How do I rearrange this power series: $$\sum_{n=1}^\infty\frac{e^{in}}{n^3}{(z^n-z^{-n})}$$ so that it can be expressed in the form of $$\sum_{n=0}^\infty\ a_n{(z-z_0)}^n $$
Asked
Active
Viewed 56 times
1 Answers
1
At least to me, this is not an easy problem since $$\sum_{n=1}^\infty\frac{e^{in}}{n^3}{(z^n-z^{-n})}=\text{Li}_3\left(e^i z\right)-\text{Li}_3\left(\frac{e^i}{z}\right)$$ Then, the expansion around $z=a$ will include a bunch of polylogarithms everywhere.
The very first terms would be $$\left(\text{Li}_3\left(a e^i\right)-\text{Li}_3\left(\frac{e^i}{a}\right)\right)+\frac{\left(\text{Li}_ 2\left(\frac{e^i}{a}\right)+\text{Li}_2\left(a e^i\right)\right) }{a}(z-a)+$$ $$\frac{ \left(-\text{Li}_2\left(\frac{e^i}{a}\right)-\text{Li}_2\left(a e^i\right)+\log \left(1-\frac{e^i}{a}\right)-\log \left(1-e^i a\right)\right)}{2 a^2}(z-a)^2+O\left((z-a)^3\right)$$
Claude Leibovici
- 260,315
-
What I was really looking for was the set of all the complex numbers for which the power series converges absolutely, but I thought it would be useful if I could rewrite the power series in the general form, but this is math way above my current level. – PTSONIC Mar 06 '20 at 08:55