Using the forward Euler method to solve $y’'(t) = f(t), y’(0)=0$ with a step size of $h$, we obtain the following values of $y$ in the first four iterations:
$0, hf (0), h(f(0) + f(h)) \text{ and }h(f(0) - f(h) + f(2h))$
$0, 0, h^2f(0)\text{ and } 2h^2 f(0) + f(h)$
$0, 0, h^2f(0) \text{ and } 3h^2f(0)$
$0, 0, hf(0) + h^2f(0) \text{ and }hf (0) + h^2f(0) + hf(h)$