Step-by-step explanation:

The graphical method has the drawback in that the straight line drawn may not

be unique but principle of least squares PROVIDES a unique set of values to the

constants and hence suggests a curve of best fit to the given data. The method of

least square is probably the most systematic procedure to fit a unique curve through

the given data points.

We will consider some of the best fitting curves of the type:

1. A straight line.

2. A second degree curve.

3. The exponential curve y = aebx.

4. The curve y — a x 71.

1. Fitting a straight line by the method of least squares:

Let (x;,yj), t = 0,1,2,....,n be the n sets of observations and let the related

relation by y = ax + b. Now we have to select a and b so that the straight line is the

best fit to the data.

As explained earlier, the residual at x = x t is

d i= y i~ f O i) = yt~ Caxi + b),i = 1,2,, 71

e = ir= i df = Xf=i [yf - (axi + b)]2

By the principle of least squares, E is minimum.

AL-Mustansirriya University

College oE Engineering

Computer & Software ENG. Dep. AiL^I Course ( l) Lecture (4)

3rd Class

dE dE — = 0 and — = 0 da db

i-e-, 2 - (axt + b)] ( - ^ ) = 0 & 2 £ [y £ - (axt + b)] ( - 1 ) = 0

i.e., - ax? - bXi) = 0 & E?=i(yi - ax< - ft) = 0

i.e., aZ?=i3C? = Z " = i^ y i .... (eq.l)

And a £"=1 x t + nb = £ f=1 y; .....(eq.2)

Since, x it y t are known, equations (1) & (2) give two equations in a & b. Solve for a

& b from (1) & (2) & obtain the best fit y= ax + b.

Note:

• Equations (1) & (2) are called normal equations.

• Dropping SUFFIX i from (1) & (2), the normal equations are

aY ,x + nb = £ y & a£x2 + b j ^ x = £xy

Which are get taking £ on both sides of y = ax + b & also taking £ on both sides

after multiplying by x both sides of y = ax + b.

• Transformation like X = ~ ~ , Y — ~ ~ reduce the linear equation y = ax + b to

the form Y = AX + B. Hence, a linear fit is another linear fit in both SYSTEMS of

coordinates.

Example 1:

By the method of least squares find the straight line to the data given below:

X 5 10 15 20 25

y 16 19 23 26 30

Solution:

Let the straight line be y=ax+b

The nor

AL-Mustansirriya University

College of Engineering

Computer & Software Eng. Dep.

Course ( l) Lecture (4)

3rd Class

X y X 2 xy

5 16 25 80

10 19 100 190

15 23 225 345

20 26 400 520

25 30 625 750

Total 75 114 1375 1885

The normal equations are 75a+5b=l 14 .....(eq.l)

1375a+75b=1885 ..... (eq.2)

Eliminate b, multiply (1) by 15

1125a+75b=1710 .....(eq.3)

(eq.2) - (eq.3) gives, 250 a=T75 or a=0.7, hence b= 1 ? ^

Hence, the best fitting line is y=0.7x+12.3

Let X = x~xmid = *~15 y = y~ymid = y~23 h 5 ’ h 5

Let the line in the new variable by Y=AX+B

X y X X2 Y XY

5 16 -2 4 -1.4 2.8

10 19 -1 1 -0.8 0.8

15 23 0 0 0 0

20 26 1 1 0.6 0.6

25 30 2 4 1.4 2.8

Total 0 10 -0.2 7

The normal equations are A £ X + 55 = £ Y .....(eq.4)

A Y X 2 + B Y , X = Z X Y .....(eq.5)

Therefore, -SB = —0.2 —> B = —0.04

10A = 7 ->A = 0.7

The equations Y=0.7X - 0.04

i.e. = 0.7 ( - 0.04 y - 23 = 0.7x - 10.5 - 0.2

i.e. y=0.7x+33.3

Which is the same equation as seen before.

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AL-Mustansirriya University

College of Engineering

Computer & Software Eng. Dep. ■Miami

Course ( l) Lecture (4)

3rd Class

Example 2:

Fit a straight line to the data given below. Also estimate the value of y at x=2.5

x 0 l 2 3 4

y l 1.8 3.3 4.5 6.3

Solution:

Let the best fit be y= ax + b .....(eq.l)

The normal equations are a £ x + 5b = £ y .....(eq.2)

a£x2 + b'Zx = 2 > y .....(eq.3)

We prepare the table for easy use.

X y X 2 xy

0 l 0 0

1 1.8 1 1.8

2 3.3 4 6.6

3 4.5 9 13.5

4 6.3 16 25.2

Total 10 16.9 30 47.1

Substituting in (eq.2) and (eq.3), we get,

10a+5b=16.9

30a+10b=47.1

Solving eq.(2)-eq.(l), we get, a=l .33, b= 0.72

Hence, the equation is y =1.33x+0.72

y (at x=2.5) =1.33 (2.5) +0.72 = 4.045