# Finding the error in a two-step finite difference numerical approximation

I got the following question in a math lecture the other day, and I'm not really sure how to go about it:

A differential equation is given in the form $$\frac{\partial y}{\partial x} = f (x, y(x))$$ where the function $f$ is known. To solve it numerically, one can use a two-step finite-difference scheme: $$y_{i+1} = \alpha y_i + \beta y_{i−1} + (\mu f_i + \nu f_{i−1}) \Delta x$$ where $$\Delta x = x_i − x_{i−1}\\ y_i = y(x_i)\\ f_i = f(x_i, y_i)$$ and $\alpha,\beta,\mu,\nu$ are constants.

Show that for $\alpha=1,\ \beta=0,\ \mu=3/2,\ \nu=–1/2$, this scheme gives errors of approximately $(\Delta x)^3$.

I'm guessing I use a Taylor Expansion of $y(x+\Delta x)$ and then compare the resulting equations, but when I do I can't see how I'm going to get an error of $(\Delta x)^3$. Any help is appreciated.

Yes, Taylor expansions are your friends here. Your local truncation error is given by \begin{align} & y_{i+1} - y_i - \Delta x (3/2 f_i - 1/2 f_{i-1}) \\ & = \left( y_i + \Delta x y_i' + \frac{\Delta x^2}{2} y_i'' + \mathcal{O}(\Delta x^3) \right) - y_i \\ & + \Delta x \left( \frac{3}{2}f_i - \frac{1}{2} \left( f_i - \Delta x f_i' + \mathcal{O}(\Delta x^2) \right) \right) \\ &= \Delta x(y_i' - f_i) + \frac{\Delta x^2}{2}(y_i'' - f_i') + \mathcal{O}(\Delta x^3). \end{align} The first two terms are zero from statement of the question so your scheme will have a local truncation error of $\mathcal{O}(\Delta x^3)$. Note that the global error is typically reduced one order.