4 Theta Functions 519
Moreover, there exists a constantA>0, depending onqbut not onnorN, such that
|cnN|≤A|q|n
2
.
For we can chooseB>0sothat|
∏m
k= 1 (^1 −q
2 k)|≥Bfor allm, we can choose
C>0sothat|
∏m
k= 1 (^1 −q
2 k)|≤Cfor allm, and we can then takeA=C/B (^2) .Since
the series
∑∞
n=−∞q
n^2 znis absolutely convergent, it follows that we can proceed to
the limit term by term in (31) to obtain (30).
In the series
∑∞
n=−∞q
n^2 znwe now put
q=eπiτ, z=e^2 πiv,
so that|q|<1 corresponds toIτ>0, and we define thetheta function
θ(v;τ)=
∑∞
n=−∞
eπiτn
2
e^2 πivn.
The functionθ(v;τ)is holomorphic invandτfor allv∈Candτ∈H (the upper
half-plane). Since initially we will be more interested in the dependence onv, withτ
just a parameter, we will often writeθ(v)in place ofθ(v;τ). Furthermore, we will still
useqas an abbreviation foreπiτ.
Evidently
θ(v+ 1 )=θ(v)=θ(−v).
Moreover,
θ(v+τ)=
∑∞
n=−∞
qn
(^2) + 2 n
e^2 πivn
=q−^1 e−^2 πiv
∑∞
n=−∞
q(n+^1 )
2
e^2 πiv(n+^1 )
=e−πi(^2 v+τ)θ(v).
It may be immediately verified that
∂^2 θ/∂v^2 =− 4 π^2 q∂θ/∂q= 4 πi∂θ/∂τ,
which becomes the partial differential equation of heat conduction in one dimension
on puttingτ= 4 πit.
By Proposition 2, we have also the product representation
θ(v)=
∏∞
n= 1
( 1 +q^2 n−^1 e^2 πiv)( 1 +q^2 n−^1 e−^2 πiv)( 1 −q^2 n).
It follows that the points
v= 1 / 2 +τ/ 2 +m+nτ(m,n∈Z)
are simple zeros ofθ(v), and that these are the only zeros.