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Ted123

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## Homework Statement

Find the general solution of Airy's equation [tex]f'' - zf=0[/tex] 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 [tex]f(z)=\sum_{n=0}^{\infty} a_n z^n[/tex] into Airy's equation and manipulating the summations I get the recurrence relation: [tex]a_2 =0[/tex] [tex]a_{n+3} = \frac{a_{n}}{(n+2)(n+3)}\;,\;\;\;\forall \;\;n=0,1,2,3,...[/tex]

Solving this:

[tex]a_{3k+2}=0\;\;\;\forall \;\;k=0,1,2,3,...[/tex] and solving separately for n=3k and n=3k+1, [tex]a_{3(k+1)} = \frac{a_{3k}}{(3k+3)(3k+2)} = \frac{a_{3k}}{9(k+1)(k+\frac{2}{3})}[/tex] [tex]a_{3k} = \frac{a_0}{9^k (1)_k (\frac{2}{3})_k}[/tex] Similarly for n=3k+1, [tex]a_{3k+1} = \frac{a_1}{9^k (1)_k (\frac{4}{3})_k}[/tex] so that the solution is [tex]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)[/tex]

f(0)=a

_{0}, f'(0)=a

_{1}

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

[tex]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!}[/tex]

How do I express this in terms of the Gauss hypergeometric series?:

[PLAIN]http://img200.imageshack.us/img200/5992/gauss.png

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