Jackson Floyd

2022-04-02

Summing up the series ${a}_{3}k$ where

$\mathrm{log}(1-x+{x}^{2})=\sum {a}_{k}{x}^{k}$

clarkchica44klt

Beginner2022-04-03Added 17 answers

Lemma: Let $f\left(x\right)=\sum {a}_{n}{x}^{n}$ be a power series. Then

$\sum {a}_{3n}{x}^{3n}=\frac{f\left(x\right)+f\left(\omega x\right)+f\left({\omega}^{2}x\right)}{3}$

where$\omega ={e}^{\frac{2\pi t}{3}}$

Proof. Ignoring convergence it suffices to prove this for a single term, and then it boils down to the identity

$\frac{1+{\omega}^{n}+{\omega}^{2n}}{3}$

This is a special case of the discrete Fourier transform. Applying the lemma, we readily obtain that the desired sum is

$\frac{\mathrm{ln}1+\mathrm{ln}(1-\omega -{\omega}^{2})+\mathrm{ln}(1-{\omega}^{2}+\omega )}{3}=\frac{1}{3}\mathrm{ln}(2{\omega}^{2}\cdot 2\omega )$

$=\frac{2}{3}\mathrm{ln}2$

where we use the fact that$1+\omega +{\omega}^{2}=0$

where

Proof. Ignoring convergence it suffices to prove this for a single term, and then it boils down to the identity

This is a special case of the discrete Fourier transform. Applying the lemma, we readily obtain that the desired sum is

where we use the fact that

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