Derivative by Interpolation

The derivative of a given function $f(t)$ can be approximated by that of its polynomial interpolation based on a set of $n+1$ points $y_i=f(x_i),\;(i=0,\cdots,n)$, such as the Lagrange polynomial interpolation considered previously:

$$f(x)\approx L_n(x)=\sum_{i=0}^n y_i l_i(x)$$ (8)

where $l_i(x)$ is the ith basis polynomial:

$$l_i(x)=\prod_{j=0,\;j\ne i}^n \frac{x-x_j}{x_i-x_j} =\frac{l(x)}{(x-x_i) l'(x)},\;\;\;\;\;(i=0,\cdots,n)$$ (9)

and $l(x)=\prod_{i=0}^n(x-x_i)$. The truncation error is:

$$R_n(x)=\frac{f^{(n+1)}(\xi)}{(n+1)!}\prod_{i=0}^n(x-x_i) =\frac{f^{(n+1)}(\xi(x))}{(n+1)!}\,l(x)$$ (10)

Taking derivative of $f(x)$ above we get:

$$D[f]=f'(x)\approx L'_n(x)=\sum_{i=0}^n y_i l'_i(x) =\sum_{i=0}^n y_i d_i(x)$$ (11)

where

$$d_i(x)=\frac{d}{dx}l_i(x) =\frac{d}{dx} \left(\prod_{j=0,\;j\ne i}^n \frac{x-x_j}{x_i-x_j}\right)$$ (12)

The error is

$$E_n(x)=R'_n(x) = \frac{1}{(n+1)!}\frac{d}{dx}\left((f^{(n+1)}(\xi(x))\,l(x)\right)$$

$$= \frac{1}{(n+1)!}\left[l'(x)f^{(n+1)}(\xi(x)) +l(x)\frac{d}{dx}f^{(n+1)}(\xi(x))\right]$$

This error is unknown in general as the function $\xi(x)\in[x_0,\,x_n]$ is not known explicitly. However, at each node $x_i$ where $w(x_i)=0$, we have

$$E_n(x)=R'_n(x)=l'(x)\frac{f^{(n+1)}(\xi(x))}{(n+1)!}$$ (13)

Consider the following two special cases:

• $n=1$, linear interpolation based on $n+1=2$ points:
  

  $$d_0 = l'_0(x)=\frac{d}{dx}\left(\frac{x-x_1}{x_0-x_1}\right) =\frac{1}{x_0-x_1}$$

  $$d_1 = l'_1(x)=\frac{d}{dx}\left(\frac{x-x_0}{x_1-x_0}\right)=\frac{1}{x_1-x_0}$$ (14)

  
  and we have

  $$f'(x)\approx L'_1(x)=y_0 d_0(x)+y_1 d_1(x) =\frac{y_1-y_0}{x_1-x_0},\;\;\;\;\;\;\;f''(x)=0$$ (15)

• $n=2$, quadratic interpolation based on $n+1=3$ points:
  

  $$d_0(x) = l'_0(x)=\frac{d}{dx}\left(\frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)}\right) =\frac{2x-x_1-x_2}{(x_0-x_1)(x_0-x_2)}$$

  $$d_1(x) = l'_1(x)=\frac{d}{dx}\left(\frac{(x-x_0)(x-x_2)}{(x_1-x_0)(x_1-x_2)}\right) =\frac{2x-x_0-x_2}{(x_1-x_0)(x_1-x_2)}$$

  $$d_2(x) = l'_2(x)=\frac{d}{dx}\left(\frac{(x-x_0)(x-x_1)}{(x_2-x_0)(x_2-x_1)}\right) =\frac{2x-x_0-x_1}{(x_2-x_0)(x_2-x_1)}$$ (16)

  
  we therefore have
  

  $$f'(x)\approx L'_2(x)=\sum_{i=1}^2 d_i y_i$$

  $$= \frac{2x-x_1-x_2}{(x_0-x_1)(x_0-x_2)}y_0 +\frac{2x-x_0-x_2}{(x_1-x_0)(x_1-x_2)}y_1 +\frac{2x-x_0-x_1}{(x_2-x_0)(x_2-x_1)}y_2$$ (17)

  
  Specially, if the 3 points are equally spaced with $x_1=x,\; x_0=x-h,\;x_2=x+h$, we have:

  $$f'(x)\approx \frac{x-(x+h)}{2h^2}y_0+\frac{2x-(x-h)-(x+h)}{-h^2}y_1 +\frac{x-(x-h)}{2h^2}=\frac{1}{2h}(y_2-y_0)$$ (18)

  which is the same as the central difference with error $O(h^2)$ found previously in Eq. (6).

  We can further approximate the 2nd order derivative:
  

  $$f''(x)\approx L''_2(x)=\frac{d}{dx}L'_2(x)$$

  $$= \frac{d}{dx}\left[\frac{2x-x_1-x_2}{(x_0-x_1)(x_0-x_2)}y_0 +\frac... ..._0-x_2}{(x_1-x_0)(x_1-x_2)}y_1+\frac{2x-x_0-x_1}{(x_2-x_0)(x_2-x_1)}y_2 \right]$$

  $$= 2\left[\frac{y_0}{(x_0-x_1)(x_0-x_2)} +\frac{y_1}{(x_1-x_0)(x_1-x_2)}+\frac{y_2}{(x_2-x_0)(x_2-x_1)}\right]$$

  
  Again, specially, if the 3 points are equally spaced with $x_1=x,\; x_0=x_1-h,\;x_2=x_1+h$, we have:

  $$f'(x)\approx \frac{2x-x_1-x_2}{2h^2}y_0 -\frac{2x-x_0-x_2}{h^2}y_1+\frac{2x-x_0-x_1}{2h^2}y_2$$ (19)

  At the three points we have
  

  $$f'(x_0) = \frac{1}{2h}(-3y_0+4y_1-y_2),\;\;\;\; E_2(x_0) =\frac{h^2}{3}f''(\xi)$$

  $$f'(x_1) = \frac{1}{2h}(-y_0+y_2),\;\;\;\; E_2(x_1) =-\frac{h^2}{6}f''(\xi)$$

  $$f'(x_2) = \frac{1}{2h}(y_0-4y_1+3y_2),\;\;\;\; E_2(x_2) =\frac{h^2}{3}f''(\xi)$$

  
  The 2nd order derivative is
  

  $$f''(x) \approx \frac{d}{dx}\left[\frac{2x-x_1-x_2}{2h^2}y_0 -\frac{2x-x_0-x_2}{h^2}y_1+\frac{2x-x_0-x_1}{2h^2}y_2\right]$$

  $$= \frac{1}{h^2}(y_0-2y_1+y_2)$$

  
  which is the same as second order difference with error $O(h^2)$ found previously in Eq. (7).