RePoSS #11: The Mathematics of Niels Henrik Abel: Continuation ...

324 Chapter 17. Steps in the process of coming to “know” elliptic functions
and he claimed and proved that the right hand side was “of the form v
2n+1 ”. In the
process, ABEL made use of the result that
φ (α) = α + Aα 3 + . . .
because φ was an odd function and φ ′ (0) = f (0) F (0) = 1.
ABEL’S next step concerned the sum of θ. He found that
and therefore, for each m,
∞
∑ (−1)
µ=n
µ θ (m, µ) = v
2n + 1
n−1
∑ (−1)
µ=0
µ ψ (m, µ) =
When he summed these, ABEL found
n−1 n−1
∑ ∑ (−1)
m=0 µ=0
m+µ n−1
ψ (m, µ) = ∑ (−1)
m=0
m
and, as ABEL wrote,
n−1
∑
m=0
vm
2n + 1
∞
∑ (−1)
µ=0
µ θ (m, µ) + vm
2n + 1 .
�
∞
∑ (−1)
µ=0
µ �
n−1
θ (m, µ) + ∑
m=0
= nv
2n + 1
= v
2 ,
vm
2n + 1
in which “v is a quantity which has zero for its limit”. 3 Consequently, ABEL had found
a way of expressing φ (α) as a double infinite sum, e.g.
φ (α) = 1
ec
∞
∑
∞
∑ (−1) m+µ
�
(2µ+1) ¯ω
m=0 µ=0
(α−(m+ 1 2)ω) 2 +(µ+ 1 2) 2 −
¯ω 2
(2µ+1) ¯ω
(α+(m+ 1 2)ω) 2 +(µ+ 1 2) 2 ¯ω 2
�
.
(17.5)
ABEL did not stop after having obtained the expansion in infinite series (17.5).
Instead, he used similar methods to search for expressions for φ (α) involving infinite
products. ABEL also invested an effort in obtaining expressions involving only one
infinite series and transcendental objects in the terms. Among the formulae which he
obtained, the following was probably the simplest:
φ (α) = 2
∞
π
ec ¯ω ∑ (−1)
k=0
k ε2k+1 − ε−(2k+1) r2k+1 − r−(2k+1) where
�
ε = exp α π
�
�
ωπ
�
and r = exp .
¯ω
2 ¯ω
Thus, as described, ABEL used his multiplication formulae to obtain rather simple
infinite representations for elliptic functions. When he passed to the infinite limit, his
arguments did not conform to the strict standards of rigor, which he had advocated
in the theory of series. Further perspectives on ABEL’S motivations for searching for
infinite expressions and his methods of obtaining them are described and discussed
in the subsequent sections.
3 (N. H. Abel, 1827b, 161).