are assigned. In that case, putting (x
- [xi]1)/(x - [xi]2) = t/(t - 1), the singular points are transformed
to 0, 1, [oo], and, as is clear, without change of indices. Still
denoting the independent variable by x, the equation then has the form

x(1 - x)y" + y'[1 - [lambda]1 - x(1 + [lambda] + [mu])] - [lambda][mu]y = 0,

which is the ordinary hypergeometric equation. Provided none of
[lambda]1, [lambda]2, [lambda] - [mu] be zero or integral about x = 0,
it has the solutions

F([lambda], [mu], 1 - [lambda]1, x), x^[lambda]1 F([lambda] + [lambda]1, [mu] + [lambda]1, 1 + [lambda]1, x);

about x = 1 it has the solutions

F([lambda], [mu], 1 - [lambda]2, 1 - x), (1 - x)^[lambda]1 F([lambda] + [lambda]2, [mu] + [lambda]2, 1 + [lambda]2, 1 - x),

where [lambda] + [mu] + [lambda]1 + [lambda]2 = 1; about x = [oo] it
has the solutions

x^-[lambda] F([lambda], [lambda] + [lambda]1, [lambda] - [mu] + 1, x^-1),
x^-[mu] F([mu], [mu] + [lambda]1, [mu] - [lambda] + 1, x^-1),

where F([alpha], [beta], [gamma], x) is the series

[alpha][beta]x [alpha]([alpha] + 1)[beta]([beta] + 1)x^2
1 + -------------- + ----------------------------------------- ...,
[gamma] 1.2.[gamma]([gamma] + 1)

which converges when |x| < 1, whatever [alpha], [beta], [gamma] may
be, converges for all values of x for which |x| = 1 provided the real
part of [gamma] - [alpha] - [beta] < 0 algebraically, and converges
for all these values except x = 1 provided the real part of [gamma] -
[alpha] -[beta] > -1 algebraically.

In accordance with our general theory, logarithms are to be expected
in the solution when one of [lambda]1, [lambda]2, [lambda] - [mu] is
zero or integral. Indeed when [lambda]1 is a negative integer, not
zero, the second solution about x = 0 would contain vanishing factors
in the denominators of its coefficients; in case [lambda] or [mu] be
one of the positive integers 1, 2, ... (-[lambda]1), vanishing factors
occur also in the numerators; and then, in fact, the second solution
about x = 0 becomes x^[lambda]1 times an integral polynomial of degree
(-[lambda]1) - [lambda] or of degree (-[lambda]1) - [mu]. But when
[lambda]1 is a negative integer including zero, and neither [lambda]
nor [mu] is one of the positive integers 1, 2 ... (-[lambda]1), the
second solution about x = 0 involves a term having the factor log x.
When