Simplifying 'b2' of Newton's divided difference interpolation

bsodmike
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Hi all,

http://www.bsodmike.com/stuff/interpolation.pdf"

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" .

\begin{equation}\label{eq:solution}\begin{split} <br /> b_2&amp;=\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]<br /> &amp;=\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)}<br /> \end{split}\end{equation}

As marked above in {\color{red}red} as three {\color{red}$\hdots$}, what algebraic manipulations are needed to arrive at the solution? <br /> <br /> The farthest I can get is,<br /> \begin{equation}\label{eq:attempt}\begin{split} <br /> b_2&amp;=\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]<br /> &amp;=\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)}<br /> \end{split}\end{equation}

I would most appreciate your comments on solving this. You can either send me a PM or an email to mike@bsodmike.com.
 
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I believe I managed to figure it out. Take the eq. (b_1) in terms of f(x_0),

\begin{equation}\label{eq:attempt}<br /> 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)<br /> \end{equation}

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

\begin{equation}\label{eq:attempt}\begin{split} <br /> b_2&amp;=\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]<br /> &amp;=\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]<br /> &amp;=\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]<br /> &amp;=\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)}<br /> \end{split}\end{equation}

\o/ :approve:
 
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