Proof that (2*M* + *T*)/3 Equals Simpson’s Rule

[by Chris N. and Andrew K., 11/14/2002]

[with minor editing by Jimmy L. and Mr. Hansen]

Prove (2*M* + *T*)/3 = Simpson’s Rule:

Let *y* = *f*(*x*) be continuous on a closed interval [*a*, *b*].

Let *n* = no. of trapezoids = no. of rectangles, with *n* Î **N**, *n* ³ 2. (Problem is uninteresting for smaller *n*.)

Let *M* = the midpoint Riemann sum using midpoints *x*_{1}, *x*_{3}, *x*_{5}, . . . , *x*_{2n–1}

Let *T* = the sum of the Trapezoidal Rule using mesh points *x*_{0}, *x*_{2}, *x*_{4}, . . . , *x*_{2n}

*Note:* Here, width (horizontal step size) of trapezoids and midpoint rectangles is 2D*x*. For Simpson’s Rule we use the more conventional mesh size of D*x*.

*M* = 2D*x*(*y*_{1} + *y*_{3 }+ *y*_{5} + . . . + *y*_{2n–1})

*T* = 2D*x*(½ *y*_{0} + *y*_{2 }+ *y*_{4} + . . . + ½ *y*_{2n}) = D*x*(*y*_{0} + 2*y*_{2} + 2*y*_{4} + . . . + *y*_{2n})

Simpson’s Rule = 1/3 · D*x*(*y*_{0} + 4*y*_{1} + 2*y*_{2} + 4*y*_{3} + 2*y*_{4} + 4*y*_{5} + . . . + *y*_{2n})

= 1/3 (D*x*(*y*_{0} + 2*y*_{2} + 2*y*_{4} + . . . + *y*_{2n}) + 4D*x*(*y*_{1} + *y*_{3} + *y*_{5} + . . . + *y*_{2n–1}))

= 1/3 (*T* + 2*M*)

= (2*M* + *T*)/3, Q.E.D.