6.2_CP1a.py

# Apply the Explicit Trapezoid method on a grid of step size h = 0.1 in [0,1] to the initial value problem in
# Exercise 1. Print a table of the t values, approximations, and global truncation error at each step.
import numpy as np
from pandas import DataFrame as DF


def trapezoid(y0, f, h, t):
    w = np.zeros(len(t))
    w[0] = y0
    for i in range(0, len(t) - 1):
        w[i + 1] = w[i] + h / 2 * (f(t[i], w[i]) + f(t[i] + h, w[i] + h * f(t[i], w[i])))
    return w


def trapes_metode(f, y_init, h, n):
    y = []
    y.append(y_init)
    for i in range(n):
        s1 = f(y[i])
        s2 = f(y[i] + h * s1)
        s = 0.5 * (s1 + s2) * h
        y.append(y[i] + s)
    return y


def exact_solution_a(t):
    return (t ** 2) / 2 + 1


def main():
    #   f = lambda t, y: t
    # h = 0.1
    # t = np.arange(0, 1 + h, h)
    #  w = trapezoid(1, f, h, t)
    #  w = trapes_metode(1, f, h, t)
    #  true_value = exact_solution_a(t)
    #  error = abs(true_value - w)
    # print(DF({"t": t, "w": w, "Error": error, 'True value': true_value}).set_index('t')[['True value', 'w', 'Error']])

    f = lambda y: -3 * y
    y = trapes_metode(f, 5, 0.1, 100)
    print("%.5f" % sum(y))



if __name__ == '__main__':
    main()