Approximating unsolvable recursion relations

1. Jul 5, 2010

Hoplite

I have a complicated recursion replation, which I'm sure is unsolvable. (By "unsolvable" I mean that there is no closed form solution expressing $$\xi_1$$, $$\xi_2$$, $$\xi_3$$, etc. in terms of $$\xi_0$$.) It goes

$$\frac{(k+4)!}{k!}\xi_{k+4} +K_1 (k+2)(k+1)\xi_{k+2}+ [ K_2 k(k-1) +K_3] \xi_{k} +K_4 \xi_{k-2} =0,$$

for $$k=0,1,2,3...$$, where $$K_1$$, $$K_2$$, $$K_3$$ and $$K_4$$ are constants. However, what I do know about the $$\xi_k$$'s is that

$$\sum_{k_{even}} \xi_{k}=\sum_{k_{odd}} \xi_k=1,$$

and that

$$\sum_{k_{even}}k \xi_{k}=\sum_{k_{odd}}k \xi_k=0.$$

$$\xi_k$$ is also zero for all $$k<0$$. What I want to do is produce an approximation of the sum for

$$S(z) =\sum_{k=0}^\infty \xi_k z^k.$$

Does anyone have any idea how to do this?

Last edited: Jul 5, 2010
2. Jul 5, 2010

Gerenuk

If I remember correctly that type of recursion relation is solvable. Somehow you take differences of consecutive equations until you have
$$\sum_1^N a_i\xi_{k+i}=0$$
and then you use a geometric series as the trial solution. Not sure if the degree of that polynomial becomes too large.

But maybe the contraints help.

3. Jul 5, 2010

Gerenuk

What do you mean? Isn't S=2 then?

4. Jul 5, 2010

Hoplite

Oops, sorry. I had the wrong equation for S. I've fixed it now.

5. Jul 5, 2010

Gerenuk

Using something which I think is called z-transform you get
$$S''''+(a+bx^2)S''+(c+dx^2)S=e_5x^5+\dotsb+e_0$$
I guess that's your initial problem in reverse :)
Anyone knows how to solve this? I've found
http://eqworld.ipmnet.ru/en/solutions/ode/ode0406.pdf
but that only works for a special contraints on the parameters.

Last edited: Jul 5, 2010
6. Jul 5, 2010

tmccullough

This series will converge really quickly - this is definitely an entire function. It's probably just as good to deal with the recursive relation. Even if you had a closed form, it wouldn't help you to evaluate the function.

Do you need a closed form, or do you just think you need it? For the purpose of it being a solution to an ODE, this is just as good.

7. Jul 6, 2010

Hoplite

That's correct. In fact my equation is

$$S''''+(a+bx^2)S''+(c+dx^2)S=0,$$

with some inhomogenious boundary conditions.

8. Jul 6, 2010

Gerenuk

Sorry for messing around and confusing people :)
But maybe the special case from that webpage helps a bit :)