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Simplifying 'b2' of Newton's divided difference interpolation

  1. Dec 1, 2008 #1
    Hi all,

    Problem Description as a PDF (36kb)

    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 linked PDF.

    [tex]\begin{equation}\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}\end{equation} [/tex]

    [tex]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,
    \begin{equation}\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}\end{equation}[/tex]

    I would most appreciate your comments on solving this. You can either send me a PM or an email to mike@bsodmike.com.
     
  2. jcsd
  3. Dec 1, 2008 #2
    I believe I managed to figure it out. Take the eq. ([tex]b_1[/tex]) in terms of [tex]f(x_0)[/tex],

    [tex]\begin{equation}\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)
    \end{equation}[/tex]

    and substitute [tex]b_1[/tex] inside. Substitute the entire [tex]f(x_0)[/tex] in,

    [tex]\begin{equation}\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}\end{equation}[/tex]

    \o/ :approve:
     
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