1. Direct Method of Interpolation
Industrial Engineering Majors
Authors: Autar Kaw, Jai Paul
Transforming Numerical Methods Education for STEM Undergraduates

2. Direct Method of Interpolation http://numericalmethods.eng.usf.edu

3. What is Interpolation ?
Given (x0,y0), (x1,y1), …… (xn,yn), find the value of ‘y’ at a value of ‘x’ that is not given.
Figure 1 Interpolation of discrete.

4. Interpolants Evaluate Differentiate, and Integrate
Polynomials are the most common choice of interpolants because they are easy to:
Evaluate
Differentiate, and
Integrate

5. Direct Method Given ‘n+1’ data points (x0,y0), (x1,y1),………….. (xn,yn),
pass a polynomial of order ‘n’ through the data as given
below:
where a0, a1,………………. an are real constants.
Set up ‘n+1’ equations to find ‘n+1’ constants.
To find the value ‘y’ at a given value of ‘x’, simply substitute the value of ‘x’ in the above polynomial.

6. Example
A curve needs to be fit through the given points to fabricate the cam. If the cam follows a straight line profile between x = 1.28 to x = 0.66, what is the value of y at x=1.1 ? Find using the direct method and linear interpolation.
Point
x (in.)
y (in.) 1
2.20
0.00 2
1.28
0.88 3
0.66
1.14 4
1.20 5
–0.60
1.04 6
–1.04
0.60 7
–1.20 1 2 3 5 6 7 X Y 4

7. Linear Interpolation Solving the above two equations gives, Hence

8. Example
A curve needs to be fit through the given points to fabricate the cam. If the cam follows a straight line profile between x = 1.28 to x = 0.66, what is the value of y at x=1.1 ? Find using the direct method and quadratic interpolation.
Point
x (in.)
y (in.) 1
2.20
0.00 2
1.28
0.88 3
0.66
1.14 4
1.20 5
–0.60
1.04 6
–1.04
0.60 7
–1.20 1 2 3 5 6 7 X Y 4

9. Quadratic Interpolation
Solving the above three equations gives

10. Quadratic Interpolation (contd)
The absolute relative approximate error obtained between the results from the first and second order polynomial is

11. Comparison Table

12. Example
A curve needs to be fit through the given points to fabricate the cam. If the cam follows a straight line profile between x = 1.28 to x = 0.66, what is the value of y at x=1.1 ? Find using the direct method and a sixth order polynomial.
Point
x (in.)
y (in.) 1
2.20
0.00 2
1.28
0.88 3
0.66
1.14 4
1.20 5
–0.60
1.04 6
–1.04
0.60 7
–1.20 1 2 3 5 6 7 X Y 4

13. Sixth Order Interpolation

14. Sixth Order Interpolation (contd)

15. Sixth Order Polynomial (contd)
Solving the above seven equations gives

16. Sixth Order Polynomial (contd)

17. Additional Resources
For all resources on this topic such as digital audiovisual lectures, primers, textbook chapters, multiple-choice tests, worksheets in MATLAB, MATHEMATICA, MathCad and MAPLE, blogs, related physical problems, please visit

18. THE END