Approximating unsolvable recursion relations

[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?

 

[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.

 

[Gerenuk]

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

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

 

[Hoplite]

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

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

 

[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.

 

[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.

 

[Hoplite]

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 :)

That's correct. In fact my equation is

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

with some inhomogenious boundary conditions.

 

[Gerenuk]

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