Chapter 58

Elliptic FunctionsThis chapter is to give a short introduction to elliptic functions. There is much more avail-able. There are books written on elliptic functions. What I am presenting here followsAlfors [3] although the material is found in many books on complex analysis. Hille, [65]has a much more extensive treatment than what I will attempt here. There are also manyreferences and historical notes available in the book by Hille. Another good source formore having much the same emphasis as what is presented here is in the book by Saks andZygmund [115]. This is a very interesting subject because it has considerable overlap withalgebra.

Before beginning, recall that an absolutely convergent series can be summed in anyorder and you always get the same answer. The easy way to see this is to think of the seriesas a Lebesgue integral with respect to counting measure and apply convergence theoremsas needed. The following theorem provides the necessary results.

Theorem 58.0.1 Suppose ∑∞n=1 |an|< ∞ and let θ ,φ :N→ N be one to one and onto map-

pings. Then ∑∞n=1 aφ(n) and ∑

∞n=1 aθ(n) both converge and the two sums are equal.

Proof: By the monotone convergence theorem,

∞

∑n=1|an|= lim

n→∞

n

∑k=1

∣∣aφ(k)∣∣= lim

n→∞

n

∑k=1

∣∣aθ(k)∣∣

but these last two equal ∑∞k=1

∣∣aφ(k)∣∣ and ∑

∞k=1

∣∣aθ(k)∣∣ respectively. Therefore, ∑

∞k=1 aθ(k)

and ∑∞k=1 aφ(k) exist (n→ aθ(n) is in L1 with respect to counting measure.) It remains to

show the two are equal. There exists M such that if n > M then

∞

∑k=n+1

∣∣aθ(k)∣∣< ε,

∞

∑k=n+1

∣∣aφ(k)∣∣< ε

∣∣∣∣∣ ∞

∑k=1

aφ(k)−n

∑k=1

aφ(k)

∣∣∣∣∣< ε,

∣∣∣∣∣ ∞

∑k=1

aθ(k)−n

∑k=1

aθ(k)

∣∣∣∣∣< ε

Pick such an n denoted by n1. Then pick n2 > n1 > M such that

{θ (1) , · · · ,θ (n1)} ⊆ {φ (1) , · · · ,φ (n2)} .

Thenn2

∑k=1

aφ(k) =n1

∑k=1

aθ(k)+ ∑φ(k)/∈{θ(1),··· ,θ(n1)}

aφ(k).

Therefore, ∣∣∣∣∣ n2

∑k=1

aφ(k)−n1

∑k=1

aθ(k)

∣∣∣∣∣=∣∣∣∣∣ ∑φ(k)/∈{θ(1),··· ,θ(n1)},k≤n2

aφ(k)

∣∣∣∣∣1817