MHB Richardson's Extrapolation for Derivative Estimation

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Richardson's extrapolation is applied to estimate the first derivative of the function y=(x^3)sinx at x=2 using centered difference schemes with step sizes h1=0.5 and h2=0.25. The central difference estimates yield approximately 7.17838 for h=0.5 and 7.48111 for h=0.25. Using Richardson's formula, the extrapolated derivative is calculated to be approximately 7.58202. The actual derivative value can be compared to assess the percentage error. This method effectively improves derivative estimates by reducing error through extrapolation.
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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
 
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CaptainBlack said:
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
 
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