How can I understand this? $-q\sum_{n=0}^{\infty}\frac{t^{p+nq}}{p+nq}=\sum_{n=0}^{q-1}\omega^{-np}\ln(1-\omega^{n}t)$ Here $p,q\in\mathbb{N}$. Let $\omega=e^{2\pi i/q}$ be a primitive $q^{\mathrm{th}}$ root of unity. I can't understand it, I hope you can help me.
Can you help me with this problem?
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0Yeah word Taylor expansion works, writing up sol. – 2012-05-24
1 Answers
I'm proving this as an equality of power series in the indeterminate $t$, where $\log(1-x)$ is defined by $\sum_{k \geqslant 0} - \frac{x^k}{k}$. Here goes (throwing out a minus sign): $\sum_{n=0}^{q-1} \omega^{-np} \sum_{k \geqslant 0} \frac{(\omega^n t)^k}{k} = \sum_{n=0}^{q-1} \sum_{k \geqslant 0} \frac{\omega^{n(k-p)} t^k}{k}$$ = \sum_{k \geqslant 0} \frac{t^k}{k} \sum_{n=0}^{q-1} \omega^{n(k-p)}$If $k-p \equiv 0 \mod q$, $\sum_0^{q-1} \omega^{n(k-p)} = \sum_0^{q-1} 1 = q$. If $k-p \not \equiv 0 \mod q$, then write $x = \omega^{k-p} = x$, another nontrivial $q^{th}$ root of unity. We have$(1 + x + \ldots + x^{q-1})(x-1) = x^q - 1 \Rightarrow (1 + x + \ldots + x^{q-1}) = \frac{x^q - 1}{x-1} = \frac{1-1}{x-1} = 0$(Note we can't divide if $x$ is the trivial root of unity). Thus this equals $q\sum_{k \equiv p \mod q} \frac{t^k}{k}$which isn't quite what you have, unless $p < q$...(otherwise you miss a few early terms).
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0:Thank you very muah! It's useful. – 2012-05-24