Eisenstein Series in Mathematica?

Question

Mathematica doesn't seem to have built-in tools to deal with the Eisenstein series:

$$\begin{align*} E_{2}(\tau)&= 1-24 \sum_{n=1}^{\infty} \frac{n e^{2 \pi i n \tau}}{1-e^{2 \pi i n \tau}}\\ E_{4}(\tau)&= 1+240 \sum_{n=1}^{\infty} \frac{n^{3} e^{2 \pi i n \tau}}{1-e^{2 \pi i n \tau}} \end{align*}$$

I'm wondering what is the best way to deal with this. Just messing around, informally, on Wolfram, it seems like these series all converge pretty fast. Can I carry out the sums manually in Mathematica including a small number of terms? Or is there a better way? I'm worried this is prone to severe inaccuracy for $\Im(\tau)$ either large or small, both cases I'm interested in.

If it simplifies anything, I only need the cases $\Re(\tau) \in \mathbb{Z}$ where the series outputs real numbers.

Answer 1

Since EllipticTheta[] is a built-in function, and since the Eisenstein series $E_4(q)$ and $E_6(q)$ are expressible in terms of theta functions (I use the nome $q$ as the argument in this answer, but you can convert to your convention by using the relation with the period ratio $\tau$: $q=\exp(2\pi i \tau)$), and since the higher-order Eisenstein series (note that they are only defined for even orders!) can be generated from $E_4(q)$ and $E_6(q)$ through a recurrence (see e.g. Apostol's book), it is relatively straightforward to write Mathematica routines for these functions:

SetAttributes[EisensteinE, {Listable, NHoldFirst}];

EisensteinE[4, q_] := (EllipticTheta[2, 0, q]^8 + EllipticTheta[3, 0, q]^8 +
                       EllipticTheta[4, 0, q]^8)/2

EisensteinE[6, q_] := With[{q2 = EllipticTheta[2, 0, q]^4,
                            q3 = EllipticTheta[3, 0, q]^4,
                            q4 = EllipticTheta[4, 0, q]^4}, 
                           (q2 + q3) (q3 + q4) (q4 - q2)/2

EisensteinE[n_Integer?EvenQ, q_] /; n > 2 := (6/((6 - n) (n^2 - 1) BernoulliB[n]))
Sum[Binomial[n, 2 k + 4] (2 k + 3) (n - 2 k - 5)
    BernoulliB[2 k + 4] BernoulliB[n - 2 k - 4]
    EisensteinE[2 k + 4, q] EisensteinE[n - 2 k - 4, q], {k, 0, n/2 - 4}]

Here are a few examples:

(* "equianharmonic case" *)
{ω1, ω3} = {1, (1 + I Sqrt[3])/2};
N[WeierstrassInvariants[{ω1, ω3}]] // Quiet // Chop
   {0, 12.825381829368068}

2 {60, 140} Zeta[{4, 6}] EisensteinE[{4, 6}, Exp[I π ω3/ω1]]/(2 ω1)^{4, 6}
// N // Chop
   {0, 12.825381829368068}

(* "lemniscatic case" *)
{ω1, ω3} = {1, I};
N[WeierstrassInvariants[{ω1, ω3}]] // Quiet // Chop
   {11.817045008077123, 0}

2 {60, 140} Zeta[{4, 6}] EisensteinE[{4, 6}, Exp[I π ω3/ω1]]/(2 ω1)^{4, 6}
// N // Chop
   {11.817045008077123, 0}

Using techniques similar to the one used in this answer, here are domain-colored plots of $E_4(q)$ (left) and $E_6(q)$ (right) over the unit disk, using the DLMF coloring scheme:

---

Now, one may ask: what about $E_2(q)$? This function is what is termed as a "quasi-modular" form, whose behavior with respect to modular transformations is completely different from the other $E_{2k}(q)$. Due to this unusual state of affairs (i.e. not expressible entirely in terms of theta functions), one needs a different formula for $E_2(q)$; one useful formula can be found hidden deep within Abramowitz and Stegun:

EisensteinE[2, q_] := With[{q3 = EllipticTheta[3, 0, q]^2}, 6/π
                           EllipticE[InverseEllipticNomeQ[q]] q3 -
                           q3^2 - EllipticTheta[4, 0, q]^4]

Test:

Series[EisensteinE[2, q], {q, 0, 12}]
   1 - 24 q^2 - 72 q^4 - 96 q^6 - 168 q^8 - 144 q^10 - 288 q^12 + O[q]^13

1 - Sum[24 DivisorSigma[1, k] q^(2 k), {k, 1, 6}]
   1 - 24 q^2 - 72 q^4 - 96 q^6 - 168 q^8 - 144 q^10 - 288 q^12

Unfortunately, altho this version is great for symbolic use, it is not too good for numerical evaluation, as can be seen from the following attempt to generate a domain-colored plot from it:

The relatively complicated branch cut structure is apparently inherited from the branch cuts of the complete elliptic integral of the second kind $E(m)$ not being canceled out by the inverse nome.

Thus, I shall present another routine for numerically evaluating $E_2(q)$, based on recursing the quasi-modular relation (note the use of $\tau$ instead of $q$)

$$E_2\left(-\frac1{\tau}\right)=\tau^2 E_2(\tau)-\frac{6i\tau}{\pi}$$

before the actual numerical evaluation of the series:

e2[zz_ /; (InexactNumberQ[zz] && Im[zz] > 0)] :=
   Block[{τ = SetPrecision[zz, 1. Precision[zz]], r = False, f, k, pr, q, qp, s},
         τ -= Round[Re[τ]]; pr = Precision[τ];
         If[7 Im[τ] < 6,
            r = True; f = e2[SetPrecision[-1/τ, pr]],
            q = SetPrecision[Exp[2 π I τ], pr]; f = s = 0; qp = 1; 
            k = 0;
            While[k++; qp *= q; f = s + k qp/(1 - qp); s != f, s = f];
            f = 1 - 24 f];
         If[r, (f/τ + 6 I/π)/τ, f] /; NumberQ[f]]

EisensteinE[2, q_?InexactNumberQ] :=
   If[q == 0, N[1, Internal`PrecAccur[q]], e2[Log[q]/(I π)]]

(Note that the subroutine e2[] actually takes the period ratio $\tau$ as the argument; if your preferred convention is to use $\tau$ instead of $q$, you can make that the main routine and skip the conversion to $q$ altogether.)

This now gives a proper-looking plot:

(Thanks to მამუკა ჯიბლაძე for convincing me to look further into this.)

---

Finally, if you prefer the function $G_{2k}(q)$, here is the corresponding formula:

EisensteinG[n_Integer?EvenQ, q_] := 2 Zeta[n] EisensteinE[n, q]

Answer 2

The above expression for EisensteinE[2, q] is a bit slow. Here is a faster version using double precision arithmetic:

EisensteinE[2, tau_Complex /; Im[tau] > 0] := ee2Upper[tau];
ee2Upper = Compile[
    {{tauIn, _Complex}},
    Block[{tau = tauIn, f, k, q, qp, s},
         tau -= Round[Re[tau]];
         If[7 Im[tau] < 6,
            f = ee2Upper[-1/tau]; (f/tau + 6 I/Pi)/tau,
            q = Exp[2 Pi I tau]; f = s = 0.0 + 0.0 I; qp = 1.0 + 0.0 I; 
            k = 0;
            While[k++; qp *= q; f = s + k qp/(1 - qp);
                  Abs[s - f] > MachinePrecision, s = f];
            1 - 24 f]],
    {{ee2Upper[_], _Complex}}];

Note that it is expressed in terms of τ instead of q.

Answer 3

Since 11.2 (2017) Mathematica does have the built-in tools to deal with the Eisenstein series:

$$G_{2k}(\tau)=\sum_{(m,n)}{'}\frac{1}{(m+n \tau)^{2k}}$$ being the functions:

WeierstrassInvariantG2
WeierstrassInvariantG3

Calculating $G_{2k}(\tau)$ can be implemented as follows:

g2[2, t_] = WeierstrassInvariantG2[{1/2, t/2}]/60;
g2[3, t_] = WeierstrassInvariantG3[{1/2, t/2}]/140;
g2[m_, t_] :=  3 Sum[(2 r - 1) (2 m - 2 r - 1) g2[r, t] g2[m - r, t], {r, 2, m - 2}]/((2 m + 1) (m - 3) (2 m - 1))

( Updated 17/6-'21 ):

nEisensteinG2K[2, t_]=nEisensteinG2K[2, {1, t}];
nEisensteinG2K[3, t_]=nEisensteinG2K[3, {1, t}];
nEisensteinG2K[2, {w1_, w3_}]=WeierstrassInvariantG2[{w1/2,w3/2}]/60;
nEisensteinG2K[3, {w1_, w3_}]=WeierstrassInvariantG3[{w1/2,w3/2}]/140;
nEisensteinG2K[m_, {w1_, w3_}]:= 3 Sum[(2 r - 1) (2 m - 2 r - 1) nEisensteinG2K[r, {w1, w3}] nEisensteinG2K[m - r, {w1, w3}], {r, 2, m - 2}]/((2 m + 1) (m - 3) (2 m - 1))

Source/Reference:

Modular Functions and Dirichlet Series in Number Theory, Tom Apostol ( Chapter 1 Par 1.9 "The Eisenstein series and the invariants g2 and g3" )