A finite difference formula was derived as , using the three points: xi+2, xi, x

Question

A finite difference formula was derived as , using the three points: xi+2, xi, xi-1, where xi+2-xi=2h and xi-xi-1=h is the step size. Generate the f (xi)' derivative approximation shown above, and describe how you combined the two Taylor series expansions for f(xi+2), f(xi-1) to do so. MAE 384 Numerical Methods Fall 2013 Dennis Dunn What is the order of error term O(h?) of the above finite difference approximation? Develop a new "smarter" formula to find f' (xi) that has a reduced error by an additional order of h, but still uses f (xi+2), f(xi-1). Hint: This will require canceling a higher order term when combining Taylor series expansions.

Explanation / Answer

Let f(x) = f(xi) + f'(xi)(x-xi) + 1/2 f''(xi)(x-xi)^2 + ..

Then, (3 f(xi+2h) - 8 f(xi) + 5 f(xi-h))/h = (3(f(xi) + f'(xi)(2h) + 1/2f''(xi)4h^2) - 8f(xi) + 5(xi - f(xi)h + 1/2f''(xi)h^2)...)/h=

(f'(xi)h + 17/2 f''xi)h^2+..)/h = f'(xi) + 17/2 f''(xi)h + ...

b) Thus, the error term is O(h^-1)

Clearly, we have to eliminate the f(xi) terms and f''(xi) terms, and the f'(xi) term has to have a constant of 1.

This creates our 3 equations with 3 unknowns.

Let x be the coefficient for the f(xi+2) term, y the coefficient for the f(xi) term, and z the coefficient for the f(xi-1) term.

Thus,

x +y+z = 0

2x -z = 1

4x + z = 0

Thus, x = -z/4

2x-z = 1, so 2(-z/4) - z = 1

-3/2z = 1

z = -2/3

x = 4/3

x+y+z = 0

-2/3 + 4/3 + y = 0

y = -2/3

The better approximation is (4/3f(xi+2) -2/3 f(xi) - 2/3 f(xi-1))/h

This will be O(h^-2)

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