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Hoplite
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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?
\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?
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