | limits question


问:
Prove that $ \Delta = (a + \bar{a} - 2)^2$ with (E) : $z^2-2(a-\bar{a})z - |a-1|^2$ in ${\displaystyle \mathbb {C} }$
Prove that $ \Delta = (a + \bar{a} - 2)^2$

with (E) : $z^2-2(a-\bar{a})z - |a-1|^2$; with a $\in {\displaystyle \mathbb {C} }$\{1}

I tried developing  $ \Delta $ but I got stuck here :
$ \Delta  = a² -2\cdot a\bar{a}  + (\bar{a})² + 4\cdot|a-1|²$

Thanks to everyone!

答:
Hint: $|a-1|^2 = (a-1)(\bar{a} -1)$.

Expand and continue.

If you're stuck, explain what else you're done.


问:
Formally writing about lists (tuples), and notation analogous to set notation
Is there any formal notation for dealing with lists, rather than sets?

e.g. if I have a set $X=\{x_1,\dots,x_n\}$ and I want to add a new item to the set, say $x_{n+1}$, I can say "Let $X = X \cup \{x_{n+1}\}$" and it is clearly understood that I want to add $x_{n+1}$ to my set.

However, if $X$ is not a set but rather a list, or tuple (i.e. the elements are ordered and duplicates are allowed), is there any way of indicating that I am adding an element to the end of the list?

e.g. given $X=(x_1,\dots,x_n)$, how do I say add an element to $X$ such that $X=(x_1,\dots,x_n,x_{n+1})$? i.e. how do I formally denote appending an element to $X$?

答:
What you call a list is formally known as sequence. There was a question which symbol is for sequence concatenation. Unfortunately there is no accepted answer. Symbols ⋅, ⌒ (commentator actually used u2322, "frown" symbol but it's resisting my attempt to copy it) and ∥ are mentioned in comments.

According the Wikipedia article ∥ is an operator for concatenation of numbers (doesn't specify which set of numbers, probably ℕ) but doesn't say much about sequences. The same symbol is in my opinion more commonly used for parallelism so it may confuse the reader.

I haven't seen ⌒symbol before but commentators agree about it.


问:
Pointwise convergence in increasing functions
If $f_{n}$ are a sequence of real valued strictly increasing functions that converges pointwise to $f$, is it necessary that $f$ be strictly increasing? Looks fine, but I can't see why. Thanks in advance!

答: No.  Take $$f_n(x)=\frac x{2^n}$$  For each $n$, $f_n$ is strictly increasing, but $\lim_{n\to\infty}f_n(x)=0$ for all $x$.  All we can say is that the limit is monotone increasing.


答:
Assume b_n converges to m.

Then

$\lim_{n} (a_n)=\lim_{n}((a_n-b_n)+b_n)=$

$\lim_{n}(a_n-b_n)+\lim_{n}(b_n)=l+m$,

a contradiction.

Note : If $\lim c_n, d_n$ exist then

$\lim (c_n+d_n)=\lim c_n +\lim d_n$.

Title: Infinitesimals | limits question

Body:
I have a question that I am not sure if I did well. can you give me your way to prove it?

Here is the question:

Suppose $\displaystyle{\lim_{n \to \infty}}{(a_n-b_n)}=l$, with $l\in\mathbb{R}$, and $\displaystyle{\lim_{n \to \infty}}{(a_n)}=\infty$ . Does $b_n$ converge?


答:
This question is part of the following "problem": if $(u_n) \to l \in \Bbb R$ and
$
(\forall n, u_n > 0)
$
does this implies that $l >0$? In general, the answer is no (take $u_n = \frac{1}{n} \to 0$). Here, you can take $f_n(x) = \frac{x}{n}$ for every $x \in \Bbb R^+$. We see that each $f_n$ is strictly increasing, $f_n \to 0$ but the $0$ function isn't strictly increasing.

Title: Pointwise convergence in increasing functions

Body: If $f_{n}$ are a sequence of real valued strictly increasing functions that converges pointwise to $f$, is it necessary that $f$ be strictly increasing? Looks fine, but I can't see why. Thanks in advance!


问:
What is the log of the beta function, how can it be simplified?
$$B(x,y) = \int_0^1 t^{x-1}(1-t)^{y-1} \, dt$$
is the beta function. What is the meaning and simplification of the log of the beta function:

$$\log\left(\int_0^1 t^{x-1}(1-t)^{y-1} \, dt\right) = \, ?$$

答:
well:
$$B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$
so:
$$\ln(B(x,y))=\ln(\Gamma(x))+\ln(\Gamma(y))-\ln(\Gamma(x+y))$$
then you can just use this instead??

Also there are identities relating to approximating $\ln(\Gamma(x))$ or finding its fourier series

Also notice that:
$$\ln\left(\prod a_i\right)=\sum\ln(a_i)$$


问:
Lebesgue points of a set
Recall that given a locally integrable function $f$ for the Lebesgue measure $\mathcal{L}^d$ we have that,
$$\lim_{r\to 0} \frac{1}{\mathcal{L}^d(B(x,r))} \int_{B(x,r)} |f(z)-f(x)|dz =0$$
for almost every $x\in \Omega$ where the open set $\Omega\subset \mathbb{R}^d.$ The points $x$ for which the above equality holds are called the Lebesgue points of the function $f$. We denote $\operatorname{Leb}(f)$ as the set of all Lebesgue points of the function $f\in L^{1}_{\operatorname{loc}}(\mathbb{R}^d).$

To define the Lebesgue points of a $\mathcal{L}^d$ measurable set $A$ we simply use the indicator function $f=1_{A}$ and thus $\operatorname{Leb}(A) = \operatorname{Leb}(1_{A})$ or in other words $x\in \operatorname{Leb}(A)$ if,
\begin{align}
\lim_{r\to 0} \frac{\mathcal{L}^d(A\cap B(x,r))}{\mathcal{L}^d(B(x,r))}=1.
\end{align}

With these two definitions, here is my question: If $\mu << \mathcal{L}^d$ and $f$ is the density function of $\mu,$
\begin{align}
f(y)=\limsup_{r\to 0} \frac{\mu(B(y,r))}{\mathcal{L}^d(B(y,r))},
\end{align}
and for some $x\in\Omega$ we have $f(x)>0$ and $x\in \operatorname{Leb}(f),$ then why is $x\in \operatorname{Leb}(A)$ where $A=\{y\in \Omega: f(y)>0\} = \{f>0\}?$

Here is what I think. To show that $x\in \operatorname{Leb}(A)$ we need to show that
\begin{align}
\lim_{r\to 0} \frac{\mathcal{L}^d(A\cap B(x,r))}{\mathcal{L}^d(B(x,r))}=1.
\end{align}
Now $x\in A$ since $f(x)>0$ and so $x\in A\cap B(x,r),$ but this not very helpful. If $x$ was a point of continuity of $f$ then the proof is trivial since I can find a ball around $x$ where $f>0$ and so limit expression would be equal to $1.$ But this is not the case here. Any comments or hints are much appreciated.

**Edit: ** I came across this fact while reading the proof of Lemma 3.5 in the following article Monge Problem in R^d

答:
This is false.

Let $A$ be any set such that $0$ is not a Lebesgue point of $A$.

Let $f(t)=|t|\chi_A(t)$, so $f\ge0$ and $A=\{f>0\}$. Then $f$ is continuous at the origin, so $0$ is a Lebesgue point of $f$.

Title: Lebesgue points of a set


问:
decidability of $\{x|W_x \text{is different from K in only finitely many elements}\}$
Is the following language decidable? Please explain your argument as I want to learn how such problems must be solved to do the rest on my own. </p>

<p>$$\{x \mid W_x \text{ is different from K in only finitely many
    elements}\},$$  where $W_x = \text{dom}(\Phi_x)$ and $K=\{x \mid \Phi_x(x) \downarrow\}$.</p>

答:
In your case you can apply Rice's Theorem.

The set you consider denotes a set of functions with a nontrivial property, so the conditions of Rice's theorem hols. If you are interested in a more elementary proof, you can carry out the (very simple) construction in the proof of Rice's Theorem. This will give you reduction to the halting problem.


问:
Change of basis matrix between orthonormal bases of different inner products
We were recently given the following exercise:

Given a vector space of real polynomials of degree 2 defined over the interval [-1,1], let the inner product be defined as $\langle f,g\rangle = \int_{-1}^1 f(t)g(t)dt$. Let the following be an orthonormal basis in terms of this inner product: $\{\frac{1}{\sqrt{2}},\sqrt{\frac{3}{2}}t,\frac{3}{2}\sqrt{\frac{5}{2}}(-\frac{1}{3}+t^2)\}$.

We were also given two polynomials (which are irrelevant to the problem I'm having) and then had to use Parseval's identity to find the "angle" between these two polynomials. Parseval's identity states that the abstract inner product between the two polynomials is the same as the euclidian inner product between their coordinate vectors in terms of this orthonormal basis.

So as a first step, I tried to find a change of basis matrix from the given orthonormal basis to the "standard" basis for polynomials, where $\begin{pmatrix}1\\0\\0\end{pmatrix}$ represents a constant polynomial, $\begin{pmatrix}0\\1\\0\end{pmatrix}$ represents $t$ and $\begin{pmatrix}0\\0\\1\end{pmatrix}$ represents $t^2$.

The coordinate vectors of the orthonormal basis in terms of the standard basis are thus $\begin{pmatrix}\frac{1}{\sqrt{2}}\\0\\0\end{pmatrix}, \begin{pmatrix}0\\\sqrt{\frac{3}{2}}\\0\end{pmatrix}, \begin{pmatrix}-\sqrt{\frac{5}{8}}\\0\\ \sqrt{\frac{45}{8}}\end{pmatrix}$.

Now, this is were I'm confused. The change of basis matrix, which takes coordinate vectors from the orthonormal basis to the standard basis, is $\begin{pmatrix} \frac{1}{\sqrt{2}} &&0&&-\sqrt{\frac{5}{8}} \\ 0&&\sqrt{\frac{3}{2}}&&0 \\ 0&&0&&\sqrt{\frac{45}{8}}\end{pmatrix}$.

This matrix is clearly not orthogonal. However, a few days ago we learned that the change of basis matrix between two orthonormal bases must always be orthogonal, and both bases I gave are clearly orthonormal. What am I missing here? Does it make a difference that one basis is orthonormal in terms of the euclidian dot product and one basis is orthonormal in terms of the newly defined inner product?

答: The result you mentioned is about a change of basis in the same $(E, \left <\,,\right>)$ inner product space. When you change the inner product $\left <\,,\right>$, distances and angles change too. To see this, take $E=\Bbb R^2$, $\left =a_1b_1 +a_2b_2$ and $a=(-1,-1), b=(1,-1)$. We have $\left=0$. Now, if $\left =2a_1 b_1 + 3a_2b_2$ (you can check that is is an inner product on $\Bbb R^2$), we have $\left =-2+3=1 \neq 0$.

Body: The result you mentioned is about a change of basis in the same $(E, \left <\,,\right>)$ inner product space. When you change the inner product $\left <\,,\right>$, distances and angles change too. To see this, take $E=\Bbb R^2$, $\left <a,b\right>=a_1b_1 +a_2b_2$ and $a=(-1,-1), b=(1,-1)$. We have $\left<a,b\right>=0$. Now, if $\left <a,b\right>=2a_1 b_1 + 3a_2b_2$ (you can check that is is an inner product on $\Bbb R^2$), we have $\left <a,b\right>=-2+3=1 \neq 0$.


问:
Residue theorem pole
How do I find poles of this function?
$$f(z) = \frac{1-\cos z}{z^4 +z^3}$$

I am unable to identify zero pole order of $f(z)$ as it includes trigonometric functions and I'm new to this concept. Please help me out to identify the pole order.

答:
Replace cos(z) with its expression as a power series:'
$cos(z)= 1+ z^2/2+ z^4/4!+ \cdot\cdot\cdot$
so $1- cos(z)= -z^2/2- z^4/4!-\cdot\cdot\cdot$.

$\frac{1- cos(z)}{z^4+ z^3}= -\frac{z^2/2+ z^4/4!+ \cdot\cdot\cdot}{z^2(z^2+ z)}$
$=-\frac{1/2+ z^2/4!+ \cdot\cdot\cdot}{z^2+z}$

Title: Residue theorem pole


问:
Bayes Theorem Probability with 3 events
I am having trouble understanding how to set this one up, I don't need the full answer just how to set it up.

Sally is always running late to campus. The probability she is late to campus for when she uses her moped is $3\%$, for her car $10\%$ and then the train is $7\%$.

Question: Say the probability that Sally uses her moped is $0.7$, car $0.2$ , or train $0.1$. What is the probability Sally ends up being late? Please use extended Bayes Theorem.

My mistake, miss typed a number

答:
Let $A_1$ be the event: Sally uses her moped

Let $A_2$ be the event: Sally uses her car

Let $A_3$ be the event: Sally uses the train

Let $B$ be the event: Sally is late

Using total probability formula:

$P(B) = \sum_{i=1}^n P(B|A_i)P(A_i) = P(B|A_1)P(A_1) + P(B|A_2)P(A_2) + P(B|A_3)P(A_3)$

Title: Bayes Theorem Probability with 3 events


问:
Two different solutions to an integral
I have $\int -\frac{x}{1-x^2}dx$. I can solve it two ways: either
$$\int -\frac{x}{1-x^2}dx = -\int \frac{x}{1-x^2}dx$$ and then letting $u = -x^2 + 1 $ resulting in $\frac{1}{2}\ln(1-x^2) +C_1$ or by doing
$$\int -\frac{x}{1-x^2}dx = \int \frac{x}{x^2 - 1}dx$$
and letting $u = x^2 - 1$, resulting in $\frac{1}{2}\ln(x^2 - 1) + C_2$

I have seen Two different solutions to integral
but cannot see how that applies in this case. I tried doing

$$\ln(1-x^2) = \ln((-1)(-1)(1-x^2)) = \ln(-1) + \ln(x^2-1)$$ but that doesn't work. Why are the solutions different and do they only differ in the constant?

答:
Depending on the value of $x$, one of the solutions is wrong because the argument of the logarithm may not be negative.

In fact you have observed that due to the singularity at $|x|=1$, you may not "cross" these points. Hence the antiderivative is made of independent sections.

Some guys will tell you that the general solution is

$$\log|x^2-1|+C.$$

But in reality, it is

$$f(x)=\begin{cases}|x|<1\to\log(1-x^2)+C_0,\\|x|>1\to\log(x^2-1)+C_1.\end{cases}$$

Title: Two different solutions to an integral


答:
The residue of $f(z)$ at $a$ is $\lim_\limits{z\to a} f(z)(z-a)$

If you have a simple pole, i.e. $f(z) = \frac {g(z)}{z-a}$ and $g(z)$ is analytic at $a$ then the residue at $a$ is $g(a).$  If you have a second order pole, $f(z) = \frac {g(z)}{(z-a)^2}$ and $g(a) = 0,$  then the residue is $g'(a).$  Which is analogous to applying L'Hoptial's rule in Calc I.

And you can keep applying higher order derivatives as necessary.

It shouldn't really matter what sort of function you have in the numerator.

In some cases it is easier to expand the function as a Laurent series.  Which is similar to a Taylor series.

That is $f(z) = \cdots \frac {c_{-2}}{(z-a)^2} + \frac {c_{-1}}{z-a} + c_0 + c_1(z-a) + \cdots$

And the residue is the $c_{-1}$ coefficient.

What do we have here?

$f(z) = \frac {1-\cos z}{z^3(z+1)}$

The function fails to be analytic at $z = -1$ and $z = 0$

The residue at $z = -1:$

$\lim_\limits{z\to -1} \frac {1-\cos z}{z^3} = \cos 1 - 1$

The residue at $z = 0:$

$\lim_\limits{z\to 0} \frac {1-\cos z}{z^2(z+1)} = \frac {d^2z}{dz^2} \frac {1-\cos z}{z+1}$

Or we could expand the numerator as a Taylor series.

$\lim_\limits{z\to 0} \frac {\frac {z^2}{2} - \frac {z^4}{4!} + \cdots}{z^2(z+1)}$

Cancel the $z^2$ terms with the denominator.

$\lim_\limits{z\to 0} \frac {\frac {1}{2} - \frac {z^2}{4!} + \cdots}{(z+1)}$

And consider the limit as $z$ approaches $0$

Title: Residue theorem pole

Body:
How do I find poles of this function?
$$f(z) = \frac{1-\cos z}{z^4 +z^3}$$

I am unable to identify zero pole order of $f(z)$ as it includes trigonometric functions and I'm new to this concept. Please help me out to identify the pole order.


问:
Does it make sense to compare sets (polygons) with different dimensions?
In the context of integer programming, I am considering 3 different linear models for a given problem. The goal is to determine which formulation is the tightest, that is, the one that gives the least fractional solution when omitting integrity constraints.

Let $X$, $Y$ and $Z$ be the polytopes defined by each of these formulations, without integrity constraints, and let $z_1, z_2, z_3$ denote the optimal values of the formulations over $X,Y,Z$, respectively. Assume we are dealing with a minimization problem. I have managed to prove that
$$
z_1 \ge z_2 \ge z_3
$$
in other words, the formulation over polytope $X$ is the tightest.

From there, do the following inclusions hold (or make any sense) ? :
$$
X\subseteq Y \subseteq Z
$$

My guess is that if the same number of variables is used for each formulation, then yes. But if these numbers are different, I am not sure the polytopes are "comparable" (my wording may be incorrect), and perhaps it makes no sense to use the symbol $\subseteq$.

答:
Afik, the question is still ill-posed, but here is what I can tell you:

Suppose, $X$ and $Y$ are (closed) convex polytopes in $R^n$ such that for every linear function $f: R^n\to R$, we have
$$
\min f|_Y\le \min f|_X.
$$
Then $X\subset Y$. Here $\min f|_Z= \min \{f(z): z\in Z\}$.

A proof of this is quite straightforward. I will prove the contrapositive: If $X$ is not contained in $Y$ then there exists a linear function function
$f: R^n\to R$, such that

$$
\min f|_X < \min f|_Y.
$$

Since $X$ is not contained in $Y$, one of the vertices $v$ of $X$ is not contained in $Y$. Let $C$ denote the convex hull of $X\cup Y$. Then $v$ is still a vertex of $C$.

Given any convex polytope $P\subset R^n$ and its vertex $p$, there exists a linear function $h: R^n\to R$ such that $v$ is the unique minimum point of $h$ on $P$: $h(p)=a$ and $h(q)>a$ for all $q\in P, q\ne p$. Applying this to the polytope $C$ and its vertex $v$, we obtain a linear function $f: R^n\to R$ such that $f(v)=0$ and $f(c)>a$ for all $c\in C$, except $c=v$. Since $v\notin Y$, we get
$$
a=\min f|_X < \min f|_Y.
$$

qed

It is, however, possible that the question that you had in mind is a bit different:

Suppose that $f_1,...,f_k$ are given defining linear functions of the polytope $X\subset R^n$  (i.e. $X$ is given by a system of linear inequalities $f_i\ge a_i, i=1,...,k$). Suppose that $Y$ is another polytope in $R^n$ such that for each $i=1,...,k$,
$$
\min f_i|_Y\le \min f_i|_X.
$$
Does it follow that $X\subset Y$? This is indeed false even when $n=2$.

The difference between the two questions is that in the first formulation, I am allowed to enlarge the given defining set of functions for $X$.


问:
Arithmetic mean vs. normalized vector length in multidimensional space
Given I have multiple values in the range of [0,1] and I want to condense them into a single value (again, in the range [0,1]). What's the semantic difference of the following two approaches?

Arithmetic mean:

Sum up the values

Divide the sum by the number of values

Example: $\frac{0.1 + 0.7 + 0.4}{3} = 0.4$

Normalized vector length:

Interpret the values as being vector coordinates in a n-dimensional space

Calculate the vector length

Normalize the length by dividing it by the maximum possible vector length, i.e., a vector with all coordinates being 1.0

Example: $\frac{\sqrt{0.1^2+0.7^2+0.4^2}}{\sqrt{1.0^2+1.0^2+1.0^2}}=0.469...$

I understand the differences in calculation, but can't make sense of why the results differ. Which approach should I use to get a sensible summary of the values?

答:
I realized that my way of calculating the "Normalized vector length" average is algebraically equivalent to the Quadratic Mean (https://en.wikipedia.org/wiki/Average#Summary_of_types), so I now have a term to look for further information on the subject.

In short, the Quadratic Mean emphasizes higher values, which the arithmetic mean doesn't.