Download - Curve Fitting and Solution of Equation - …newagepublishers.com/samplechapter/001938.pdf · UNIT V Curve Fitting and Solution of Equation 5.1 CURVE FITTING In many branches of applied

UNIT V

Curve Fitting and Solution of Equation

5.1 CURVE FITTING

In many branches of applied mathematics and engineering sciences we come across experiments andproblems, which involve two variables. For example, it is known that the speed v of a ship varies with

the horsepower p of an engine according to the formula = + 3.p a bv Here a and b are the constants tobe determined. For this purpose we take several sets of readings of speeds and the correspondinghorsepowers. The problem is to find the best values for a and b using the observed values of v and .p

Thus, the general problem is to find a suitable relation or law that may exist between the variables xand y from a given set of observed values =( , ), 1, 2,.........., .i ix y i n Such a relation connecting x and y

is known as empirical law. For above example, x = v and y = p.The process of finding the equation of the curve of best fit, which may be most suitable for

predicting the unknown values, is known as curve fitting. Therefore, curve fitting means an exactrelationship between two variables by algebraic equations. There are following methods for fitting acurve.

I. Graphic methodII. Method of group averages

III. Method of momentsIV. Principle of least square.

Out of above four methods, we will only discuss and study here principle of least square.

5.2 PRINCIPLE OF LEAST SQUARES

The graphical method has the drawback in that the straight line drawn may not be unique but principleof least squares provides a unique set of values to the constants and hence suggests a curve of best fit tothe given data. The method of least square is probably the most systematic procedure to fit a uniquecurve through the given data points.

CURVE FITTING AND SOLUTION OF EQUATION 383

Let the curve −= + + +2 1................. my a bx cx kx ...(1)

be fitted to the set of n data points 1 1( , )x y , 2 2( , )x y , 3 3( , )x y ,……, ( , ).n nx y At =( )ix x the observed

(or experimental) value of the ordinate is yi = PiMi and the corresponding value on the fitting curve ( )i

is + + + =2 ........ mi i i i ia bx cx kx L M which is the expected or calculated value. The difference of the

observed and the expected value is − = (say)i i i i iP M L M e this difference is called error at =( )ix x

clearly some of the error 1 2 3, , ,....., , .......,i ne e e e e will be positive and other negative. To make all errors

positive we square each of the errors i.e. = + + + + + +2 2 2 2 21 2 3 ........ ......i nS e e e e e the curve of best fit is

that for which e's are as small as possible i.e. S, the sum of the square of the errors is a minimum this

is known as the principle of least square. The theoretical values for x1, x2, ..., xn may be λ λ λ1 2, , ...... .n

y y y

5.3 FITTING OF STRAIGHT LINE

Let a straight line = +y a bx ...(1)

which is fitted to the given data points 1 1( , )x y , 2 2( , )x y , 3 3( , )x y ,……, ( , )n nx y .

Let λ1y be the theoretical value for x1 then λ= −

11 1e y y

⇒ = − +1 1 1( )e y a bx

⇒ = − −2 21 1 1( )e y a bx

Now we have = + + + +2 2 2 21 2 3 ............ nS e e e e

== ∑ 2

1

n

ii

S e

== − −∑ 2

1

( )n

i ii

S y a bx

By the principle of least squares, the value of S is minimum therefore,

∂ =∂

0S

a...(2)

384 A TEXTBOOK OF ENGINEERING MATHEMATICS–III

and∂ =∂

0S

b...(3)

On solving equations (2) and (3), and dropping the suffix, we have

= +∑ ∑y na b x ...(4)

= +∑ ∑ ∑ 2xy a x b x ...(5)

The equation (3) and (4) are known as normal equations.

On solving equations (3) and (4), we get the value of a and .b Putting the value of a and b inequation (1), we get the equation of the line of best fit.

5.4 FITTING OF PARABOLA

Let a parabola = + + 2y a bx cx ...(1)

which is fitted to a given data 1 1( , )x y , 2 2( , )x y , 3 3( , )x y ,……, ( , )n nx y .

Let λy be the theoretical value for 1x then λ= −1 1e y y

⇒ = − + + 21 1 1 1( )e y a bx cx

⇒ = − − −2 2 21 1 1 1( )e y a bx cx

Now we have=

= ∑ 2

1

n

ii

S e

== − − −∑ 2 2

1

( )n

i i ii

S y a bx cx

By the principle of least squares, the value of S is minimum, therefore

∂ =∂

0S

a,∂ =∂

0S

b and

∂ =∂

0S

c...(2)

Solving equation (2) and dropping suffix, we have

= + +∑ ∑ ∑ 2y na b x c x ...(3)

= + +∑ ∑ ∑ ∑2 3xy a x b x c x ...(4)

= + +∑ ∑ ∑ ∑2 2 3 4x y a x b x c x ...(5)

The equation (3), (4) and (5) are known as normal equations.On solving equations (3), (4) and (5), we get the values of a,b and c. Putting the values of a, b

and c in equation (1), we get the equation of the parabola of best fit.

5.5 CHANGE OF SCALE

When the magnitude of the variable in the given data is large number then calculation becomes verymuch tedious then problem is further simplified by taking suitable scale when the value of x are givenat equally spaced intervals.


Let h be the width of the interval at which the values of x are given and let the origin of x and

y be taken at the point 0 0,x y respectively, then putting

−

= 0( )x xu

h and = − 0v y y

If m is odd then, u = )( interval

term) (middle

h

x −

But if m is even then, u = (interval)

21

term) middle two of (middle−x

Example 1: Find the best-fit values of a and b so that = +y a bx fits the data given in the table.

: 0 1 2 3 4

: 1 1.8 3.3 4.5 6.3

x

y

Sol. Let the straight line is = +y a bx ...(1)

= = = =∑ ∑ ∑ ∑

2

2

0 1 0 0

1 1.8 1.8 1

2 3.3 6.6 4

3 4.5 13.5 9

4 6.3 25.2 16

10 16.9 47.1 30

x y xy x

x y xy x

Normal equations are, = +∑ ∑y na b x ...(2)

= +∑ ∑ ∑ 2xy a x b x ...(3)

Here = 5,n =∑ 10,x =∑ 16.9,y =∑ 47.1,xy =∑ 2 30x

Putting these values in normal equations, we get

= +16.9 5 10a b

= +47.1 10 30a bOn solving these two equations, we get

= 0.72a , = 1.33.b

So required line = +0.72 1.33 .y x Ans.


Example 2: Fit a straight line to the given data regarding x as the independent variable.

1 2 3 4 5 6

1200 900 600 200 110 50

x

y

Sol. Let the straight line obtained from the given data by = +y a bx ...(1)

Then the normal equations are = +∑ ∑y na b x ...(2)

= +∑ ∑ ∑ 2xy a x b x ...(3)

2

2

1 1200 1 1200

2 900 4 1800

3 600 9 1800

4 200 16 800

5 110 25 550

6 50 36 300

21 91 64503060

x y x xy

x x xyy= = ==∑ ∑ ∑∑

Putting all values in the equations (2) and (3), we get

= +3060 6 21a b

= +6450 21 91a bSolving these equations, we get

= 1361.97a and = −243.42b

Hence the fitted equation is = −1361.97 243.42 .y x Ans.Example 3: Fit a straight line to the following data:

71 68 73 69 67 65 66 67

69 72 70 70 68 67 68 64

x

y

Sol. Here we from the following table:

x y xy x2

71 69 4899 5041

68 72 4896 4624

73 70 5110 5329

69 70 4830 4761

67 68 4556 4489

65 67 4355 4225

66 68 4488 4356

67 64 4288 4489

�x = 546 �y = 548 �xy = 37422 �x2 = 37314


Let the equation of straight line to be fitted bey = a + bx ...(1)

And the normal equations are

yΣ an b x= + Σ ...(2)

xyΣ 2a x b x= Σ + Σ ...(3)⇒ 8a + 546b = 548

546a + 37314b = 37422Solving these equations, we get

a = 39.5454, b = 0.4242

Hence from (1)

y = 39.5454 + 0.4242x. Ans.

Example 4: Find the least square polynomial approximation of degree two to the data.

− −0 1 2 3 4

4 1 4 11 20

x

y

Also compute the least error.

Sol. Let the equation of the polynomial be = + + 2y a bx cx ...(1)

2 2 3 4

2 2 3 4

0 4 0 0 0 0 0

1 1 1 1 1 1 1

2 4 8 4 16 8 16

3 11 33 9 99 27 81

4 20 80 16 320 64 256

10 30 120 30 434 100 354

x y xy x x y x x

x y xy x x y x x

−− − −

= = = = = = =∑ ∑ ∑ ∑ ∑ ∑ ∑The normal equations are,

= + +∑ ∑ ∑ 2y na b x c x ...(2)

= + +∑ ∑ ∑ ∑2 3xy a x b x c x ...(3)

= + +∑ ∑ ∑ ∑2 2 3 4x y a x b x c x ...(4)

Here n = 5, =∑ 10,x =∑ 30,y =∑ 120,xy =∑ 2 30,x =∑ 2 434,x y =∑ 3 100,x

=∑ 4 354.x

Putting all these values in (2), (3) and (4), we get

= + +30 5 10 30a b c ...(5)

= + +120 10 30 100a b c ...(6)

= + +434 30 100 354a b c ...(7)


On solving these equations, we get = −4a , = 2b , = 1.c Therefore required polynomial is

= − + + 24 2y x x , errors = 0. Ans.

Example 5: Fit a second degree curve of regression of y on x to the following data:

1 2 3 4

6 11 18 27

x

y

Sol. We form the following table:

x y x2 x3 x4 xy x2y

1 6 1 1 1 6 6

2 11 4 8 16 22 44

3 18 9 27 81 54 162

4 27 16 64 256 108 432

�x = 10 �y = 62 �x2 = 30 �x3 = 100 �x4 = 354 �xy = 190 �x2y = 644

The equation of second degree parabola is given byy = a + bx + cx2 ...(1)

And the normal equations are

yΣ 2an b x c x= + Σ + Σ ...(2)

xyΣ 2 3a x b x c x= Σ + Σ + Σ ...(3)

2x yΣ 2 3 4a x b x c x= Σ + Σ + Σ ...(4)

�

4 10 30 62

10 30 100 190 3, 2, 1

30 100 354 644

a b c

a b c a b c

a b c

+ + = + + = ⇒ = = =+ + =

Hence y = 3 + 2x + x2. Ans.

Example 6: By the method of least squares, find the straight line that best fits the following data:

1 2 3 4 5 6 7

14 27 40 55 68 77 85

x

y

Sol. The equation of line isy = a + bx ...(1)

The normal equations are y an b xΣ = + Σ ...(2)

and xyΣ 2a x b x= Σ + Σ ...(3)


Now we from the following table:

x y xy x2

1 14 14 1

2 27 54 4

3 40 120 9

4 55 220 16

5 68 340 25

6 77 462 36

7 85 595 49

�x = 28 �y = 356 �xy = 1805 �x2 = 140

� From equations (2) and (3), we get

7 28a b+ = 356

28 140a b+ = 1805

On solving these equations, we get

a = – 3.5714

b = 13.6071

� y = – 3.5714 + 13.6071x. Ans.

Example 7: Find the least squares fit of the form 20 1y a a x= + to the following data

1 0 1 2

2 5 3 0

x

y

−

Sol. We have y 20 1a a x= +

By principle of least squares

s { }2

20 1(i i

i

y a a x= − +∑

s

a

∂∂

{ }2 20 12 ( ( ) 0i i i

i

y a a x x= − + − =∑

⇒ 2x yΣ 2 40 1a x a x= Σ + Σ (Drop suffix) ...(1)


and0

s

a

∂∂ { }2

0 12 ( ) 0i iy a a x= − − + =∑

⇒ yΣ 20 1a n a x= + Σ (Drop suffix) ...(2)

Now, we form the following table:

x y x2 x2y x4

–1 2 1 2 1

0 5 0 0 0

1 3 1 3 1

2 0 4 0 16

�x = 2 �y = 10 �x2 = 6 �x2y = 5 �x4 = 18

From equations (1) and (2), we get

0 16 18a a+ 5= ...(3)

and 0 14 6a a+ 10= ...(4)

Solving the equations (3) and (4), we get

a1 = – 1.111, a0 = 4.166

� The equation is given by

y = 4.166 – 1.111x2. Ans.

Example 8: Fit a second-degree parabola to the following data taking x as the independentvariable.

1 2 3 4 5 6 7 8 9

2 6 7 8 10 11 11 10 9

x

y

Sol. The equation of second-degree parabola is given by = + + 2y a bx cx and the normal equations

are:

2

2 3

2 2 3 4

y na b x c x

xy a x b x c x

x y a x b x c x

= + += + +

= + +

∑ ∑ ∑∑ ∑ ∑ ∑

∑ ∑ ∑ ∑...(1)

Here = 9n . The various sums are appearing in the table as follows:


= = = = = = =∑ ∑ ∑ ∑ ∑ ∑ ∑

2 2 3 4

2 2 3 4

1 2 2 1 2 1 1

2 6 12 4 24 8 16

3 7 21 9 63 27 81

4 8 32 16 128 64 256

5 10 50 25 250 125 625

6 11 66 36 396 216 1296

7 11 77 49 539 343 2401

8 10 80 64 640 512 4096

9 9 81 81 729 729 6561

45 74 421 284 2771 2025 15333

x y xy x x y x x

x y xy x x y x x

Putting these values of ∑ ,x ∑ ,y ∑ 2,x ∑ ,xy ∑ 2 ,x y ∑ 3x and ∑ 4x in equation (1) and

solving the equations for a, b and c, we get

= −0.923a ; = 3.520b ; = −0.267c .Hence the fitted equation is

= − + − 20.923 3.53 0.267 .y x x Ans.

Example 9: Show that the line of fit to the following data is given by = +0.7 11.28.y x

0 5 10 15 20 25

12 15 17 22 24 30

x

y

Sol. Here n = 6 (even)

Let = = =0 012.5, 5, 20 (say)x h y

Then, −= 12.5

2.5

xu and = − 20,v y we get

− −− −− −

= = = =∑ ∑ ∑ ∑

2

2

0 12 5 8 40 25

5 15 3 5 15 9

10 17 1 3 3 1

15 22 1 2 2 1

20 24 3 4 12 9

25 30 5 10 50 25

0 0 122 70

x y u v uv u

u v uv u

The normal equations are,

0 6 0 0a b a= + ⋅ ⇒ =

122 0 70a b= ⋅ + ⇒ = 1.743b


Thus line of fit is = 1.743 .v u

or12.50

20 (1.743) 0.69 8.7152.5

xy x

− − = = −

or = +0.7 11.285y x . Ans.

Example 10: Fit a second-degree parabola to the following data by least squares method.

1929 1930 1931 1932 1933 1934 1935 1936 1937

352 356 357 358 360 361 361 360 359

x

y

Sol. Taking =0 1933,x =0 357y then −

= 0( )x xu

hHere h = 1

Taking = − 0u x x and = − 0 ,v y y therefore, = −1933u x and = − 357v y

2 2 3 41933 357

1929 4 352 5 20 16 80 64 256

1930 3 356 1 3 9 9 27 81

1931 2 357 0 0 4 0 8 16

1932 1 358 1 1 1 1 1 1

1933 0 360 3 0 0 0 0 0

1934 1 361 4 4 1 4 1 1

1935 2 361 4 8 4 16 8 16

1936 3 360 3 9 9 27 27 81

1937 4 359 2 8 16 32 64 256

0 11Total

x u x y v y uv u u v u u

u v

= − = −− − − −− − − −− −− − −

= =∑ ∑ 2 2 3 451 60 9 0 708uv u u v u u= = = − = =∑ ∑ ∑ ∑ ∑

Then the equation = + + 2y a bx cx is transformed to = + + 2v A Bu Cu ...(1)

Normal equations are:

= + +∑ ∑ ∑ 29v A B u C u ⇒ = +11 9 60A C

= + +∑ ∑ ∑ ∑2 3uv A u B u C u ⇒ = 17 / 20B

= + +∑ ∑ ∑ ∑2 2 3 4u v A u B u C u ⇒ − = +9 60 708A C

On solving these equations, we get 694

3231

A = = , 17

0.8520

B = = and 247

0.27924

C = − = −

∴ 23 0.85 0.27v u u= + −

⇒ 2357 3 0.85( 1933) 0.27( 1933)y x x− = + − − −

⇒ 21010135.08 1044.69 0.27 .y x x= − + − Ans.


Example 11: Fit second degree parabola to the following:

0 1 2 3 4

1 1.8 1.3 2.5 6.3

x

y

Sol. Here = 5n (odd) therefore =0 2x , h = 1, y0 = 0 (say)

Now let = − 2u x , =v y and the curve of fit be = + + 2 .v a bu cu

− − −− − −

2 2 3 4

0 1 2 1 2 4 4 8 16

1 1.8 1 1.8 1.8 1 1.8 1 1

2 1.3 0 1.3 0 0 0 0 0

3 2.5 1 2.5 2.5 1 2.5 1 1

4 6.3 2 6.3 12.6 4 25.2 8 16

Total 0 12.9 11.3 10 33.5 0 34

x y u v uv u u v u u

Hence the normal equations are,

= + +∑ ∑ ∑ 25v a b u c u

= + +∑ ∑ ∑ ∑2 3uv a u b u c u

= + +∑ ∑ ∑ ∑2 2 3 4u v a u b u c u

On putting the values of ∑u , ∑v etc. from the table in these, we get

= +12.9 5 10 ,a c =11.3 10 ,b = +33.5 10 34 .a cOn solving these equations, we get

a = 1.48, b = 1.13 and c = 0.55

Therefore the required equation is = + + 21.48 1.13 0.55 .v u u

Again substituting = − 2u x and v = y, we get

= + − + − 21.48 1.13( 2) 0.55( 2)y x x

or = − + 21.42 1.07 0.55 .y x x Ans.

5.6 FITTING OF AN EXPONENTIAL CURVE

Suppose an exponential curve of the form

= bxy ae

Taking logarithm on both the sides, we get

= +10 10 10log log logy a bx e

i.e., = +Y A Bx ...(1)

where = = =10 10 10log , log and log .Y y A a B b e


The normal equations for (1) are,

= +∑ ∑Y nA B x

= +∑ ∑ ∑ 2xY A x B x

On solving the above two equations, we get A and B

then a = antilog A, =10log

Bb

e

5.7 FITTING OF THE CURVE y = ax + bx2

Error of estimate for ith point ( , )i ix y is

= − − 2( )i i i ie y ax bx

We have, =

= ∑ 2

1

n

ii

S e

== − −∑ 2 2

1

( )n

i i ii

y ax bx

By the principle of least square, the value of S is minimum

∴∂ =∂

0S

a and

∂ =∂

0S

b

Now∂ =∂

0S

a

⇒ =

− − − =∑ 2

1

2( )( ) 0n

i i i ii

y ax bx x

or= = =

= +∑ ∑ ∑2 3

1 1 1

n n n

i i i ii i i

x y a x b x ...(1)

and∂ =∂

0S

b

⇒=

− − − =∑ 2 2

1

2( )( ) 0n

i i i ii

y ax bx x

or= = =

= +∑ ∑ ∑2 3 4

1 1 1

n n n

i i i ii i i

x y a x b x ...(2)

Dropping the suffix i from (1) and (2), then the normal equations are,

= +∑ ∑ ∑2 3xy a x b x

= +∑ ∑ ∑2 3 4x y a x b x


5.8 FITTING OF THE CURVE xb

axy +=


= − −( )i i ii

be y ax

x

We have,=

= ∑ 2

1

n

ii

S e

== − −∑ 2

1

( )n

i iii

by ax

x


∴ ∂ =∂

0S

a and

∂ =∂

0S

b

Now∂ =∂

0S

a

⇒ =

− − − =∑1

2( )( ) 0n

i i iii

by ax x

x

or = =

= +∑ ∑ 2

1 1

n n

i i ii i

x y a x nb ...(1)

and∂ =∂

0S

b

⇒ =

− − − =∑1

12( )( ) 0

n

i ii ii

by ax

x x

or= =

= +∑ ∑ 21 1

1n ni

ii i i

yna b

x x ...(2)


= +∑ ∑ 2xy nb a x

= +∑ ∑ 2

1yna b

x xwhere n is the number of pair of values of x and y.


5.9 FITTING OF THE CURVE 01

cy = + c x

x


= − −01( )i i i

i

ce y c x

x

We have, =

= ∑ 2

1

n

ii

S e

== − −∑ 20

11

( )n

i iii

cy c x

x


∴∂ =∂ 0

0S

c and

∂ =∂ 1

0S

c

Now∂ =∂ 0

0S

c

⇒ =

− − − =∑ 01

1

12( )( ) 0

n

i ii ii

cy c x

x x

or= = =

= +∑ ∑ ∑0 121 1 1

1 1n n ni

ii i ii i

yc c

x x x ...(1)

and∂ =∂ 1

0S

c

⇒ =

− − − =∑ 01

1

2( )( ) 0n

i i iii

cy c x x

x

or = = =

= +∑ ∑ ∑0 11 1 1

1n n n

i i ii i ii

y x c c xx ...(2)


= +∑ ∑ ∑0 12

1 1yc c

x x x

0 11

.y x c c xx

= +∑ ∑ ∑

Example 12: Find the curve of best fit of the type = bxy ae to the following data by the methodof least squares:

1 5 7 9 12

10 15 12 15 21

x

y

Sol. The curve to be fitted is = bxy ae or = +Y A Bx , where = 10log ,Y y = 10log ,A a and

= 10log .B b e


Therefore the normal equations are:

= +∑ ∑5Y A B x

= +∑ ∑ ∑ 2xY A x B x

=

= = = =∑ ∑ ∑ ∑

210

2

log

1 10 1.0000 1 1

5 15 1.1761 25 5.8805

7 12 1.0792 49 7.5544

9 15 1.1761 81 10.5849

12 21 1.3222 144 15.8664

34 5.7536 300 40.8862

x y Y y x xY

x Y x xY

Substituting the values of ∑ ,x etc. and calculated by means of above table in the normal

equations, we get

= +5.7536 5 34A B

and = +40.8862 34 300A BOn solving these equations, we obtain,

= 0.9766A ; = 0.02561B

Therefore = =10antilog 9.4754a A ; = =10

0.059log

Bb

e

Hence the required curve is = 0.0599.4754 xy e . Ans.Example 13: For the data given below, find the equation to the best fitting exponential curve of

the form = .bxy ae

1 2 3 4 5 6

1.6 4.5 13.8 40.2 125 300

x

y

Sol. = bxy ae

On taking log both the sides, = +log log logy a bx e which is of the form = + ,Y A Bx where

= log ,Y y = logA a and = log .B b e

=

= = = =∑ ∑ ∑ ∑

2

2

log

1 1.6 0.2041 1 0.2041

2 4.5 0.6532 4 1.3064

3 13.8 1.1399 9 3.4197

4 40.2 1.6042 16 6.4168

5 125 2.0969 25 10.4845

6 300 2.4771 36 14.8626

21 8.1754 91 36.6941

x y Y y x xY

x Y x xY


Normal equations are: = +∑ ∑6Y A B x

= +∑ ∑ ∑ 2xY A x B x

Therefore from these equations, we have

= +8.1754 6 21A B = +36.6941 21 91A B

⇒ = − =0.2534, 0.4617A B

Therefore, = = − = =antilog antilog( 0.2534) antilog(1.7466) 0.5580.a A

and = = =0.46171.0631

log 0.4343

Bb

e

Hence required equation is = 1.06310.5580 xy e . Ans.Example 14: Given the following experimental values:

0 1 2 3

2 4 10 15

x

y

Fit by the method of least squares a parabola of the type = + 2.y a bx

Sol. Error of estimate for ith point ( , )i ix y is = − − 2( )i i ie y a bx

By the principle of least squares, the values of a and b are such that

2 2 2

1 1

( )n n

i i ii i

S e y a bx= =

= = − −∑ ∑ is minimum.

Therefore normal equations are given by

∂ = ⇒ = +∂ ∑ ∑ 20S

y na b xa

...(1)

and∂ = ⇒ = +∂ ∑ ∑ ∑2 2 40S

x y a x b xb

...(2)

= = = =∑ ∑ ∑ ∑

2 2 4

2 2 4

0 2 0 0 0

1 4 1 4 1

2 10 4 40 16

3 15 9 135 81

31 14 179 98Total

x y x x y x

y x x y x

Here n = 4.


From (1) and (2), = +31 4 14a b and = +179 14 98a b

Solving for a and b, we get

= 2.71a and = 1.44b

Hence the required curve is = + 22.71 1.44 .y x Ans.

Example 15: By the method of least square, find the curve = + 2y ax bx that best fits the following

data:

1 2 3 4 5

1.8 5.1 8.9 14.1 19.8

x

y

Sol. Error of estimate for ith point ( , )i ix y is = − − 2( )i i i ie y ax bx

By the principle of least squares, the values of a and b are such that

2 2 2

1 1

( )n n

i i i ii i

S e y ax bx= =

= = − −∑ ∑ is minimum.

Therefore normal equations are given by

2 3

1 1 1

0n n n

i i i ii i i

Sx y a x b x

a = = =

∂ = ⇒ = +∂ ∑ ∑ ∑ and

2 3 4

1 1 1

0n n n

i i i ii i i

Sx y a x b x

b = = =

∂ = ⇒ = +∂ ∑ ∑ ∑

Dropping the suffix i, normal equations are

= +∑ ∑ ∑2 3xy a x b x ...(1)

and = +∑ ∑ ∑2 3 4x y a x b x ...(2)

= = = = =∑ ∑ ∑ ∑ ∑

2 3 4 2

2 3 4 2

1 1.8 1 1 1 1.8 1.8

2 5.1 4 8 16 10.2 20.4

3 8.9 9 27 81 26.7 80.1

4 14.1 16 64 256 56.4 225.6

5 19.8 25 125 625 99 495

55 225 979 194.1 822.9Total

x y x x x xy x y

x x x xy x y

Substituting these values in equations (1) and (2), we get

= +194.1 55 225a b and = +822.9 225 979a b

⇒ = ≈83.851.52

55a

and = ≈317.40.49

664b

Hence required parabolic curve is = + 21.52 0.49y x x . Ans.


Example 16: Fit an exponential curve of the form = xy ab to the following data:

1 2 3 4 5 6 7 8

1.0 1.2 1.8 2.5 3.6 4.7 6.6 9.1

x

y

Sol. = xy ab takes the form = +Y A Bx , where = logY y ; = logA a and = log .B b

Hence the normal equations are given by

= +∑ ∑Y nA B x and = +∑ ∑ ∑ 2xY A x x .

=

= = = = =∑ ∑ ∑ ∑ ∑

2

2

log

1 1.0 0.0000 0.000 1

2 1.2 0.0792 0.1584 4

3 1.8 0.2553 0.7659 9

4 2.5 0.3979 1.5916 16

5 3.6 0.5563 2.7815 25

6 4.7 0.6721 4.0326 36

7 6.6 0.8195 5.7365 49

8 9.1 0.9590 7.6720 64

36 30.5 3.7393 22.7385 204

x y Y y xY x

x y Y xY x

Putting the values in the normal equations, we obtain

= +3.7393 8 36A B and = +22.7385 36 204A B

⇒ 0.1407B = and 0.1656A =

⇒ = =antilog 1.38b B and = =antilog 0.68.a A

Thus the required curve of best fit is (0.68)(1.38) .xy = Ans.

Example 17: Fit a curve = xy ab to the following data:

2 3 4 5 6

144 172.8 207.4 248.8 298.5

x

y

Sol. Given equation = xy ab reduces to = +Y A Bx where = log ,Y y = logA a and = log .B b

The normal equations are,

= +∑ ∑log log logy n a b x

= +∑ ∑ ∑ 2log log logx y a x b x

The calculations of ∑ ,x ∑ log ,y ∑ 2x and ∑ logx y are substitute in the following tabular

form.


2 log log

2 144 4 2.1584 4.3168

3 172.8 9 2.2375 6.7125

4 207.4 16 2.3168 9.2672

5 248.8 25 2.3959 11.9795

6 298.5 36 2.4749 14.8494

20 90 11.5835 47.1254

x y x y x y

Putting these values in the normal equations, we have= +11.5835 5log 20loga b

47.1254 20 log 90log .a b= +Solving these equations and taking antilog, we have a = 100, b = 1.2 approximate. Therefore

equation of the curve is y = 100(1.2)x. Ans.

Example 18: Derive the least square equations for fitting a curve of the type = +2 ( / )y ax b x to

a set of n points. Hence fit a curve of this type to the data.

−1 2 3 4

1.51 0.99 3.88 7.66

x

y

Sol. Let the n points are given by 1 1( , ),x y 2 2( , ),x y 3 3( , ),x y ……, ( , ).n nx y The error of estimate

for the ith point ( , )i ix y is = − −2[ ( / )].i i i ie y ax b x

By the principle of least square, the values of a and b are such so that the sum of the square oferror S, viz.,

=

= = − −

∑ ∑

22 2

1

n

i i iii

bS e y ax

xis minimum.

Therefore the normal equations are given by∂ ∂= =∂ ∂

0, 0S S

a b

or= = =

= +∑ ∑ ∑2 4

1 1 1

n n n

i i i ii i i

y x a x b x and = = =

= +∑ ∑ ∑ 21 1 1

1n n ni

iii i i i

ya x b

x x

These are the required least square equations.

2 4 22

1 1

1 1.51 1 1 1 1 1.51 1.51

2 0.99 4 16 0.5 0.25 3.96 0.495

3 3.88 9 81 0.3333 0.1111 34.92 1.2933

4 7.66 16 256 0.25 0.0625 122.56 1.0943

10 354 1.4236 159.93 1.1933

yx y x x yx

x xx− − −


Putting the values in the above least square equations, we get

= +159.93 354 10a b and 1.1933 10 1.4236 .a b= +Solving these, we get a = 0.509 and b = –2.04.

Therefore, the equation of the curve fitted to the above data is = −2 2.040.509y x

x. Ans.

Example 19: Fit the curve γ =pv k to the following data:

2(kg/cm ) 0.5 1 1.5 2 2.5 3

(litres) 1620 1000 750 620 520 460

p

v

Sol. Given γ =pv k

γγ − γ

= =

1/1/ 1/k

v k pp

Taking log both the sides, we get

= −γ γ1 1

log log logv k p

which is of the form = +Y A BX

where = log ,Y v = log ,X p =γ1

logA k and = −γ1

.B

− −

=∑

2

0.5 1620 0.30103 3.20952 0.96616 0.09062

1 1000 0 3 0 0

1.5 750 0.17609 2.87506 0.50627 0.03101

2 620 0.30103 2.79239 0.84059 0.09062

2.5 520 0.39794 2.716 1.08080 0.15836

3 460 0.47712 2.66276 1.27046 0.22764

1.05115Total

p v X Y XY X

X Y = = =∑ ∑ ∑ 217.25573 2.73196 0.59825XY X

Here n = 6Normal equations are,

= +17.25573 6 1.05115A B = +2.73196 1.05115 0.59825A B

On solving these, we get = 2.99911A and = −0.70298B


∴ γ = − = =1 11.42252

0.70298B

Again = γ =log 4.26629k A

∴ antilog(4.26629) 18462.48k = =

Hence, the required curve is =1.42252 18462.48.pv Ans.

Example 20: For the data given below, find the equation to the best fitting exponential curve of

the form = .bxy ae

1 2 3 4 5 6

1.6 4.5 13.8 40.2 125 300

x

y

Sol. Given y = aebx, taking log we get = + 10log log logy a bx e which is of the = + ,Y A Bx

where = logY y , = logA a and = 10log .B e

Put the values in the following tabular form, also transfer the origin of x series to 3, so that

= − 3u x .

2log 3

1 1.6 0.204 2 0.408 4

2 4.5 0.653 1 0.653 1

3 13.8 1.140 0 0 0

4 40.2 1.604 1 1.604 1

5 125.0 2.094 2 4.194 4

6 300 2.477 3 7.431 9

Total 8.175 3 12.168 19

x y y Y x u uY u= − =− −− −

In case . ,Y A B u= + then normal equations are given by

Y nA B u= +∑ ∑ ⇒ 8.175 6 3A B= + ...(1)

2uY A u B u= +∑ ∑ ∑ ⇒ 12.168 3 19A B= + ...(2)

Solving (1) and (2), we get

1.13A = and 0.46B =Thus equation is = +1.13 0.46 ,Y u i.e. = + −1.13 0.46( 3)Y x

or = −0.46 0.25Y x

Which gives log 0.25a = − i.e. antilog( 0.25) antilog (1.75) 0.557− = =

10

.461.06

log 0.4343

Bb

e= = =

Hence, the required equation of the curve is 1.06(0.557) .xy e= Ans.


PROBLEM SET 5.1

1. Fit a straight line to the given data regarding x as the independent variable:

1 2 3 4 6 8

2.4 3.1 3.5 4.2 5.0 6.0

x

y = +[ 2.0253 0.502 ]y xAns.

2. Fit a straight line = +y a bx to the following data by the method of least square:

0 1 3 6 8

1 3 2 5 4

x

y +[ 1.6 0.38 ]xAns.

3. Find the least square approximation of the form = + 2y a bx for the following data:

0 0.1 0.2 0.3 0.4 0.5

1 1.01 0.99 0.85 0.81 0.75

x

y = − 2[ 1.0032 1.1081 ]y xAns.

4. Fit a second degree parabola to the following data:

0.0 1.0 2.0

1.0 6.0 17.0

x

y = + + 2[ 1 2 3 ]y x xAns.

5. Fit a second degree parabola to the following data:

1.0 1.5 2.0 2.5 3.0 3.5 4.0

1.1 1.3 1.6 2.0 2.7 3.4 4.1

x

y = − + 2[ 1.04 0.193 0.243 ]y x xAns.

6. Fit a second degree parabola to the following data by the least square method:

1 2 3 4 5

1090 1220 1390 1625 1915

x

y = + +2[ 27.5 40.5 1024]y x xAns.

7. Fit a parabola = + + 2y a bx cx to the following data:

2 4 6 8 10

3.07 12.85 31.47 57.38 91.29

x

y = − + 2[ 0.34 0.78 0.99 ]y x xAns.

8. Determine the constants a and b by the method of least squares such that = bxy ae fits the

following data:

2 4 6 8 10

4.077 11.084 30.128 81.897 222.62

x

y = 0.50001[ 1.49989 ]xy eAns.

9. Fit a least square geometric curve = by ax to the following data:

1 2 3 4 5

0.5 2 4.5 8 12.5

x

y = 1.9977[ 0.5012 ]y xAns.


10. A person runs the same race track for five consecutive days and is timed as follows:

( ) 1 2 3 4 5

( ) 15.3 15.1 15 14.5 14

Day x

Time y

Make a least square fit to the above data using a function + +2

b ca

x x.

2

6.7512 4.4738 13.0065y

x x

= + + Ans.

11. Use the method of least squares to fit the curve = +01

cy c x

x to the following table of

values:

0.1 0.2 0.4 0.5 1 2

21 11 7 6 5 6

x

y1.97327

3.28182y xx

= + Ans.

12. Using the method of least square to fit a parabola = + + 2y a bx cx in the following data:

−( , ) : ( 1,2),(0,0),(0,1),(1,2)x y21 3

2 2

y x = + Ans.

13. The pressure of the gas corresponding to various volumes V is measured, given by thefollowing data:

3

2

(cm ) 50 60 70 90 100

(kgcm ) 64.7 51.3 40.5 25.9 78

V

p −

Fit the data to the equation γ =pV c . [Ans. =0.28997 167.78765pV ]

14. Employ the method of least squares to fit a parabola y = a + bx + cx2 in the followingdata: (x, y): (–1, 2), (0, 0), (0, 1), (1, 2) [Ans. y = 0.5 + 1.5x2 ]

15. Fit a second degree parabola in the following data: [U.T.U. 2008]

0.0 1.0 2.0 3.0 4.0

1.0 4.0 10.0 17.0 30.0

x

y [Ans. y = 1 + 2x + 3x2]

16. Fit at least square quadratic curve to the following data:

1 2 3 4

1.7 1.8 2.3 3.2

x

y , estimate y(2.4)

[Ans. y = 2 – 0.5x + 0.2x2 and y(2.4) = 1.952]


17. Fit an exponential curve by least squares

1 2 5 10 20 30 40 50

98.2 91.7 81.3 64.0 36.4 32.6 17.1 11.3

x

y

Estimate y when x = 25. [Ans. y = 100(0.96)x, y(25) = 33.9]

18. Fit the curve b

y ax

= + to the following data

1 2 3 4

3 1.5 6 7.5

x

y

Estimate y when x = 2.25.1.7

1.3 , (2.25) 4.625y yx

= + = Ans.

5.10 POLYNOMIAL

If 1 20 1 2 1( ) ..............n n n

n nf x a x a x a x a x a− −−= + + + + +

Then the above relation is called the polynominal of nth order in x.

5.10.1 Degree of Polynomial

The highest power of x occurring in the given polynomial is called degree of polynomial.

The constant = 0c cx is called a polynomial of degree zero.

The polynomial = + ≠( ) , 0f x ax b a is of degree one and is called a linear polynomial.

The polynomial = + + ≠2( ) , 0f x ax bx c a is of degree two and is called a quadratic polynomial.

The polynomial = + + + ≠3 2( ) , 0f x ax bx cx d a is of degree three and is called a cubic polynomial.

The polynomial = + + + + ≠4 3 2( ) , 0f x ax bx cx dx e a is of degree four and is called biquadratic

polynomial.

5.11 DESCARTE’S RULE OF SIGNS

The number of positive roots of the equation f (x) = 0 cannot exceed the number of changes of sign inf (x), and the number of negative roots cannot exceed the number of changes of sign of f (–x).

Existence of imaginary roots: If an equation of the nth degree has at most p positive roots and

at most q negative roots, then it has at least − +( )n p q imaginary roots.

Example 1: Apply Descarte’s Rule of signs to discuss the nature of the roots of the equation

+ + − =4 215 7 11 0.x x x

Sol. The given equation is = + + − =4 2( ) 15 7 11 0f x x x x

Signs of ( )f x are + + + – [from + to –]


It has one change of sign and hence it must have one +ve root.

− = − + − + − − =4 2( ) ( ) 15( ) 7( ) 11 0f x x x x

or − = + − −4 2( ) 15 7 11f x x x x = 0

Signs of −( )f x are + + – – [from + to –]

It has only one change in sign and hence it must have one –ve root.Thus the equation has two real roots , one +ve and one –ve and hence the other two roots must be

imaginary.

Example 2: Show that − + − =7 4 33 2 1 0x x x has at least four imaginary roots.

Sol. The given equation is

( )f x = − + − =7 4 33 2 1 0x x x [from + to – or – to +]

Signs of ( )f x + – + –

∴ =( ) 0f x has 3 changes of sign. Therefore, it cannot have more than three positive roots.

Also −( )f x = − − − + − − =7 4 3( ) 3( ) 2( ) 1 0x x x

or − = − − − − =7 4 3( ) 3 2 1 0f x x x x

Signs of ( )f x are – – – –

∴ − =( ) 0f x has no changes in sign. Therefore the given equation has no negative root.

Thus the given equation cannot have more than 3 + 0 = 3 real roots. But the given equation has7 roots. Hence the given equation has 7 – 3 = 4 imaginary roots.

Example 3: Find the least positive number of imaginary roots of the equation

= − + + + =9 5 4 2( ) 1 0f x x x x x

Sol. The given equation is = − + + + =9 5 4 2( ) 1 0f x x x x x

Signs of ( )f x are + – + + +

∴ =( ) 0f x has two changes of signs, and hence 2 is the max. number of +ve root.

− = − − − + − + − + =9 5 4 2( ) ( ) ( ) ( ) ( ) 1 0f x x x x x

or − = − + + + + =9 5 4 2( ) 1 0f x x x x x

Signs of −( )f x are – + + + +

∴ − =( ) 0f x has only one changes of sign and hence it has only one –ve root or =( ) 0f x has only

one –ve root.Thus the max. number of real roots is 2 + 1=3 and the equation being of 9th degree and it will

have at least 9 – 3 = 6 imaginary roots.


5.12 CARDON’S METHOD

Case 1. When Cubic is of the Form 3x + qx +r = 0

The given cubic is + + =3 0x qx r ...(1)

Let x = u + v be a root of (1)

Cubing , = + + + = + +3 3 3 3 33 ( ) 3x u v uv u v u v uvx

or − − + =3 3 33 ( ) 0x uvx u v ...(2)

Comparing (1) and (2), 1

3uv q= − or = −3 3 31

27u v q and + = −3 3u v r

∴ 3 3andu v are the roots of the equation

− + + =2 3 3 3 3( ) 0t u v t u v or + − =2 310

27t rt q ...(3)

Solving (3), − ± +

=

32 4

272

qr r

t

Let− + + − − +

= =

3 32 2

3 3

4 427 27;

2 2

q qr r r r

u v

Now, the three cube roots of u3 are 2, ,u u uω ω and those of 3v are 2, , ,v v vω ω where 2andω ωare imaginary cube root of unity.

Since = +x u vTo find x, we have to add a cube root of u3 and a cube of v3 in such a manner that their product

is real.

∴The three values of x are 2 2, ,u v u v u v+ ω + ω ω + ω 3( 1)∴ω =Example 4: Use Cardon’s method to solve

− + =3 27 54 0.x x

Sol. Let = +x u v

Cubing, = + = + + + = + +3 3 3 3 3 3( ) 3 ( ) 3x u v u v uv u v u v uvx

⇒ − − + =3 3 33 ( ) 0x uvx u vComparing with the given equation,we get

= ⇒ =3 39 729uv u v (on cubing)

and + = −3 3 54u v


� u3 and v3 are the root of − + + =2 3 3 3 3( ) 0t u v t u v

⇒ + + = ⇒ + =2 254 729 0 ( 27) 0t t t

⇒ = − −27, 27t

Let = − = −3 327 and 27u v

So that, 2 23, 3 , 3 and 3, 3 , 3u v= − − ω − ω = − − ω − ωTo find x, we have to add a cube root of u3 and a cube root of v3 in such a way that their product

is real.

� 2 2( 3 3), ( 3 3 ), ( 3 3 )x = − − − ω − ω − ω − ω 3( 1)ω =�

2 26, 3( ), 3( ) 6,3,3= − − ω + ω − ω + ω = − 2( 1 0)+ ω + ω =�

Hence the required roots are –6, 3, 3. Ans.

Example 5: Solve by Cardon’s method 3 15 126 0.x x− − =Sol. Let = +x u v

Cubing, = + = + + + = + +3 3 3 3 3 3( ) 3 ( ) 3x u v u v uv u v u v uvx

⇒ − − + =3 3 33 ( ) 0x uvx u vComparing with the given equation, we get

= ⇒ =3 35 125uv u v (on cubing)

and + =3 3 126u v

∴ 3 3andu v are the roots of − + + =2 3 3 3 3( ) 0t u v t u v

⇒ − + =2 126 125 0t t ⇒ − − =( 1)( 125) 0t t ⇒ = 1,125t

Let = =3 31 and 125u v

So that, 2 21, , and 5, 5 , 5u v= ω ω = ω ωTo find x, we have to add a cube root of u3 and a cube root of v3 in such a way that their product

is real.

� 2 21 5, 5 , 5x = + ω + ω ω + ω ( )3 1ω =�

− + − − − − − + = + +

1 3 1 3 1 3 1 36, 5 , 5

2 2 2 2

i i i i

= − − − + = − ±6, 3 2 3 , 3 2 3 6, 3 2 3i i i

Hence, the required roots are − ±6, 3 2 3 .i Ans.


Example 6: Solve − − =3 6 9 0x x by Cardon’s method.

Sol. Let = +x u v

Cubing, = + = + + +3 3 3 3( ) 3 ( )x u v u v uv u v

= + +3 3 3u v uvx

⇒ − − + =3 3 33 ( ) 0x uvx u vComparing, we get

= ⇒ =3 32 8uv u v (on cubing)

and + =3 3 9u v

∴ 3 3,u v are the roots of − + + =2 3 3 3 3( ) 0t u v t u v

⇒ − + =2 9 8 0t t ⇒ − − = ⇒ =( 1)( 8) 0 1,8t t t

Let = =3 31and 8u v so that

2 21, , and 2,2 ,2u v= ω ω = ω ωTo find x, we have to add a cube root of u3 and a cube root of v3 in such that their product is real.

∴ 2 21 2, 2 , 2x = + ω + ω ω + ω 3( 1)ω =�

− + − − − − − + = + +

1 3 1 3 1 3 1 33, 2 , 2

2 2 2 2

i i i i

− − − + − ±= ⇒ =3 3 3 3 3 33, , 3, .

2 2 2

i i ix Ans.

Example 7: Solve the cubic equation − − =3 18 35 0x x

Sol. Let = +x u v

Cubing, = + = + + +3 3 3 3( ) 3 ( )x u v u v uv u v

= + +3 3 3u v uvx

⇒ − − + =3 3 33 ( ) 0x uvx u v

Comparing, we get = ⇒ =3 36 216uv u v (on cubing)

and + =3 3 35u v

∴ 3 3and u v are the roots of − + + =2 3 3 3 3( ) 0t u v t u v

⇒ − + = ⇒ − − = ⇒ =2 35 216 0 ( 8)( 27) 0 8,27t t t t t

Let = =3 38 & 27u v so that

2 22,2 ,2 and 3,3 ,3u v= ω ω = ω ω


To find x, we have to add a cube root of u3 and a cube root of v3 in such that their product is real.

∴ 2 22 3, 2 3 , 2 3x = + ω + ω ω + ω ( 3 1ω =� )

1 3 1 3 1 3 1 35, 2 3 , 2 3

2 2 2 2

i i i i − + − − − − − + = + +

5 3 5 35, ,

2 2

i i− − − +=

⇒ − ±= 5 3

5, .2

ix Ans.

Case 2. When the Cubic Equation is of the Form 3 20 1 2 3a + a + a + a = 0x x x

Then, first of all we remove the term containing x2.

This is done by diminishing the roots of the given equation by − 1

0

a

na or Sum of root/No. of

roots. Where n is 3. We proceed with the help of following examples:

Example 8: Solve by Cardon’s method + + + =3 26 9 4 0x x x

Sol. Equationing + + + =3 26 9 4 0x x x ...(1)

Equating with + + + =3 20 1 2 3 0,a x a x a x a we get ...(2)

= =0 11, 6a a

Then = − = − = −1

0

62

3 3

ah

a

= −2.hNow remove the x2 terms ,we have

−− − −

− −−

−

2 1 6 9 4

2 8 2

1 4 1 (2)

2 4

1 2 ( 3)

2

1 (0)

The transformed equation is − + =3 3 2 0y y ...(3)

where = + 2.y x

Let = +y u v be a solution of (3), then we get

3 3 33 ( ) 0y uvy u v− − + = ...(4)

Equating (3) and (4), we get = 1uv

⇒ = + = −3 3 3 31 and 2u v u v


Let us consider an equation whose roots are 3 3andu v

− + =2 0t st p s = sum of roots, p = product of roots

+ + = ⇒ + =2 22 1 0 ( 1) 0t t t

⇒ = − −1, 1t

Let = − = −3 31 and 1u v so that

2 21, , and 1, ,u v= − −ω −ω = − −ω −ω

� 2 2( 1 1), ( ),( )y u v= + = − − −ω − ω −ω − ω

= −2,1,1 2( 1 0)+ ω + ω =�

But = − 2x y

= − − − − = − − −2 2,1 2,1 2 4, 1, 1

∴ = − − −4, 1, 1x . Ans.

Example 9: Solve by Cardon’s method − − + =3 215 33 847 0.x x x

Sol. The given equation − − + =3 215 33 847 0x x x ...(1)

Equating with + + + =3 20 1 2 3 0,a x a x a x a we get

= = −0 11, 15a a

� = − = =1

0

155

3 3

ah

a

Now remove the x2 term, using synthetic division, we have

( )

( )

( )

− −− −

− −−

− −

5 1 15 33 847

5 50 415

1 10 83 432

5 25

1 5 108

5

1 0

Transformed equation is

− + =3 108 432 0y y ...(2)

where = − 5.y x

Let = +y u v

� − − + =3 3 33 ( ) 0y uvy u v ...(3) (on cubing)


Comparing (2) and (3),

= ⇒ =3 3 636 (6)uv u v and + = −3 3 432u v

∴ 3u and 3v are the roots of + + =2 6432 (6) 0t t

⇒ + + = ⇒ + = ⇒ = − −2 3 6 22. (6) (6) 0 ( 216) 0 216, 216t t t t

Let = − = −3 3216 and 216u v so that

2 26, 6 , 6 and 6, 6 , 6u v= − − ω − ω = − − ω − ω

∴ 2 2( 6 6),( 6 6 ),( 6 6 )y u v= + = − − − ω − ω − ω − ω

= −12, 6, 6

2

2

1 0

1

ω + ω + = ω + ω = −

�

But = + = − + + +5 12 5, 6 5, 6 5x y

� = −7,11,11.x Ans.

Example 10: Solve 3 23 12 16 0x x x− + + =

Sol. Here 10 1

0

31, 3 1

3 3

aa a h

a= = − ∴ = − = =

Now remove the term x2, using synthetic division

1 1 –3 12 16 1 –2 10

1 –2 10 (26) 1 –1

1 –1 (9) 1

1 (0)

The transformed equation is3( 1) 9( 1) 26x x− + − + = 0

Let y = x – 1, then y3 + 9y + 26 = 0 ...(1)

Let y = u + v ...(2)

⇒ 3 3 33 ( ) ( )y uv y u v− − + = 0

Comparing (1) and (2), we get

uv 3 33 and 26u v= − + = −

⇒ u3v3 = –27


Now let us have an equation whose roots are u3 and v3

2 ( 26) ( 27)t t− − + − 20 26 27 0t t= ⇒ + − =

⇒ ( 27)( 1)t t+ − = 0

⇒ t 3 31, 27 . ., 1 and 27i e u v= − = = −

So that u 2 31, , and 3, 3 , 3v= ω ω = − − ω − ω

y u v= + 2 2(1 3), ( 3 ), ( 3 )= − ω − ω ω − ω

1 3 1 3 1 3 1 32, 3 , 3

2 2 2 2

i i i i − + − − − − − + = − − −

{ } { }2, (1 2 3) , (1 2 3)i i= − + −

� x = y + 1

⇒ x ( 2 1), (1 2 3 1), (1 2 3 1)i i= − + + + − +

1, (2 2 3), (2 2 3)i i= − + −

1, 2(1 3).i= − ± Ans.

Example 11: Solve − + − =3 26 6 5 0x x x by Cardon’s method.

Sol. The given equation − + − =3 26 6 5 0x x x ...(1)

Comparing with + + + =3 20 1 2 3 0,a x a x a x a we get

= = −0 11, 6a a

� = − = =1

0

62

3 3

ah

a

Now remove the x2 term, using synthetic division

− −− −

− − −−

− −

2 1 6 6 5

2 8 4

1 4 2 ( 9)

2 4

1 2 ( 6)

2

1 (0)

Transformed equation is − − =3 6 9 0y y ...(2)


where = − 2y x

Let y = u + v be the solution of (2), then

= + + +3 3 3 3 ( )y u v uv u v

= + +3 3 3 3 ( )y u v uv y

− − + =3 3 33 ( ) ( ) 0y uv y u v ...(3)

Comparing (2) and (3), we have

= ⇒ = ⇒ =3 33 6 2 8uv uv u v

and + =3 3 9u v

Now let us have an equation whose roots are 3 3andu v

− + =2 0t st p ; s = sum of roots, p = product of the roots.

− + =2 9 8 0t t

= 1,8t i.e., = =3 31and 8u v so that

2 21, , and 2,2 ,2u v= ω ω = ω ω� = +y u v

2 2(1 2),( 2 ),( 2 )= + ω + ω ω + ω

1 3 1 3 1 3 1 33, 2 , 2

2 2 2 2

i i i i − + − − − − − + = + +

− − − += 3 3 3 33, , .

2 2

i i

But = + 2x y

− += 1 3 1 35, ,

2 2

i i

= ±15, (1 3).

2i Ans.

Example 12: Solve the equation − + − =3 23 3 1 0x x x by Cardon’s method.

Sol. Here, = = −0 11, 3a a

� = − = =1

0

31

3 3

ah

a


Now remove the 2x term, using synthetic division, we have

− −−

−−

−

1 1 3 3 1

1 2 1

1 2 1 (0)

1 1

1 1 (0)

1

1 (0)

Transformed equation is =3 0y ...(1)

where = −1.y x

From (1), = 0,0,0y

∴ = + =1 1,1,1x y

Hence the required roots are 1, 1, 1. Ans.

Example 13: Solve the cubic equation + − + =3 2 16 20 0.x x x

Sol. Given equation is + − + =3 2 16 20 0x x x

Here, = =0 11, 1a a

� = − = −1

0

1

3 3

ah

a


( )

− −

− −

−

− −−

−

1/ 3 1 1 16 20

1461/ 3 2 / 9

27146 686

1 2 / 39 27

1/ 3 1/ 9

1471 1/ 3

9

1/ 3

1 0

Transformed equation is − + =3 147 6860

9 27y y ...(1)

where = + 1.

3y x

Let = +y u v


Cubing, = + +3 3 3 3y u v uvy

⇒ − − + =3 3 33 ( ) 0y uvy u v ...(2)

Comparing with (1), we get

= ⇒ = + = −

63 3 3 349 7 686

and 9 3 27

uv u v u v

∴ 3 3andu v are the roots of + + =

62 686 7

027 3

t t

+ + =

3 62 7 7

2 03 3

t t

⇒ + = ⇒ = − −

23 3 37 7 7

0 ,3 3 3

t t

Let = − = −

3 33 37 7

and 3 3

u v

So that 2 27 7 7 7 7 7

, , and , ,3 3 3 3 3 3

u v−= − ω − ω = − − ω − ω

∴2 27 7 7 7 7 7 14 7 7

, , , ,3 3 3 3 3 3 3 3 3

y u v = + = − − − ω − ω − ω − ω = −

Now,−= − = − − − = −1 14 1 7 1 7 1

, , 5,2,23 3 3 3 3 3 3

x y

Hence required roots are –5, 2, 2. Ans.

Example 14: Solve the cubic equation + − + =3 26 12 32 0.x x x

Sol. Given equation is + − + =3 26 12 32 0x x x

Here, = =0 11, 6a a

� = − = − = −1

0

62

3 3

ah

aNow remove the x2 term, using synthetic division method, we have

− −− −

−− −

−−

2 1 6 12 32

2 8 40

1 4 20 (72)

2 4

1 2 ( 24)

2

1 (0)


Transformed equation is

− + =3 24 72 0y y ...(1)

where = + 2y x

Let = +y u v

Cubing, = + +3 3 3 3y u v uvy

⇒ − − + =3 3 33 ( ) 0y uvy u v ...(2)

Comparing (1) and (2), we get

= ⇒ = + = −3 3 3 38 512 and 72uv u v u v

∴ 3 3 and u v are the roots of + + =2 72 512 0t t

⇒ = − −8, 64t

Let = − = −3 38 and 64u v

So that 2 22, 2 , 2 & 4, 4 , 4u v= − − ω − ω = − − ω − ω

∴ 2 2( 2 4),( 2 4 ),( 2 4 )y u v= + = − − − ω − ω − ω − ω

= –6,3 + i −3,3 3i

Now, = − = − + −2 8,1 3,1 3x y i i

Hence the required roots are − ±8,1 3.i Ans.

Example 15: Solve x3 + 3ax2 +3(a2 – bc)x + a3 + b3 + c3 – 3abc = 0 by Cardon’s method.

Sol. Given equation is x3 + 3ax2 + 3(a2 – bc)x + a3 + b3 + c3 –3abc = 0Here a0 = 1, a1 = 3a

� h = 1

0

3

. 3.1

a aa

n a− = − = −


2 3 3 3

2 3

2 3 3

2

1 3 3( ) 3

2 3

1 2 3 ( )

1 3

1 (0)

a a a bc a b c abc

a a a abc

a a bc b c

a a

a bc

a

− − + + −

− − − +

− +

− −−

−

Download - Curve Fitting and Solution of Equation - …newagepublishers.com/samplechapter/001938.pdf · UNIT V Curve Fitting and Solution of Equation 5.1 CURVE FITTING In many branches of applied

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