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364 pages, Paperback
First published January 1, 2008
F(x)=(1/(1-x-x^2)
It is convenient to factor the denominator as follows:
F(x)=1/[(1-((-2)/(1-sqrt(5)))x:]*[1-((-2)/(1+sqrt(5)))x):]
A little more algebraic manipulation yields that
F(x)=(5+sqrt(5)/10)[1/(1+(2/(1-sqrt(5)))x):]+(5-sqrt(5)/10)[1/(1+(2/(1+sqrt(5)))x):]
Now we want to apply the formula in Eq. (6.2) from Sec. 6.4 to each of the fractions in brackets ([:]). For the first fraction, we think of -(2/(1-sqrt(5)))x as lambda. Thus the first expression in brackets equals
(sigma(j=0->infinity)(-(2/(1-sqrt(5)))x))^j
Likewise the second sum equals
(sigma(j=0->infinity)(-(2/(1+sqrt(5)))x))^j
All told, we find that
F(x)=((5+sqrt(5))/(10))(sigma(j=0->infinity)(-(2/(1-sqrt(5)))x))^j+ ((5-sqrt(5))/(10))(sigma(j=0->infinity)(-(2/(1-sqrt(5)))x))^j
Grouping terms with like powers of x, we finally conclude that
F(x)=(sigma(j=0->infinity)[((5+sqrt(5))/(10))(-(2/(1-sqrt(5))))^j+((5-sqrt(5))/(10))(-(2/(1+sqrt(5))))^j:])x^j