[Edu-sig] RE: Integration correction

[Kirby Urner]

> 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

And with a decorator:

def simpson(f):
   def defint(a,b,n=1000):
       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
   return defint

@simpson
def g(x):  return x*x

 >>> g(0, 3)  
 9.0000000000000036
 >>> g(0, 4)
 21.333333333333329
 >>> g(1, 7)
 113.99999999999983
 >>> g(1, 7, n=10000)
 114.0

Kirby