Chapter 20, Problem 41P

Ohm's law states that the voltage drop V across an ideal resistor is linearly proportional to the current i flowing through the resistor as in $V=iR$, where R is the resistance. However, real resistors may not always obey Ohm's law. Suppose that you performed some very precise experiment stomeasure the voltaged rop and corresponding current for a resistor. The results, as listed in Table P20.41, suggest a curvilinear relationship rather than the straight line represented by Ohm's law. In order to quantify this relation-ship, a curve must be fit to these data. Because of measurement error, regression would typically be the preferred method of curve fitting for analyzing such experimental data. However, the smoothness of the relationship, as well as the precision of the experimental methods, suggests that interpolation might be appropriate. Use Newton's interpolating polynomial to fit these data and compute V for $i=0.10$. What is the order of the polynomial that was used to generate the data?

TABLE P20.41

Experimental data for voltage drop across a resistor subjected to various levels of current.

i	–2	–1	–0.5	0.5	1	2
V	–637	–96.5	–20.5	20.5	96.5	637

To determine

To calculate: The value of V at $i=0.10$ by the use of Newton’s interpolating polynomial and the order of the polynomial for generation of the data where the provided data is,

i	$-2$	$-1$	$-0.5$	0.5	1	2
V	$-637$	$-96.5$	$-20.5$	20.5	96.5	637

Answer to Problem 41P

Solution:

The value of V at $i=0.10$ by the third-order Newton’s interpolation polynomial is 2.324.

Explanation of Solution

Given Information:

The provided data is,

i	$-2$	$-1$	$-0.5$	0.5	1	2
V	$-637$	$-96.5$	$-20.5$	20.5	96.5	637

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:

Assume $x=i$,

First, order the provided value as close to 0.10 as below,

$x_0=0.5,x_1=-0.5,x_2=1,x_3=-1,x_4=2$ and $x_5=-2$.

Therefore,

$\begin{array}{c}f\left(x_0\right)=20.5\\ f\left(x_1\right)=-20.5\\ f\left(x_2\right)=96.5\\ f\left(x_3\right)=-96.5\end{array}$

And,

$\begin{array}{l}f\left(x_4\right)=637\\ f\left(x_5\right)=-637\end{array}$

The first divided difference is,

$\begin{array}{c}f\left[x_1,x_0\right]=\frac{f\left(x_1\right)-f\left(x_0\right)}{x_1-x_0}\\ =\frac{-20.5-20.5}{-0.5-0.5}\\ =41\end{array}$

Thus, the first degree polynomial value can be calculated as,

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

Put $b_0=20.5,b_1=41,x=0.1\text{and}{x}_0=0.5$ in above equation,

$\begin{array}{c}{f}_1\left(x\right)=20.5+\left(41\right)\left(0.1-0.5\right)\\ =4.1\end{array}$

Solve for other values as,

$\begin{array}{c}f\left[x_2,x_1\right]=\frac{f\left(x_2\right)-f\left(x_1\right)}{x_2-x_1}\\ =\frac{96.5+20.5}{1+0.5}\\ =\frac{117}{1.5}\\ =78\end{array}$

And,

$\begin{array}{c}f\left[x_3,x_2\right]=\frac{f\left(x_3\right)-f\left(x_2\right)}{x_3-x_2}\\ =\frac{-96.5-96.5}{-1-1}\\ =96.5\end{array}$

Similarly,

$\begin{array}{l}f\left[x_4,x_3\right]=244.5\\ f\left[x_5,x_4\right]=318.5\end{array}$

The second divided difference is,

$\begin{array}{c}f\left[x_2,x_1,x_0\right]=\frac{f\left[x_2,x_1\right]-f\left[x_1,x_0\right]}{x_2-x_0}\\ =\frac{78-41}{1-0.5}\\ =74\end{array}$

Thus, the second degree polynomial value can be calculated as,

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

Put $b_0=20.5,b_1=41,b_2=74,x_1=-0.5,x=0.1\text{and}{x}_0=0.5$ in above equation,

$\begin{array}{c}{f}_2\left(x\right)=20.5+\left(41\right)\left(0.1-0.5\right)+74\left(0.1-0.5\right)\left(0.1+0.5\right)\\ =-13.66\end{array}$

And,

$\begin{array}{c}f\left[x_3,x_2,x_1\right]=\frac{f\left[x_3,x_2\right]-f\left[x_2,x_1\right]}{x_3-x_1}\\ =\frac{96.5-78}{-1+0.5}\\ =-37\end{array}$

And,

$\begin{array}{c}f\left[x_4,x_3,x_2\right]=\frac{f\left[x_4,x_3\right]-f\left[x_3,x_2\right]}{x_4-x_2}\\ =\frac{244.5-96.5}{2-1}\\ =148\end{array}$

The third divided difference is,

$\begin{array}{c}f\left[x_3,x_2,x_1,x_0\right]=\frac{f\left[x_3,x_2,x_1\right]-f\left[x_2,x_1,x_0\right]}{x_3-x_0}\\ =\frac{-37-74}{-1-0.5}\\ =74\end{array}$

Thus, the third degree polynomial value can be calculated as,

$f_3\left(x\right)=b_0+b_1\left(x-x_0\right)+b_2\left(x-x_0\right)\left(x-x_1\right)+b_3\left(x-x_0\right)\left(x-x_1\right)\left(x-x_2\right)$

Put $b_0=20.5,b_1=41,b_2=74,b_3=74,x_2=1,x_1=-0.5,x=0.1\text{and}{x}_0=0.5$ in above equation,

$\begin{array}{c}{f}_3\left(x\right)=20.5+\left(41\right)\left(0.1-0.5\right)+74\left(0.1-0.5\right)\left(0.1+0.5\right)+\left(74\right)\left(0.1-0.5\right)\left(0.1+0.5\right)\left(0.1-1\right)\\ =2.324\end{array}$

And, the error is calculated as,

$\begin{array}{c}\text{Error}=4.1-20.5\\ =-16.4\end{array}$

Similarly the other dividend can be calculated as shown above,

Therefore, the difference table can be summarized as,

Order	$f\left(0.1\right)$	Error
0	20.5	$-16.4$
1	4.1	$-17.76$
2	$-13.66$	15.984
3	2.324	0
4	2.324	0
5	2.324

Since the error after order becomes zero, therefore, it can be concluded that the data is generated with a cubic polynomial.

Hence, the value of V at $i=0.10$ by the third-order Newton’s interpolation polynomial is 2.324.