Question 8.5: Using Richardson's extrapolation in differentiation. Use Ric......
Using Richardson’s extrapolation in differentiation.
Use Richardson’s extrapolation with the results in Example 8-4 to calculate a more accurate approximation for the derivative of the function f(x)=\frac{2^{x}}{x} at the point x = 2.
Compare the results with the exact (analytical) derivative.
In Example 8-4 two approximations of the derivative of the function at x = 2 were calculated using the central difference formula in which the error is O(h²). In one approximation h = 0.2, and in the other h = 0.1. The results from Example 8-4 are:
for h = 0.2, f ”(2) = 0.577482. The error in this approximation is 0.5016 %.
for h = 0.1, f ”(2) = 0.575324. The error in this approximation is 0.126 %.
Richardson’s extrapolation can be used by substituting these results in Eq. (8.45)
D=\frac{1}{3}\left(4 D\left(\frac{h}{2}\right)-D(h)\right)+O\left(h^4\right) (8.45)
(or Eq. (8.53)):
f^{\prime}\left(x_i\right)=\frac{1}{3}\left[4 \frac{f\left(x_i+h / 2\right)-f\left(x_i-h / 2\right)}{h} \frac{f\left(x_i+h\right)-f\left(x_i-h\right)}{2 h}\right]+O\left(h^4\right) (8.53)
D=\frac{1}{3}\left(4 D\left(\frac{h}{2}\right)-D(h)\right)+O\left(h^{4}\right)=\frac{1}{3}(4 \cdot 0.575324-0.577481)=0.574605
The error now is \quad error =\frac{0.574605-0.5746}{0.5746} \cdot 100=0.00087 \%
This result shows that a much more accurate approximation is obtained by using Richardson’s extrapolation.