[math-fun] sqrt(binomial)=octonomial, sort of

I said >Still no luck with sqrt(binomial) = hexanomial. 1/4 sqrt(113 %i - 84 2 ) = 3/4 1/4 3 %i 3/4 1/4 3 - 2 %i + 4 2 %i + ------- + 6 %i + 2 - 4 2 + ------- + 6, sqrt(2) sqrt(2) but half of this is from sqrt(i)=(i+1)/sqrt(2). --rwg