# 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 (x1, y1), (x2, y2), …, (xn , yn) of the two variables x and y are given.

These values are plotted on the xy-plane.

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.

The method of least squares is used to find a curve which passes through the maximum number of points.

Let P (xi, yi) be a point on the scatter diagram as shown in figure (ii).

Figure (ii)

Let the ordinate at P meet the curve y = f(x) at Q and the x-axis at M.

Distance QP = MP-MQ
= yi – f(xi)

The distance QP is known as deviation, error, or residual and is denoted by di

It may be positive, negative, or zero depending upon whether P lies above, below, or on the curve.

Similar residuals or errors corresponding to the remaining (n-1) points may be obtained.

The sum of squares of residuals, denoted by E, is given as

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 + bx

∑xy = a∑x + b∑x2

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 0.6 0.7 0.8 0.85

Solution:

Let the straight line be fitted to the data be

y = a + bx

The normal equations are

y = na + bx

∑xy = a∑x + b∑x2

 X y x2 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 ∑x2 = 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:

The normal equations for the curve y = a+bx+cx² are

∑y = na + b∑x + c∑x2

∑xy = a∑x + b∑ x2 + c∑x3

∑x2y = a∑x2+ b∑x3 + c∑x4

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∑x2

∑xy = a∑x + b∑ x2 + c∑x3

∑x2y = a∑x2+ b∑x3 + c∑x4

Here n = 5,

 X y x2 x3 x4 xy x2y 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 ∑x2 = 55 ∑x3 = 225 ∑x4 = 979 ∑xy = 832 ∑x2y = 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 + 5.7143x2

HOW TO FIND THE REQUIRED EQUATION OF PARABOLA USING fx82MS CALCULATOR:

## FITTING OF EXPONENTIAL AND LOGARITHMIC CURVES

For the curve y = abx

Taking Logarithm on both sides of the equation y = abx,

logey = logea + xlogeb

Putting logey = Y, logea = A, logeb = B

Y = A + BX

The normal equations are

Y = nA + BX

∑XY = a∑X + b∑X2

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 = abx.

Question:

Fit a curve of the form y = abx 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 = abx

Taking Logarithm on both sides of the equation y = abx,

logey = logea + xlogeb

Putting logey = Y, logea = A, logeb = B

Y = A + BX

The normal equations are

Y = nA + BX

∑XY = a∑X + b∑X2

Here n = 7,

 x Y X Y X2 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 ∑X2 = 140 ∑XY = 142.2551

Substituting these values in the normal equations,

34.4718   = 7A + 28B                            … (i)

142.2551 = 28A + 140B                       … (ii)

Solving equations (i) and (ii),

A = 4.3006   and   B = 0.156

logea = A

logea = 4.3006

a = 73.744

logeb = B

logeb = 0.156

b = 1.1688

Hence, the required curve is y = 73.744 (1.1688)

HOW TO FIND THE REQUIRED EQUATION OF EXPONENTIAL CURVE USING fx82MS CALCULATOR: