Let f(x)=(x+1)P(x)-x. So f(k)=0 for k=0,1,2, ...,n. -->
f(x)=cx(x-1)(x-2)...(x-n) with f(-1)=(-1)^(n+1)*c*(n+1)!=1
So f(x)=(-1)^(n+1)*x(x-1)(x-2)...(x-n)/(n+1)!
f(n+1)=(-1)^(n+1)=(n+2)P(n+1)-(n+1) -->
P(n+1)=( (-1)^(n+1) + (n+1) )/(n+2)
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Posted by Dennis
on 2007-02-06 13:34:19 |