What is three-point backward difference?

Question

I fully understand two-point, forward/backward/central difference. I know them mathematically and graphically but I'm not quite sure what the three-point backward difference is. I know that we get the derivative using the previous two points but I don't really get the concept of it. Can someone explain it preferably with the aid of diagram or a graph please?

Answer 1

As mentioned in the comments, one way to derive the 3-point backwards finite difference would be to perform polynomial interpolation. However, I’ll derive the 3-point backwards difference using Taylor series.

Let $h>0$ denote our working spatial increment. For a smooth function $f(x)$, Taylor’s theorem gives $$f(x-h) = f(x) - f’(x)h + f’’(x)\frac{h^2}{2} + \mathcal{O}(h^3),$$ $$f(x-2h) = f(x) - 2f’(x)h + 2f’’(x)h^2+ \mathcal{O}(h^3).$$ Therefore, $$f(x-h) - \frac{1}{4}f(x-2h)= \frac{3}{4}f(x) - \frac{1}{2}f’(x)h + \mathcal{O}(h^3).$$ Solving for the derivative $f’(x)$ then gives $$f’(x) = \frac{\frac{3}{2}f(x) - 2f(x-h)+\frac{1}{2}f(x-2h)}{h} + \mathcal{O}(h^2).$$ This is the second order accurate three point backward finite difference approximation to the derivative. There are also other first order discretizations as well.

Answer 2

You want the first-order derivative approximated to second error order. We already know that the central difference quotient is second order. To apply this here we need to move the center backwards. Use Taylor for that, $f'(t)=f'(t-h)+hf''(t-h)+O(h^2)$. Now insert the difference quotients centered at $t-h$, $$ f'(t)=\frac{f(t)-f(t-2h)}{2h}+\frac{f(t)-2f(t-h)+f(t-2h)}{h}+O(h^2) $$