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Since sin(0)=0, we can set a<sub>ij</sub>=0 for all i>0, or simply disregard these coefficients. So here are some of the coefficients apparently solved as a<sub>0j</sub> (noting that binomial(i,j) is the binomial coefficient for integers i and j where, if i<j, binomial(i,j)=0):
a<sub>0n</sub> = 1 for all n;
a<sub>0(n-2)</sub> = (-1)binomial(n,3) for all n;
a<sub>0(n-4)</sub> = (1/3)(5n - 22)binomial(n,5) for all n>4.
Last edited by phukraut; 04-11-2005 at 03:19 PM..
Reason: Correction on the a<sub>ij</sub>=0. This is in fact not true. What I wanted to say was that we can disregard these coefficients since the term is identically zero.
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