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Example

Learn how to estimate the derivative of a function at x=0.5 using Richardson extrapolation and how to calculate the corresponding error.

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Example

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  1. Example Estimate the derivative of f(x) at x=0.5 using h=0.5, the true value is -0.9125 F(x)=(-0.1*x^4 )-(0.15*x^3)-(0.5*x^2)-(0.25*x)+1.2 Solution: X(i-1)=0 f(x(i-1))=1.2 X(i)=0.5 f(x(i))=0.925 X(i+1)=1.0 f(x(i+1))=0.2 F’(x(i))= ((f(x(i+1))- f(x(i))) / h) + 0(h)  F’ (0.5)=~ ( (0.2 – 0.925) ) 0.5 = -1.45  |E(t)|=58.9 0(h)= halring (h) halres the error (approximately) 0(h^2)= halring (h) quarters the error (approximately)

  2. Richardson Extrapolation There are three ways to improve derivatives estimates Decrease h Use a higher order formula that employs more points Richardson extrapolation: Uses two derivatives estimates to compute a third , more accurate approximation. If “D” is Derivative: D=~ ( (4/3)*D(h2) ) - ( (1/3)*D(h1) )

  3. Derivatives of unequally spaced data F’(x)= f(x(i-1)) * ( (2x - x(i) - x(i+1) ) ) / (x(i-1) - x(i) ) * (x(i-1) - x(i+1)) + f(x(i)) * ( (2x- x(i-1) – x(i+1)) / (x(i) – x(i-1))* (x(i) – x(i+1)) + F(x(i+1)) * ( (2x- x(i-1) - x(i) ) / (x(i+1) – x(i-1) ) * (x(i+1) – x ) This is the derivative of a 2nd-order language interpolating polynomial that fits three points.

  4. Chapter 25 Ordinary Differential Equations Runge-Kutta Methods Euler’s Method Y(i+1)= y(i)+ f(x(i) , y(i) ) * h h=step size , f(x(i),y(i))= DE evaluated It also called Euler-Cauchy or the point slope method.

  5. Example Solve using Euler’s: (Dy/dx)= -2x^3 + 12x^2 - 20x + 8.5 From x=0 to x=4  y=1  y(0) h=0.5 Exact solution is y= -0.5^4 + 4*x^3 - 10x^2 +8.5 Solution: y(i+1)= y(i) + f(x(i),y(i))* h Y(0.5)=y(0) + f(0,1)* 0.5 Slope at x=0 =f(0,1) dy/dx | (x=0,y=1) = 8.5 y(0.5)= 1 + 8.5(0.5)= 5.25

  6. True solution at x=0.5 is 3.2188  E(t)= true – approximation = 3.2188 – 5.25 = -2.0312 E(t)= - 63.1 % 2nd step: y(1)= y(0.5) + f(0.5,5.25)*0.5 y(1)=5.25 + [-2*(0.5)^3 + 12*(0.5)^2 - 20 *(0.5) + 8.5 ] * 0.5 y(1)= 5.875 True solution at x=1 is 3  E(t)= -95.8%

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