624 Series
converge to f(x) and g(x) when jxj < r, then the series in (3) will converge to
f(x)g(x) when ixi < r provided the constants co, ci, c2, - are determined by
the formulas
co = aobo
c, = aob, + aibo
c2 = aob2 + aibi + a2bo
c3 = aoba + aib2 + a2bi + aabo
etcetera. Observe that this is precisely the way we would write the product of
the right members of (1) and (2) if they were polynomials. To obtain a bit of
experience with these formulas, write the formulas to which (1), (2), and (3)
reduce when ak = bk = I for each k. Check your work by obtaining the third
formula from the first in another way.
9 Prove that if the series uo + u1 + u2 + - has partial sums so, s3, s2,
and if the series in
f(x) = u0 + ulx + u2x2 + u3x3 +
converges to f(x) when lxi < 1, then
f(x) _ (1 - X) (SO + slx + J2x2 + 13x3 +
when Jxj < 1. Hint: Use the information given at the start of the preceding
problem, putting bk = 1 for each k.
10 Write two more terms of each of the series
a
es=1-x+23+
1
1+x= 1 -x+x2 - x3+.
-: 2
+x=1-2x+ 2
11 It can be proved that if the series in the first of the formulas
(1) f(x) = ao + aix + a2x2 + aax3 + a4xa + ...
(2) 1
f(z )
=bo+bIx+b2x2+bax3+b4x4+
converges to f(x) when jxj < r1 and if ao 0 0, then there exist numbers r2, bo,
bi, b2,--- such that the series in (2) converges to 1/f(x) when jxi < r2. Since
the product of the left members of (1) and (2) has the power series expansion
1+0x+0x2+0x3+0x4 + .. ,
the coefficients bo, bi, b2,- can be calculated from the formulas
1 = aobo
0 = aoba + albo
0 = aob2 + aibi + a2bo
0 = aoba + aib2 + a2b1 + a3bo.