.. _algorithms_polylogarithms:

Algorithms for polylogarithms
===============================================================================

The polylogarithm is defined for `s, z \in \mathbb{C}` with `|z| < 1` by

.. math ::

    \operatorname{Li}_s(z) = \sum_{k=1}^{\infty} \frac{z^k}{k^s}

and for `|z| \ge 1` by analytic continuation, except for the singular
point `z = 1`.

Computation for small `z`
-------------------------------------------------------------------------------

The power sum converges rapidly when `|z| \ll 1`.
To compute the series expansion with respect to `s`, we substitute
`s \to s + x \in \mathbb{C}[[x]]` and obtain

.. math ::

    \operatorname{Li}_{s+x}(z) = \sum_{d=0}^{\infty} x^d
        \frac{(-1)^d}{d!} \sum_{k=1}^{\infty} T(k)

where

.. math ::

        T(k) = \frac{z^k \log^d(k)}{k^s}.

Let `U(k)` be a nonincreasing bound for the magnitude of the term ratio

.. math ::

    \frac{T(k+1)}{T(k)} = z \left(\frac{k}{k+1}\right)^s 
        \left( \frac{\log(k+1)}{\log(k)} \right)^d.

Then the remainder after the `k = N-1` term is bounded by

.. math ::

    \left| \sum_{k=N}^{\infty} T(k) \right|
        \le |T(N)| \sum_{k=0}^{\infty} U(N)^k = \frac{|T(N)|}{1 - U(N)}

whenever `U(N) < 1`.

If `s = \sigma + i \tau` where `\sigma, \tau \in \mathbb{R}`, we can take

.. math ::

    U(k) = |z| \; B(k, \max(0, -\sigma)) \; B(k \log(k), d)

wherein `B(m,n) = (1 + 1/m)^n`. This follows from the bounds

.. math ::

    \left| \left(\frac{k}{k+1}\right)^s \right|
    = \left(\frac{k}{k+1}\right)^{\sigma}
    \le \begin{cases}
        1                    & \text{if }         \sigma \ge 0 \\
        (1 + 1/k)^{-\sigma}  & \text{if }         \sigma < 0.
        \end{cases}

and

.. math ::

    \left( \frac{\log(k+1)}{\log(k)} \right)^d \le
        \left(1 + \frac{1}{k \log(k)}\right)^d.

We can replace `\sigma` with any lower bound, conveniently
the nearest integer in the direction of `-\infty`, and `|z|` with any
upper bound.

To bound `B(m,n)` when `m` is large, it is useful to note that
`B(m,n) = \exp(n \log(1+1/m)) \le \exp(n/m)`).

Expansion for general `z`
-------------------------------------------------------------------------------

For general complex `s, z`, we write the polylogarithm as a sum of
two Hurwitz zeta functions

.. math ::

    \operatorname{Li}_s(z) = \frac{\Gamma(v)}{(2\pi)^v}
        \left[
            i^v
            \zeta \left(v, \frac{1}{2} + \frac{\log(-z)}{2\pi i}\right)
            + i^{-v}
            \zeta \left(v, \frac{1}{2} - \frac{\log(-z)}{2\pi i}\right)
        \right]

in which `s = 1-v`.
With the principal branch of `\log(-z)`, we obtain the conventional
analytic continuation of the polylogarithm with a branch
cut on `z \in (1,+\infty)`.

To compute the series expansion with respect to `v`, we substitute
`v \to v + x \in \mathbb{C}[[x]]` in this formula
(at the end of the computation, we map `x \to -x` to
obtain the power series for `\operatorname{Li}_{s+x}(z)`).
The right hand side becomes

.. math ::

    \Gamma(v+x) [E_1 Z_1 + E_2 Z_2]

where `E_1 = (i/(2 \pi))^{v+x}`, `Z_1 = \zeta(v+x,\ldots)`,
`E_2 = (1/(2 \pi i))^{v+x}`, `Z_2 = \zeta(v+x,\ldots)`.

When `v = 1`, the `Z_1` and `Z_2` terms become Laurent series with
a leading `1/x` term. In this case,
we compute the deflated series `\tilde Z_1, \tilde Z_2 = \zeta(x,\ldots) - 1/x`.
Then

.. math ::

    E_1 Z_1 + E_2 Z_2 = (E_1 + E_2)/x + E_1 \tilde Z_1 + E_2 \tilde Z_2.

Note that `(E_1 + E_2) / x` is a power series, since the constant term in
`E_1 + E_2` is zero when `v = 1`. So we simply compute one extra derivative
of both `E_1` and `E_2`, and shift them one step.
When `v = 0, -1, -2, \ldots`, the `\Gamma(v+x)` prefactor has a pole.
In this case, we proceed analogously and formally multiply
`x \, \Gamma(v+x)` with `[E_1 Z_1 + E_2 Z_2] / x`.

Note that the formal cancellation only works when the order `s` (or `v`)
is an exact integer: it is not currently possible to use this method when
`s` is a small ball containing any of `0, 1, 2, \ldots` (then the
result becomes indeterminate).

The Hurwitz zeta method becomes inefficient when `|z| \to 0` (it
gives an indeterminate
result when `z = 0`). This is not a problem since we just use the defining series
for the polylogarithm in that region.
It also becomes inefficient when `|z| \to \infty`, for which an asymptotic
expansion would better.