Difference between revisions of "Identity written as a sum of Möbius functions"
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==Theorem== | ==Theorem== | ||
The following formula holds for $|x|<1$: | The following formula holds for $|x|<1$: | ||
| − | $$\displaystyle\sum_{k=1}^{\infty} \dfrac{\mu(k)x^k}{1-x^k} | + | $$x=\displaystyle\sum_{k=1}^{\infty} \dfrac{\mu(k)x^k}{1-x^k},$$ |
where $\mu$ denotes the [[Möbius function]]. | where $\mu$ denotes the [[Möbius function]]. | ||
| Line 7: | Line 7: | ||
==References== | ==References== | ||
| − | * {{BookReference|Handbook of mathematical functions|1964|Milton Abramowitz|author2=Irene A. Stegun|prev=Reciprocal of Riemann zeta as a sum of Möbius function for Re(z) greater than 1|next=}}: 24.3.1 I.B. | + | * {{BookReference|Handbook of mathematical functions|1964|Milton Abramowitz|author2=Irene A. Stegun|prev=Reciprocal of Riemann zeta as a sum of Möbius function for Re(z) greater than 1|next=Möbius function is multiplicative}}: $24.3.1 \mathrm{I}.B.$ |
[[Category:Theorem]] | [[Category:Theorem]] | ||
[[Category:Unproven]] | [[Category:Unproven]] | ||
Latest revision as of 01:33, 22 June 2016
Theorem
The following formula holds for $|x|<1$: $$x=\displaystyle\sum_{k=1}^{\infty} \dfrac{\mu(k)x^k}{1-x^k},$$ where $\mu$ denotes the Möbius function.
Proof
References
- 1964: Milton Abramowitz and Irene A. Stegun: Handbook of mathematical functions ... (previous) ... (next): $24.3.1 \mathrm{I}.B.$
