# richardson's extrapolation

#### CaptainBlack

##### Well-known member
From Randy Blythe's Yahoo Answers Question:

Use Richardson’s extrapolation, using a centred difference scheme for the initial estimates, to estimate the first derivative of y=(x^3)sinx at x=2 with step sizes h1=0.5 and h2=0.25. Calculate the actual derivative and the percentage error

#### CaptainBlack

##### Well-known member
From Randy Blythe's Yahoo Answers Question:

Use Richardson’s extrapolation, using a centred difference scheme for the initial estimates, to estimate the first derivative of y=(x^3)sinx at x=2 with step sizes h1=0.5 and h2=0.25. Calculate the actual derivative and the percentage error
First you need to find the central difference estimate of $$y'(2)$$ with $$h=0.5$$ and $$h=0.25$$.

As the central difference estimate of the derivative at $$x$$ with step size $$h$$ is:

$y'_h(x)= \frac{y(x+h/2)-y(x-h/2)}{h}$

we heve

$y'_{0.5}(2) = \frac{ y(2+0.25) - y(2-0.25) }{ 0.5} \approx 7.17838$

and

$y'_{0.25}(2) = \frac{ y(2+0.125) - y(2-0.125) }{ 0.25} \approx 7.48111$

Now we know that the central differences approximation for the derivative is or order $$h^2$$, so our Richardson extrapolation $$y'_R(x)$$ for $$\lim_{h\to 0} y'_h(x)$$ is:

$y'_R(x)= \frac{ 2^n y'_{h/2}(x) - y'_{h}(x) }{ 2^n-1 }$

with $$n=2$$.

Hence:

$y'_R(2) = \frac{ 2^2 y'_{0.25}(2) - y'_{0.5}(2) }{ 2^2-1 } \approx 7.58202$

I will leave the last part, the comparison with the true value to you.

CB

• Chris L T521