> But you never use the trapezoidal rule in real life! > Simpson gives much smaller error for the same amount > of computation. In fact, it is exact for Kirby's example! > > Henrik
Here's my implementation of Simpson's, except it divides the interval into 2n segments. >>> def simpson(f,a,b,n): h = float(b-a)/(2*n) sum1 = sum(f(a + 2*k *h) for k in range(1,n)) sum2 = sum(f(a + (2*k-1)*h) for k in range(1,n+1)) return (h/3)*(f(a)+f(b)) + (2*h/3)*sum1 + (4*h/3)*sum2 >>> def g(x): return x*x >>> simpson(g,0,3,10000) 8.9999999999999893 >>> simpson(g,0,3,5000) 8.9999999999999787 Kirby _______________________________________________ Edu-sig mailing list Edu-sig@python.org http://mail.python.org/mailman/listinfo/edu-sig