CURVE FITTING  HOW TO FIND EQUATION OF STRAIGHT LINE, PARABOLA AND EXPONENTIAL CURVE USING fx82 MS CALCULATOR
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
Comments
Post a Comment