CURVE FITTING
WHAT IS CURVE FITTING?
Curve fitting is the process of finding the best fit curve for a given set of data. The relationship between two variables is represented by means of an algebraic equation.
Suppose a set of n points of values (x_{1}, y_{1}), (x_{2}, y_{2}), …, (x_{n }, y_{n}) of the two variables x and y are given.
These values are plotted on the xyplane.
Figure (i): Scatter Diagram
The resulting set of points is known as a scatter diagram as shown in figure (i).
The scatter diagram exhibits the trend and it is possible to visualize a smooth curve approximating the data which is known as an approximating curve.
LEAST SQUARE METHOD
From a scatter diagram, generally, more than one curve may be seen to be appropriate to the given set of data.
Let P (x_{i}, y_{i}) be a point on the scatter diagram as shown in figure (ii).
Distance QP = MPMQ
= y_{i }– f(x_{i})
The distance QP is known as deviation, error, or residual and is denoted by d_{i}.
If E= 0 then all the n points will lie on y = f(x). If E ≠ 0, f(x) is chosen such that E is minimum, i.e., the best fitting curve to the set of points is that for which E is minimum.
This method is known as the least square method.
This method does not attempt to determine the form of the curve y = f(x) but it determines the values of the parameters of the equation of the curve.
FITTING OF LINEAR CURVES
The normal equations for the straight line y = a + bx are
∑y = na + b∑x
∑xy = a∑x + b∑x^{2}
These equations can be solved simultaneously to give the best values of a and b.
The best fitting straight line is obtained by substituting the values of a and b in the equation y = a + bx.
Question:
Fit a straight line to the following data:
x 
100 
120 
140 
160 
180 
200 
y 
0.45 
0.55 
0.60 
0.70 
0.80 
0.85 
Solution:
Let the straight line be fitted to the data be
y = a + bx
The normal equations are
∑y = na + b∑x
∑xy = a∑x + b∑x^{2}
X 
y 
x^{2} 
xy 
100 
0.45 
10000 
45 
120 
0.55 
14400 
66 
140 
0.60 
19600 
84 
160 
0.70 
25600 
112 
180 
0.80 
32400 
144 
200 
0.85 
40000 
170 
∑x = 900 
∑y = 3.95 
∑x^{2} = 142000 
∑xy = 621 
Substituting these values in normal equations
3.95 = 6a + 900b …(i)
621 = 900a + 14200b …(ii)
Solving equations (i) and (ii),
a = 0.0476 and b = 0.0041
Hence, the required equation of the straight line is
y = 0.0476 + 0.0041x
HOW TO FIND THE REQUIRED EQUATION OF STRAIGHT LINE USING fx82MS CALCULATOR:
FITTING OF QUADRATIC CURVES
The normal equations for the curve y = a+bx+cx² are
∑y = na + b∑x + c∑x^{2}
∑xy = a∑x + b∑ x^{2 }+ c∑x^{3}
∑x^{2}y = a∑x^{2}+ b∑x^{3} + c∑x^{4}
These equations can be solved simultaneously to give the best values of a, b, and c.
The best fitting parabola is obtained by substituting the values of a, b, and c in the equation y = a+bx+cx².
Question:
Fit a parabola to the following data:
x 
1 
2 
3 
4 
5 
y 
5 
12 
26 
60 
97 
Solution:
Let the equation of the parabola be a+bx+cx²
The normal equations are
∑y = na + b∑x + c∑x^{2}
∑xy = a∑x + b∑ x^{2 }+ c∑x^{3}
∑x^{2}y = a∑x^{2}+ b∑x^{3} + c∑x^{4}
Here n = 5,
X 
y 
x^{2} 
x^{3} 
x^{4} 
xy 
x^{2}y 
1 
5 
1 
1 
1 
5 
5 
2 
12 
4 
8 
16 
24 
48 
3 
26 
9 
27 
81 
78 
234 
4 
60 
16 
64 
256 
240 
960 
5 
97 
25 
125 
625 
485 
2425 
∑x = 15 
∑y = 200 
∑x^{2} = 55 
∑x^{3} = 225 
∑x^{4} = 979 
∑xy = 832 
∑x^{2}y = 3672 
Substituting these values in the normal equations,
200 = 5a + 15b + 55c …(i)
832 = 15a + 55b + 225c …(ii)
3672 = 55a + 225b + 979c …(iii)
Solving equations (i), (ii) and (iii) we get
a = 10.4, b = 11.0857 and c = 5.7143
Hence, the required equation of the parabola is
y = 10.4 – 11.0857x + 5.7143x^{2}
HOW TO FIND THE REQUIRED EQUATION OF PARABOLA USING fx82MS CALCULATOR:
FITTING OF EXPONENTIAL AND LOGARITHMIC CURVES
For the curve y = ab^{x}
Taking Logarithm on both sides of the equation y = ab^{x},
log_{e}y = log_{e}a + xlog_{e}b
Putting log_{e}y = Y, log_{e}a = A, log_{e}b = B
Y = A + BX
The normal equations are
∑Y = nA + B∑X
∑XY = a∑X + b∑X^{2}
Solving these equations, A and B, and, hence, a and b can be found.
The best fitting exponential curve is obtained by substituting the values of a and b in the equation y = ab^{x}.
Question:
Fit a curve of the form y = ab^{x} to the following data by the method of least squares:
x 
1 
2 
3 
4 
5 
6 
7 
y 
87 
97 
113 
129 
202 
195 
193 
Solution:
y = ab^{x}
Taking Logarithm on both sides of the equation y = ab^{x},
log_{e}y = log_{e}a + xlog_{e}b
Putting log_{e}y = Y, log_{e}a = A, log_{e}b = B
Y = A + BX
The normal equations are
∑Y = nA + B∑X
∑XY = a∑X + b∑X^{2}
Here n = 7,
x 
Y 
X 
Y 
X^{2} 
XY 
1 
87 
1 
4.4659 
1 
4.4659 
2 
97 
2 
4.5747 
4 
9.1494 
3 
113 
3 
4.7274 
9 
14.1822 
4 
129 
4 
4.8598 
16 
19.4392 
5 
202 
5 
5.3083 
25 
26.5415 
6 
195 
6 
5.2730 
36 
31.6380 
7 
193 
7 
5.2627 
49 
36.8389 


∑X = 28 
∑Y = 34.4718 
∑X^{2} = 140 
∑XY = 142.2551 
34.4718 = 7A + 28B … (i)
142.2551 = 28A + 140B … (ii)
Solving equations (i) and (ii),
A = 4.3006 and B = 0.156
log_{e}a = A
log_{e}a = 4.3006
a = 73.744
log_{e}b = B
log_{e}b = 0.156
b = 1.1688