$f'(0)=g'(0)=1⟹f(g(x))-g(f(x))∼f^{-1}(g^{-1}(x))-g^{-1}(f^{-1}(x))$

hbghlyj

点A的横坐标为$x$，则$AB$的长为$f(g(x))-g(f(x))$，$D'E'$的长为$f^{-1}(g^{-1}(x))-g^{-1}(f^{-1}(x))$. 由于$DE$与$D'E'$关于$y=x$对称，$DE=D'E'$. $$\frac{f(g(x))-g(f(x))}{f^{-1}(g^{-1}(x))-g^{-1}(f^{-1}(x))}=\frac{AB}{DE}=\frac{AB}{BC}\frac{BC}{DE}\to1\cdot 1=1$$ A点趋于0的时候，割线AC趋于切线，于是∠ACB趋于45°，$\triangle$ACB趋于等腰直角三角形，于是$AB/BC\to1$. 　　同理，割线BD和CE的斜率也分别趋向于1，所以BD和CE趋于平行，另外BC与DE平行，于是BCED趋于平行四边形，进而对边$BC/DE\to1$ （参见阿尔诺德所著《惠更斯与巴罗，牛顿与胡克：数学分析与突变理论的起步，从渐伸线到准晶体》(Huygens and Barrow, Newton and Hooke: Pioneers in Mathematical Analysis and Catastrophe Theory from Evolvements to Quasicrystals)一书第一章第5节与附注8。） 　　这个证明看上去确实很漂亮，颇有欧几里得遗风，而且清晰明了。 WolframAlpha
Arnold-Trivium-1991的第2题
弄个大新闻，把阿尔诺德批判一番！ – 超理论坛

hbghlyj

Problem. Compute
\[
\lim _{x \rightarrow 0} \frac{\sin \tan x-\tan \sin x}{\arcsin \arctan x-\arctan \arcsin x} .
\]
Solution. It is convenient to do the computation in the general form. Let
$f(x)=x+a x^3+b x^5+c x^7+o\left(x^7\right)$ and $\phi(x)=x+\alpha x^3+\beta x^5+\gamma x^7+o\left(x^7\right)$
be the Taylor expansions of two odd infinitely differentiable functions with the derivatives at $x=0$ equal to 1 . Then
\[
\begin{aligned}
&f(\phi(x)) \\
&= x+\alpha x^3+\beta x^5+\gamma x^7+a\left(x+\alpha x^3+\beta x^5\right)^3+b\left(x+\alpha x^3\right)^5+c x^7+o\left(x^7\right)\\
&=x+(a+\alpha) x^3+(b+3 a \alpha+\beta) x^5+\left(c+3 a \beta+3 a \alpha^2+5 b \alpha+\gamma\right) x^7+o\left(x^7\right)
\end{aligned}
\]
Since the coefficients at $x^3$ and $x^5$ are symmetric with respect to exchanging Latin and Greek letters, the expansion of $f(\phi(x))-\phi(f(x))$ starts only with $x^7$. This is a bad news, but the good news is that the coefficient at $x^7$ depends on the Taylor coefficients of $f$ and $\phi$ only up to order $x^5$ :
\[
f(\phi(x))-\phi(f(x))=[3 a \alpha(\alpha-a)+2(b \alpha-\beta a)] x^7+o\left(x^7\right) .
\]
On the other hand, if $\phi=f^{-1}(x)=x+A x^3+B x^5+o\left(x^5\right)$, then
\[
x=f\left(f^{-1}(x)\right)=x+(a+A) x^3+(b+3a A+B) x^5+o\left(x^5\right)
\]
i.e. $A=-a$, and $B=3 a^2-b$. Substituting into $3 a \alpha(\alpha-a)+2(b \alpha-\beta a)$ respectively: $-a$ for $a,-\alpha$ for $\alpha, 3 a^2-b$ for $b$, and $3 \alpha^2-\beta$ for $\beta$, we find:
\[
\begin{aligned}
&f^{-1}\left(\phi^{-1}(x)\right)-\phi^{-1}\left(f^{-1}(x)\right)\\
&= \left[3 a \alpha(a-\alpha)+2\left(a\left(3 \alpha^2-\beta\right)-\alpha\left(3 a^2-b\right)\right)\right] x^7+o\left(x^5\right) \\
&= [3 a \alpha(\alpha-a)+2(b \alpha-\beta a)] x^7+o\left(x^7\right) .
\end{aligned}
\]
Thus
\[
\lim _{x \rightarrow 0} \frac{f(\phi(x))-\phi(f(x))}{f^{-1}\left(\phi^{-1}(x)\right)-\phi^{-1}\left(f^{-1}(x)\right)}=1
\]
Could you explain (predict) this result without much computation?

kuing

经典极限： lim.pdf (22.98 KB, 下载次数: 202)

2021-6-4 17:50 上传

点击文件名下载附件

isee

hbghlyj 发表于 2022-9-24 20:08
Problem. Compute
\[
\lim _{x \rightarrow 0} \frac{\sin \tan x-\tan \sin x}{\arcsin \arctan x-\arctan ...

重新手打录入，还是找到源文件了？

kuing

isee 发表于 2022-10-4 00:09
重新手打录入，还是找到源文件了？

也可能是用 mathpix 之类的识别工具。

hbghlyj

\[\text{Compute expansion of series }

\frac{f(g(x))-g(f(x))}{f^{-1}\left(g^{-1}(x)\right)-g^{-1}\left(f^{-1}(x)\right)}

\text{, given }

f'(0)=g'(0)=1

:\]

1. serf=x+c1 x^2+c2 x^3+c3 x^4+c4 x^5+O[x]^6;
2. serg=x+d1 x^2+d2 x^3+d3 x^4+d4 x^5+O[x]^6;
3. serCommute[s1_,s2_]:=ComposeSeries[s1,s2]-ComposeSeries[s2,s1]
4. serCommute[serf,serg]/serCommute[InverseSeries[serf],InverseSeries[serg]]//Simplify

复制代码

$$\frac{f(g(x))-g(f(x))}{f^{-1}(g^{-1}(x))-g^{-1}(f^{-1}(x))}=1+3 x (f''(0)+g''(0))+O\left(x^2\right)$$