Proof that (2M + 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 (2M + 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 x1, x3, x5, . . . , x2n–1
Let T = the sum of the Trapezoidal Rule using mesh points x0, x2, x4, . . . , x2n
Note: Here, width (horizontal step size) of trapezoids and midpoint rectangles is 2Dx. For Simpson’s Rule we use the more conventional mesh size of Dx.
M = 2Dx(y1 + y3 + y5 + . . . + y2n–1)
T = 2Dx(½ y0 + y2 + y4 + . . . + ½ y2n) = Dx(y0 + 2y2 + 2y4 + . . . + y2n)
Simpson’s Rule = 1/3 · Dx(y0 + 4y1 + 2y2 + 4y3 + 2y4 + 4y5 + . . . + y2n)
= 1/3 (Dx(y0 + 2y2 + 2y4 + . . . + y2n) + 4Dx(y1 + y3 + y5 + . . . + y2n–1))
= 1/3 (T + 2M)
= (2M + T)/3, Q.E.D.