Difference between revisions of "Series for log(z) for Re(z) greater than 1/2"
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==Theorem== | ==Theorem== | ||
The following formula holds for $\mathrm{Re}(z) \geq \dfrac{1}{2}$: | The following formula holds for $\mathrm{Re}(z) \geq \dfrac{1}{2}$: | ||
| − | $$\log(z) = -\displaystyle\sum_{k=1}^{\infty} \left(\dfrac{z-1}{z} \right)^k \dfrac{ | + | $$\log(z) = -\displaystyle\sum_{k=1}^{\infty} \left(\dfrac{z-1}{z} \right)^k \dfrac{1}{k},$$ |
where $\log$ denotes the [[logarithm]]. | where $\log$ denotes the [[logarithm]]. | ||
| Line 7: | Line 7: | ||
==References== | ==References== | ||
| − | * {{BookReference|Handbook of mathematical functions|1964|Milton Abramowitz|author2=Irene A. Stegun|prev=Taylor series of log(1+z)|next=Series for log(z) for | + | * {{BookReference|Handbook of mathematical functions|1964|Milton Abramowitz|author2=Irene A. Stegun|prev=Taylor series of log(1+z)|next=Series for log(z) for absolute value of (z-1) less than 1}}: $4.1.25$ |
| + | |||
| + | [[Category:Theorem]] | ||
| + | [[Category:Unproven]] | ||
Latest revision as of 17:28, 27 June 2016
Theorem
The following formula holds for $\mathrm{Re}(z) \geq \dfrac{1}{2}$: $$\log(z) = -\displaystyle\sum_{k=1}^{\infty} \left(\dfrac{z-1}{z} \right)^k \dfrac{1}{k},$$ where $\log$ denotes the logarithm.
Proof
References
- 1964: Milton Abramowitz and Irene A. Stegun: Handbook of mathematical functions ... (previous) ... (next): $4.1.25$
