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richardson's extrapolation

CaptainBlack

Well-known member
Jan 26, 2012
890
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
Jan 26, 2012
890
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