9.5.3 Other iterative procedures  9.5 Roots of equations  9.5.1 Newton's method

9.5.2 Interpolation

Roots of equations can also be found by interpolation if the identifications of dependent and independent variables are interchanged: From known values of the function, one interpolates to the point where the value of the function, rather than the independent variable, is zero. Lagrange interpolation is particularly suited to this approach because the values of the function will usually not form an evenly spaced array, and Lagrange interpolation is still convenient with unequal spacing between data values.

To illustrate the procedure, consider the equation

f(x)=x-e^-x=0  (9.42)

 

To start the problem, plot the function or make some guesses spanning the root:
 

	x	f(x)
	0.1	0.804837
	0.5	0.1065306
	0.7	-0.2034147
	0.9	-0.4934303

The answer lies between 0.5 and 0.7. A fairly good estimate of the answer can be obtained by using f(x) as the independent variable and interpolating among the tabulated values to find x corresponding to f(x)=0. Lagrange interpolation with the above points produces x=0.567, while the true root is 0.56714$\ldots$ . To improve the estimate, values at 0.555, 0.560, 0.565, 0.570, and 0.575 were taken; when the interpolation procedure was repeated, the result was 0.56714.

Similar accuracy can be reached via Newton's method in four steps. Lagrange interpolation can be very useful, however, when only tabular values are available. In the preceding example, the initial result was close to the true root even though the tabulated values were widely spaced about the root.

   
 9.5.3 Other iterative procedures  9.5 Roots of equations  9.5.1 Newton's method