Revision history of "Log(1+z) as continued fraction"

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  • (cur | prev) 20:04, 25 June 2017Tom (talk | contribs). . (516 bytes) (+516). . (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}...")