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

18.1. Transformation theory 335
Obviously, only finitely many roots could exist, and the value of α could be obtained
from (18.1),
α = µω + µ ′ ω ′
(18.3)
in which µ, µ ′ were rational constants. However, the degree of the equation y = ψ (x)
might surpass the number of different values produced in this fashion and a new
group corresponding to a new value α2 of α might be necessary. ABEL found that a
certain number ν (which he did not describe in any detail) existed such that all the
roots of the equation y = ψ (x) would be representable by the different values of the
expression
λ
�
θ +
�
ν
∑ knαn
n=1
when k1, . . . , kν took all integer values. However, the different values could also be
represented as (possibly changing the values of α1, α2, . . . )
λ (θ) , λ (θ + α1) , . . . , λ (θ + αm−1)
in which α1, . . . , αm−1 were still rational linear combinations of ω and ω ′ (as in 18.3).
ABEL then wrote the rational function ψ (x) = p(x)
q(x)
obtained 12
m−1
p (x) − q (x) y = A
∏
n=0
with no common divisors and
(x − λ (θ + αn)) .
The constant A was of the form A = f − gy with f , g constants. ABEL’S next step
was to find an expression for A and he did so by first imposing a limiting assumption
and gradually relaxing it. First, ABEL considered the case in which both p and q were
polynomials of the first degree. In this case (e.g. by Euclidean division),
and ABEL found
y = f ′ + f x
g ′ + gx ,
dy = f g′ − f ′ g
(g ′ dx.
2
+ gx)
When he inserted this into the original differential equation, its dependence on dy
disappeared. Consequently, ABEL could conclude that the differential equation in this
case implied either of the three solutions
I.y = ax, c 2 1
II.y = a 1
ec
= c2
a 2 ,e2 1
x , c2 c2
1 =
III.y = m 1 − x√ ec
1 + x √ ec , c1 = 1
m
12 Here I write α0 = 0 for brevity.
a 2 ,e2 1
e2
= , or
a2 = e2
a
√ c − √ e
√ c + √ e ,e1 = 1
m
2 , or
√ √
c + e
√ √ , a =
c − e im
(c − e) .
2