Difference between revisions of "Series for log(z) for absolute value of (z-1) less than 1"
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(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...") |
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==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== | ||
==References== | ==References== | ||
| − | * {{BookReference|Handbook of mathematical functions|1964|Milton Abramowitz|author2=Irene A. Stegun|prev=Series for log(z) for Re(z) greater than 1/2|next=}}: 4.1.26 | + | * {{BookReference|Handbook of mathematical functions|1964|Milton Abramowitz|author2=Irene A. Stegun|prev=Series for log(z) for Re(z) greater than 1/2|next=Series for log(z) for Re(z) greater than 0}}: $4.1.26$ |
| + | |||
| + | [[Category:Theorem]] | ||
| + | [[Category:Unproven]] | ||
Latest revision as of 17:28, 27 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) ... (next): $4.1.26$
