MHB Richardson's Extrapolation for Derivative Estimation

  • Thread starter Thread starter CaptainBlack
  • Start date Start date
  • Tags Tags
    Extrapolation
AI Thread Summary
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.
CaptainBlack
Messages
801
Reaction score
0
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
 
Mathematics news on Phys.org
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
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Thread 'Unit Circle Double Angle Derivations'
Here I made a terrible mistake of assuming this to be an equilateral triangle and set 2sinx=1 => x=pi/6. Although this did derive the double angle formulas it also led into a terrible mess trying to find all the combinations of sides. I must have been tired and just assumed 6x=180 and 2sinx=1. By that time, I was so mindset that I nearly scolded a person for even saying 90-x. I wonder if this is a case of biased observation that seeks to dis credit me like Jesus of Nazareth since in reality...

Similar threads

Back
Top