Solution of Airy's Equation in Terms of Gauss Hypergeometric Series

[Ted123]

Homework Statement

Find the general solution of Airy's equation $$f'' - zf=0$$ satisfying the initial conditions f(0)=1, f'(0)=0 as a power series expansion at z=0. Express the result in terms of the Gauss hypergeometric series.

The Attempt at a Solution

After subbing $$f(z)=\sum_{n=0}^{\infty} a_n z^n$$ into Airy's equation and manipulating the summations I get the recurrence relation: $$a_2 =0$$ $$a_{n+3} = \frac{a_{n}}{(n+2)(n+3)}\;,\;\;\;\forall \;\;n=0,1,2,3,...$$
Solving this:
$$a_{3k+2}=0\;\;\;\forall \;\;k=0,1,2,3,...$$ and solving separately for n=3k and n=3k+1, $$a_{3(k+1)} = \frac{a_{3k}}{(3k+3)(3k+2)} = \frac{a_{3k}}{9(k+1)(k+\frac{2}{3})}$$ $$a_{3k} = \frac{a_0}{9^k (1)_k (\frac{2}{3})_k}$$ Similarly for n=3k+1, $$a_{3k+1} = \frac{a_1}{9^k (1)_k (\frac{4}{3})_k}$$ so that the solution is $$f(z) = a_0 \left( \sum_{k=0}^{\infty} \frac{1}{9^k (1)_k (\frac{2}{3})_k} z^{3k} \right) + a_1 \left( \sum_{k=0}^{\infty} \frac{1}{9^k (1)_k (\frac{4}{3})_k} z^{3k+1} \right)$$

f(0)=a₀, f'(0)=a₁

Hence the solution to the initial value problem f(0)=1, f'(0)=0 is:

$$f(z) = \sum_{k=0}^{\infty} \frac{1}{9^k (1)_k (\frac{2}{3})_k} z^{3k} = \sum_{k=0}^{\infty} \frac{1}{9^k (\frac{2}{3})_k} \frac{z^{3k}}{k!}$$

How do I express this in terms of the Gauss hypergeometric series?:
[PLAIN]http://img200.imageshack.us/img200/5992/gauss.png

 

[mighty2000]

Great job on finding the general solution to Airy's equation and satisfying the initial conditions! To express the solution in terms of the Gauss hypergeometric series, we can use the identity:
_2F_1(a,b;c;z) = \sum_{k=0}^{\infty} \frac{(a)_k (b)_k}{(c)_k} \frac{z^k}{k!}
where (x)_k is the Pochhammer symbol.
In this case, we can rewrite the solution as:
f(z) = a_0 \left( \sum_{k=0}^{\infty} \frac{(1)_k (\frac{2}{3})_k}{(9)_k} z^{3k} \right) + a_1 \left( \sum_{k=0}^{\infty} \frac{(1)_k (\frac{4}{3})_k}{(9)_k} z^{3k+1} \right)
= a_0 \left( \sum_{k=0}^{\infty} \frac{(1)_k (\frac{2}{3})_k}{(9)_k} z^{3k} \right) + a_1 \left( \sum_{k=0}^{\infty} \frac{(1)_k (\frac{4}{3})_k}{(9)_k} z^{3k} \frac{z}{k+1} \right)
= a_0 \left( \sum_{k=0}^{\infty} \frac{(1)_k (\frac{2}{3})_k}{(9)_k} z^{3k} \right) + a_1 \left( \sum_{k=1}^{\infty} \frac{(1)_{k-1} (\frac{4}{3})_{k-1}}{(9)_{k-1}} z^{3k} \frac{z}{k} \right)
= a_0 \left( \sum_{k=0}^{\infty} \frac{(1)_k (\frac{2}{3})_k}{(9)_k} z^{3k} \right) + a_1 \left( \sum_{k=0}^{\infty