Question

Recall from Lecture the fourth order Runge-Kutta method: x_{n+1} = x_n + 1/6 (k_1 + 2k_2 + 2k_3 + k_4) k_1 = hf(x_n, t_n) k_2 = hf(x_n + 1/2 k_1, t_n + 1/2 h) k_3 = hf(x_n + 1/2 k_2, t_n + 1/2 h) k_4 = hf(x_n + k_3, t_n + h) Apply it, with h = 1, to the initial value problem of the previous Question to find a (better) approximation to ?2 (recall ?2 ? 1.41421). Provide a numeric answer rounded to five decimal places.

          Recall from Lecture the fourth order Runge-Kutta method:

x_{n+1} = x_n + 1/6 (k_1 + 2k_2 + 2k_3 + k_4)

k_1 = hf(x_n, t_n)
k_2 = hf(x_n + 1/2 k_1, t_n + 1/2 h)
k_3 = hf(x_n + 1/2 k_2, t_n + 1/2 h)
k_4 = hf(x_n + k_3, t_n + h)

Apply it, with h = 1, to the initial value problem of the previous Question to find a (better) approximation to ?2 (recall ?2 ? 1.41421). Provide a numeric answer rounded to five decimal places.

Recall from Lecture the fourth order Runge-Kutta method:

xn+1 = xn + 1/6 (k1 + 2k2 + 2k3 + k4)

k1 = hf(xn, tn)
k2 = hf(xn + 1/2 k1, tn + 1/2 h)
k3 = hf(xn + 1/2 k2, tn + 1/2 h)
k4 = hf(xn + k3, tn + h)

Apply it, with h = 1, to the initial value problem of the previous Question to find a (better) approximation to ?2 (recall ?2 ? 1.41421). Provide a numeric answer rounded to five decimal places.

Added by Kristina A.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Supreeta N

04/27/2023

Step 1

We are given the initial value problem (IVP) from the previous question. Let's assume it is: \[y'(t) = f(t, y(t)), \quad y(t_0) = y_0\] Show more…

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Recall from Lecture the fourth-order Runge-Kutta method: k1 = hf(r,u,t) k2 = hf(r+0.5k1,u+0.5k1,t+0.5h) k3 = hf(r+0.5k2,u+0.5k2,t+0.5h) k4 = hf(r+k3,u+k3,t+h) Δu = (k1 + 2k2 + 2k3 + k4)/6 u = u + Δu Apply with t0 the Initial value problem from the previous question to find a better approximation to √2 (recall √2 ≈ 1.41421). Provide a numeric answer rounded to five decimal places.