It could be counted as 9 multiplies, actually:Originally Posted by Chumbucket843
with intermediate results a=x^2 and b=x^5:
x - (x*x*x)* (1/3!) + (b=((a=x*x)*a*x)) * (1/5!) - (a*b) * (1/7!)
It could be counted as 9 multiplies, actually:
with intermediate results a=x^2 and b=x^5:
x - (x*x*x)* (1/3!) + (b=((a=x*x)*a*x)) * (1/5!) - (a*b) * (1/7!)
Last edited by SEA; 08-25-2010 at 11:38 AM.
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