# Simplifying 'b2' of Newton's divided difference interpolation

1. Dec 1, 2008

### bsodmike

Hi all,

http://www.bsodmike.com/stuff/interpolation.pdf" [Broken]

I am going through some of my notes and quite a few books; they all skip the over the point I have marked with 3 red dots in the http://www.bsodmike.com/stuff/interpolation.pdf" [Broken].

$$\label{eq:solution}\begin{split} b_2&=\dfrac{f(x_2)-b_0-b_1(x_2-x_0)}{(x_2-x_0)(x_2-x_1)}=\dfrac{f(x_2)-f(x_0)-\dfrac{f(x_1)-f(x_0)}{x_1-x_0}(x_2-x_0)}{(x_2-x_0)(x_2-x_1)}={\color{red}\hdots}= \\[10px] &=\dfrac{\dfrac{f(x_2)-f(x_1)}{x_2-x_1}-\dfrac{f(x_1)-f(x_0)}{x_1-x_0}}{(x_2-x_0)} \end{split}$$

$$As marked above in {\color{red}red} as three {\color{red}\hdots}, what algebraic manipulations are needed to arrive at the solution? The farthest I can get is, \label{eq:attempt}\begin{split} b_2&=\dfrac{f(x_2)-f(x_0)-\dfrac{f(x_1)-f(x_0)}{x_1-x_0}(x_2-x_0)}{(x_2-x_0)(x_2-x_1)}\\[10px] &=\dfrac{f(x_2)-f(x_0)-\left[\left(\dfrac{f(x_1)}{x_1-x_0}+\dfrac{f(x_0)}{x_0-x_1}\right)(x_2-x_0)\right]}{(x_2-x_0)(x_2-x_1)} \end{split}$$

I would most appreciate your comments on solving this. You can either send me a PM or an email to mike@bsodmike.com.

Last edited by a moderator: May 3, 2017
2. Dec 1, 2008

### bsodmike

I believe I managed to figure it out. Take the eq. ($$b_1$$) in terms of $$f(x_0)$$,

$$\label{eq:attempt} f(x_0)=f(x_1)-b_1(x_1-x_0)=f(x_1)-\dfrac{f(x_1)-f(x_0)}{x_1-x_0}(x_1-x_0)$$

and substitute $$b_1$$ inside. Substitute the entire $$f(x_0)$$ in,

$$\label{eq:attempt}\begin{split} b_2&=\dfrac{f(x_2)-f(x_0)-\dfrac{f(x_1)-f(x_0)}{x_1-x_0}(x_2-x_0)}{(x_2-x_0)(x_2-x_1)}\\[10px] &=\dfrac{f(x_2)-\left(f(x_1)-\dfrac{f(x_1)-f(x_0)}{x_1-x_0}(x_1-x_0)\right)-\dfrac{f(x_1)-f(x_0)}{x_1-x_0}(x_2-x_0)}{(x_2-x_0)(x_2-x_1)}\\[10px] &=\dfrac{\dfrac{f(x_2)-f(x_1)}{x_2-x_1}-\left(\dfrac{f(x_1)-f(x_0)}{(x_1-x_0)(x_2-x_1)}((x_0-x_1)+(x_2-x_0))\right)}{(x_2-x_0)}\\[10px] &=\dfrac{\dfrac{f(x_2)-f(x_1)}{x_2-x_1}-\dfrac{f(x_1)-f(x_0)}{x_1-x_0}}{(x_2-x_0)} \end{split}$$

\o/