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Next: (2.2.2)Newton Divided Difference Table: Up: Curve Fitting : Interpolation, Previous: Lagrange Interpolation:

Newton Interpolation polynomial:

Suppose that we are given a data set $ (x_{i},f_{i}),i=0,1...n-1$. Let us assume that these are interpolating points of Newton form of interpolating polynomial $ p_{n}(x)$ of degree i.e

$\displaystyle p_{n}(x_{i})=f_{i}, \qquad i=0,1,..n-1$ (1)

The Newton form of the interpolating polynomial $ p_{n}(x)$ is given by

$\displaystyle p_{n}(x)=a_{0}+a_{1}(x-x_{0})+a_{2}(x-x_{0})(x-x_{1})+ a_{3}(x-x_{0})(x-x_{1})(x-x_{2})+.......+ a_{n}(x-x_{0})(x-x_{1})....(x-x_{n-1}).$ (2)

For i=0, from (1)-(2) we get

$\displaystyle f_{0}=P_{n}(x_{0})=a_{0}$ (3.1)

For $ i=1$ from (1)-(2) we get

$\displaystyle f_{1}=p_{n}(x_{1})=a_{0}+a_{1}(x_{1}-x_{0})$

$\displaystyle a_{1}=\frac{f_{1}-f_{0}}{x_{1}-x_{0}}$ (3.2)

For i=2 from(1)-(2) We get

$\displaystyle f_{2}=p_{n}(x_{2})=a_{0}+a_{1}(x_{2}-x_{0})+a_{2}(x_{2}-x_{0})(x_{2}-x_{1})$

using (3.1)-(3.2), we get

$\displaystyle a_{2}=\frac{[(f_{2}-f_{1})/(x_{2}-x_{1})]-[(f_{1}-f_{0})/(x_{1}-x_{0})]}{(x_{2}-x_{0})}$ (3.3)

Similarly we can find $ a_{3}.....a_{n-1}$. To express $ a_{i},i=0......n-1$ in a compact manner let us first define the following notation called divided differences:

$\displaystyle f[x_{k}]=f_{k}$ (4.1)

$\displaystyle \hspace{0.5in}f[x_{k},x_{k+1}]=\frac{f[x_{k+1}]-f[x_{k}]}{x_{k+1}-x_{k}}$ (4.2)

$\displaystyle \hspace{0.9in}f[x_{k},x_{k+1},x_{k+2}]=\frac{f[x_{k+1},x_{k+2}]-f[x_{k},x_{k+1}]}{x_{k+2}-x_{k}}$ (4.3)

$\displaystyle \hspace{0.5in}f[x_{k},x_{k-1}....x_{i},x_{i+1}]=\frac{f[x_{k+1}...x_{i+1}]-f[x_{k}...x_{i}]}{x_{i+1}-x_{k}}$ (4.4)

Now the co-efficients $ a_{i}, i=_{0}...n-1$ can be expressed in terms of divided differences as fellows:

$\displaystyle a_{0}=f_{0}=f[x_{0}]$ (5.1)

$\displaystyle a_{1}=\frac{f_{1}-f_{0}}{x_{1}-x_{0}}=f[x_{1},x_{0}]$ (5.2)

$\displaystyle a_{2}=\displaystyle
{\frac{\frac{f_{2}-f_{1}}{x_{2}-x_{1}}-\frac{...
...}{x_{1}-x_{0}}}{x_{2}-x_{0}}}=\frac{f[x_{1},x_{2}]-f[x_{0},x_{1}]}{x_{2}-x_{0}}$

$\displaystyle =f[x_{0},x_{1},x_{2}]$ (5.3)

$\displaystyle a_{n}=f[x_{0},x_{1}...x_{n}]$ (5.4)

Note that $ a_{1}$ is called as the first divided difference, $ a_{2}$ as the second divided difference and so on. Now the polynomial (2) can be rewritten as:

$\displaystyle p_{n}(x)=f[x_{0}]+f[x_{0},x_{1}](x-x_{0})+f[x_{0},x_{1},x_{2}](x-x_{0})(x-x_{1})+......+f[x_{0}x_{1}......x_{n}](x-x_{0})(x-x_{1})...(x-x_{n-1})$

i.e.$\displaystyle \hspace{1in}p_{n}(x)=\sum\limits_{k=0}^{n}f[x_{0}......x_{k}]\prod\limits_{i=0}^{k-1}(x-x_{i})$ (6)

This is called as Newton's Divided Difference interpolation polynomial.

Example:
Given the following data table, evaluate $ f(2.4)$ using $ 3^{rd}$ order Newton's Divided Difference interpolation polynomial.
i 0 1 2 3 4
$ x_{i}$ 0 1 2 3 4
$ y_{i}=f(x_{i})$ 1 2.25 3.75 4.25 5.81
Solution:
Here $ n=5$. For constructing $ 3^{rd}$ order Newton Divided Difference polynomial we need only four points. Let us use the first four points. The $ 3^{rd}$ Newton Divided Difference polynomial is given by:
$ p_{3}(x)=\sum\limits_{k=0}^{3}a_{k}\prod\limits_{j=0}^{k-1}(x-x_{j})\\ =\sum\limits_{k=0}^{3}f[x_{0}...x_{k}]\prod\limits_{j=0}^{k-1}(x-x_{j})$
$ =f[x_{0}]+f[x_{0},x_{1}](x-x_{0})+
f[x_{0},x_{1},x_{2}](x-x_{0})(x-x_{1})+
f[x_{0},x_{1},x_{2},x_{3}](x-x_{0})(x-x_{1})(x-x_{3})$

% latex2html id marker 2479
$ \therefore a_{0}=f[x_{0}]=1$

$ a_{1}=f[x_{0},x_{1}]=\displaystyle
{\frac{f(x_{1})-f(x_{0})}{(x_{1}-x_{0})}}=\frac{2.25-1}{1-0}=1.25$

$ \displaystyle
{f[x_{1}x_{2}]=\frac{f[x_{2}]-f[x_{1}]}{x_{2}-x_{1}}=\frac{3.75-2.25}{2-1}=1.5}$


% latex2html id marker 2485
$ \displaystyle{\therefore
a_{2}=f[x_{0},x_{1},x_{2}]=\frac{f[x_{1},x_{2}]-f[x_{0},x_{1}]}{x_{2}-x_{0}}=\frac{1.5-1.25}{2-0}=0.125}$

$ \displaystyle{f[x_{2},x_{3}]=\frac{f[x_{3}]-f[x_{2}]}{x_{3}-x_{2}}=\frac{4.25-3.75}{3-2}=\frac{0.5}{1}=0.5}$

$ \displaystyle{f[x_{1},x_{2},x_{3}]=\frac{f[x_{2},x_{3}]-f[x_{1},x_{2}]}{x_{3}-x_{1}}=\frac{0.5-1.5}{3-1}=-0.5}$


% latex2html id marker 2491
$ \displaystyle{\therefore
a_{3}=f[x_{0},x_{1},x_{2},x_{3}]=\frac{f[x_{1},x_{2},x_{3}]-f[x_{0},x_{1},x_{2}]}{x_{3}-x_{0}}}$

$\displaystyle =\frac{-0.5-0.125}{3-0}=\frac{-0.625}{3}=-0.20833$


% latex2html id marker 2495
$ \therefore
p_{3}(x)=1+1.25(x-0)+0.125(x-0)(x-1)+(-0.20833)(x-0)(x-1)(x-2)$

% latex2html id marker 2497
$ \therefore f(2.4)=p_{3}(2.4)$
$ =1+1.25(2.4-0)+0.125(2.4-0)(2.4-1)+(-0.20833)(2.4-0)(2.4-1)(2.4-2)$
$ =1+(1.25)(2.4)+0.125(2.4)(1.4)-0.20833(2.4)(1.4)(0.4)$
$ =4.2200032$

In this example it may be noted that for calculating the $ 3rd$ order polynomial, we first start with $ P_{0}=f[x_{0}]=1$. To it we add $ a_{1}(x-x_{0})$ to get $ P_{1}$ and to $ P_{1}$ we add $ a_{2}(x-x_{0})(x-x_{1})$ to get $ P_{2}$. Finally on adding $ a_{3}(x-x_{0})(x-x_{1})(x-x_{2})$ to $ P_{2}$ we get $ P_{3}$.
Some remarks on the Error in the interpolation approach:
Suppose that the given data points $ (x_{i},y_{i})$, $ i=0,1,..n$ correspond to a real valued function $ f(x)$ defined on the internal $ I=[a,b]$. Let $ p_{n}(x)$ be the interpolating polynomial of degree $ \leq n.$ $ i.e.$ $ P_{n}(x_{i})=f(x_{i})=y_{i}$     $ i=0,1...n$. Then the interpolation error $ e_{n}(x)$ due to interpolation by $ p_{n}(x)$ is given by

$\displaystyle e_{n}(x)=f(x)-p_{n}(x).$ (6.2)

An estimate of the error is provided in the following theorem.

Theorem: Let $ f(x)$ be a real-valued function define on $ [a,b]$ and $ n+1$ times differentiable on $ (a,b).$ If $ p_{n}(x)$ is the polynomial of degree $ \leq n$ which interpolates $ f(x)$ at the (n+1) distinct points $ x_{0}..x_{n}\epsilon[a,b]$, then for all $ \overline{x}\epsilon[a,b]$, there exits $ \xi=\xi(\overline{x})\epsilon (a,b) s.t.$

$\displaystyle e_{n}(\overline{x})=f(\overline{x})-P_{n}(\overline{x})$    

$\displaystyle =\frac{f^{n+1}\vert\xi\vert}{(n+1)!}\prod \limits_{j=0}^{n}(\overline{x}-x_{j})$ (6.3)

Note:
i) $ \xi=\xi(\overline{x})$ i.e. $ \xi$ depends on the point $ \overline{x}$ at which the error estimate is required.
ii) since $ f^{n+1}(x)$ i.e. $ (n+1)^{th}$ derivative is seldom known the error formula $ (6.3)$ in the above theorem is of limited practical value. But when a bound on $ \vert f^{n+1}(x)\vert$ is known over the entire internal $ [a,b]$, then the formula $ (6.3)$ may be used to get a bound on the interpolation error.
next up previous
Next: (2.2.2)Newton Divided Difference Table: Up: Curve Fitting : Interpolation, Previous: Lagrange Interpolation:
root 2006-02-14