Skip to main content

numerics - Eisenstein Series in Mathematica?


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



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:


domain-colored plots of the Eisenstein series




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:



wrong domain-colored plot of E2


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:


domain-colored plot of E2


(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]

Comments

Popular posts from this blog

plotting - Filling between two spheres in SphericalPlot3D

Manipulate[ SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, Mesh -> None, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], {n, 0, 1}] I cant' seem to be able to make a filling between two spheres. I've already tried the obvious Filling -> {1 -> {2}} but Mathematica doesn't seem to like that option. Is there any easy way around this or ... Answer There is no built-in filling in SphericalPlot3D . One option is to use ParametricPlot3D to draw the surfaces between the two shells: Manipulate[ Show[SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], ParametricPlot3D[{ r {Sin[t] Cos[1.5 Pi], Sin[t] Sin[1.5 Pi], Cos[t]}, r {Sin[t] Cos[0 Pi], Sin[t] Sin[0 Pi], Cos[t]}}, {r, 1, 2 - n}, {t, 0, Pi}, PlotStyle -> Yellow, Mesh -> {2, 15}]], {n, 0, 1}]

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]

plotting - Mathematica: 3D plot based on combined 2D graphs

I have several sigmoidal fits to 3 different datasets, with mean fit predictions plus the 95% confidence limits (not symmetrical around the mean) and the actual data. I would now like to show these different 2D plots projected in 3D as in but then using proper perspective. In the link here they give some solutions to combine the plots using isometric perspective, but I would like to use proper 3 point perspective. Any thoughts? Also any way to show the mean points per time point for each series plus or minus the standard error on the mean would be cool too, either using points+vertical bars, or using spheres plus tubes. Below are some test data and the fit function I am using. Note that I am working on a logit(proportion) scale and that the final vertical scale is Log10(percentage). (* some test data *) data = Table[Null, {i, 4}]; data[[1]] = {{1, -5.8}, {2, -5.4}, {3, -0.8}, {4, -0.2}, {5, 4.6}, {1, -6.4}, {2, -5.6}, {3, -0.7}, {4, 0.04}, {5, 1.0}, {1, -6.8}, {2, -4.7}, {3, -1.