如何比较两个对数值的大小

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睡神 发表于 2024-4-4 17:44
还是战巡大佬的更精确！！！拜服！！！

@战巡 用的是$\log \left(\frac{1+z}{1-z}\right)$，它的幂级数比$\log(1+z)$的幂级数收敛更快。
$\log \left(\frac{1+z}{1-z}\right)$也有连分式。猜测$\log \left(\frac{1+z}{1-z}\right)$的连分式也比$\log(1+z)$的连分式收敛更快？
Log continued fraction identity 17
$$\log \left(\frac{1+z}{1-z}\right)=\frac{2 z}{1+\mathop{\Large\mathrm{K}}\limits_{k=1}^{\infty} \dfrac{-k^2 z^2}{2 k+1}}$$
渐近分数为\[ \frac{{2z}}{{1 - \frac{{z^2}}{3}}} \]
\[ \frac{{2z}}{{1 - \frac{{z^2}}{{3 - \frac{{4z^2}}{5}}}}} \]
\[ \frac{{2z}}{{1 - \frac{{z^2}}{{3 - \frac{{4z^2}}{{5 - \frac{{9z^2}}{7}}}}}}} \]
\[\vdots\]
Reference

• Oskar Perron. Die Lehre von den Kettenbrüchen. p. 370, 1913. (in German)
• Bruce Carl Berndt. Ramanujan's Notebooks, Part II. p. 133, 1989.
• Lisa Lorentzen and Haakon Waadeland. Continued Fractions, Vol. 1: Convergence Theory, 2nd ed. p. 271, 2008.

睡神

不得不服战巡大佬！当取4项时，

$\hat{\ln(x)}=2[\dfrac{x-1}{x+1}+\dfrac{1}{3}(\dfrac{x-1}{x+1})^3+\dfrac{1}{5}(\dfrac{x-1}{x+1})^5+\dfrac{1}{7}(\dfrac{x-1}{x+1})^7]$

结果得0.01696023602，这精确度高得离谱，达到7位小数吻合，跪拜！