I am trying to show that a fourth order Runge-Kutta method is equivalent to Simpson's method for approximating an integral over an interval [a,b]. I've read that it is, but I'm unsure how to demonstrate this. I would appreciate any help in figuring out how to show this equivalency.
2026-03-26 11:06:45.1774523205
Demonstrating equivalency between Runge-Kutta and Simpson's Rule
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The Runge-Kutta method I think you are talking about is \begin{align} &k_1 = f(t_n, y_n) \\ &k_2 = f(t_n + h/2, y_n + hk_1/2) \\ &k_3 = f(t_n + h/2, y_n + hk_2/2) \\ &k_4 = f(t_n + h, y_n + hk_3) \\ &y_{n + 1} = y_n + \frac{h}{6}(k_1 + 2k_2 + 2k_3 + k_4). \end{align} If the ode is just integration, i.e. $y'(t) = f(t)$, this reduces to \begin{align} &k_1 = f(t_n) \\ &k_2 = f(t_n + h/2) \\ &k_3 = f(t_n + h/2) \\ &k_4 = f(t_n + h) \\ &y_{n + 1} = y_n + \frac{h}{6}(k_1 + 2k_2 + 2k_3 + k_4) = y_n + \frac{h}{6}(f(t_n) + 4f(t_n + h/2) + f(t_n + h)) \end{align} which is identical to the 3 point Simpsons rule.