Chapter 18, Problem 10P

Use Newton's interpolating polynomial to determine y at $x=8$ to the best possible accuracy. Compute the finite divided differences as in Fig. 18.5 and order your points to attain optimal accuracy and convergence.

x	0	1	2	5.5	11	13	16	18
y	0.5	3.134	5.3	9.9	10.2	9.35	7.2	6.2

To determine

To calculate: The value of $y$ at $x=8$ by the use of Newton’s interpolating polynomial where the provided data are,

x	0	1	2	5.5	11	13	16	18
y	0.5	3.134	5.3	9.9	10.2	9.35	7.2	6.2

Answer to Problem 10P

Solution:

Thevalue of $y$ at $x=8$ by zero order Newton’s interpolating polynomial is $9.9$, by first order Newton’s interpolating polynomial is $10.036$, by second order Newton’s interpolating polynomial is $10.516$, by third order Newton’s interpolating polynomial is $10.775$ and by fourth order Newton’s interpolating polynomial is $10.812$.

Explanation of Solution

Given Information:

The provided data are,

x	0	1	2	5.5	11	13	16	18
y	0.5	3.134	5.3	9.9	10.2	9.35	7.2	6.2

Formula used:

The zero-order Newton’s interpolation formula:

$f_0\left(x\right)=b_0$

The first-order Newton’s interpolation formula:

$f_1\left(x\right)=b_0+b_1\left(x-x_0\right)$

The second- order Newton’s interpolating polynomial is given by,

$f_2\left(x\right)=b_0+b_1\left(x-x_0\right)+b_2\left(x-x_0\right)\left(x-x_1\right)$

The n th-order Newton’s interpolating polynomial is given by,

$f_n\left(x\right)=b_0+b_1\left(x-x_0\right)+b_2\left(x-x_0\right)\left(x-x_1\right)+\cdots +b_n\left(x-x_0\right)\left(x-x_1\right)\cdots \left(x-x_{n-1}\right)$

Where,

$\begin{array}{c}{b}_0=f\left(x_0\right)\\ b_1=f\left[x_1,x_0\right]\\ b_2=f\left[x_2,x_1,x_0\right]\\ ⋮\\ b_2=f\left[x_n,\cdots ,x_2,x_1,x_0\right]\end{array}$

The first finite divided difference is,

$f\left[x_i,x_j\right]=\frac{f\left(x_i\right)-f\left(x_j\right)}{x_i-x_j}$

And, the n th finite divided difference is,

$f\left[x_n,x_{n-1},\mathrm{...},x_1,x_0\right]=\frac{f\left[x_n,x_{n-1},\mathrm{...},x_1\right]-f\left[x_{n-1},\mathrm{...},x_1,x_0\right]}{x_n-x_0}$

Calculation:

First, order the provided value as close to 8 as below,

$x_0=5.5,x_1=11,x_2=13,x_3=2,x_4=1,x_5=16,x_6=0$ and $x_6=18$.

Therefore,

$\begin{array}{c}f\left(x_0\right)=9.9\\ f\left(x_1\right)=10.2\\ f\left(x_2\right)=9.35\\ f\left(x_3\right)=5.3\end{array}$

And,

$\begin{array}{l}f\left(x_4\right)=3.134\\ f\left(x_5\right)=7.2\\ f\left(x_6\right)=0.5\\ f\left(x_7\right)=6.2\end{array}$

The first divided difference is,

$\begin{array}{c}f\left[x_1,x_0\right]=\frac{f\left(x_1\right)-f\left(x_0\right)}{x_1-x_0}\\ =\frac{10.2-9.9}{11-5.5}\\ =\frac{0.3}{5.5}\\ =0.054545\end{array}$

And,

$\begin{array}{c}f\left[x_2,x_1\right]=\frac{f\left(x_2\right)-f\left(x_1\right)}{x_2-x_1}\\ =\frac{9.35-10.2}{13-11}\\ =\frac{-0.85}{2}\\ =-0.425\end{array}$

And,

$\begin{array}{c}f\left[x_3,x_2\right]=\frac{f\left(x_3\right)-f\left(x_2\right)}{x_3-x_2}\\ =\frac{5.3-9.35}{2-13}\\ =\frac{-4.05}{-11}\\ =0.368182\end{array}$

Similarly,

$\begin{array}{l}f\left[x_4,x_3\right]=2.166\\ f\left[x_5,x_4\right]=0.271067\\ f\left[x_6,x_5\right]=0.41875\\ f\left[x_7,x_6\right]=0.31667\end{array}$

The second divided difference is,

$\begin{array}{c}f\left[x_2,x_1,x_0\right]=\frac{f\left[x_2,x_1\right]-f\left[x_1,x_0\right]}{x_2-x_0}\\ =\frac{-0.425-0.054545}{13-5.5}\\ =\frac{-0.479545}{7.5}\\ =-0.06394\end{array}$

And,

$\begin{array}{c}f\left[x_3,x_2,x_1\right]=\frac{f\left[x_3,x_2\right]-f\left[x_2,x_1\right]}{x_3-x_1}\\ =\frac{0.368182-\left(-0.425\right)}{2-11}\\ =\frac{0.793182}{-9}\\ =-0.08813\end{array}$

And,

$\begin{array}{c}f\left[x_4,x_3,x_2\right]=\frac{f\left[x_4,x_3\right]-f\left[x_3,x_2\right]}{x_4-x_2}\\ =\frac{2.166-0.368182}{1-13}\\ =\frac{1.797818}{-12}\\ =-0.14982\end{array}$

Similarly,

$\begin{array}{c}f\left[x_5,x_4,x_3\right]=-0.13535\\ f\left[x_6,x_5,x_4\right]=-0.147683\\ f\left[x_7,x_6,x_5\right]=-0.05104\end{array}$

The third divided difference is,

$\begin{array}{c}f\left[x_3,x_2,x_1,x_0\right]=\frac{f\left[x_3,x_2,x_1\right]-f\left[x_2,x_1,x_0\right]}{x_3-x_0}\\ =\frac{\left(-0.08813\right)-\left(-0.06394\right)}{2-5.5}\\ =\frac{-0.02419}{-3.5}\\ =0.00691143\end{array}$

And,

$\begin{array}{c}f\left[x_4,x_3,x_2,x_1\right]=\frac{f\left[x_4,x_3,x_2\right]-f\left[x_3,x_2,x_1\right]}{x_4-x_1}\\ =\frac{\left(-0.14982\right)-\left(-0.08813\right)}{1-11}\\ =\frac{-0.06169}{-10}\\ =0.006169\end{array}$

Similarly,

$\begin{array}{c}f\left[x_5,x_4,x_3,x_2\right]=0.004822\\ f\left[x_6,x_5,x_4,x_3\right]=0.0061665\\ f\left[x_7,x_6,x_5,x_4\right]=0.005513\end{array}$

The fourth divided difference is,

$\begin{array}{c}f\left[x_4,x_3,x_2,x_1,x_0\right]=\frac{f\left[x_4,x_3,x_2,x_1\right]-f\left[x_3,x_2,x_1,x_0\right]}{x_4-x_0}\\ =\frac{\left(0.006169\right)-\left(0.00691143\right)}{1-5.5}\\ =\frac{-0.00074243}{-4.5}\\ =0.000165\end{array}$

And,

$\begin{array}{c}f\left[x_5,x_4,x_3,x_2,x_1\right]=\frac{f\left[x_5,x_4,x_3,x_2\right]-f\left[x_4,x_3,x_2,x_1\right]}{x_5-x_1}\\ =\frac{\left(0.004822\right)-\left(0.006169\right)}{16-11}\\ =\frac{-0.001347}{5}\\ =-0.0002694\end{array}$

And,

$\begin{array}{c}f\left[x_6,x_5,x_4,x_3,x_2\right]=\frac{f\left[x_6,x_5,x_4,x_3\right]-f\left[x_5,x_4,x_3,x_2\right]}{x_6-x_2}\\ =\frac{\left(0.0061665\right)-\left(0.004822\right)}{0-13}\\ =\frac{0.0013445}{-13}\\ =0.0001034\end{array}$

And,

$\begin{array}{c}f\left[x_7,x_6,x_5,x_4,x_3\right]=\frac{f\left[x_7,x_6,x_5,x_4\right]-f\left[x_6,x_5,x_4,x_3\right]}{x_7-x_3}\\ =\frac{\left(0.005513\right)-\left(0.0061665\right)}{18-2}\\ =\frac{-0.0006535}{16}\\ =-0.00004084\end{array}$

The fifth divided difference is,

$\begin{array}{c}f\left[x_5,x_4,x_3,x_2,x_1,x_0\right]=\frac{f\left[x_5,x_4,x_3,x_2,x_1\right]-f\left[x_4,x_3,x_2,x_1,x_0\right]}{x_5-x_0}\\ =\frac{\left(-0.0002694\right)-\left(0.000165\right)}{16-5.5}\\ =\frac{-0.0004344}{10.5}\\ =-0.0000414\end{array}$

And,

$\begin{array}{c}f\left[x_6,x_5,x_4,x_3,x_2,x_1\right]=\frac{f\left[x_6,x_5,x_4,x_3,x_2\right]-f\left[x_5,x_4,x_3,x_2,x_1,x_0\right]}{x_6-x_1}\\ =\frac{\left(0.0001034\right)-\left(0.0002694\right)}{0-11}\\ =\frac{-0.000166}{-11}\\ =0.0000151\end{array}$

And,

$\begin{array}{c}f\left[x_7,x_6,x_5,x_4,x_3,x_2\right]=\frac{f\left[x_7,x_6,x_5,x_4,x_3\right]-f\left[x_6,x_5,x_4,x_3,x_2\right]}{x_7-x_2}\\ =\frac{\left(-0.00004048\right)-\left(0.0001034\right)}{18-13}\\ =\frac{-0.00014388}{5}\\ =0.000028776\end{array}$

The sixth divided difference is,

$\begin{array}{c}f\left[x_6,x_5,x_4,x_3,x_2,x_1,x_0\right]=\frac{f\left[x_6,x_5,x_4,x_3,x_2,x_1\right]-f\left[x_5,x_4,x_3,x_2,x_1,x_0\right]}{x_6-x_0}\\ =\frac{\left(0.0000151\right)-\left(-0.0000414\right)}{0-5.5}\\ =\frac{0.0000565}{-5.5}\\ =-0.0000103\end{array}$

And,

$\begin{array}{c}f\left[x_7,x_6,x_5,x_4,x_3,x_2,x_1\right]=\frac{f\left[x_7,x_6,x_5,x_4,x_3,x_2\right]-f\left[x_6,x_5,x_4,x_3,x_2,x_1\right]}{x_7-x_1}\\ =\frac{\left(0.0000288776\right)-\left(0.0000151\right)}{18-11}\\ =\frac{0.00001378}{7}\\ =0.000002\end{array}$

The seventh divided difference is,

$\begin{array}{c}f\left[x_7,x_6,x_5,x_4,x_3,x_2,x_1,x_0\right]=\frac{f\left[x_7,x_6,x_5,x_4,x_3,x_2,x_1\right]-f\left[x_6,x_5,x_4,x_3,x_2,x_1,x_0\right]}{x_7-x_0}\\ =\frac{\left(0.000002\right)-\left(-0.0000103\right)}{18-5.5}\\ =\frac{0.0000123}{12.5}\\ =0.000001\end{array}$

Therefore, the difference table can be summarized as,

$i$	$x_i$	$f\left(x_i\right)$	First	Second	Third	Fourth	Fifth	Sixth	7th
0 1 2 3 4 5 6 7	5.5 11 13 2 1 16 0 18	9.9 10.2 9.35 5.3 3.134 7.2 0.5 6.2	0.054545-0.425 0.3682 2.166 0.2711 0.4188 0.3167	$-0.06394$ $-0.08813$ $-0.14982$ $-0.13535$ $-0.14768$ $-0.05104$	0.0069 0.0062 0.0048 0.0062 0.0055	0.0002 $-0.0003$ 0.0001 $-0.00004$	$-0.00004$ 0.00002 0.00003	$-0.00001$ 0.000002	0.00

Since, the divided difference of fifth order is nearly equals to zero. So, the fourth-order polynomial is the optimal.

Therefore, the zero-order Newton’s interpolation polynomial is,

$f_0\left(x\right)=9.9$

Thus, the value of y at $x=8$ is,

$y=9.9$

The first-order Newton’s interpolation polynomial is:

$f_1\left(x\right)=9.9+0.054545\left(x-5.5\right)$

Thus, the value of y at $x=8$ is,

$\begin{array}{c}y=9.9+0.054545\left(8-5.5\right)\\ =9.9+0.054545\times 2.5\\ =9.9+0.1363625\\ =10.036\end{array}$

The second- order Newton’s interpolating polynomial is,

$f_2\left(x\right)=9.9+0.054545\left(x-5.5\right)-0.06394\left(x-5.5\right)\left(x-11\right)$

Thus, the value of y at $x=8$ is,

$\begin{array}{c}y=9.9+0.054545\left(8-5.5\right)-0.06394\left(8-5.5\right)\left(8-11\right)\\ =10.036-0.06394\times 2.5\times \left(-3\right)\\ =10.036+0.47955\\ =10.516\end{array}$

The third-order Newton’s interpolating polynomial is,

$f_3\left(x\right)=\left\{\begin{array}{l}9.9+0.054545\left(x-5.5\right)-0.06394\left(x-5.5\right)\left(x-11\right)\\ +0.0069\left(x-5.5\right)\left(x-11\right)\left(x-13\right)\end{array}\right\}$

Thus, the value of y at $x=8$ is,

$\begin{array}{c}y=\left\{\begin{array}{l}9.9+0.054545\left(8-5.5\right)-0.06394\left(8-5.5\right)\left(8-11\right)\\ +0.0069\left(8-5.5\right)\left(8-11\right)\left(8-13\right)\end{array}\right\}\\ =10.516+0.0069\times 2.5\times \left(-3\right)\times \left(-5\right)\\ =10.775\end{array}$

The fourth-order Newton’s interpolating polynomial is,

$f_4\left(x\right)=\left\{\begin{array}{l}9.9+0.054545\left(x-5.5\right)-0.06394\left(x-5.5\right)\left(x-11\right)\\ +0.0069\left(x-5.5\right)\left(x-11\right)\left(x-13\right)+0.000165\left(x-5.5\right)\left(x-11\right)\left(x-13\right)\left(x-2\right)\end{array}\right\}$

Thus, the value of y at $x=8$ is,

$\begin{array}{c}y=\left\{\begin{array}{l}9.9+0.054545\left(8-5.5\right)-0.06394\left(8-5.5\right)\left(8-11\right)\\ +0.0069\left(8-5.5\right)\left(8-11\right)\left(8-13\right)+0.000165\left(8-5.5\right)\left(8-11\right)\left(8-13\right)\left(8-2\right)\end{array}\right\}\\ =10.775+0.000165\times 2.5\times \left(-3\right)\times \left(-5\right)\times 6\\ =10.812\end{array}$

Hence, the value of $y$ at $x=8$ by zero order Newton’s interpolating polynomial is $9.9$, by first order Newton’s interpolating polynomial is $10.036$, by second order Newton’s interpolating polynomial is $10.516$, by third order Newton’s interpolating polynomial is $10.775$ and by fourth order Newton’s interpolating polynomial is $10.812$