I'm trying to understand why if $ \displaystyle \sum_{n=0}^{\infty} a_{n}x^{n} = f(x) $, then $$ \sum_{n=0}^{\infty} a_{p+nq} x^{p+nq} = \frac{1}{q} \sum_{j=0}^{q-1} \omega^{-jp} f(\omega^{j} x)$$ where $\omega$ is a primitive $q$-th root of unity.

I'm assuming it has something to do with the $q$-th roots of unity summing to zero.

It is used in a proof of Gauss' digamma theorem.


  • $\begingroup$ This has a number of duplicates but I can't find them immediately. $\endgroup$
    – anon
    Commented Aug 31, 2013 at 17:34

1 Answer 1


In the right hand side of your expresion, remplace $f(\omega^j x)$ by $\sum_{k=0}^{\infty} a_k \omega^{jk}x^k $.

You will get : $$\frac{1}{q}\sum_{j=0}^{q-1} \omega^{-jp}\sum_{k} a_k x^k \omega^{jk}$$

By interverting the sums (formally at least it has to be justified : see Jyrki Lahtonen's comment) you get : $$\sum_{k} a_k x^k \frac{1}{q}\sum_{j=0}^{q-1} \omega^{j(k-p)}$$

And then you can use the result you are refering to in order to get the left hand side :

If $(k-p)=nq$ : $\sum_{j=0}^{q-1} \omega^{jnq} =\sum_{j=0}^{q-1} (\omega^q)^{jn}= q$ ($\omega$ is a $q$-th root I guess ?) Otherwise $(k-p)=nq+r$ ($0< r<q$) and $$\sum_{j=0}^{q-1} \omega^{j(k-p)}=\sum_{j=0}^{q-1} \omega^{jnq+jr}=\sum_{j=0}^{q-1} \omega^{jr}=0$$

  • 1
    $\begingroup$ As this is a power series, it will converge on a disk (possibly one with infinite radius) about the origin. Possibly with exceptions along the boundary the substitution $x\to\omega^jx$ will thus not change the convergence. In any compact disk contained in the disk of convergence, the convergence will be absolute, and thus you can reorder the terms at will. +1 - obviously :-) $\endgroup$ Commented Aug 31, 2013 at 17:21

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