Difference between revisions of "Log(1+z) as continued fraction"
From specialfunctionswiki
(Created page with "==Theorem== The following formula holds for any $z \in \mathbb{C} \setminus (-\infty,-1)$: $$\log(1+z)=\cfrac{z}{1+\cfrac{z}{2+\cfrac{z}{3+\cfrac{4z}{4+\cfrac{4z}{5+\cfrac{9z}...") |
(No difference)
|
Latest revision as of 20:04, 25 June 2017
Theorem
The following formula holds for any $z \in \mathbb{C} \setminus (-\infty,-1)$: $$\log(1+z)=\cfrac{z}{1+\cfrac{z}{2+\cfrac{z}{3+\cfrac{4z}{4+\cfrac{4z}{5+\cfrac{9z}{6+\ddots}}}}}},$$ where $\log$ denotes the logarithm.
Proof
References
- 1964: Milton Abramowitz and Irene A. Stegun: Handbook of mathematical functions ... (previous) ... (next): $4.1.39$
