问:
Relation between Rogers Ramanujan continued fraction and $j$-invariant
While going through this answer I found an interesting but slightly complicated relation between Rogers-Ramanujan continued fraction and the j-invariant. I would like to know an elementary proof of the same.

Before proceeding let me define all the necessary terms and symbolism to set the proper context. Let $\tau$ be a complex number with positive imaginary part and $q=\exp(2\pi i\tau) $ so that $|q|<1$. Below I define functions and the relations which I am aware of. The Rogers-Ramanujan continued fraction is given by $$R(q) =\cfrac{q^{1/5}}{1+\cfrac{q}{1+\cfrac{q^2}{1+\cfrac{q^3}{1+\dots}}}}\tag{1}$$ Ramanujan studied this function in great detail and obtained the following fundamental identities $$\frac{1}{R(q)}-1-R(q)=\frac{\eta(q^{1/5})} {\eta(q^5)}\tag{2}$$ and $$\frac{1}{R^5(q)}-11-R^5(q)=\left(\frac{\eta(q)}{\eta(q^5)}\right)^6\tag{3}$$ where $\eta(q) $ is Dedekind eta function defined by $$\eta(q) =q^{1/24}\prod_{n=1}^{\infty} (1-q^n)\tag{4}$$ To define the $j$-invariant we need to introduce Ramanujan's version of Eisenstein series denoted by $L, M, N$ (symbols $P, Q, R$ are typically used but we want to avoid conflict with Rogers-Ramanujan continued fraction $R(q) $)
\begin{align}
L(q) &= 1-24\sum_{n=1}^{\infty}\frac{nq^n}{1-q^n}\tag{5a}\\
M(q)&=1+240\sum_{n=1}^{\infty} \frac{n^3q^n}{1-q^n}\tag{5b}\\
N(q) &=1-504\sum_{n=1}^{\infty} \frac{n^5q^n}{1-q^n}\tag{5c}
\end{align}
It should be observed that $L$ is related to $\eta$ via $$L(q) =24q\frac{d}{dq}(\log\eta(q))\tag{6}$$ The $j$-invariant is defined as $$j(q) =\frac{1728M^3(q)}{M^3(q)-N^2(q)}\tag{7}$$
Ramanujan obtained a system of differential equations connecting $L, M, N$:
\begin{align}
q\frac{dL(q) } {dq} &=\frac{L^2(q)-M(q)}{12}\tag{8a}\\
q\frac{dM(q)}{dq}&=\frac{L(q)M(q)-N(q)}{3}\tag{8b}\\
q\frac{dN(q)} {dq} &=\frac{L(q) N(q) - M^2(q)}{2}\tag{8c}
\end{align}
Using $(6)$ and $(8a)$ it is evident that $M(q) $ can also be expressed in terms of $\eta(q) $. On the other hand the above differential equations allow us to prove that $$M^3(q)-N^2(q)=1728\eta^{24}(q)\tag{9}$$ and thus we have some expression for $j(q) $ in terms of $\eta(q) $.

The following complicated relation holds between Rogers Ramanujan continued fraction $R(q)$ and $j(q) $ : $$ R^5 (R^{10}+11 R^5-1)^5j+(R^{20}-228 R^{15}+494 R^{10}+228 R^5+1)^3 = 0\tag{10}$$ I checked Wikipedia and found that this is derived from another identity $$j(q) =\frac{(x^2+10x+5)^3} {x} \tag{11}$$ where $$x=125\left(\frac {\eta(q^5)}{\eta(q)}\right)^6\tag{12}$$ Using $(11),(12)$ and $(3)$ we can deduce $(10)$ with a little algebra.

Thus the problem boils down to a proof of equation $(11)$. I don't know if this can be derived using algebraic manipulation of the identities given above or does it need some specific modular equation.

Any proofs or suggestions for proof are welcome. I don't understand the machinery of modular forms properly and would prefer an approach more in the spirit of Ramanujan. The question is however tagged "modular-forms" to get the attention of experts from that tag.

Update: I have finally managed to give a proof based on modular equation of degree $5$ given by Ramanujan and posted it as an answer. The proof is more of a verification and a more natural proof utilizing some transformation formula of eta function is desired.

There is a related question which assumes $(10),(11)$ and proves $(3)$, but the approach uses Mathematica.

答:
An approach based on modular equations of degree $5$ works out fine with some tedious algebra. The identity $(11) $ in question will be proved by replacing $q$ with $q^2$.

Let $k, l$ denote the elliptic modulus and $K, L$ the complete elliptic integrals of first kind corresponding to nomes $q, q^5$ respectively and further let $\alpha=k^2,\beta =l^2$. Also let $m=K/L$ denote the multiplier.

Then we have the standard formulas
\begin{align}
M(q^2)&=\left(\frac{2K}{\pi}\right)^4(1-\alpha(1-\alpha))\tag{1}\\
\eta(q^2)&=2^{-1/3}\sqrt {\frac {2K}{\pi}}(\alpha(1-\alpha))^{1/12}\tag{2}\\
\eta(q^{10})&=2^{-1/3}\sqrt{\frac{2L}{\pi}}(\beta(1-\beta))^{1/12}\tag{3}
\end{align}
Dr. Bruce C. Berndt and his collaborators analyzed Ramanujan's modular equations of degree $5$ and gave the following relations between $\alpha, \beta, m$:
\begin{align}
(\alpha^3\beta)^{1/8}&=\frac{\rho+3m-5}{4m}\tag{4}\\
(\alpha\beta^3)^{1/8}&=\frac{\rho+m^2-3m}{4m}\tag{5}\\
\rho&=\sqrt{m^3-2m^2+5m}\tag{6}
\end{align}
From the above equations we get $$\alpha=\frac{(\alpha^3\beta)^{3/8}}{(\alpha\beta^3)^{1/8}}=\frac{(\rho+3m-5)^3}{16m^2(\rho+m^2-3m)}$$ Multiplying numerator and denominator by $\rho-m^2+3m$ and using some algebra we get $$\alpha=\frac{(\rho+3m-5)^2(2m+\rho)} {16m^3(m-1)} $$ Next we replace $\alpha, \beta, m$ by $1-\beta,1-\alpha,5/m$ to get $$((1-\alpha)(1-\beta)^3)^{1/8}=\frac{\rho+3m-m^2}{4m}$$ and $$((1-\alpha)^3(1-\beta))^{1/8}=\frac{\rho+5-3m}{4m}$$ and thus $$1-\alpha=\frac{(\rho+5-3m)^2(2m-\rho)} {16m^3(m-1)} $$ and hence $$\alpha(1-\alpha)=\frac{(m-1)(5-m)^5}{256m^5} \tag{7}$$ A similar calculation leads to the relation $$\left(\frac{\beta(1-\beta)}{\alpha(1-\alpha)} \right)^{1/2}=\frac{m^2(m-1)^2}{(5-m)^2}\tag{8}$$ Let us now observe that if $$x=125\left(\frac{\eta (q^{10})}{\eta(q^2)}\right)^6$$ then by the identities $(2),(3),(8)$ we get $$x=\frac{125(m-1)^2}{m(5-m)^2}$$ On the other hand we have $$j(q^2)=\frac{M^3(q^2)}{\eta^{24}(q^2)}=\frac{(1-\alpha(1-\alpha)) ^3}{2^{-8}(\alpha(1-\alpha))^2}=\frac{(256m^5-(m-1)(5-m)^5)^3}{m^5(m-1)^2(5-m)^{10}}$$ And the expression $(x^2+10x+5)^3/x$ equals $$\frac{(3125(m-1)^4+250m(m-1)^2(5-m)^2+m^2(5-m)^4)^3}{m^5(m-1)^2(5-m)^{10}}$$ Our job is thus complete if we can show that $$3125(m-1)^4+250m(m-1)^2(5-m)^2+m^2(5-m)^4=256m^5-(m-1)(5-m)^5$$ Both sides are polynomials of degree $6$ and hence we need to verify the identity for at most $7$ values of $m$. The verification is trivial for $m=0,1,5$ and one may try to verify it for small values like $m=-1,-2, 2,3$.

There is another approach to rewrite the desired identity so that both sides can be factorized easily. Thus the desired identity is equivalent to $$250m(m-1)^2(5-m)^2+3125(m-1)^4=256m^5+(m-1)(m-5)^5-m^2(m-5)^4$$ or $$250m(m-1)^2(m-5)^2+3125(m-1)^4=3125-6250m+3750m^2-1000m^3+125m^4+250m^5$$ or $$(m-1)^2(2m(m-5)^2+25(m-1)^2)=25-50m+30m^2-8m^3+m^4+2m^5$$ It can be checked now that each side factors as $$(m-1)^2(2m^3+5m^2+25)$$


问:
Solving very large matrices in "pieces"
Say you have a very dense matrix that is 30000x30000 elements. The very dense matrix comes from the radiosity equation, which I discussed here.

Say you have Ax = B.  You have B, and A is 30000x30000 elements.  Solving:  You need to find x given B.

Solving a matrix like this in C code actually introduces an interesting problem:  you cannot actually store it all in memory at once.

So, I want to know about ways to solve this problem "in pieces" -- I want to mathematically decompose the 30000x30000 matrix into smaller sub-matrices (perhaps in chunks of 30000x20?), and somehow solve that way.

What algorithms exist to break down / solve a matrix "in pieces" or steps, that can be solved regardless of memory restrictions?  It can be an iterative technique.

答:
Searching for "Out of core LU" on google might help you.

Essentially, there are two simple choices:

An iterative method that merely depends on repeatedly computing A*v and perhaps A'*v for a given vector 'v'. Such a computation can be done without having the entire matrix A in RAM

A direct method, for this the "out of core LU" might help.

In general, for both choices above, searching for external memory algorithms will be useful.

ContentLicense: CC BY-SA 2.5

Title: Solving very large matrices in "pieces"


问:
Series of Vectors
In $\mathbb{R}^n$ we define sequences of it's elements in a very natural manner, we say that a sequence is a function $x : \mathbb{N} \to \mathbb{R}^n$ and we denote it by $(x_k)$ as in the $n=1$ case. Then defining limit of a sequence and all of that is pretty straightforward and acts like one extension of everything done in $\mathbb{R}$. My doubt is: is there something that stops us from defining the sum of a sequence in $\mathbb{R}^n$?

All the books I've seem until now define sequences in $\mathbb{R}^n$ but not series. I thought it was possible to define it in an analogous manner: we consider the sequence of partial sums of some sequence $(x_k)$, this will give us another sequence defined in $\mathbb{R}^n$ and we could say that the series converges if the sequence of partial sums converges. Is it possible to do this or there is some inconsistency I'm failing to see?

Thanks very much in advance.

答: It is possible. The sum of a series of vectors has as its kth component the sum of the series of kth components of vectors in the series, each of which is a series in $\mathbb{R}$, and converges iff all of the component series converge.</p>

ContentLicense: CC BY-SA 3.0

Title: Series of Vectors


问:
How to prove that if $S_2$ is linearly independent then $S_1$ is also linearly independent, where $S_2$ is superset of $S_1$?
Ok, so this is my attempt to prove.

Let $S_2$ be linearly independent (L.I)

This implies that $a_1 v_1 + a_2 v_2+ \dots + a_n = 0$ if $a_1 = a_2 = \dots = a_n = 0$

Now assume that $S_1$ is linearly dependent (L.D)

This implies that

$a_1v_1 + a_2v_2 + \dots + a_n v_n = 0$ for some non-zero scalars.

Then, $v_1 = -(a_2/a_1)v_2 - \dots - (a_n/a_1)v_n$

This implies that $v_1$ is L.D

Now, since $S_1$ is subset of $S_2$
implies that $v_1$ belongs to $S_2$.
But $S_2$ is L.I

Therefore, our assumption that $S_1$ is L.D wrong.

This implies that $S_1$ is L.I

答:
The idea feels right, but be careful. Note that you wrote

This implies that $a_1v_1+a_2v_2+......+a_nv_n=0$ if
$a_1=a_2=....=a_n=0$,

but that's obvious.

If $S_2$ is L.I, what you can deduce is that IF $a_1v_1+a_2v_2+......+a_nv_n=0$, THEN $a_1=a_2=....=a_n=0$. Don't mess up with necessity/sufficiency.

ContentLicense: CC BY-SA 4.0


问:
show that if $f: \Bbb R^k \rightarrow \Bbb R$ when $ k \geq 2 $ is partial continuous function, than its not injective
I told on a different question that I study calculus 3, and the teacher likes to give us difficult questions. here's one of them: "let $ k \geq 2 $, let $f :\Bbb R^k \rightarrow \Bbb R $ be a partial function who is continuous in a punctured neighborhood of a point $ a \in \Bbb R^k $. prove that f is not injective". as a clue, he wrote "in every punctured neighborhood there exists a loop. if the function is injective, you will get a contradiction to the intermediate value theorem on one part of the loop". I don't really know how this is helping me in any way. maybe I'm missing something which is obvious? I could really use some help here.

答:
There is a circle $C\subset dom(f).$ That is, there is a continuous surjection $g:[0,2\pi]\to C$ such that $g(0)=g(2\pi)$ and the restriction of $g$ to the domain $[0,2\pi)$ is injective.

For $x\in [0,2\pi]$ let $h(x)=f(g(x)).$ We have $h(0)=h(2\pi).$ Now if $f$ is continuous then so is $h$.

Exercise: Prove that if $h:[0,2\pi]\to\Bbb R$ is continuous and $h(0)=h(2\pi)$ then there exist $x_1,x_2$ with $0<x_1<x_2<2\pi$ and $h(x_1)= h(x_2).$

Hence if $f$ is continuous then $h$ cannot be injective on the domain $[0,2\pi).$ But $g$ $is$ injective on $[0,2\pi),$ so a continuous $f$ cannot be injective on $C.$

Proof of Exercise: There exists $x_1\in (0,\pi)$ with $h(x_1)=\frac {h(0)+h(\pi)}{2}$ and there exists $x_2\in (\pi,2\pi)$ with $h(x_2)=\frac {h(2\pi)+h(\pi)}{2}=\frac {h(0)+h(\pi)}{2}=h(x_1).$


问:
$\operatorname{lcm}(a, b, c) = 100$ What are all the triads $a, b, c $ that happen? lcm-least common multiple
$$\operatorname{lcm}(a, b, c) = 100$$

What are all the triads $a, b, c $ that happen?

How many answers are there?

We've the canonical form $100=2^2 \cdot 5^2$ therefore $\tau(100)=9$ Assuming $a,b,c $ to be distinct integers there are $\binom{18}{3}$ tuples $(a,b,c)$ satisfying the desired conditions
Sidenote : OP must specify the nature of $a,b,c$, i.e. whether $a,b,c \in \mathbb{Z}$or $\mathbb{Z}^{+}$ and whether they're all distinct or not  ???   or  I think the answer is $\binom{3}{1}^2 \cdot 9^2 $.

Who knows if it is right or not please help prove the solution ?

答:
The number of ordered triples of nonnegative integers less than or equal to $2$ with at least one coordinate equal to $2$ (and hence having a maximum of $2$) is $3^3-2^3=27-8=19$ (all ordered triples, minus those with all coordinates less than $2$).

Hence, since LCMs are all about taking the maximum of the exponents of each prime factor (and likewise, the minimum for the GCD), the answer is $19^2=361$, assuming that only solutions with $a, b,$ and $c$ all positive are allowed. If one allows at least one of $a, b,$ and $c$ to be negative, then one must multiply the answer by $8$ to get $361 \cdot 8=2888$.


问:
Convergence of conditional expectations given convergence of random variables
Say $X_n$ is an $\mathcal{F}_n$-adapted sequence of random variables converging to $X_{\infty}$ in $L^1$. Clearly, fixed $n\leq m$ one has $E[X_m\mid \mathcal{F}_n] \to E[X_{\infty}\mid \mathcal{F}_n]$ in $L^1$. What's the easiest way to prove this? I know that $E[X_{\infty}\mid \mathcal{F}_n]\in L^1$ but I can't use the dominated convergence because I don't have P-a.s. convergence of the random variables, nor monotone convergence because I don't have non-negativity.

答:
This amounts to saying that the mapping $X \mapsto E[X|\mathcal{G}]$ is continuous for the $L^1$ norm (here $\mathcal{G}$ is any sub-$\sigma$-field.)

The mapping is clearly linear and satisfies
$$\lVert E[X|\mathcal{G}]\rVert_{L^1}= E\left|E[X|\mathcal{G}]\right|\leq E\left[E[|X||\mathcal{G}]\right]=E|X|=\lVert X\rVert_{L^1}.$$


问:
Normal line to the border of a set

$D=\left\{(x, y): x^{2}+y^{2}<4, x y>\sqrt{3}, x>0\right\}$. Find the normal line to $\partial D$ that passes through $(\sqrt{2}, \sqrt{3 / 2})$.

I don't really know how to approach this problem. I've found the derivates of the circle for the tangent line, but I don't think it's correct, because I have to consider the hyperbola too.

答:
The point $\left(\sqrt2,\sqrt{\frac32}\right)$ belongs indeed to the boundary of $D$. But it is an interior point of the circle. On the other hand, it is a point of the hyperbola $xy=\sqrt3$. So, consider the function$$f(x)=\frac{\sqrt3}x.$$Then $f'(x)=-\frac{\sqrt3}{x^2}$ and $f'\left(\sqrt2\right)=-\frac{\sqrt3}2$. So, the slope of the tangent at $\left(\sqrt2,\sqrt{\frac32}\right)$ is $-\frac{\sqrt3}2$ and therefore the slope of the normal line at that point is $\frac2{\sqrt3}$. So, the normal line is$$y=\frac2{\sqrt3}\left(x-\sqrt2\right)+\sqrt{\frac32}.\tag1$$

You can reach the same conclusion using the gradient. The hyperbola is the level curve $xy=\sqrt3$. So, let $g(x,y)=xy$. Then$$\nabla g\left(\sqrt2,\sqrt{\frac32}\right)=\left(\sqrt{\frac32},\sqrt2\right)$$and this vector has the same direction as the normal line. Therefore, the slope of the normal line is $\frac{\sqrt2}{\sqrt{3/2}}=\frac2{\sqrt3}$. And this is all you need to deduce that the normal line is the line $(1)$.

Title: Normal line to the border of a set


问:
Examples of a non-finitely generated $\mathbb Z$-module
I am looking for a couple of examples of a $\mathbb Z$-module $M \neq 0$ with
$$
\mathrm{Hom}_{\mathbb Z}(M, \mathbb Z) = 0 \\
\mathrm{Ext}^1_{\mathbb Z}(M, \mathbb Z) = 0
$$

If $M$ was finitely generated this would not be possible since $M$ would be the direct sum of a free part and a torsion part.

To put this into context: I would like to realize $M$ as the first homology group of a CW complex $X$. My ultimate goal would be to construct a complex $X$ such that its first homology group is nonzero but all cohomology groups (except in degree 0) vanish. The above constraints arise by looking at the exact sequence of the universal coefficient theorem in cohomology.

答: One may ask if $\mathrm{Ext}^1(M,\mathbb{Z})=0$ implies that $M$ is free, this is the famous Whitehead problem. Shelah has shown that it is independent from ZFC. It is false in certain models of ZFC, but true for example in Gödel's constructible universe L. But if $M$ is free with basis $B \neq \emptyset$, then of course $\hom(M,\mathbb{Z}) \cong \mathbb{Z}^B \neq 0$. This means that for example in L there is no example.


问:
Show $T$ is compact operator if $\langle Te_n,e_n \rangle$ tend to zero.
Suppose $\mathcal{H}$ is a Hilbert space, and $T\in B(\mathcal{H})$. If for each orthonormal (norm 1) basis $\{e_n\}\subseteq \mathcal{H}$, we have $\langle Te_n, e_n \rangle \rightarrow 0$. Can we deduce $T$ is compact?

I guess it may use spectra decomposition. Take adjoin, use the fact that $\langle Ae_n,e_n\rangle \rightarrow 0$ iff $\langle (A+A^*)e_n,e_n \rangle \rightarrow 0$ and $\langle (A-A^*)e_n,e_n\rangle \rightarrow 0$ , we can assume $T$ is self-adjoint. So $T$ can be viewed as a multiple operator(multiple by a $L^{\infty}(X,d\mu)$ function) on $L^2(X,d\mu)$, via unitary equivalence. Here $d\mu$ is an abstract $\sigma$-finite Borel measure on X. But I don’t know how to use the condition.. And, another way, if we use the spectra decomposition via $T=\int z dE$, E is the corresponding spectra measure. To show $T$ is compact, it is sufficient to prove the projection $E(-\infty,-\epsilon)$ and $E(\epsilon, +\infty)$ are all finite rank, for every $\epsilon>0$. And I feel the condition may can be use together with something like dominate converge theorem? But I fail.
Any help or hint? Thanks.

答:
Lemma.  If $T$ is a bounded operator on $H$,  the following are equivalent:

$\langle T(e_n), e_n\rangle \to 0$, for every orthonormal basis $\{e_n\}_n$,

$\langle T(e_n), e_n\rangle \to 0$, for every orthonormal sequence $\{e_n\}_n$.

Proof.

(1) $\Rightarrow$ (2) Obvious.

(2) $\Rightarrow$ (1)  Given an orhonormal sequence $\{e_n\}_n$, choose another orthonormal set $\{f_i\}_{i\in I}$, such that
$$
  B:= \{e_n:n\in {\mathbb N}\}\cup   \{f_i:i\in I\}
  $$
is an orthonormal basis.  Assuming, as we supposedly are, that $H$ is separable, it is possible to order $B$
into a sequence admiting   $\{e_n\}_n$ as a subsequence.  So the proof follows since a subsequence of a convergent
sequence converges to the same limit.
QED

As already noticed,  in the complex case we may assume without loss of generality that $T$ is self-adjoint.

Let $\mathfrak B(\mathbb R)$ denote the $\sigma $-algebra of all Borel subsets of ${\mathbb R}$, and let
$$
  E:\mathfrak B(\mathbb R)\to B(H)
  $$
be the spectral resolution for $T$.

We then claim that
$$
  P:= E(\varepsilon ,\infty )
  $$
is a finite rank projection for every $\varepsilon >0$.  To see this,
suppose otherwise, so there exists an orthonormal sequence $\{e_n\}_n$ contained in the range of $P$.  Observing that
$TP\geq \varepsilon P$, we then have that
$$
  \langle T(e_n), e_n\rangle  =  \langle TP(e_n), e_n\rangle  \geq  \varepsilon    \langle e_n, e_n\rangle  = \varepsilon \|e_n\|^2 = \varepsilon ,
  $$
contradicting the hypothesis.

In a similar way one proves that  $E(-\infty , -\varepsilon )$ is finite rank.

Finally the compactness of $T$ follows from the fact that
$$
  T = \lim_{\varepsilon \to 0} TE (-\infty , -\varepsilon  )+TE (\varepsilon , \infty  )
  $$


问:
Evaluating $\lim_{x → 4}\frac{\sqrt{x^2-7}-3}{4-\sqrt{3x+4}}$

How can I find
$$\lim_{x → 4}\frac{\sqrt{x^2-7}-3}{4-\sqrt{3x+4}}$$

I know the domain of the function is any real number except 4, but how can I eliminate the zero denominator?

答:
Note that there are no singularities in the denominator, therefore you can do direct substitution.
\begin{align*}
\lim_{x\to 4}\frac{\sqrt{(x^2-7)-3}}{4-\sqrt{3x-4}}&=\frac{\sqrt{4^2-7-3}}{4-\sqrt{3(4)-4}}\\
&=\frac{\sqrt{6}}{4-\sqrt{10}}\\
&=\frac{2\sqrt{6}+\sqrt{15}}{3}
\end{align*}


问:
ODE - getting reciprocal of correct answer
Consider the first order differential equation
$$(1+x^2)\frac{dy}{dx}=x(1-y^2), \quad y(0)=2.$$

I've tried to solve this by separating variables, and I get
$$y=\frac{3(1+x^2)-1}{3(1+x^2)+1}.$$

However, Wolfram Alpha gives the reciprocal of this.

During my solution, I ended up with
\begin{equation}\log|1+y|-\log|1-y|=\log|3(1+x^2)|,\end{equation}
which I simplified to
$$\log\left|\frac{1+y}{1-y}\right|=\log |3(1+x^2)|,$$
and subsequently got my answer.

However, if I had simplified this to
$$\log\left|\frac{1+y}{y-1}\right|=\log |3(1+x^2)|,$$
I get the same answer as WA.

What's going on here? What's wrong with my initial approach?

EDIT

After arriving at
$$\log\left|\frac{1+y}{1-y}\right|=\log |3(1+x^2)|,$$
I wrote
$$e^{\log\left|\frac{1+y}{1-y}\right|}=e^{\log |3(1+x^2)|},$$
and then
$$\frac{1+y}{1-y}=3(1+x^2).$$

Perhaps this is where the issue arises?

答:
It is often helpful to determine the constants only after getting rid of the absolute value. That is, first find
$$
\frac{1+y}{1-y}=C(1+x^2), ~~C\in\Bbb R\cup\{\infty\}
$$
then use the initial condition to find the constant with the correct sign, $C=-3$. Continue transforming as you did to find the reciprocal of your first solution.


问:
Simple conceptual Fundamental Thm of Calculus question
When applying the Fundamental Thm of Calculus in complex analysis, what does it mean for an open connected set to contain a loop? For example, does my red-color open annulus contain the black colour loop? I think so, but I'm struggling with understanding this:

When asked to integrate $\frac 1{(z-5)}$ around a circle centred at 5, why can't I use the Fundamental Thm of Calculus? After all, the function is continuous in the red annulus , and furthermore, its primitive $\ln z$ is analytic throughout the annulus which does not touch the negative real axis.

答:
The fundamental theorem of calculus is applied to a special contour: intervals of the real line.  And it applies to functions that have an antiderivative, or primitive.  In the complex plane, we have Cauchy's Theorem which states that such nicely behaved functions ("analytic") may be integrated by evaluating its antiderivative at the endpoints of a contour, no matter the shape of the contour.

In your case, the function $1/(z-5)$ has a pole in the interior of the black contour. By Cauchy's theorem, the value of the integral is $i 2 \pi$.  You can see this from a parametrization $z=5+r e^{i \phi}$.  Now, suppose you evaluate the integral via a Fundamental Theorem of calculus mindset: let's say we want to evaluate $\log{(z-5)}$ at the beginning and end of the contour.   Parametrize by $z=5 + r e^{i \phi}$, and so we will be subtracting the value of the antiderivative at $\phi=0$ from the value at $\phi=2 \pi$:

$$\log{(r e^{i 2 \pi})} - \log{(r e^{i 0} )} = \log{r} + i 2 \pi - \log{r} - i 0 = i 2 \pi$$

It is the multivaluedness of the log function that causes the nonzero value.  But you see that it all agrees and provides for a consistent basis for computation.

A word about your annulus: yes, $1/(z-5)$ is analytic throughout your annulus.  The integral about the boundary of the annulus is indeed zero: you also have to integrate about the inner circle in the opposite direction, so the integral is $i 2 \pi - i 2 \pi = 0$.


问:
Where does $f_n$ converge pointwise, is it uniformly convergent?
Let $f_{n}:[0,1] \rightarrow \mathbb{R}$ be a sequence of function defined by
$$
f_{n}(x)=\left\{\begin{array}{cc}
n(1-n x), & \text { if } 0<x<\frac{1}{n} \\
0, & \text { if } x=0 \text { or } \frac{1}{n} \leq x \leq 1
\end{array}\right.
$$

The graph is a right triangle and when $n\to \infty$, triangle's basis vanishes. So can I say that this function converges pointwisely to $f(x)=0$?

If I can say that I can choose $$x=\dfrac{n-1}{n^2}\in (0,1/n)$$

For $\epsilon=1/2$, $\forall n\in \mathbb N^+$ $|f_n(x)-0|=1>\epsilon$

If I cannot assume the above pointwise assumption, how can I show it is not a Cauchy sequence?

答:
Hint.

You have the intuition for the pointwise convergence. But in order to give a proof, you need to show that for every $x\in[0,1]$,
$$
\lim_{n\to 0}f_n(x)=0
$$

One way to see that the sequence is not uniformly convergent is by observing that
$$
\int_0^1 f_n(x)dx=\frac12
$$
for all $n$ and that $\int_0^1f(x)\,dx=0$, where $f(x)\equiv 0$.

If you want to follow the definition of uniform convergence, you need to show that there exists $\epsilon>0$ such that for any $N$, there exist $x\in[0,1]$ and $n>N$ such that
$$
|f_n(x)-0|\ge \epsilon
$$


问:
Integration of $\int_0^{\infty}\frac{e^{(\lambda +is)t}-e^{-(\lambda +is)t}}{e^{\pi t}-e^{-\pi t}}dt$
Integration of $$\int_0^{\infty}\frac{e^{(\lambda +is)t}-e^{-(\lambda +is)t}}{e^{\pi t}-e^{-\pi t}}dt$$

I am thinking about this integral from last half and hour but still can't figure it out how to solve this. what i was think is to make this in form

$$\int_{0}^{\infty}\frac{\sinh(\lambda+is)t}{\sinh (\pi t)} dt$$ but still this will get me nowhere. please just give me a hint how to proceed this.

Thank you.

答:
By 3.511.2 in Gradshteyn and Ryzhik,
$$
\int_0^{ + \infty } {\frac{{\sinh (zt)}}{{\sinh (\pi t)}}dt}  = \frac{1}{2}\tan \left( {\frac{z}{2}} \right)
$$
provided $\left| z \right| < \pi$. Take $z=\lambda+is$.


问:
What is the measure of the set of numbers in $[0,1]$ whose decimal expansions do not contain $5$?
I just came across an exercise that asks to show that set of numbers in $[0,1]$ which possess decimal expansion not containing the digit $5$ has measure zero.

How do I approach this?

I tried thinking of relating this to Cantor set but, I see that numbers whose decimal expansion contains the digit $5$ are both contained in the Cantor set as well in its complement too, I thought if it was in the Cantor set and since measure of Cantor set is zero, so it's measure will be zero but this does not happen.

In fact how is this set measurable?

Any idea?

答:
Describing the set

There are actually a couple of little subtleties here which are worth mentioning, though they end up not really causing problems at the end of the day.  To fix notation, let $A$ denote the set of numbers in $[0,1]$ possessing a decimal expansion which does not contain a $5$.

Part of the question, which is not addressed by the other answers, is "Is $A$ measurable?"  There is also the question of showing that $A$ is of measure zero.  What is missing here is a description of the measure with respect to which we should answer the questions.  The "obvious" answer is that the Lebesgue measure is the measure in question, but it doesn't hurt to specify this.

Decimal expansions are not necessarily unique.  For example,
$$ \frac{1}{2} = 0.5 = 0.4\overline{9}. $$
The first decimal representation of one half contains a $5$, while the second does not.  Similarly, $3/5 = 0.6 = 0.5\overline{9}$.  Because both $1/2$ and $3/5$ possess decimal expansions which do not contain a $5$, both numbers are in $A$.  The number of such "ambiguous points" is small (countable, therefore measure zero), but recognizing that there is a little wrinkle here ensures that the set can be described accurately, and may have implications with respect to other questions (e.g. about the topology of the set being considered).

It is correct to note that this set is similar to "the" Cantor set.  One common description of the Cantor set is that it consists of all of the numbers in $[0,1]$ possessing a ternary expansion which does not contain a $1$.  In fact, the set being considered in this question is a Cantor set, if not the usual middle-thirds or ternary Cantor set.  If one already knows how to show that the Cantor set has measure zero, then adjusting that argument to the current context should be doable.

With that in mind, we can give a more precise statement of the problem, as well as a description of the set which is a little easier to work with:

Define the functions $\varphi_{j} : [0,1] \to [0,1]$ by
$$ \varphi_j(x) = \frac{1}{10} x + \frac{j}{10}, $$
where $j \in \{0, 1, 2, 3, 4, 6, 7, 8, 9 \}$.  Let $A_0 = [0,1]$, and for each natural number $n$ define
$$ A_{n+1} = \bigcup_{j=0}^{4} \varphi_j(A_n) \cup \bigcup_{j=6}^{9} \varphi(A_n). $$
Then
$$ A = \bigcap_{n=0}^{\infty} A_n. $$

Is $A$ Lebesgue measurable?

What is the Lebesgue measure of $A$?

This looks gross, but watching what happens in the first iteration may help make clear what is going on:  the initial set $A_0$ is mapped by the functions $\varphi_j$ into nine closed intervals, which are then unioned up:
\begin{align}
A_1
&= \bigcup_{j=0}^{4} \varphi_j(A_0) \cup \bigcup_{j=6}^{9} \varphi_j(A_0) \\
&= \bigcup_{j=0}^{4} \left[ \frac{j}{10}, \frac{j+1}{10} \right] \cup \bigcup_{j=6}^{9} \left[ \frac{j}{10}, \frac{j+1}{10} \right] \\
&= [0.0, 0.1] \cup [0.1,0.2] \cup [0.2,0.3] \cup [0.3,0.4] \\
&\qquad\cup [0.6,0.7] \cup [0.7,0.8] \cup [0.8,0.9] \cup [0.9, 1.0] \\
&= [0,1] \setminus (0.5,0.6).
\end{align}
Hence $A_1$ consists precisely of those numbers in $[0,1]$ which possess a decimal expansion non containing a $5$ in the tenths place (again, note that $0.5 = 0.4\overline{9}$, which does not contain a $5$ in the tenths place).

Following a similar, though much more tedious, argument, it is possible to show that $A_2$ consists of all of those points in $[0,1]$ which possess a decimal expansion that does not contain a $5$ in either the tenths or hundredths place.  Indeed, $A_n$ is precisely the set of numbers in $[0,1]$ which possess a decimal expansion which does not contain a $5$ at or before the $n$-th place of the decimal expansion.  Intersecting all of these sets gives the set of points possessing a decimal expansion with no $5$s at all.

Is $A$ Lebesgue measurable?

Constructing a workable description of the set is the hard part.  With this description in hand, answering the question of measurability should be reasonably straight forward:

Note that each $A_n$ is the union of $9^n$ closed intervals ($A_0$ is one closed interval, $A_1$ is the union of nine closed intervals, $A_2$ is the union of 81 closed intervals, and so on).  Closed intervals are measurable sets, and the $\sigma$-algebra of Lebesgue measurable sets is closed under countable unions, so each $A_n$ is Lebesgue measurable.

Each $A_n$ is measurable and $A$ is the intersection of the $A_n$.  The $\sigma$-algebra of Lebesgue measurable sets is closed under countable intersections, hence $A$ is Lebesgue measurable, which gives an affirmative answer to the first question.

What is the measure of $A$?

As noted above, $A_n$ consists of $9^n$ closed intervals.  Each one of these intervals has Lebesgue measure $10^{-n}$ ($A_0$ consists of the closed interval $[0,1]$, which has measure $1$; $A_1$ consists of the $9$ intervals listed above, each of which has measure $1/10$; and so on).  The intervals overlap a little, so we can't directly use the additivity of the measure with respect to disjoint unions, we can use subadditivity:

\begin{align}
m(A_n)
&= m( \text{$9^n$ closed intervals of measure $10^{-n}$} ) \\
&\le m( \text{$9^n$ disjoint closed intervals of measure $10^{-n}$} ) \\
&=  9^n \cdot 10^{-n} \\
&= \left( \frac{9}{10} \right)^n.
\end{align}

Recall that if $m(A_0) < \infty$ and $A_n \supseteq A_{n+1}$ for all $n$, then $$m\left( \bigcap_{n=0}^{\infty} A_n \right) = \lim_{n\to\infty} m(A_n)$$
(this is continuity from above).  It is not too difficult to show that the $A_n$ are nested in this way, and so
$$ m(A)
= m\left( \bigcap_{n=0}^{\infty} A_n \right)
= \lim_{n\to \infty} m(A_n)
\le \lim_{n\to \infty} \left( \frac{9}{10} \right)^n
= 0.
$$


问:
Explanation of $\frac{sgn(x) + 1}{2} \approx sigmoid(x)$
I saw an approximation in my textbook which states

$$ \frac{sgn(x) + 1}{2} \approx sigmoid(x)$$

Why is this true? Is there any famous reference to where this approximation came from or even some kind of derivation/proof? As I cannot find a relationship, one is continuous and one is not.

Here is a plot of sgn(x)

And here is a plot of sigmoid(x). I can see they are similar at $y=0,1$ but not in between, as there is a jump.

答: For positive $x\gg1$, $e^{-x}\ll1$ so $\sigma(x)\approx1-e^{-x}\approx1$. For positive $-x\gg1$, use $\sigma(-x)=1-\sigma(x).$ Small $x$ satisfy $\sigma(x)\approx\sigma(0)=\tfrac12$, in line with the Gibbs phenomenon for a Fourier treatment of step functions.


问:
An isoperimetric problem
How can I find a triangle which encloses the largest area with given (total) length using caculus of variation? I know the direct method but I can't find one using variational method.</p>

答:
Let the sides be of length

$$ (2x,2y,2L) $$

where $L$ is a constant and $x,y$ are variables.

Semi-perimeter constraint function is
$$ p=(x+y+L) $$

Square of largest area objective function is
$$ q=A^2=(-L+x+y)(L+x+y)(L-x+y)(L+x-y) $$

Apply Euler-Lagrange equation and simplify algebraically using symmetry in variables ($x,y),$

Area square $q$
$$-L^4 + 2 L^2 x^2 - x^4 + 2 L^2 y^2 + 2 x^2 y^2 - y^4$$
resulting in relation between partial derivatives
$$\dfrac{p_x}{p_y} =\dfrac{q_x}{q_y} \text{ i.e.,}$$

$$\frac{\frac{\partial p}{ \partial x}}{\frac{\partial p}{ \partial y}}=
\frac{\frac{\partial q}{ \partial x}}{\frac{\partial q}{ \partial y}}$$

$$\dfrac{4L^2-4x^3+4xy^2}{4L^2y+4x^2y-4y^3}=1$$

Simplifying, $(x-y)$ is found to be a factor. So

$$ x=y=L,$$

i.e., maximum area is enclosed for an equilateral triangle.

Title: An isoperimetric problem


问:
What is the probability that it was the two-headed coin?
There are four coins in a box. One is a two-headed coin, two are fair coins, and the third is a biased coin that comes up tails 65 percent of the time. When one of the four coins is selected at random and flipped, it shows heads.

Please help me to solve it. Advance thanks

答:
The probability you're asking for is the conditional probability of having thrown the two-headed coin after knowing the result is heads. It's value can be calculated by the following expression (calling $X$ the event of throwing the two headed coin, and $H$ the event of getting heads after the throw):
$$P[X|H]=\frac{P[X\cap H]}{P[H]},$$
so we calculate $P[H]$, which clearly is:
$$P[H]=\frac{1}{4}+\frac{1}{4}\cdot\frac{1}{2}+\frac{1}{4}\cdot\frac{1}{2}+\frac{35}{100}\cdot\frac{1}{4}=\frac{47}{80}.$$
We already know that $P[X\cap H]=P[X]=\frac{1}{4}$ because it will always give you heads as result and the coins are selected with equal chance. So the final probability is:
$$P[X|H]=\frac{P[X\cap H]}{P[H]}=\frac{\frac{1}{4}}{\frac{47}{80}}=\boxed{\frac{20}{47}}$$


问:
Why $2n = 6k - 1$ doesn't have a solution?
Can't we do something like this:

$2n = 6k - 1$ where k is an integer and n is a normal number

We treat $k' = 2k$ which means

$2n = 3k' - 1$

And by using modulo arithmetic

$2n = -1 (mod 3)$

Which is $n = 1 (mod 3)$ since 2 and 3 are co-prime.

Then the solution is that $n = 3p + 1$ where p is an integer.

The book says that it has no solution but isn't this a solution?

I'm definitely wrong but I applied this solution to many problems before, I dunno why I can't apply it here?

EDIT: I'm so sorry, I mean $n$ is a natural number not normal.

答: The left hand side of your equation is even. The right hand side is odd. It's going to be very tough to find a number that's both even and odd.</p>


问:
Sum of two orthogonal projections - equivalent properties
I am trying to prove the following equivalences.

Let $P_1$ and $P_2$ be the orthogonal projections on the closed subspaces $M_1$ and $M_2$ respectively. The following are equivalent:

(i) $P_1+P_2$ is an orthogonal projection

(ii) $P_1P_2 = 0$

(iii) $M_1\perp M_2$

I have proved $(ii) \to (iii)$, any hint on the other implications would be precious.

答:
Let me prove (i) $\Rightarrow$ (ii), while giving you some more time to think about (iii) $\Rightarrow$ (i), which is
not so hard.

By hypothesis  $$
  P_1+P_2=
(P_1+P_2)^2 =
  P_1^2+P_1P_2+P_2P_1+P_2^2 =
  P_1+P_1P_2+P_2P_1+P_2.
  $$
so $P_2P_1=-P_1P_2$.  Left multiplying the above by  $P_2$ we get
$$
  P_2P_1=-P_2P_1P_2.
  $$
Since $P_2P_1P_2$ is self-adjoint, then so is $P_2P_1$.  But then
$$
  P_1P_2=  P_1^*P_2^*= (P_2P_1)^* = P_2P_1 = -P_1P_2,
  $$
whence $P_1P_2=0$.


问:
Check my proof that $(ab)^{-1} = b^{-1} a^{-1}$
The following question is problem Pinter's Abstract Algebra. And to put things in context: $G$ is a group and $a, b$ are elements of $G$.

I want to show $(ab)^{-1}$ = $b^{-1}a^{-1}$.

I originally thought of proving the fact in the following manner:
\begin{align*}
(ab)^{-1}(ab) &= e \newline
(ab)^{-1}(ab)(b^{-1}) &= (e)(b^{-1}) \newline
(ab)^{-1}(a)(bb^{-1}) &= (b^{-1}) \newline
(ab)^{-1}(a)(e) &= (b^{-1}) \newline
(ab)^{-1}(a) &= (b^{-1}) \newline
(ab)^{-1}(a)(a^{-1}) &= (b^{-1})(a^{-1}) \newline
(ab)^{-1}(e) &= (b^{-1})(a^{-1}) \newline
(ab)^{-1} &= (b^{-1})(a^{-1}) \newline
\end{align*}

I know this may seem extremely inefficient to most, and I know there is a shorter way. But would this be considered a legitimate proof?

Thanks in advance!

Body:
For a direct proof systematically using the associative property and the fact that $x = xe = xyy^{-1}$ for all $y\in G$  proceed as follows.
\begin{align}
(ab)^{-1}
=
&
(ab)^{-1}e
\\
=
&
(ab)^{-1}[aa^{-1}]
\\
=
&
(ab)^{-1}\big[(ae)a^{-1}\big]
\\
=
&
(ab)^{-1}\Big[\big(a[bb^{-1}]\big)a^{-1}\Big]
\\
=
&
(ab)^{-1}\Big[([ab]b^{-1})a^{-1}\Big]
\\
=
&
(ab)^{-1}\Big[[ab]\big(b^{-1}a^{-1}\big)\Big]
\\
=
&
\Big[(ab)^{-1}[ab]\Big]\big(b^{-1}a^{-1}\big)
\\
=
&
e\big(b^{-1}a^{-1}\big)
\\
=
&
b^{-1}a^{-1}
\end{align}


问:
How to calculate the length of arcs in the given periodic function?
I have this function
$$f(x)=-11 \cos x+11 \cos 2 x+6$$
with period $2\pi$. The plot of function is on the left below. Then, I want to know how I can calculate the lengths of the four arcs for which $-2<f(x)<6$  that is the plot on the right. Any comments or hints are appreciated.

答:
If we have a curve, $y=f(x)$, then the length of the curve between $x=a$ and $x=b$ is equal
$$\int_a^b\sqrt{\left(\frac{dy}{dx}\right)^2+1}~dx$$
Does that help?


问:
Why is $d(x^2)= 2xdx$?
This seems a bit obvious, but I'm not clear on one part. My teacher had told us $d(x^2) =2xdx$.
However on asking why, we were told $d(x^2)= (x+dx)^2-x^2= 2xdx+(dx)^2$. We were told that we can ignore the $(dx)^2$ because its small.
For obvious reason I feel this step is wrong, why can we ignore $(dx)^2$, is their any other way of getting the result?

答:
by definition:
$$\frac{df(x)}{dx}=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}\tag{1}$$ if we let $f(x)=x^2$ then we get:
$$\frac{d(x^2)}{dx}=\lim_{h\to0}(2x+h)=2x$$
now remember that whilst $dx$ represents a small change, it is non-zero and with the chain rule we can say:
$$df=\frac{df}{dx}dx\tag{2}$$
which for us means:
$$d(x^2)=2xdx$$

I think where you are getting confused with "ignoring" this $(dx)^2$ term is in finding the limit to define the derivative. There are some other interesting cases that we can derive where we end up with terms like:
$$\sin(dx)$$
where we say that this is equal to $dx$ but this is not the same situation. Hope this helps

One interesting way of visualising the derivative is using equal sided shapes (i.e. squares, cubes etc.) and looking at what the change in area/volume/hyper-volume is as we make incremental changes to the lengths of sides. For example, lets take a normal square of side length $x$. The area of this square will be $x^2$. Now lets increase the length of all sides by a small amount $\delta x$. The new length of the sides will be $x+\delta x$, which makes the new area $(x+\delta x)^2$. Now if we expand this out we get:
$$A(x)=x^2$$
$$A(x+\delta x)=(x+\delta x)^2+x^2+2x\delta x+(\delta x)^2$$
The change in the area, which we will denote $\delta A$ will be as follows:
$$\delta A=2x\delta x+(\delta x)^2$$
teachers often use this to define the derivative by saying "the differential of the area is equal to the change in area divided by the change in side length" and whilst this is true we get given the following expression:
$$\frac{\delta A}{\delta x}=2x+\delta x$$
Which can confuse many people as we clearly have this additional $\delta x$ term which does not agree with the limit definition of the derivative at first appearance. However, if you look at the limit we took we did in fact have a term of $h$ left over but since $h\to 0$ it does not change our value of $\frac{df}{dx}$. The same is true when we talk of it in this form, whilst $\delta x$ has a value which is not close to zero we cannot negate it from our equations, but for small enough values the difference in the order of magnitudes between $x$ and $\delta x$ means that as $\delta x$ gets smaller it has an unnoticeable impact on our expression

Title: Why is $d(x^2)= 2xdx$?


问:
Let $\{p,q\}$ be linearly independent polynomials. Show that $\{p,q,pq\}$ is linearily independent if and only if $1\leq deg(p)$ and $1\leq deg(q)$
Let $\{p,q\}$ be linearly independent polynomials. Show that $\{p,q,pq\}$ is linearily independent if and only if $1\leq deg(p)$ and $1\leq deg(q)$

My Proof Attempt

$[\rightarrow]$ If $\operatorname{deg}(q) = \operatorname{deg}(p)=0$, then they are all constants. Hence, by finding a common factor I can find $pa_1+qa_2+pqa_3=0$ such that not all $a_i = 0$. Hence $1 \leq \operatorname{deg}(q)$ and $1 \leq \operatorname{deg}(p)$.

$[\leftarrow]$ if $1\leq \operatorname{deg}(p)$ and $1\leq \operatorname{deg}(q)$ then $ \operatorname{deg}(p)\operatorname{deg}(q)< \operatorname{deg}(pq)$. A higher degree polynomial can not be written as a sum of lower degrees. Hence $\{p,q,pq\}$ must be linearly independent. QED

My Proof Attempt II (from suggestions edited, second part)

$[\rightarrow]$ Contrapositive: $\operatorname{deg}(q)<
 1 $ OR $\operatorname{deg}(p)<1$. WLOG, $p$ and $q$ are constants. Hence, by finding a common factor I can find $pa_1+qa_2+pqa_3=0$ such that not all $a_i = 0$. Hence $1 \leq \operatorname{deg}(q)$ and $1 \leq \operatorname{deg}(p)$.

$[\leftarrow]$ if $1\leq \operatorname{deg}(p)$ and $1\leq \operatorname{deg}(q)$ then $ \operatorname{deg}(p)< \operatorname{deg}(pq)$ and $\operatorname{deg}(q)< \operatorname{deg}(pq)$. A higher degree polynomial can not be written as a sum of lower degrees. Hence $\{p,q,pq\}$ must be linearly independent, since we know $\{p,q\}$ is L.I. and pq can not be written as a sum of $p$ and $q$. QED

Is this approach correct?

答:
Let's first look at your forward direction (which I think you're handling via contraposition):

Namely, the negation of "$1 \le \deg(p)$ and $1 \le \deg(q)$" is "$1 > \deg(p)$ or $1 > \deg(q)$". So the start of this proof should be "WOLOG, $p$ is a constant." I think you'll find that the fix is minor though.

Now, the backward direction:

It concerned me that you didn't use the hypothesis that $\{p, q\}$ are linearly independent. Moreover, your inequality $\deg(p) \deg(q) < \deg(pq)$ isn't correct: take $p = x^2$ and $q = x^3$, so that $pq = x^5$ has degree strictly smaller than 6! I am happy to provide hints on how to incorporate these ideas.

Edit: Backwards direction looks great. (You showed in effect that $pq$ is outside of the span of $\langle p, q \rangle$ due to degree considerations.) However, for the forward direction the WLOG is that only one polynomial is a constant.


问:
How to interpret Variation Inflation factor of $1$, when tolerance is equal to $1$ and there are only two predictors.
So, I need to determine multicollinearity of predictors, but I have only two. So, if VIF = $\frac{1}{1-R_j^2}$, then in case there are no other predictors VIF will always equal $1$? So, maybe it's not even possible to gauge multicollinearity if there are only two predictors?

答: Let's say you have $Y$ and two predictors $X_1$ and $X_2$, then the $R_j^2$ in the VIF index is $R_1^2$ obtained from the model $X_1 = \alpha_0 + \alpha_1X_2 + \xi$. That is, unless $X_1$ and $X_2$ are uncorrelated, $R^2_1 > 0$, hence VIF $>1$.


问:
Properties of Continuous Functions
Prove that there is no continuous function $f:\mathbb{R} \rightarrow \mathbb{R}$ such that for $c \in \mathbb{R}$ the equation $f(x)=c$ has exactly two solutions.
This is what I have so far.

Proof by contradiction, suppose $f:\mathbb{R} \rightarrow \mathbb{R}$ is continuous and the equation $f(x)=c$ has exactly two solutions, $a$ and $b$ with $a<b$.  On the interval $[a,b], f(t)\leq c$ for all $t$, or $f(t) \geq c$ for all $t$.  Assume the latter.  Then $f(t)<c$ for $t<a$ and $f(t)<c$ for $t>b$.  $f$ has a maximum value on $[a,b]$ at exactly two points, $d$ and $g$, assume $d<g$.  Now choose any $t$ such that $d<t<g$.  Then $f(a)=c<f(t)<f(d)$ and $f(b)=c<f(t)<f(g)$.  So the equation $f(x)=f(t)$ has three solutions, $t, c_{1}, c_{2}$ where $a<c_{1}<d$ and $g<c_{2}<b$ by the intermediate value theorem.

答: Let $x_1(c), x_2(c)$ be two distinct solutions of $f(x) = c$ chosen so that both $x_1$ and $x_2$ are continuous functions. $x_1(c) \neq x_2(c)$ for any $c \in \mathbb{R}$, assume w.l.o.g. that $x_1(c) < x_2(c)$. Thus, the image of the function $x \mapsto (x, f(x))$ is not connected. But $\mathbb{R}$ is connected and $f$ is continuous - contradiction.</p>

Title: Properties of Continuous Functions


答:
So if $n=3p+1$ then $$6p+2=6k-1$$ or $$6(p-k)=3$$

Now both sides of this are divisible by $3$, so this is a valid equation modulo $3$, but $3$ is not a multiple of $6$.

Working modulo $3$ only captures part of the information in the original equation, so doesn't guarantee to find a solution of the original.

There is an interesting development in more advanced mathematics when one does look to solve equations modulo $p$ for primes $p$ (called working locally). Then one looks for conditions under which the existence of a local solution for every prime $p$ guarantees a global solution - ie a finite integer solution to the original. Note that even finding a solution for every prime does not guarantee a global solution (to more complex kinds of problems).

Title: Why $2n = 6k - 1$ doesn't have a solution?

Body:
Can't we do something like this:

$2n = 6k - 1$ where k is an integer and n is a normal number

We treat $k' = 2k$ which means

$2n = 3k' - 1$

And by using modulo arithmetic

$2n = -1 (mod 3)$

Which is $n = 1 (mod 3)$ since 2 and 3 are co-prime.

Then the solution is that $n = 3p + 1$ where p is an integer.

The book says that it has no solution but isn't this a solution?

I'm definitely wrong but I applied this solution to many problems before, I dunno why I can't apply it here?

EDIT: I'm so sorry, I mean $n$ is a natural number not normal.


问:
Why is this not an exponential distribution?
On an exercise sheet I found the following question:

You monitor planes landing at an international airport. X measures the time intervals between consecutive landings. Is X exponentially distributed (in good approximation)?

To me it seemed that this is a clear case of an exponential distribution. However, the correct answer, supposedly, is no.

Can you explain to me why X is not exponentially distributed?

答:
It slightly depends on how busy the airport is, but note

(a) the mode of an exponential distribution is $0$ while (ignoring emergencies) there is usually a substantial gap between planes landing so the first can clear the runway, and

(b) in a busy airport with restricted landing slots the controllers try to make planes tend to land with close to equal gaps between them to balance efficiency and safety, so a low standard deviation of gaps compared to their average, while an exponential distribution has a standard deviation equal to its average


问:
Need Greatest common divisor explaination
In the book the problem says:
What is the GCD of two natural numbers m and n if after increasing the number m by 6 the GCD of two numbers increased 9 times.

And the solution it gives is 2,3 or 6.
Can someone explain how is it. And what values of m and n satisfy the equation. I can't find any. May be i couldn't understand the question or the book is miss printed. Idk. Please help me understand this with some examples. Thanks

答:
Hint $\, $ By below, with $\,j=2,\, k = 3\,\Rightarrow\, 3\mid (m,n)\mid 6$

Lemma $\, \ k,(m,n)\mid (m,\:\!n\!+\!jk)\!\iff\! k\mid (m,n)\mid jk$

Proof $\,\ $ Note $\ (m,n)\mid n,\,n\!+\!jk\,\iff (m,n)\mid jk,\, $

and $\ k\mid m,n\!+\!jk\! \iff k\mid m,n\iff k\mid (m,n)$


答:
You have:

$$2n+1=6k$$
so:

$$2n+1=2(3k)$$
if you put $3k=h$
you have:

$$2n+1=2h$$
So an odd number shoud be equal to an odd number which is impossible


问:
When is the pullback functor on sheaves faithful?
For a flat finite surjective morphism of smooth varieties $f : X \rightarrow Y$ we have the pushforward functor $f_* : \mathcal{S}h (X) \rightarrow \mathcal{S}h (Y)$ and its left adjoint $f^* : \mathcal{S}h (Y) \rightarrow \mathcal{S}h (X)$ between coherent sheaves of $\mathcal{O}_X$-modules and $\mathcal{O}_Y$-modules. Is it true that $f^*$ is a faithful functor? It doesn't seem obvious to me...I don't mind if it is full or not. Thanks!</p>

答: If $f : X \to Y$ is a flat and surjective, i.e. faithfully flat morphism, then $f^* : \mathsf{Qcoh}(Y) \to \mathsf{Qcoh}(X)$ is faithful. In fact, it is exact since $f$ is flat, so that it remains to prove $f^* M = 0 \Rightarrow M = 0$. But this can be checked locally, and is one of the well-known characterizations of faithfully flat ring homomorphisms: $A \to B$ is faithfully flat iff it is flat and $M \otimes_A B = 0 \Rightarrow M=0$.</p>


问:
Summing $1+\cos(\theta)+\cos(2\theta) +\cdots + \cos(n\theta)$
First of all, I'd like to acknowledge that there are already solutions to this question on this forum. However, I'm repeating the question here because my solution doesn't quite get me what those other solutions are and I'm wondering if I'm doing something wrong.

I know that taking $z\in \mathbb{C}$ s.t. $|z|=1$ and $\arg(z) = \theta$, we get $z=\cos(\theta)+i\sin(\theta)$. Then by DeMoivre's Theorem we get $z^n= (\cos(\theta)+i\sin(\theta))^n=\cos(n\theta)+i\sin(n\theta)$, and we see that the sum
$1+\cos(\theta)+\cos(2\theta) +\cdots + \cos(n\theta)$ is equal to $\operatorname{Re}(z^0+z^1+\cdots+z^n)$ (call it $S$).

From there, using geometric sums, $S=\frac{z^{n+1}-1}{z-1}$. In order to find $\operatorname{Re}(S)$, I am multiplying the top and the bottom by $\overline{z-1}$ to get $S=\frac{(z^{n+1}-1)*(\overline{z-1})}{(z-1)*(\overline{z-1})}$.

This is where I'm running into trouble. Expanding $z$ into $\cos(\theta)+i\sin(\theta)$ I get that the denominator is equal to $2-2\cos(\theta)$ (and whatever other variants of this using trig identities). However, this doesn't seem to line up with any of the other standard answers I'm finding online (which mostly involve a $\sin(\frac{\theta}{2})$ in the denominator).

I was wondering if anyone could point out if/where I've gone wrong in my process. Is there another way to handle this sum (while still using DeMoivre's Theorem)?

答:
$$\frac{z^{n+1}-1}{z-1}=\frac{(z^{n+1}-1)\overline{(z-1)}}{|z-1|^2}\tag{$\ast$}$$

Denominator is real, so you need to separate numerator into real and imaginary parts.

$$(z^{n+1}-1)=(\cos((n+1)\theta)-1)+i\sin((n+1)\theta)\tag1$$$$\overline{(z-1)}=(\cos\theta-1)-i\sin\theta\tag2$$

Multiply (1) and (2) and extract real part.

The real part of the expression in $(\ast)$ is

$$\frac{(\cos((n+1)\theta)-1)(\cos\theta-1)+\sin((n+1)\theta)\sin\theta}{(\cos\theta-1)^2+\sin^2\theta}$$

The solution you saw with $\theta/2$ in bottom most likely simplified the denom as $2(1-\cos\theta)$ and used the identity $\cos(x)=\cos^2(x/2)-\sin^2(x/2)=1-2\sin^2(x/2)=2\cos^2(x/2)-1$

The numerator might also be simplified by expanding out and using trig identities which express $\cos x\cos y$ and $\sin x\sin y$ as sums of $\sin$ and $\cos$


问:
Is real segment [0; 1] enumerable?
Cantor's diagonal argument shows that there is no bijection between the real segment [0; 1] and natural numbers. Does that mean that we can't enumerate this set ?

If we take an application f(n) from N to R that, for any number, reverse it and place it after the comma, like it:

```<code>f(0) = 0.0
f(1) = 0.1
...
f(1234) = 0.4321
...
</code>```

We clearly see that the application enumerate any given finite real in [0; 1]. What's wrong ?

答:
Well it makes sense that you can attempt to "construct" such a bijection since those are all rational numbers (ones of the form: $x_Nx_{N-1}...x_0.y_1y_2...y_M$) and the field of rational numbers $\mathbb{Q}$ is countable, so $\mathbb{Q}\cap I$ is countable for any interval $I\subset \mathbb{R}$. You can give an ordering to any interval by just taking the well ordering theorem (by assuming the axiom of choice), but that ordering will not be indexed by the natural numbers (since that would contradict the fact that intervals are uncountable, unless the interval is empty or a singleton).

Here is an intuition for what is going on: any subset of an interval I indexed by the natural numbers, is a sequence in I, this sequence is visually a (one-dimensional) "lattice" of points in I, and the actual interval I itself is too "large" to be "filled in" by a sequence.

More clearly, $[0,1]$ has the same cardinality as $J:= (-\pi/2, \pi/2)$ (take a dilation bijection from J to the unit interval $(0,1)$ and then another bijection to $[0,1]$), but $tan(x)\mid _{(-\pi/2, \pi/2)}$ is a bijection from J to the real line (perhaps it's more intuitive to see that $\mathbb{R}$ does not have an enumeration by $\mathbb{N}$ than the closed unit interval).

Title: Is real segment [0; 1] enumerable?


问:
Integrate:$ \int_{0}^{1} x^{1/2}(1-x^{2})^{3/2}dx$
Integrate:
$\int_{0}^{1} x^{1/2}(1-x^{2})^{3/2}dx$

I tried substituting $x=\sin \theta$ and ended up with $\int_{0}^{\pi/2} \sqrt{\sin\theta}\cos ^{4}\theta d\theta$. Further substitution made it more complex.
After this what should I do?

答:
$$I=\int_0^1 x^{1/2}(1-x^2)^{3/2}dx$$
$u=x^2\Rightarrow dx=\frac{du}{2x}$ now sub in:
$$I=\int\limits_{u=0}^1u^{1/4}(1-u)^{3/2}\frac{du}{2u^{1/2}}=\frac12\int_0^1u^{-1/4}(1-u)^{3/2}du$$
Now the definition of the beta function is:
$$B(x,y)=\int_0^1t^{x-1}(1-t)^{y-1}dt\tag{1}$$
$$B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}\tag{2}$$

where $\Gamma$ is the Gamma function, an analytic continuation of the factorial function.


问:
How to show $x^4 - 1296 = (x^3-6x^2+36x-216)(x+6)$
How to get this result: $x^4-1296 = (x^3-6x^2+36x-216)(x+6)$?

It is part of a question about finding limits at mooculus.

答:
A bit more roundabout way:

$x^4 - 1296 \ $ is a difference of two squares, as is $ \ x^2 - 36 \ $, so

$$(x^4 - 1296) \ = \ [ \ (x^2)^2 \ - \ (36^2) \ ] \ = \ (x^2 + 36) \ \cdot (x^2 - 36 ) $$

$$= \ (x^2 + 36) \cdot (x - 6) \cdot (x + 6) \ = \ (x^3 \ + \ 36x \ - \ 6x^2 - 216 ) \cdot (x + 6) \ . $$

EDIT: I missed that Πάρτη Κοηλί had done this going down to complex roots; this version "keeps things real"...

All of the answers I see at the present are fine -- there isn't any getting around that one needs to recognize that 1296 is $ \ 6^4 \ $ , or otherwise needs to find enough factors of 1296 to see a way to "break down" the polynomial.


答: None of your inputs map to $\frac13 =  0.\overset{.}{3}.$

Title: Is real segment [0; 1] enumerable?

Body:
Cantor's diagonal argument shows that there is no bijection between the real segment [0; 1] and natural numbers. Does that mean that we can't enumerate this set ?

If we take an application f(n) from N to R that, for any number, reverse it and place it after the comma, like it:

```<code>f(0) = 0.0
f(1) = 0.1
...
f(1234) = 0.4321
...
</code>```

We clearly see that the application enumerate any given finite real in [0; 1]. What's wrong ?


问:
Laplace transform of $\int_t^{+\infty}\,\psi(\tau)\,d\tau$
Why this equality?

$$\mathcal{L}\left(\int_t^{+\infty}\,\psi(\tau)\,d\tau\right)=\frac{1-\psi(s)}{s}$$

Can you help me with the steps?

答:
Denoting $\Psi(t) = \int_t^\infty \psi(\tau) \, d\tau$, we have by definition of $\mathcal L$, that
\begin{align*}
  \mathcal L(\Psi)(s) &= \int_0^\infty \exp(-st)\Psi(t)\, dt\\
      &= \int_0^\infty \exp(-st)\int_t^\infty \psi(\tau)\, d\tau\, dt\\
      &= \int_0^\infty \int_t^\infty \exp(-st)\psi(\tau)\, d\tau\, dt\\
      &= \int_0^\infty \int_0^\tau \exp(-st)\psi(\tau)\, dt\, d\tau\\
      &= \int_0^\infty \int_0^\tau\exp(-st)\, dt\cdot \psi(\tau)\, d\tau\\
      &= \int_0^\infty \left(\frac 1s -\frac{1}s\exp(-s\tau)\right)\psi(\tau)\, d\tau\\
      &=\frac{\int_0^\infty \psi(\tau)\, d\tau - \mathcal L(\psi)(s)}{s}
\end{align*}
So your equation is true, if $\int_0^\infty \psi(\tau)\, d\tau = 1$ and $\mathcal L(\psi) = \psi$.</p>


问:
Why an internal map, maps an internal set to an internal set?
In this paper page 129, assumed as a fact that an internal map, maps internal subsets to internal subsets but I can't see why this happens.....</p>

答:
Any function $f$ in $M$ trivially has the property that the image of every set under $f$ is a set. We can express the property in the formal language of $M$ (since said language is higher-order, consult page 125, paragraph 3), as e.g. $\forall S. \exists S'. \forall x. x \in S' \leftrightarrow \exists y \in S. f(y) = x$.

By the Extension Principle (Section 2.2. of page 126), every property of $f$ which is expressible in the language of $M$ also holds for its extension in $\!^{\star}M$, with the caveat that the quantifiers now range over internal entities (page 125, bullet point iii). So we obtain the property that the image of every internal set under the standard function $f$ is itself an internal set.

A similar argument (using an additional $\forall f$ and appealing to the permanence principle instead of extension) allows us to conclude that generally the image of every internal set under an internal function is itself internal.


问:
redistributing value according to proportion
If 1/2 of a population receive £400 each, and the other 1/2 £400 too, then if we redistribute the value from the second population to the first the first will get £800 each.

However I would like to know the general principle for formulating this type of problem. For example I cannot figure out how to restribute from second to first if

1st population: 2/3 get £400

2nd population: 1/3 get £400

How might I do this?

答:
After the redistribution, the first population get

$$400+\frac{3\times 400}{2\times 3}=400+200=600$$

each person.

alternative reasoning

Assume you have a total population of $9$

Before the redistribution you have

$6$ persons with $400$ each that is $6\times 400=2,400$ total income Pop #1

$3$ persons with $400$ each that is $3\times 400=1,200$ total income Pop #2

The total income is now $3,600$ to be divided by 6 persons....

that means $600$ each

Of course you can choose a total number you want as a population. If you choose $n=1$ you get the first formula I showed


问:
Combinatorics - Letter Ordering
Considering the letters STUVWXYZ, how many strings can be formed with the letters Y, U, T and S appearing in that order but not necessarily together?</p>

答:
First, place Y, U, T, S. There will be 8 possible locations in the string $x_1x_2x_3x_4x_5x_6x_7x_8$ to place those letters. In other words, every placement will correspond to a 4-element subset of positions chosen from the set $\{1, 2, \dots, 8\}$. You probably know that there are $\binom{8}{4}$ such subsets.

Having done that, we're left with 4 positions, say $i_1, i_2, i_3, i_4$, into which we have to place the remaining letters, V, W, X, Z. We have 4 choices for the letter in place $i_1$. Having filled that slot, there are 3 letters that can go in position $i_2$, then 2 remaining letters to go in position $i_3$ and of course only one letter to place in position $i_4$. The total number of arrrangements, then, will be
$$
\binom{8}{4}\cdot4\cdot3\cdot2\cdot1
$$
as Quixotic indicated.

Title: Combinatorics - Letter Ordering


问:
Help solving $\int \:\frac{dx}{x+\sqrt{9x^2-9x+2}}$
Reading an article online I came across this integral I've been trying to solve since yesterday morning and I literally tried every integration method I know and I still can't solve it. I am a beginner to integrals so help will be appreciated!

I tried substitution, integration by parts and I can't seem to find the correct way!

$$\int \:\frac{dx}{x+\sqrt{9x^2-9x+2}}$$

Does anyone know how to proceed from here? What to do? Because I am clueless. Thanks!

Body:
The Euler substitution $\sqrt{9x^2-9x+2} =t+3x$ is better suited here, resulting in $x=\frac{2-t^2}{3(2t+3)} $, $dx=-\frac23 \frac{t^2+3t+2}{(2t+3)^2}dt$
\begin{align}
&\int \:\frac{1}{x+\sqrt{9x^2-9x+2}}dx\\
=& -2 \int \frac{t^2+3t+2}{(2t+3)(2t^2+9t+8)}dt\\
=&-\frac12 \int\left(\frac1{2t+3}+\frac t{2t^2+9t+8} \right)dt\\
=& -\frac14\ln|2t+3| -\frac1{8}\ln| 2t^2+9t+8|+\frac9 {8\sqrt{17}}\ln\bigg|\frac{4t+9-\sqrt{17}}{4t+9+\sqrt{17}} \bigg|
\end{align}


问:
How to prove $x^{(nk)} - 1$ can be divided by $x^k - 1$
Here $x$ is the polynomial and $n,k$ are both integers. Why is this true or is not true?

I tried to prove it by induction.

Base case: $(q^{k} - 1) / (q^{k} - 1) = 1$.

Inductive step: $(q^{ck} - 1) / (q^{k} - 1) = f(x)$, then prove that $(q^{(c+1)k} - 1) / (q^{k} - 1) = f'(x)$.
And here is where I stuck, how can I express such $f'(x)$, that is to say, I need to solve $(q^k - 1) * f(x) + q^{k} = (q^k - 1) * g(x) = f'(x) $.

答:
One way to prove it is by using the ciclotomic equation (you can see a prove to this equality in this same site [here][1]): that given $x,y\in\mathbb{R}$ and $n\in\mathbb{N}$, then
$$x^n-y^n = (x-y)(x^{n-1}+x^{n-2}y+\cdots xy^{n-2}+y^{n-1}).$$
If we pick $x$ as your $x^k$, and $y$ as $1$, then we get that
$$x^{nk}-1=\left(x^k\right)^{n}-\left(1\right)^n = \left(x^k-1\right)\left(\left(x^k\right)^{n-1}+\left(x^k\right)^{n-2}+\cdots + x + 1\right),$$
so from this we conclude that $x^{kn}-1$ is divisible by $x^k-1$.

[1]: Proving $x^n - y^n = (x-y)(x^{n-1} + x^{n-2} y + ... + x y^{n-2} + y^{n-1})$)


问:
Is Bayes' Theorem really that interesting?
I have trouble understanding the massive importance that is afforded to Bayes' theorem in undergraduate courses in probability and popular science.

From the purely mathematical point of view, I think it would be uncontroversial to say that Bayes' theorem does not amount to a particularly sophisticated result. Indeed, the relation
$$P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{P(B\cap A)P(A)}{P(B)P(A)}=\frac{P(B|A)P(A)}{P(B)}$$
is a one line proof that follows from expanding both sides directly from the definition of conditional probability. Thus, I expect that what people find interesting about Bayes' theorem has to do with its practical applications or implications. However, even in those cases I find the typical examples being used as a justification of this to be a bit artificial.

To illustrate this, the classical application of Bayes' theorem usually goes something like this: Suppose that

1% of women have breast cancer;

80% of mammograms are positive when breast cancer is present; and

10% of mammograms are positive when breast cancer is not present.

If a woman has a positive mammogram, then what is the probability that she has breast cancer?

I understand that Bayes' theorem allows to compute the desired probability with the given information, and that this probability is counterintuitively low. However, I can't help but feel that the premise of this question is wholly artificial. The only reason why we need to use Bayes' theorem here is that the full information with which the other probabilities (i.e., 1% have cancer, 80% true positive, etc.) have been computed is not provided to us. If we have access to the sample data with which these probabilities were computed, then we can directly find
$$P(\text{cancer}|\text{positive test})=\frac{\text{number of women with cancer and positive test}}{\text{number of women with positive test}}.$$
In mathematical terms, if you know how to compute $P(B|A)$, $P(A)$, and $P(B)$, then this means that you know how to compute $P(A\cap B)$ and $P(B)$, in which case you already have your answer.

From the above arguments, it seems to me that Bayes' theorem is essentially only useful for the following reasons:

In an adversarial context, i.e., someone who has access to the data only tells you about $P(B|A)$ when $P(A|B)$ is actually the quantity that is relevant to your interests, hoping that you will get confused and will not notice.

An opportunity to dispel the confusion between $P(A|B)$ and $P(B|A)$ with concrete examples, and to explain that these are very different when the ratio between $P(A)$ and $P(B)$ deviates significantly from one.

Am I missing something big about the usefulness of Bayes' theorem? In light of point 2., especially, I don't understand why Bayes' theorem stands out so much compared to, say, the Borel-Kolmogorov paradox, or the "paradox" that $P[X=x]=0$ when $X$ is a continuous random variable, etc.

答:
First see the comments following this answer, especially the last few comments.  I was totally unaware that Bayes Theorem is simply a consequence of axioms around the definition of Conditional Probability.  Based on this assertion, I can't refute the idea that the following problem can be solved without Bayes Theorem.

Hard to imagine attacking a conditional probability problem without it.  Imagine traveling back in time 1000 years.  You are the captain of a ship.  You have two sailors, A and B that you independently use to predict rain.

A is right 90% of the time and B is right 80% of the time.

A says it will rain today, and B says it won't rain today.

Absent Bayes Theorem, and absent any info on how often (in general) it rains, how do you (intuitively) determine the chance that it will rain today?  Clearly, the problem is well defined, so the problem has a meaningful answer.  Absent Bayes Theorem, or anything like it, how do you compute the answer?


答:
The simplest method uses the exponential notation: this sum is the real part of
$$S=1+\mathrm e^{i\theta}+\mathrm e^{2i\theta}+\dots++\mathrm e^{ni\theta},$$
which is the sum of a geometric progression with common ratio $+\mathrm e^{i\theta}$, hence
\begin{align}
S&=\frac{\mathrm e^{(n+1)i\theta}-1}{\mathrm e^{i\theta}-1}=\frac{\mathrm e^{\tfrac{(n+1)i\theta}2}\Bigl(\mathrm e^{\tfrac{(n+1)i\theta}2}-\mathrm e^{-\tfrac{(n+1)i\theta}2}\Bigr)}{\mathrm e^{\tfrac{i\theta}2}\Bigl(\mathrm e^{\tfrac{i\theta}2}-\mathrm e^{-\tfrac{i\theta}2}\Bigr)}\\
&=\mathrm e^{\tfrac{ni\theta}2}\,\frac{\mathrm e^{\tfrac{(n+1)i\theta}2}-\mathrm e^{-\tfrac{(n+1)i\theta}2}}{\mathrm e^{\tfrac{i\theta}2}-\mathrm e^{-\tfrac{i\theta}2}}=\mathrm e^{\tfrac{ni\theta}2}\,\frac{\sin\frac{(n+1)\theta} 2}{\sin\frac\theta 2}
\end{align}

and the real part should not be too hard to determine now…

Title: Summing $1+\cos(\theta)+\cos(2\theta) +\cdots + \cos(n\theta)$

Body:
First of all, I'd like to acknowledge that there are already solutions to this question on this forum. However, I'm repeating the question here because my solution doesn't quite get me what those other solutions are and I'm wondering if I'm doing something wrong.

I know that taking $z\in \mathbb{C}$ s.t. $|z|=1$ and $\arg(z) = \theta$, we get $z=\cos(\theta)+i\sin(\theta)$. Then by DeMoivre's Theorem we get $z^n= (\cos(\theta)+i\sin(\theta))^n=\cos(n\theta)+i\sin(n\theta)$, and we see that the sum
$1+\cos(\theta)+\cos(2\theta) +\cdots + \cos(n\theta)$ is equal to $\operatorname{Re}(z^0+z^1+\cdots+z^n)$ (call it $S$).

From there, using geometric sums, $S=\frac{z^{n+1}-1}{z-1}$. In order to find $\operatorname{Re}(S)$, I am multiplying the top and the bottom by $\overline{z-1}$ to get $S=\frac{(z^{n+1}-1)*(\overline{z-1})}{(z-1)*(\overline{z-1})}$.

This is where I'm running into trouble. Expanding $z$ into $\cos(\theta)+i\sin(\theta)$ I get that the denominator is equal to $2-2\cos(\theta)$ (and whatever other variants of this using trig identities). However, this doesn't seem to line up with any of the other standard answers I'm finding online (which mostly involve a $\sin(\frac{\theta}{2})$ in the denominator).

I was wondering if anyone could point out if/where I've gone wrong in my process. Is there another way to handle this sum (while still using DeMoivre's Theorem)?


问:
Find Total number of ways out of N Number taking K numbers every M interval
I have been stuck in a problem, that has thrown my brain out of the coding. This problem is at very high priority and I need the solution as early as possible. Problem is as :

There are exactly N advertising boards on the highway. Now a company want to advertise on some of these advertising boards(each advertising board costs some money) .

Company strategy is that, they want at least 'K' advertisement should be there among M consecutive advertising boards. But at the same time Company want to pay minimum for its advertisement .

Now, what is the total number of ways, different ways Company can advertise meeting its minimum cost strategy.

As for Example: N= 3, M = 2, K=1 ==> there is only one way for minimum cost, ie. 0C0 , where '0' denotes No company advertisement, and 'C' denotes company advertisement board.

Similarly, for N =4, M =2, K =1 ==> there is 3 possible ways, ie. C0C0, 0C0C, 0CC0.

Hope to get a reply from someone soon. Thanks.

答:
For $K=1$ there is a closed form.  The minimum number of billboards is $n=\lfloor \frac NM \rfloor$ and that number suffices if $N$ is increased to $(n+1)M-1$  Imagine a run of $M-1$ o's, a C, $M-1$ o's, a C, ... a C, and $M-1$ o's.  This is the only arrangement for $N=(n+1)M-1$.  Now if $N$ is lower, we can just erase the right number of o's.  These can be chosen n ${N-M \choose (M+1)\lfloor \frac NM \rfloor-N-1}$ ways.

I think the same approach works if $K$ divides $M$ but am not sure.


问:
Why is the exponential distribution specified with parameter $X$ instead of $T$
One thing that always confused me about the exponential distribution was why the probability distribution function or even the expected value is modeled as a function of $X$ instead of a variable like $T$ for time--note capitalization means a random variable. This is not meant to be a nit-picky notation question, but really just trying to understand if I am missing something about the distribution and why $X$ is used.

The exponential is used for waiting time or inter-arrival times in general.

If you look at the Wikipedia article for example, the pdf is specified as:
$$
f(x;\lambda) = \lambda e^{-\lambda x}
$$

The exponential distribution models the probability that an event happens within some continuous interval $[0, t]$, given an intensity parameter $\lambda$. Now other distributions such as the poisson distribution, explicitly reference time as the variable of interest: $P(k, t; \lambda) = \frac{(\lambda t)^k e^{-\lambda t}}{k!}$

Using $x$, the notation for the pdf of the exponential becomes something awkward like $ P(X \leq T; \lambda)$ or equivalently $P(x \in [0, T])$. Wouldn't it seem more intuitive to specify the pdf as $P(T \leq t; \lambda)$, where the random variable is specified as $T$?

答:
I guess so, but actually we use the letter $X$, $Y$ and so on because it is a standard idea to use such letters. Of course "standard" depends on the context. In probability theory or in general in statistics, we mostly use $t$ when talking about $t$-student distribution.

In any case you are entitled to use the letter you like the most, as long as you specify  it: nobody can tell you anything if you write:
$$T\sim Exp(\lambda)$$
Notice moreover that what really matters is how you write the variable you referr to:
Exp will always denote an exponential, $\textit{Po}$ a Poisson, $\mathcal{N}$ a normal and so on.


问:
Difficult matrix equation $A(x-x_0) + \alpha\left(\frac{\Sigma x}{\mu^\top x}\right) = 0$
Let $x,x_0,\theta,\mu \in \mathbb{R}^n$ vectors, $\alpha \in\mathbb{R}$ a scalar, and $\Sigma, A\in \mathbb{R}^{n\times n}$ Positive Semi-Definite matrices. Also, $\mu$ is a mean vector and $\Sigma$ is a covariance matrix of a Random Variable $Z$ (I don't know if this information may help). Is there a way to solve in closed form the following matrix equation for $x$?

\begin{equation}
A(x-x_0) + \alpha \left(\frac{\Sigma x}{\mu^\top x}\right) = 0.
\end{equation}

Main difficulty: I find difficulties in handling the denominator. If the equation would have been simply

$$A(x-x_0) + \alpha \Sigma x = 0,$$ the solution would be $x = (A+\alpha\Sigma)^{-1}Ax_0$. I am looking for a similar solution for the first equation but the denominator complicates things and I don't know how to proceed.

What I tried:

\begin{aligned}
& \mu^TxA(x-x_0) + \alpha \Sigma x = 0
\\ &\mu^TxAx - \mu^TxAx_0 - \alpha \Sigma x = 0
\\ &\mu^TxA(x-x_0) = \alpha \Sigma x
\end{aligned}

Is this way of proceeding correct? If yes, how is it possible to solve this matrix equation for $x$? I am not an expert in solving matrix equations and any help would be greatly appreciated. Thanks!

答:
To extend Hyperplane's comments about the definition of the term "closed-form", suppose you defined a function which, given the parameters $(A,\Sigma,\mu,x_0)$ computes the following iteration until it converges
$$\eqalign{
x_{k+1} &= x_0 - \left(\frac{\alpha A^{-1}\Sigma x_k}{\mu^Tx_k}\right) \\
x_* &= \lim_{k\to\infty} x_k\\
}$$
and returns the matrix
$$\eqalign{
B &= \frac{x_* x_0^T}{x_0^Tx_0}
}$$
This would allow you to write a concise solution to your problem in the desired form, i.e.
$$x = Bx_0$$
Would that meet your definition of "closed form"? If not, then why do you accept
$$\eqalign{Ax = b \quad\implies\quad x=A^+b}$$
as a closed form solution of an over-determined system of equations? If you peek inside the algorithm for the Moore-Penrose inverse you will see similar iterative techniques in use.


问:
A generalization of reflection
This is a problem from Feller's Introduction to probability theory and its aplications, Vol 1. The complete statement of the problem goes like:

Let $a$ and $b$ be positive, and $-b < c < a$. The number of paths to the point $(n,c)$ wich meet neither the line $x = -b$ nor $x = a$ is given by the series:
$$\tag1 \sum_{k={-\infty}}^{\infty} (N_{n,4k(a+b)+c}-N_{n,4k(a+b)+2a-c})$$

Where $N_{n,c}$ is the number of paths from the origin to the point $(n,c)$. The only hint of the problem is to use repeated reflection but i don't get how to do that. I sincerely don't have much of a clue of how to do this problem, any help would be apreciated.

答:
I gave a sketch of a proof in the special case $a=b$ in this answer. It is an inclusion-exclusion type argument; take all $N(n,c)$ paths, then subtract the "bad" paths which cross either $x=a$ or $x=-b$. But you must then add back in the doubly subtracted paths which cross both lines, which can be done by applying the reflection principle twice. However, there are still triple crossing to account for, etc, etc.

Here is a second argument which is more direct.

If you repeatedly reflect the point $(n,c)$ across the lines $x=a$ and $x=-b$, then the points you reach are of the form
$$
(n,2k(a+b)+c)\qquad \text{and}\qquad (n,2k(a+b)+2a-c)
$$
Note the difference between this and the problem statement; Feller is in error.

Let the points in the first form be colored blue, and the second category be colored red. If you reflect a red point though either $x=a$ or $x=-b$, you get a blue point, and vice versa.

For any point $(t,x)$, let $N^*(t,x)$ be the set of "bad" paths to $(t,x)$ which cross either the line $x=a$ or $x=-b$ at some point. The quantity you are trying to calculate is $$N(n,c)-N^*(n,c),\tag1$$ so you just need to find $N^*(n,c)$. This is done with the following lemma:

Lemma:  $$
\sum_{k} N^*(n,2k(a+b)+c)=\sum_k N^*(n,2k(a+b)+2a-c)\tag2
$$

Proof: The left hand side counts the set of bad paths to any blue point, while the RHS counts bad paths to any red point. A bijection from the LHS to the RHS is given by taking a bad path to a blue point, and reflecting everything after the first point it hits either of the lines $x=a$ or $x=-b$. $\square$

The proof is completed by solving $(2)$ for $N^*(n,c)$, substituting that expression into $(1)$, and noting that $N^*(n,c')=N(n,c')$ for any red or blue point $(n,c')$ (except when $c=c'$).

Title: A generalization of reflection


问:
Prove if $f : \mathbb{R}\to\mathbb{R}$ is continuous and $S \subset \mathbb{R}$ is compact, then the image is also compact.
I am not 100% sure about my proof, ideally I would like some hints/confirmation on my proof.

Proof:

Since I want to show $f(S)$ is compact I need to show it is closed and bounded.

Closure

Consider a sequence $(x_n) \subset S$ where  $x_n \to L \in S$ (by compactness). Since $L \in S$ we have that since the function is continuous $f(L) \in f(S)$.

Then using sequential continuity on this sequence  $f(x_n) \to f(L) \in f(S)$. Since we can represent any sequence in $f(S)$ as the image of any given sequence in $S$ the set $f(S)$ is closed.

Bounded

Since $S$ is bounded and $f$ continuous, suppose $s^*$ maximises the function $f$ on $S$ - if $f(s^*) \to \infty$ $f$ is not continuous. Since $max(S)$ and $min(S)$ are finite numbers, we cannot have divergence to infinity of the function over the elements of $s$, so $f(s^*)$ is the upper bound of $f(S)$. Repeating for the lower bound I have that S is bounded.

Is my proof correct. I have concerns that my proof for bounded is not very well worded, also I feel like my proof for closure may miss certain cases.

答:
The idea for the proof is fine as long as you are only concerned about finite dimensional cases. However in general closed and bounded is not a sufficient condition for compactness (in particular in infinite dimensional spaces).

The proof itself also makes some errors, for example what @Max already pointed out.


问:
Why $2n = 6k - 1$ doesn't have a solution?
Can't we do something like this:

$2n = 6k - 1$ where k is an integer and n is a normal number

We treat $k' = 2k$ which means

$2n = 3k' - 1$

And by using modulo arithmetic

$2n = -1 (mod 3)$

Which is $n = 1 (mod 3)$ since 2 and 3 are co-prime.

Then the solution is that $n = 3p + 1$ where p is an integer.

The book says that it has no solution but isn't this a solution?

I'm definitely wrong but I applied this solution to many problems before, I dunno why I can't apply it here?

EDIT: I'm so sorry, I mean $n$ is a natural number not normal.

答:
$\begin{align} {\bf Hint}\quad\   6\mid 2n\!+\!1 &\iff\ \ \ \ 3\mid 2n\!+\!1\ \text{ is false  (your assumption)}\\[.2em]
    6\mid 2n\!+\!1 &\iff \color{#c00}2,3\mid 2n\!+\!1\ \text{ is true (by lcm or CRT)}
\end{align}$

Renark $ $ It is generally true that congruences persist mod factors of the modulus (i.e. that the direction $(\Rightarrow)$ above holds true), and this arrow doesn't reverse for proper factors of the modulus (because it doesn't capture the divisibility information implied by the deleted factors of the modulus, e.g. deleting the factor $\,\color{#c00}2\,$ above loses parity information, which is crucial there).


问:
Compact sets are closed?
I feel really ignorant in asking this question but I am really just don't understand how a compact set can be considered closed.

By definition of a compact set it means that given an open cover we can find a finite subcover the covers the topological space.

I think the word "open cover" is bothering me because if it is an open cover doesn't that mean it consists of open sets in the topology? If that is the case how can we have a "closed compact set"?

I know a topology can be defined with the notion of closed sets rather than open sets but I guess I am just really confused by this terminology. Please any explanation would be helpful to help clear up this confusion. Thank you!

答:
For anyone who comes across this question in the future, here is a proof:

Theorem: Compact subsets of metric spaces are closed.

Proof: Let $K$ be a compact subset of a metric space $X$ and to show that $K$ is closed we will show that its complement $K^c$ is open.

Let $p \in K^c$. Now if $q_\alpha \in K$, let $r_\alpha = \frac{1}{2}d(p,q_\alpha)$ and we will denote the neighbourhood of radius $r_\alpha$ around $q_\alpha$ to be $B_{r_\alpha}(q_\alpha) = \{x \in X \mid d(q_\alpha,x) < r_\alpha\}$ and the neighbourhood of radius $r_\alpha$ around $p$ to be $B_{r_\alpha}(p) = \{x \in X \mid d(p,x) < r_\alpha\}$. Then the collection of open sets $\{B_{r_\alpha}(q_\alpha)\}_\alpha$ is an open cover of $K$. As $K$ is compact there exists a finite subcover of $\{B_{r_\alpha}(q_\alpha)\}_\alpha$ such that

$$K \subset B_{r_1}(q_1) \cup \cdots \cup B_{r_n}(q_n) = U.$$

I now make the following claim:

Claim: $(B_{r_1}(p) \cap \cdots \cap B_{r_n}(p)) \cap U = \emptyset$.

Proof: Assume that $x \in (B_{r_1}(p) \cap \cdots \cap B_{r_n}(p)) \cap U$. Then we must have that $x\in B_{r_i}(p)$ for $1 \leq i \leq n$ and $x\in U$. As $x\in U$, then there exists an $i (1 \leq i \leq n)$ such that $x\in B_{r_i}(q_i)$ and, without any loss of generality, we assume $x \in B_{r_1}(q_1)$. In particular, we must also have that $x\in B_{r_1}(p)$. Therefore, by the triangle inequality we have,

$$d(p,q_1) \leq d(p,x) + d(x,q_1) < r_1 + r_1 = d(p,q_1).$$

This however is a contradiction. Therefore, $p \in B_{r_1}(p) \cap \cdots \cap B_{r_n}(p) \subset K^c$ which means that $p$ is an interior point of $K^c$. As $p$ was arbitrary, $K^c$ is open and therefore, $K$ is closed as desired. $_\Box$

Hope this may help someone.

Title: Compact sets are closed?


答:
Then using sequential continuity on this sequence $f(x_n) \to f(L) \in S$.

For closure what you want to show is that if $f(x_n) \to L$ then $L \in f(S)$.

$f(s^*) \to \infty$

This is meaningless to me. $f(s^*)$ is constant it does not "tend to" anything.

Since $\max(S)$ and $\min(S)$ are finite numbers, we cannot have divergence to infinity of the function over the elements of $S$

Why not, exactly? This is, after all, what you are trying to prove and just stating that it's true is not a proof.

I think the better way to look at this is through the Bolzano-Weierstrass theorem. To show $f(S)$ is compact you need to show that every sequence $y_n = f(x_n)$ has a convergent subsequence and here you can use a convergent subsequence of $x_n$ to reach that conclusion.

Title: Prove if $f : \mathbb{R}\to\mathbb{R}$ is continuous and $S \subset \mathbb{R}$ is compact, then the image is also compact.

Body:
I am not 100% sure about my proof, ideally I would like some hints/confirmation on my proof.

Proof:

Since I want to show $f(S)$ is compact I need to show it is closed and bounded.

Closure

Consider a sequence $(x_n) \subset S$ where  $x_n \to L \in S$ (by compactness). Since $L \in S$ we have that since the function is continuous $f(L) \in f(S)$.

Then using sequential continuity on this sequence  $f(x_n) \to f(L) \in f(S)$. Since we can represent any sequence in $f(S)$ as the image of any given sequence in $S$ the set $f(S)$ is closed.

Bounded

Since $S$ is bounded and $f$ continuous, suppose $s^*$ maximises the function $f$ on $S$ - if $f(s^*) \to \infty$ $f$ is not continuous. Since $max(S)$ and $min(S)$ are finite numbers, we cannot have divergence to infinity of the function over the elements of $s$, so $f(s^*)$ is the upper bound of $f(S)$. Repeating for the lower bound I have that S is bounded.

Is my proof correct. I have concerns that my proof for bounded is not very well worded, also I feel like my proof for closure may miss certain cases.


问:
Integration of $\int_0^{\infty}\frac{e^{(\lambda +is)t}-e^{-(\lambda +is)t}}{e^{\pi t}-e^{-\pi t}}dt$
Integration of $$\int_0^{\infty}\frac{e^{(\lambda +is)t}-e^{-(\lambda +is)t}}{e^{\pi t}-e^{-\pi t}}dt$$

I am thinking about this integral from last half and hour but still can't figure it out how to solve this. what i was think is to make this in form

$$\int_{0}^{\infty}\frac{\sinh(\lambda+is)t}{\sinh (\pi t)} dt$$ but still this will get me nowhere. please just give me a hint how to proceed this.

Thank you.

答:
An amusing way get the answer: first extend the integration range to $-\infty, \infty$. The integrand is even in $t$, so we get a $1/2$ factor up front. I'm taking $a=\lambda+is$.

$$ I(a)=\frac{1}{2}\int_{-\infty}^{\infty} dt \ \frac{e^{at}-e^{-at}}{e^{\pi t}-e^{-\pi t}}$$

The integrand has poles at $t=in$, $n \in Z $. The pole at $t=0$ is a removable singularity. An anticlockwise contour$^\dagger$ in the complex $t$ plane will enclose all the residues for $n=1,2,3,..$, which are in the upper half plane. Courtesy of the residue theorem, we have

$$ I(a)= \sum_{n=1}^{\infty}R_n $$

$$ R_n = 2\pi i \lim_{t \rightarrow  in} (t-in) \frac{1}{2} \frac{e^{at}-e^{-at}}{e^{\pi t}-e^{-\pi t}} $$

$$ R_n =-(-1)^n \sin{(na)} $$

The sum for $I(a)$ may be performed by writing $\sin$ in exponential form and doing the two geometric sums$^\ddagger$.

$$ R_n=-\frac{1}{2i}\left[ (-e^{ia})^n- (-e^{-ia})^n \right]$$

For which we get

$$ I(a)= -\frac{1}{2i} \left[ \frac{-e^{ia}}{1+e^{ia}} - \frac{-e^{-ia}}{1+e^{-ia}} \right]$$

After simplification, this yields

$$ I(a)= \frac{1}{2} \tan \left( \frac{a}{2} \right)$$

$$\tag*{$\blacksquare$}$$

Clearly, this converges for $-\pi <a<\pi$ due to the nearest singularities in $\tan$ at $\pm \pi/2$.

$\dagger$ Along the real axis then closed in the upper plane.

$\ddagger$ As mentioned in the comments, these individual sums are divergent. Don't worry! Either we can sum $-(-1)^n \sin(na)$ directly, or regularize the individual sums. For instance, using Euler summation

$$ E_1(x)= - \sum_{n=1}^{\infty} x^n (-e^{ia})^n=\frac{xe^{ia}}{1+xe^{ia}}$$

$$ E_2(x)= - \sum_{n=1}^{\infty} x^n (-e^{-ia})^n=\frac{x}{x+e^{ia}}$$

Which converge for $x$ sufficiently small. We then have

$$I(a)= \lim_{x \rightarrow1} (E_1(x)-E_2(x))$$

Which leads to the same answer.


问:
power series solution of $(x-3)y'+2y=0$

Using power series, Find  solution of

differential equation $(x-3)y'+2y=0$

What i try :: Let $\displaystyle y=\sum^{\infty}_{n=0}a_{n}x^{n}.$ Then $\displaystyle y'=\sum^{\infty}_{n=1}na_{n}x^{n-1}$

So we have $$(x-3)\sum^{\infty}_{n=1}na_{n}x^{n-1}+2\sum^{\infty}_{n=0}a_{n}x^{n}=0$$

$$\sum^{\infty}_{n=1}\bigg(na_{n}+2a_{n}\bigg)x^n+2a_{0}-3\sum^{\infty}_{n=1}na^{n}x^n=0$$

How do i solve it, Help me please, Thanks

Title: power series solution of $(x-3)y'+2y=0$

Body:
$$(x-3)y'+2y=0~~~~(1)$$
Its direct solution by separation of variables is $$y=\frac{C}{(x-3)^2}=\frac{C}{9}[1+2x/3+x^2/3+4x^3/27+5x^4/81+...]~~~~(2)$$
Here we have used binomial series $(1-z)^{-2}.$
Let us go for series solution near $x=0$ in powers of $x$, take
$$y(x)=\sum_{n=0}^{\infty} A_n x^n~~~~(3)$$
Putting it in (1), we get (name of the summation idex is dummy)
$$\sum_{n=0}^{\infty}(n+2)A_n x^n-3\sum_{m=0}^{\infty}m A_m x^{m-1}~~~~(4)$$
Let $m-1=n$ in the second sum above, then (4) becomes
$$y(x)=\sum_{n=0}^{\infty} (n+2) A_n x^n-3\sum_{n=-1} (n+1) A_{n+1} x^n~~~~(5)$$
$$\implies y(x)=\sum_{n=0}^{\infty} (n+2) A_n x^n-3\sum_{n=0} (n+1) A_{n+1} x^n~~~~(6)$$
$$\implies y(x)=\sum_{n=0}^{\infty} [(n+2)A_n-3A_{n+1}] x^n=0=\sum_{n=0}^{\infty} 0 x^n~~~~(7)$$
$$\implies (n+2) A_{n}-3(n+1) A_{n+1}=0$$
$$\implies A_{n+1}=\frac{n+2}{3(n+1)}A_n, n=0,1,2,3,...~~~~(8)$$
$$\implies A_1=(2/3)A_0, A_2=(1/2)A_1,A_3=(4/9)A_2, A_4=(5/12)A_3,....~~~~~(9)$$
$$\implies A_1=(2/3)A_0, A_2=(1/3) A_0, A_3=(4/27)A+0,A_4=(5/81) A_0 ~~~~(10)$$
So finally from (3), we get
$$y(x)=A_0[[1+2x/3+x^2/3+4x^3/27+5x^4/81+...]~~~~(11)$$
Here, $A_0$ may be set as $C/9$. Finally (2) and (11) coincide.


问:
Product rule for integration???
Knowing that there is a product rule for differentiating things, how would a product rule work in reverse?

For instance, how would you go around integrating $8x^3(3x-1)^2$?

At school, we haven't learn a rule to solve this at all from my teachers, and coming across a question like this, I couldn't answer it.

Thanks

答:
$$\int fdg=fg-\int gdf$$
where $df$ denotes the derivative of $f$, so $df(x)=f'(x)dx$. Howerever in this exercise the best idea is just to do the multiplication, as you are simply integrating polynomials.

Title: Product rule for integration???


问:
Picture of l'Hopital's rule at infinity
I know that when L'Hopital's rule is taken for the $\frac{0}{0}$ case, we get that the functions start looking at lines, like in this figure from Stewart's Calculus:

But what is the visual when the limit is the $\frac{\infty}{\infty}$ case?

The only thing that comes to mind that is easily visualized is a rational function's asymptotes (and that comes from dividing numerator and denominator by the largest exponent in the denominator)

答:
Well, you actually CAN do it.

There is one simple way, which sounds a bit like cheating: taking the limit means looking at the function in a neighbourhood of infinity, so you could just end your graph at a really big number and see how the slope behaves.

However, there is also a more insightful point of view, but it is really long and pretty much useless. Once you have the right intuition for the finite case, you just use that.

The idea for what you're asking would be "squeezing" the real line $\mathbb{R}$ to an interval. For example: $atg([0,\infty))=[0,\frac{\pi}{2})$, where you could consider the closed interval $[0,\frac{\pi}{2}]$ regarding $\frac{\pi}{2}$ as the image of $\infty$.

Then you could draw your function using these coordinates: what you get is a deformed & squeezed graph which is contained in the square $[0,\frac{\pi}{2}]\times[0,\frac{\pi}{2}]$.

Meanwhile, you should also write down the rule for how to convert a slope according to the squeezed graph into the right slope (i.e. according to the original graph).

Finally, you should be able to see that the slope given by the squeezed graph is actually that given by De L'hopital's rule (adequately transformed).

This is a little example I tried to carry out on the spot: here is the squeezed graph of the function $x^2$ (I'm thinking about computing the limit $\lim_{x\rightarrow \infty} \frac{x^2}{x}$, which should obviously be $\infty$, but I wish to use De L'hopital for understanding purpose)

You can see that at infinity (the upper right part) the slope is zero. the transformation rule for slope of this change of variables is probably something like

$ real\_slope = \frac{1}{squeezed\_slope} $

(I din't check it though).

So you correctly get that the real slope is $\infty$ as you would expect by applying De L'hopital's rule.

As I said this is long and not necessary. Also, beware that I omitted the details and you should go check them, if you want a full understanding. I actually never got further than this, cause I didn't feel the urge. I also had the same question and this is an answer I gave myself. If you like this kind of questions, you should like measure theory.

Title: Picture of l'Hopital's rule at infinity


问:
Deriving the equation of the polar axis-aligned ellipsoid
The polar of a convex body $C$ is defined as:

For a convex body $C$, its polar $C^*$ is given by $$C^* = \{x\in\mathbb{R}^n: \langle x,c\rangle \le 1, \forall c\in C\}$$

If I start with an orthonormal basis $(e_i)_1^n$, positive constants $(\alpha_j)_1^n > 0$ and the ellipsoid $$\varepsilon = \left\{x\in\mathbb{R}^n: \sum\limits_{j=1}^n \frac{\langle x,e_j\rangle^2}{\alpha_j^2} \leq 1  \right\}$$

how do I show that its polar ellipsoid is $$\varepsilon^* = S_1 = \left\{y\in\mathbb{R}^n: \sum_{j=1}^n \alpha_j^2 \langle y,e_j\rangle^2 \le 1\right\}$$

Using the definition, I saw that $$\varepsilon^* = S_2 = \{z\in\mathbb{R}^n: \langle z,x\rangle \le 1, \text{for all }x\in \varepsilon\}$$
Also, $\langle z,x\rangle = \sum_i\langle z,e_i\rangle\langle x,e_i\rangle$ I used Cauchy Schwartz to do one side of the proof, i.e. $S_1\subseteq S_2$:
$$\left(\sum_i\langle y,e_i\rangle\langle x,e_i\rangle\right)^2 \leq \left(\sum_i\alpha_i^2 \langle y,e_i\rangle^2\right)\left(\sum_i \frac{\langle x,e_i\rangle^2}{\alpha_i^2}\right) \le 1 \implies \langle y,x\rangle \le 1$$

I need help with the other direction, i.e. $S_2\subseteq S_1$. Which other inequality might help?

答:
This is a great deal easier to analyze in the case where you take $C=B_2^n$, the closed Euclidean unit ball. The basic idea is that you take the largest vector contained in the body along the same direction as a given point in the polar. The following proof is in the general case.

For the sake of brevity, denote $\langle x,e_i\rangle$ as $x_i$ for any $x$. Let $z\in S_2$. Define
$$v=(\alpha_1 z_1,\alpha_2z_2,\ldots,\alpha_nz_n), v^\circ=\frac{v}{\|v\|},\text{ and }z^\circ = (\alpha_1 v^\circ_1,\alpha_2 v^\circ_2,\ldots,\alpha_n v^\circ_n).$$
It is easily checked that $z^\circ\in C$ (because $v^\circ\in B_2^n$). Applying the fact that $z\in S_2$, we get
$$\langle z,z^\circ\rangle\leq 1\implies\langle v,v^\circ\rangle\leq 1 \implies \|v\|\leq 1.$$
Writing this out gives
$$\sum_{i} (\alpha_i z_i)^2 \leq 1,$$
which is exactly what we want.


问:
Missing factor equal to 0 after factoring

The equation $x^{10}+(13x-1)^{10}=0$ has $10$ complex roots, $r_i$ for integers $1\le i\le5$. Find $\sum^5_{i=1}\frac{1}{r_i\overline{r_i}}.$

My solution was simply to factor the given polynomial into $(r^2+(13r-1)^2)(r^8-\dots+(13r-1)^8)=0.$ We suppose the left factor is $0,$ such that $170r^2-26r+1=0.$ By Vieta's, the product of the roots if $1/170.$ Since this product is the product of the conjugate pairs, we can multiply $170\times5$ to find our answer, and $850$ turns out to be correct.

However, how come we don't need to account for the right factor here? We need not consider it if it were never equal to $0$, so how do we prove that?

Body:
We solve equation $x^{10}+(13x-1)^{10}=0$.

$$\left(13-\frac{1}{x}\right)^{10}=-1$$
$$13-\frac{1}{x_k}=\cos\frac{(2k+1)\pi}{10}+i\sin\frac{(2k+1)\pi}{10},$$
$$\frac{1}{x_k}=13-\cos\frac{(2k+1)\pi}{10}-i\sin\frac{(2k+1)\pi}{10},\quad k=0\ldots9.$$
Then
$$\frac{1}{x_k\bar x_k}=170-26\cos\frac{(2k+1)\pi}{10}$$
$$\sum_{k=0}^4\frac{1}{x_k\bar x_k}=850-26\left(\cos\frac{\pi}{10}+\cos\frac{3\pi}{10}+\cos\frac{7\pi}{10}+\cos\frac{9\pi}{10}\right)=850$$


问:
Testing if a function derivative is continuous
How can I methodically test if

$f(x)= \log(x)/x^3$

Has a continuous derivative?

答: First of all, notation. I guess you actually meant $f(x)=\frac{\ln x}{x^{3}}$ which is a product of differentiable functions (although you should also provide the set over which you defined $f$). Morover, $\ln x$ and $\frac{1}{x^{3}}$ are smooth, i.e. they admit derivatives of all orders, so each derivative must be continuous. Notice that again what I just said makes sense only if you provide an appropriate set: on $(0,1]$ what I said is true, but on $[0,1]$ is not.


问:
Simplified expression for centers of period-three Mandlebrot bulbs
The Mandelbrot set contains three regions (two bulbs and a cardioid) with periods of three. These regions each contain a fixed point which is a root of the expression $x^3 + 2x^2 + x + 1$.

The fixed point at the center of the period-three cardioid has the value
$$-\frac{1}{3} \left(2 + \sqrt[3]{\frac{25 - 3 \sqrt{69}}{2}} + \sqrt[3]{\frac{25 + 3 \sqrt{69}}{2}}\right)$$

This simplifies to $-ρ^2$, where $ρ$ is the plastic constant equal to $\sqrt[3]{\frac{9 + \sqrt{69}}{18}} + \sqrt[3]{\frac{9 - \sqrt{69}}{18}}$.

The fixed points at the centers of the two period-three bulbs are
$$-\frac{2}{3} + \frac{1 + i \sqrt{3}}{6} \sqrt[3]{\frac{25 - 3 \sqrt{69}}{2}} + \frac{1 - i \sqrt{3}}{3 \sqrt[3]{2}^2 \sqrt[3]{25 - 3 \sqrt{69}}}$$ and $$-\frac{2}{3} + \frac{1 - i \sqrt{3}}{6} \sqrt[3]{\frac{25 - 3 \sqrt{69}}{2}} + \frac{1 + i \sqrt{3}}{3 \sqrt[3]{2}^2 \sqrt[3]{25 - 3 \sqrt{69}}}$$

The real part of these two points simplifies to $\frac{ρ^2 - 2} {2}$. The imaginary parts are approximately $\pm$ 0.744861766619744236593170... $i$, but I can't figure out a closed-form expression for that. I'm hopeful that I can find something in terms of $ρ$, since that worked for the real part and for the cardioid point, but so far I've been stumped.

答:
Given that I know the real root of the function $x^3 + 2x^2 + x + 1$ is $-ρ^2$, I can factor that out to get $(x+ρ^2)(x^2 + (2 - ρ^2) x + (2 - ρ^6 + 3ρ^4 - 3ρ^2))$.

Using the quadratic formula on that second term, I get $\frac{ρ^2 ± \sqrt{-4 + 8ρ^2 - 11ρ^4 + 4ρ^6} - 2} {2}$.

Because $ρ^3 = ρ + 1$ and $ρ^4 = ρ^2 + ρ$, the term inside the square root simplifies to $ρ^2 - 3ρ$. That is negative, so factoring out $ί$ gives $\sqrt{3ρ - ρ²} ί$.

Rearranging, that gives me the two complex roots as $\frac{ρ^2 - 2} {2} ± \frac{\sqrt{3ρ - ρ²}} {2} ί$.


问:
Solving a Third order Differential Equation
How would I solve:

$$
v_{yyy} + vv_y + \frac{1}{3}yv_y + \frac{2}{3}v = 0
$$

where $v$ is a function of $y$.

Is there a way to reduce it to a solveable second order equation?

This equation is found when looking for a solution along one of the Lie Symmetries of the ODE:
$$
u_t - uu_x - u_{xxx}
$$

From what I have read, there is a transformation discovered by Miura, for the equation.  We are to set:
$$
v = \frac{-1}{6}w^2 + \frac{dw}{dy}
$$

I now need to find $v_{yyy}, v_y$ in terms of $\frac{dw}{dy}$ and sub that into the initial equation.

I have:
\begin{align}
v &= \frac{-1}{6}w^2 + w_y\\
v_y &= \frac{-1}{3}w w_y + w_{yy}\\
v_{yy} &= \frac{-1}{3}(w_y)^2 - \frac{1}{3}ww_{yy}+w_{yyy}\\
v_{yyy}&=-w_yw_{yy}-\frac{1}{3}w_yw_{yy}-\frac{1}{3}ww_{yyy}+w_{yyyy}\\
\end{align}

When I substitute this into the third order equation I get:

$$
w_{yyyy}-\frac{1}{3}ww_{yyy}-\frac{1}{3}w(w_y)^2-\frac{1}{6}w^2w_{yy}+\frac{1}{18}w^3w_y+\frac{1}{3}yw_{yy}-\frac{1}{9}yww_y+\frac{2}{3}w_y-\frac{1}{9}w^2=0
$$

According to what I have read, this can now be factored and then integrated.
The example I am following and using to apply to my equation is on page 195 of Olver.

In following the examplefrom Olver and applying the technique to my equation I get:
$$
(D_y - \frac{1}{3}w)(w_{yyy} - \frac{1}{6}w^2w_y + \frac{1}{3}yw_y + \frac{1}{3}w) = 0
$$

Now I want to integrate and solve:
$$
w_{yyy} - \frac{1}{6}w^2w_y + \frac{1}{3}yw_y + \frac{1}{3}w = 0
$$

When I integrate I get:
$$
w_{yy} - \frac{1}{18}w^3 + \frac{1}{3}yw + k = 0
$$

How do I go about solving this second order NonLinear Differential Equation?

The example in Olver, which differs slightly from my equation, has a positive $w^3$ term and states that the equation is a from of the Second Painleve Transcendent.

答:
A particular solution is :
$$v=-\frac{12}{y^2}$$

Of course this is far to be the general solution. Hopping that help.

Obviously $\quad v=0\quad$ and $\quad v=-y\quad$ are  trivial solutions.