Difference between revisions of "Log((1+z)/(1-z)) as continued fraction"
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(Created page with "==Theorem== The following formula holds for $z \in \mathbb{C} \setminus \left[ (-\infty,-1) \bigcup (1,\infty) \right]$: $$\log \left( \dfrac{1+z}{1-z} \right) = \cfrac{2z}{1-...") |
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The following formula holds for $z \in \mathbb{C} \setminus \left[ (-\infty,-1) \bigcup (1,\infty) \right]$: | The following formula holds for $z \in \mathbb{C} \setminus \left[ (-\infty,-1) \bigcup (1,\infty) \right]$: | ||
$$\log \left( \dfrac{1+z}{1-z} \right) = \cfrac{2z}{1-\cfrac{z^2}{3-\cfrac{4z^2}{5-\cfrac{9z^2}{7-\cfrac{16z^2}{9-\cfrac{25z^2}{11-\cfrac{36z^2}{13-\ddots}}}}}}},$$ | $$\log \left( \dfrac{1+z}{1-z} \right) = \cfrac{2z}{1-\cfrac{z^2}{3-\cfrac{4z^2}{5-\cfrac{9z^2}{7-\cfrac{16z^2}{9-\cfrac{25z^2}{11-\cfrac{36z^2}{13-\ddots}}}}}}},$$ | ||
| − | where $log$ denotes the [[logarithm]]. | + | where $\log$ denotes the [[logarithm]]. |
==Proof== | ==Proof== | ||
Latest revision as of 04:56, 21 December 2017
Theorem
The following formula holds for $z \in \mathbb{C} \setminus \left[ (-\infty,-1) \bigcup (1,\infty) \right]$: $$\log \left( \dfrac{1+z}{1-z} \right) = \cfrac{2z}{1-\cfrac{z^2}{3-\cfrac{4z^2}{5-\cfrac{9z^2}{7-\cfrac{16z^2}{9-\cfrac{25z^2}{11-\cfrac{36z^2}{13-\ddots}}}}}}},$$ where $\log$ denotes the logarithm.
Proof
References
- 1964: Milton Abramowitz and Irene A. Stegun: Handbook of mathematical functions ... (previous) ... (next): $4.1.40$
