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5.2 Methods of Curve Fitting
Data Linearization Method for y = C e Ax
Suppose that we are given the points (x1 , y1 ), (x2 , y2 ), . . . , (x N , y N ) and want to fit an exponential curve of the form (1) y = Ce Ax .

The first step is to take the logarithm of both sides: (2) ln(y) = Ax + ln(C).

Then introduce the change of variables: (3) Y = ln(y), X =x, and B = ln(C).

This results in a linear relation between the new variables X and Y : (4) Y = AX + B.

The original points (xk , yk ) in the x y-plane are transformed into the points (X k , Yk ) = (xk , ln(yk )) in the X Y -plane. This process is called data linearization. Then the leastsquares line (4) is fit to the points {(X k , Yk )}. The normal equations for finding A and B are
N 2 Xk NN

A+
k=1

Xk

B=
k=1 N

X k Yk , Yk .
k=1

(5)

k=1 N

Xk
k=1

A+

NB

=

After A and B have been found, the parameter C in equation (1) is computed: (6) C = eB.

Example 5.4. Use the data linearization method and find the exponential fit y = Ce Ax for the five data points (0, 1.5), (1, 2.5), (2, 3.5), (3, 5.0), and (4, 7.5). Apply the transformation (3) to the originalpoints and obtain (7) {(X k , Yk )} = {(0, ln(1.5), (1, ln(2.5)), (2, ln(3.5)), (3, ln(5.0)), (4, ln(7.5))} = {(0, 0.40547), (1, 0.91629), (2, 1.25276), (3, 1.60944), (4, 2.01490)}.

These transformed points are shown in Figure 5.4 and exhibit a linearized form. The equation of the least-squares line Y = AX + B for the points (7) in Figure 5.4 is (8) Y = 0.391202X + 0.457367.

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Y2.0 1.5 1.0 0.5

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Y = AX + B

X 0 1 2 3 4

Figure 5.4 The transformed data points {(X k , Yk )}.

Table 5.4 Obtaining Coefficients of the Normal Equations for the Transformed Data Points {(X k , Yk )}
xk 0.0 1.0 2.0 3.0 4.0 yk 1.5 2.5 3.5 5.0 7.5 Xk 0.0 1.0 2.0 3.0 4.0 10.0 = Xk Yk = ln(yk ) 0.405465 0.916291 1.252763 1.609438 2.014903 6.198860 = Yk =
2Xk

X k Yk 0.000000 0.916291 2.505526 4.828314 8.059612 16.309743 = X k Yk

0.0 1.0 4.0 9.0 16.0 30.0 2 Xk

Calculation of the coefficients for the normal equations in (5) is shown in Table 5.4. The resulting linear system (5) for determining A and B is (9) 30A + 10B = 16.309742 10A + 5B = 6.198860.

The solution is A = 0.3912023 and B = 0.457367. Then C is obtained with the calculation C= e0.457367 = 1.579910, and these values for A and C are substituted into equation (1) to obtain the exponential fit (see Figure 5.5): (10) y = 1.579910e0.3912023x (fit by data linearization).

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y 8 6 4 2

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y = Ce Ax

x 0 1 2 3 4

Figure 5.5 The exponential fit y = 1.579910e0.3912023x obtained by using the data linearization method.

NonlinearLeast-Squares Method for y = C e Ax
Suppose that we are given the points (x1 , y1 ), (x2 , y2 ), . . . , (x N , y N ) and want to fit an exponential curve: (11) y = Ce Ax .

The nonlinear least-squares procedure requires that we find a minimum of
N

(12)

E(A, C) =
k=1

(Ce Axk − yk )2 .

The partial derivatives of E(A, C) with respect to A and C are (13) and (14) ∂E =2 ∂C
N

∂E =2 ∂A

N(Ce Axk − yk )(C xk e Axk )
k=1

(Ce Axk − yk )(e Axk ).
k=1

When the partial derivatives in (13) and (14) are set equal to zero and then simplified, the resulting normal equations are
N N

C (15)
k=1

xk e

2Axk

−

xk yk e Axk = 0,
k=1 N

N

C
k=1

e Axk −
k=1

yk e Axk = 0.

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The equations in (15) are nonlinear inthe unknowns A and C and can be solved using Newton’s method. This is a time-consuming computation and the iteration involved requires good starting values for A and C. Many software packages have a built-in minimization subroutine for functions of several variables that can be used to minimize E(A, C) directly, For example, the Nelder-Mead simplex algorithm can be used to minimize (12) directly...