Question

3. Assume we have Simpson's Rule: to = a, 13 = , h = (b-a)/2 = a +h. (20) + 47(01) + f(x)]- ()where do < < Let fe .b), be even, h= (b-a)/n, and = a + jh, for each j = 0,1...... Show that there exists a l E (a,b) for which the Composite Simpson's rule for n subintervals can be written with its crror term as n/2 bar (n/2) - 1 f(a) +2 =1 (12) + 4 (12-1) + f(b) - - 180 / 11

Answer 1

Let y = f(x) denotes a continuous function in the interval (xo, xor2h) and posseses continuous deravahous. of all orders up to and including the fourth oredere in that interval and also let there existe a fune F(x) such that Fox orzh Then, I dx = F(20+2h)-f(x) Also Toy Simson's one third scule, we have north Is - de -₂ [Yo tay, +Y] - he [f(xv) + 4 f(x+h) + f (co+3h)] There force the Error Es is given by Es = Ir to a F. (xo + 2h) – F(x0) – 2/3 [(26) + If (worth) + flautes) =f()+241F'(x) + (2) F"(76)+(24) F" (2)4 (ory coas + &nds f(20) + --F(X) – bu [tems) 44 ffera) hf (6) & her f(x0) or the software.in] & flash enf'cks) + b)* 7" (96) + ) ? frukos) + (29) * " 4) --> 13 [Since F'(x) = f(x) F"(x) = f (x), FM (wafetal When, Eg~ to fiv (6) z – the five) where, no LeL22 Thus the Simsons one-third reuke with error terem is

Fore Composite simsone rule the correspond erereor es is given by Es - C f 'Ceme) f condes- tf "(*^1)] if tham) denotes the greatest of any of the & quantities f (xo), f ( ...) men, then =-62 f"Cu), no zuen By Simson's one third rule we get. noll s fra du = 6/3 [fla) + 2 f(x23) + 4 f (x287) + f(b)] - bara ho fa (14) j24 j24