Difference between revisions of "Series for log(z) for absolute value of (z-1) less than 1"
From specialfunctionswiki
(Created page with "==Theorem== The following formula holds for $|z-1| \leq 1$ and $z \neq 0$: $$\log(z) = \displaystyle\sum_{k=1}^{\infty} \dfrac{(-1)^k(z-1)^k}{k},$$ where $\log(z)$ denotes the...") |
|||
| Line 1: | Line 1: | ||
==Theorem== | ==Theorem== | ||
The following formula holds for $|z-1| \leq 1$ and $z \neq 0$: | The following formula holds for $|z-1| \leq 1$ and $z \neq 0$: | ||
| − | $$\log(z) = \displaystyle\sum_{k=1}^{\infty} \dfrac{(-1)^k(z-1)^k}{k},$$ | + | $$\log(z) = -\displaystyle\sum_{k=1}^{\infty} \dfrac{(-1)^k(z-1)^k}{k},$$ |
where $\log(z)$ denotes the [[logarithm]]. | where $\log(z)$ denotes the [[logarithm]]. | ||
==Proof== | ==Proof== | ||
Revision as of 07:41, 4 June 2016
Theorem
The following formula holds for $|z-1| \leq 1$ and $z \neq 0$: $$\log(z) = -\displaystyle\sum_{k=1}^{\infty} \dfrac{(-1)^k(z-1)^k}{k},$$ where $\log(z)$ denotes the logarithm.
Proof
References
- 1964: Milton Abramowitz and Irene A. Stegun: Handbook of mathematical functions ... (previous): 4.1.26
