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Interpolate the values of $ \ f \ $ at $ \ x_0-h , \ \ x_0, \ \ x_0+2h \ $. Use the Interpolant to find an approximation of $ \ f'(x_0+\frac{h}{2}) \ $.

Answer:

Then the Lagrange interpolant $ \ L(x) \ $ is given by

$L(x)=\frac{(x-x_0)(x-(x_0+2h))}{((x_0-h)-x_0)((x_0-h)-(x_0+2h))} f(x_0-h)+\frac{(x-(x_0-h))(x-(x_0+2h))}{(x_0-(x_0-h))(x_0-(x_0+2h))}f(x_0)+\frac{(x-(x_0-h))(x-x_0)}{((x_0+2h)-x_0)((x_0+2h)-(x_0-h))}f(x_0+2h) \ $

This is the interpolant of $ f \ $ at $ \ x_0-h, \ x_0, \ x_0+2h \ $.

Now I can not approximate $ \ f'(x_0+h/2) \ $ using this interpolant.

I need help doing this.

Also I am not quite sure whether my approach so far is correct or not.

Help me out

MAS
  • 10,638

1 Answers1

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It seems that the idea is to approximate $f'(x_0+h/2)$ by $L'(x_0+h/2)$.

As: $$L'(x)=\frac{(2x-(2x_0+2h))}{3 h^2}f(x_0-h)+\frac{(2x-(2x_0+h))}{2 h^2}f(x_0)+\frac{(2x-(2x_0-h))}{6 h^2}f(x_0+2h)$$ you have: $$L'\left(x_0+\frac{h}{2} \right)=\frac{-h}{3 h^2}f(x_0-h)+\frac{0}{2 h^2}f(x_0)+\frac{2h}{6 h^2}f(x_0+2h)=\frac{1}{3h}\left(f(x_0+2h)-f(x_0-h) \right)$$

Remark: actually the result is a centered approximation: $$\frac{1}{3h}\left(f(x_0+2h)-f(x_0-h) \right)=\frac{1}{2}\frac{f\left(x_0+\frac{h}{2}+\frac{3}{2}h\right)-f\left(x_0+\frac{h}{2}-\frac{3}{2}h\right) }{\frac{3}{2}h}$$

Delta-u
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