This is an offshoot of the question: Series Expansions in Mathematica. In that question, I thought I would simplify my problem and ask the simplest version of it. While I have gotten that answered, it seems there is still an issue with the real Mathematica code that I have.
I'm studying the Hurwitz Lerch Transcendant $\Phi(z,s,k)$ and am particularly interested only with $\Phi(z,n,\frac{1}{2})$ for $n\in {\mathbb Z}$. This function satisfies the following property. Define
f[z_, n_] := 2^(n - 1)/Sqrt[z] (PolyLog[n, Sqrt[z]] - PolyLog[n, -Sqrt[z]])
Then,
HurwitzLerchPhi[z, n, 1/2] == f[z, n]
Now, Mathematica knows this fact. For instance, if I write
Plot[(HurwitzLerchPhi[z, n, 1/2] - f[z, n]) /. n -> 9, {z, 0.0000001,1}]
Plot[(HurwitzLerchPhi[z, n, 1/2] - f[z, n]) /. n -> 13, {z, 0.0000001,1}]
Plot[(HurwitzLerchPhi[z, n, 1/2] - f[z, n]) /. n -> 21, {z, 0.0000001,1}]
Plot[(HurwitzLerchPhi[z, n, 1/2] - f[z, n]) /. n -> 3, {z, 0.0000001,1}]
the plots are identically 0. I chose some random integers above. You can check this with other ones as well. We can also check this numerically with arbitrary accuracy.
However, if I consider the following code into Mathematica
FullSimplify[Series[HurwitzLerchPhi[1 - Sqrt[z], 3, 1/2] - f[1 - Sqrt[z], 3], {z, 0, 6}], 0 < z < 1]
I get $$ \frac{z^2}{4}+\frac{3 z^{5/2}}{8}+\frac{55 z^3}{96}+\frac{131 z^{7/2}}{192}+\frac{19219 z^4}{23040} +\frac{42493 z^{9/2}}{46080}\\\qquad \qquad \qquad \qquad +\frac{268843 z^5}{258048}+\frac{957181 z^{11/2}}{860160}+\frac{107031761 z^6}{88473600}+O\left(z^{13/2}\right) $$ which is demonstrably non-vanishing. The problem seems to occur if the argument is $1-\sqrt{z}$. For instance, there is no issue if I take the argument to be $1-z$ or $\sqrt{1-z}$, but you do get a non-zero answer with $\sqrt{1-\sqrt{z}}$.
I don't understand at all what Mathematica is doing here. Further, the issue does not seem to be one of convergence (As suggested in the answer to Series Expansions in Mathematica).
What is going on?
f
written in formatted inline code. I believe you will get more help if you do so. $\endgroup$f
by @JackLVigne. That said, some numerical checks point to the possibility of a bug in theSeries
expansion. Possibly related to the issue reported in the earlier MSE note (same OP) reArcTanh
. $\endgroup$HurwitzLerchPhi
. $\endgroup$