Derivation of fourth-order accurate formula for the second derivative

Question

I am trying to derive / prove the fourth order accurate formula for the second derivative:

$f''(x) = \frac{-f(x + 2h) + 16f(x + h) - 30f(x) + 16f(x - h) - f(x -2h)}{12h^2}$.

I know that in order to do this I need to take some linear combination for the Taylor expansions of $f(x + 2h)$, $f(x + h)$, $f(x - h)$, $f(x -2h)$. For example, when deriving the the centered-difference formula for the first derivative, the Taylor expansion of $f(x + h)$ minus $f(x-h)$ can be computed to give the desired result of $f'(x)$, in that case.

In what way would I have to combine these Taylor expansions above to obtain the required result?

Answer 1

Exactly as Gammatester says, Taylor expand the terms upto order $4$ and verify. \begin{eqnarray*} -f(x+2h) &=& -f(x) &-& 2h f'(x) &-& 2h^2 f''(x) &-& \frac{4}{3} h^3 f'''(x) &-& \frac{2}{3} h^4 f''''(x) &+& O(h^5) \\ 16f(x+h) &=& 16 f(x)&+& 16h f'(x) &+& 8h^2 f''(x) &+& \frac{8}{3} h^3 f'''(x) &+& \frac{2}{3} h^4 f''''(x) &+& O(h^5) \\ -30f(x) &=& -30f(x) & & & & & & & & & & \\ 16f(x-h) &=& 16 f(x)&-& 16h f'(x) &+& 8h^2 f''(x) &-& \frac{8}{3} h^3 f'''(x) &+& \frac{2}{3} h^4 f''''(x) &+& O(h^5) \\ -f(x-2h) &=& -f(x) &+& 2h f'(x) &-& 2h^2 f''(x) &+& \frac{4}{3} h^3 f'''(x) &-& \frac{2}{3} h^4 f''''(x) &+& O(h^5) \\ \end{eqnarray*}

Answer 2

$$ f(x+h) = f(x) + h f'(x) + \frac{h^2}{2} f''(x) + \frac{h^3}{6} f'''(x) + O(h^4) $$

$$ f(x-h) = f(x) - h f'(x) + \frac{h^2}{2} f''(x) - \frac{h^3}{6} f'''(x) + O(h^4) $$

$$ f(x+2h) = f(x) + 2h f'(x) + 2 h^2 f''(x) + \frac{4 h^3}{3} f'''(x) + O(h^4) $$

$$ f(x-2h) = f(x) - 2h f'(x) + 2 h^2 f''(x) - \frac{4 h^3}{3} f'''(x) + O(h^4) $$

Calculate: $$ -f(x + 2h) + 16f(x + h) - 30f(x) + 16f(x - h) - f(x -2h) $$

Which is $$ \begin{aligned} & - \left[ f(x) + 2h f'(x) + 2 h^2 f''(x) + \frac{4 h^3}{3} f'''(x) \right] \\ & +16 \left[ f(x) + h f'(x) + \frac{h^2}{2} f''(x) + \frac{h^3}{6} f'''(x) \right] \\ & -30 f(x) \\ & +16 \left[ f(x) - h f'(x) + \frac{h^2}{2} f''(x) - \frac{h^3}{6} f'''(x) \right] \\ & - \left[ f(x) - 2h f'(x) + 2 h^2 f''(x) - \frac{4 h^3}{3} f'''(x) \right] \\ & + O(h^4) \end{aligned} $$

Which evaluates to $ 12 h^2 $ to give the required result.

Answer 3

You can easily derive the formula, if you do not know it, as a derivative of the Lagrange polynomial

D[D[InterpolatingPolynomial[{(-2*h,y0),(-1*h,y1),(0*h,y2),(1*h,y3),(2*h,y4)},x],x],x] /. x=0

Try at Wolfram Alpha

The other answers show how to prove the order of accuracy of an already-known formula.