CODE IN PYTHON

* 5(1) dt = T. + 2h +0(14) where C2 = (f'(a) f'(b))/12. Now if we double the number of intervals, n, we halve h, so = 2 1 * f(t) dt = T2n + C2 h 2 +0(h) = T2n + Ch? + O(^) Taking four times the second formula and subtracting the first, we get 4 = 4T2n +cah2 +O(h4) 41 5 f(t) dt = -L 31 Ss(e)dt = 4T3n T. f(t)dt = -Tn -ch2 -O(h4) +0 +O(h4) That gives us 4T2n | () at 47, 7, +O(h) 3 OPTIONAL EXTRA CREDIT TOWARD ASSIGNMENT SCORES: Write a program (you must create this code yourself) that numerically approximates an integral of f(x) over an interval [a, b] that uses a Romberg estimate with m = 1. See Theorem 4.6 on page 80 and see the formula on page 78 just above Example 4.11. Try your program out on a function for which you can compute the exact value, then commment on the error of this Romberg (m = 1). Does the error seem to support the error estimate given in Theorem 4.6? Explain. Theorem 4.6: (Romberg Integration) If: f has 2m + 4 continuous derivatives, and the numbers Rm,n are defined recursively so that 4" Rm-1,5 Rm-1,n-1 Ronn = T2r, and Rm,n = = 4m - 1 then for h = (b-a)/2" and some constant C2m+2 (depending on the derivatives of f, but independent of n) 5 $(t) dt = Rmn + o2m+2h2m +2 + 0 (12m+a) +4 + All these subscripts and estimates can become confusing. One way to organize them is to make a table. n 0 1 2" 1 2 4 8 Ron R1,n R2,n R3,n R0.0 R01 R0,2 R1,2 R2,2 R0,3 R1,3 R2,3 R3,3 R1,1 2 3 Since higher order Rs are based on lower order R's, we fill out this table by calculating each column starting from the left