3.2.5 Use of Higher Order Polynomials

In some cases the use of second order polynomials does not give satisfying results for the fit of the data values. Higher order polynomials can then be be used for the model function. But these functions need more factors than e.g., a quadratic polynomial. In Table 3.10 the number of required factors k for several types of polynomials are listed.

Table 3.10: Formulas for the number of required factors for different types of polynomials.

Type of polynomial	Number of variables
Linear	1 + n
Quadratic	1 + n + n
Quadratic with cross-terms3.4	$1 + n + \frac{n \cdot (n+1)}{2}$
Cubic with second order cross-terms	$1 + n + \frac{n \cdot (n+1)}{2} + n$
Cubic with second and third order cross-terms3.5	$1 + n + \frac{n \cdot (n+1)}{2} + \frac{n \cdot (n + 1) (n + 2)}{2 \cdot 3}$

Using the formulas from Table 3.10 for different numbers of input parameters n and different model functions like linear, quadratic, and cubic polynomials with and without cross-terms the required factors are listed in Table 3.11.

Table 3.11: Number of required factors for several types of polynomials and dimension of independent parameters.

Parameters	Linear	Quadratic	Quadratic with cross-terms	Cubic with quadratic cross-terms	Cubic with cross-terms
2	3	5	7	6	10
3	4	7	10	13	20
4	5	9	15	19	35
5	6	11	21	26	56
6	7	13	28	34	84
7	8	15	36	43	120
8	9	17	45	53	165
9	10	19	55	64	220
10	11	21	66	76	286
11	12	23	78	89	364
12	13	25	91	103	455

There are two disadvantages of higher order polynomials that have to be considered:

1. Higher order polynomials have a larger number of factors than quadratic polynomials. This implies that the dimension of the linear system that has to be solved increases, and more experiments and results are required to solve it. For a response surface of third order the number of required factors dependent on the number of input parameters is listed in the last column of Table 3.11. For less than 7 input parameters and for the experimental designs listed in Table 3.9 only the Three Level Factorial can be used to create enough experimental points for solving the linear system build by the fit function. For more parameters, a Central Composite type design should be chosen.

2. In most cases with a good transformation function for the controls and responses, gives a second order model can give sufficient results.

Footnotes

... cross-terms^3.4

The number of quadratic cross-terms is ${n \choose 2} = n \cdot (n-1)$

... cross-terms^3.5

The number of cubic cross-terms is ${n + 3 - 1 \choose 3} - n = \frac{n \mult (n + 1) (n + 2)}{2 \mult 3} - n$