Page - 140 - in Programming for Computations – Python - A Gentle Introduction to Numerical Simulations with Python 3.6, Volume Second Edition

140 6 ComputingIntegralsandTestingCode Implementation Without Function Definitions Basically, we use a for loop to compute the sum. Each term withf(x) in the formula (6.17) is replaced by 3t2et 3 , x by t, and h byΔt.1 A first version of this special implementation, without any functiondefinition,could read from math import exp a = 0.0; b = 1.0 n = int(input(’n: ’)) dt = (b - a)/n # Integral by the trapezoidal method v_sum = 0 for i in range(1, n, 1): t = a + i*dt v_sum = v_sum + 3*(t**2)*exp(t**3) numerical = dt*(0.5*3*(a**2)*exp(a**3) + v_sum + 0.5*3*(b**2)*exp(b**3)) exact_value = exp(1**3) - exp(0**3) error = abs(exact_value - numerical) rel_error = (error/exact_value)*100 print(’n={:d}: {:.16f}, error: {:g}’.format(n, numerical, error)) Theproblemwith theabovestrategy is at least three-fold: 1. To write the code, we had to reformulate (6.17) for our special problem with a differentnotation.Errorscomeeasy then. 2. Towrite thecode,wehadtoinsert theintegrand3t2et 3 severalplacesin thecode, whichquickly leads toerrors. 3. Ifwelaterwant tocomputeadifferentintegral, thecodemustbeeditedinseveral places.Sucheditsare likely to introduceerrors. Thepotentialerrorsrelated topoint2serve to illustratehowimportant it is todefine anduse appropriatefunctions. Implementation with Function Definitions An improved second version of the special implementation, now with functions for the integrand v and the anti- derivativeV,might then read from math import exp v = lambda t: 3*(t**2)*exp(t**3) # Define integrand a = 0.0; b = 1.0 n = int(input(’n: ’)) dt = (b - a)/n # Integral by the trapezoidal method 1 Replacing h by Δt is not strictly required as many use h as interval also along the time axis. Nevertheless,Δt is an even more popular notation for a small time interval, so we use thathere.