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

322 Chapter 17. Steps in the process of coming to “know” elliptic functions
was a polynomial in x 2 and so was Q2n+1. Thus, with
P2n+1 (x) = ax (2n+1)2 + · · · + bx and
Q2n+1 (x) = cx (2n+1)2 −1 + · · · + d,
ABEL found that the sum of the roots of the equation (17.1) would equal c aφ ((2n + 1) β).
Furthermore, he already knew the complete solution of the equation (17.1), and thus
he obtained
φ ((2n + 1) β) = A
n
∑
m=−n
n
∑
µ=−n
Similarly for the products of the roots of (17.1),
φ ((2n + 1) β) = B
n
∏
m=−n
(−1) m+µ �
φ β +
n
∏
µ=−n
�
φ β +
�
mω + µ ¯ωi
. (17.2)
2n + 1
�
mω + µ ¯ωi
.
2n + 1
These formulae invited the limit process n → ∞, and it is interesting to see how ABEL
carried it out.
17.1.1 Determination of the coefficient A
In order to determine the constant A of (17.2), ABEL wanted to insert a particular value
for β and he chose β = ω 2 + ¯ω 2 i. However, this is a singular value (a pole) of φ, and
ABEL applied a limit argument in the following form. First, using the relation
�
φ α + ω ¯ω
+
2 2 i
�
= − i 1
ec φ (α)
derivable from the addition formulae, ABEL found that for (m, µ) �= (0, 0),
�
mω + µ ¯ωi ω ¯ω
φ
+ +
2n + 1 2 2 i
� �
mω + µ ¯ωi ω ¯ω
+ φ − + +
2n + 1 2 2 i
�
= 0.
(17.3)
Consequently, the sum reduced to the term corresponding to (m, µ) = (0, 0). For the
last term, ABEL found that
A = lim
β→ ω 2 + ¯ω 2 i
φ ((2n + 1) β)
φ (β)
=
by (17.3) lim
α→0
φ (α)
φ ((2n + 1) α)
φ
= lim
α→0
� (2n + 1) � ω
2 + ¯ω 2 i + α��
φ � ω
2 + ¯ω 2 i + α�
= 1
2n + 1
where tacit applications of the differentiability of φ and of the Rule of l’Hospital are
involved.