Difference between revisions of "Series for log(z) for Re(z) greater than 0"
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(Created page with "==Theorem== The following formula holds for $\mathrm{Re}(z) \geq 0, z \neq 0$: $$\log(z) = 2 \displaystyle\sum_{k=1}^{\infty} \left( \dfrac{z-1}{z+1} \right)^k \dfrac{1}{k},$$...") |
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==References== | ==References== | ||
− | * {{BookReference|Handbook of mathematical functions|1964|Milton Abramowitz|author2=Irene A. Stegun|prev=Series for log(z) for absolute value of (z-1) less than 1|next=Laurent series for log((z+1)/(z-1)) for absolute value of z greater than 1}}: 4.1.27 | + | * {{BookReference|Handbook of mathematical functions|1964|Milton Abramowitz|author2=Irene A. Stegun|prev=Series for log(z) for absolute value of (z-1) less than 1|next=Laurent series for log((z+1)/(z-1)) for absolute value of z greater than 1}}: $4.1.27$ |
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+ | [[Category:Theorem]] | ||
+ | [[Category:Unproven]] |
Latest revision as of 17:28, 27 June 2016
Theorem
The following formula holds for $\mathrm{Re}(z) \geq 0, z \neq 0$: $$\log(z) = 2 \displaystyle\sum_{k=1}^{\infty} \left( \dfrac{z-1}{z+1} \right)^k \dfrac{1}{k},$$ where $\log$ denotes the logarithm.
Proof
References
- 1964: Milton Abramowitz and Irene A. Stegun: Handbook of mathematical functions ... (previous) ... (next): $4.1.27$