Solved by Mike Skirvin

Question:

Encoding Fibonacci

Find a closed form for the Maclaurin series

(f_1)*x + (f_2)*x^2 + (f_3)*x^3 + ...

Where f_n is the n-th Fibonacci number, that is f_1 = 1, f_2 = 1, and for
n >= 3, f_n = f_(n-1) + f_(n-2).

For example, a closed for the Maclaurin series for 1 + x + x^2 + x^3 + ... would be 1/(1 - x) (|x| < 1).

Answer:

Let g(x) = f_1 * x + (f_2)*x^2 + (f_3)*x^3 + ...

Then x*g(x) = f_1 * x^2 + (f_2)*x^3 + (f_4)*x^4 + ...

Adding the two, g(x) + x*g(x) = x + (f_1 + f_2)x^2 + (f_3 + f_2)x^3 + ...

= x + (f_3)x^2 + (f_4)x^3 + ... = x + (1/x)((f_3)x^3 + (f_4)x^4 + ...)  (x != 0)

= x + (1/x)(g(x) - x - x^2)

therefore,
g(x) + x*g(x) = x + (1/x)(g(x) - x - x^2)  (x != 0)

or

g(x) = x/(1 - x - x^2). (x != 0)

if x = 0 then g(0) = 0 as well as f_1 * x + (f_2)*x^2 + ... so

g(x) = x/(1 - x - x^2)

Now we need to define the interval of convergence for this series.  We can rewrite g(x) = -x/(x^2 + x -1).  Factoring the bottom by completing the square gives g(x) = -x/[(x + 1/2)^2 - 5/4].  Comparing this to our standard geometric series 1/(1-x), we see that it will converge when (x + 1/2)^2 < 5/4 (alternatively we could have used the ratio test).  Therefore |x + 1/2| < sqrt(5)/2.  Finally we are left with (-1 - sqrt(5))/2 < x < (-1 + sqrt(5))/2.

-Mike Skirvin