Newton’s & Lagrange polynomials | Engineering Mathematics - Civil Engineering (CE) PDF Download

 Page 1


    
  
 
 
               
 
 
   
 
 
 
      
 
     
     
 
 
Geometrical Interpretation of 
Newton–Raphson Formula
y
f(x
0
)
f(x
f(x
2
)
x
0
x
x
1
x
2
y = f(x)
O
1
)
The geometrical meaning of Newton–Raphson method is a 
tangent is drawn at the point [x
0
, f (x
0
)] to the curve y = f (x). 
It cuts the x-axis at x
1
 which will be a better approximation 
of the root. Now drawing another tangent at [x
1
, f (x
1
)] which 
cuts the x-axis at x
2
 which is a still better approximation and 
the process can be continued till the desired accuracy has 
been achieved.
Convergence of Newton–Raphson Method
The order of convergence of Newton–Raphson method is 2 
or the convergence is quadratic. It converges if |f (x). f ?(x)| < 
|f '(x) |
2
. Also this method fails if f '(x) = 0
Newton’s Iterative Formula to Find bth Root 
of a Positive Real Number a
The iterative formula is given by x
n+1
=- +
?
?
?
?
?
?
?
?
?
?
-
1
1
1
b
bx
a
x
n
n
b
()
Newton’s Iterative Formula to Find a 
Reciprocal of a Number N
The iterative formula is given by 
x
i+1
 = x
i
 (2 - x
i
N)
Example
Find a real root of the equation -4x + cos x + 2 = 0, by 
Newton–Raphson method upto four decimal places assum-
ing x
0
 = 0.5
Solution
Let f (x) = -4x + cos x + 2 and
 f '(x) = -4 - sin x
Chapter 06.indd   125 5/31/2017   10:59:31 AM
Page 2


    
  
 
 
               
 
 
   
 
 
 
      
 
     
     
 
 
Geometrical Interpretation of 
Newton–Raphson Formula
y
f(x
0
)
f(x
f(x
2
)
x
0
x
x
1
x
2
y = f(x)
O
1
)
The geometrical meaning of Newton–Raphson method is a 
tangent is drawn at the point [x
0
, f (x
0
)] to the curve y = f (x). 
It cuts the x-axis at x
1
 which will be a better approximation 
of the root. Now drawing another tangent at [x
1
, f (x
1
)] which 
cuts the x-axis at x
2
 which is a still better approximation and 
the process can be continued till the desired accuracy has 
been achieved.
Convergence of Newton–Raphson Method
The order of convergence of Newton–Raphson method is 2 
or the convergence is quadratic. It converges if |f (x). f ?(x)| < 
|f '(x) |
2
. Also this method fails if f '(x) = 0
Newton’s Iterative Formula to Find bth Root 
of a Positive Real Number a
The iterative formula is given by x
n+1
=- +
?
?
?
?
?
?
?
?
?
?
-
1
1
1
b
bx
a
x
n
n
b
()
Newton’s Iterative Formula to Find a 
Reciprocal of a Number N
The iterative formula is given by 
x
i+1
 = x
i
 (2 - x
i
N)
Example
Find a real root of the equation -4x + cos x + 2 = 0, by 
Newton–Raphson method upto four decimal places assum-
ing x
0
 = 0.5
Solution
Let f (x) = -4x + cos x + 2 and
 f '(x) = -4 - sin x
Chapter 06.indd   125 5/31/2017   10:59:31 AM
    
Also f (0) = 1 + 2 = 3 > 0 and 
f (1) = -4 + cos 1 + 2 = -1.4596 < 0
So, a root lies between 0 and 1. 
Given x
0
 = 0.5
\ The first approximation
xx
fx
fx
10
0
0
=-
'
()
()
 
=-
-+ +
--
05
40 50 52
40 5
.
[( .) cos( .)
sin( .)
 
=-
-+ +
--
(. )
[cos(. )
sin.
05
22 05
40 5
 
=-
-
05
0 8775
4 4794
.
.
.
  = 0.5 + 0.1958 = 0.6958.
Example 
Obtain the cube root of 14 using Newton–Raphson method, 
with the initial approximation as 2.5.
Solution
We know that, the iterative formula to find a
b
 is
x
b
bx
a
x
nn
n
b + -
=- +
?
?
?
?
?
?
?
?
?
?
1 1
1
1 ()
Here b = 3 and a = 14 and let x
0
 = 2.5
?= +
?
?
?
?
?
?
?
?
?
?
xx
x
10
0
2
1
3
2
14
x
1 2
1
3
22 5
14
25
=+
?
?
?
?
?
?
(. )
(. )
=+
?
?
?
?
?
?
=+ =
1
3
5
14
625
1
3
52 24 2 413
.
{. }.
x
2 2
1
3
22 413
14
2 413
=+
?
?
?
?
?
?
(. )
(. )
=+
?
?
?
?
?
?
1
3
4 826
14
5 822569
.
.
.
.
=+ =
1
3
4 826 2 4044 2 410 {. .} .
\ The approximate cube root of 14 is 2.41.
Example 
Find the reciprocal of 24 using Newton–Raphson method 
with the initial approximation as 0.041.
Solution
The iterative formula to find 
1
N
 is,
x
n+1
 = x
n
(2 - x
n
N)
Let x
0
 = 0.041. Then x
1
 = x
0
(2 - x
0
 (24))
? x
1
 = (0.041) (2 - (24) (0.041))
= 0.04165
x
2
 = (0.0416) {2 - (24) (0.04165)} = 0.04161, similarly 
proceeding we get x
3
 = 0.041666
\ The reciprocal of 24 is 0.04166.
Curve Fitting 
In engineering applications, many a times, we need to find 
a suitable relation or law that may exist between the vari-
ables x and y from a given set of observed values (x
1
, y
1
), 
(x
2
, y
2
), …, (x
n
, y
n
), The relation connecting x and y is called 
as empirical law. 
The process of finding the equation of the curve of best 
fit which may be most suitable for predicting the value of y 
for a given value of x is known as curve fitting.
Least Squares Approximation
Least squares approximation method is one of the best 
methods available for curve fitting. 
Let (x
1
, y
1
), (x
2
, y
2
), …, (x
n
, y
n
) be the pairs of observed 
set of values of x and y. Let y = f (x) be the functional rela-
tionship sought between the variables x and y where f (x) 
consists of some unknown parameters. We need to find the 
relationship y = f (x) by using the observed values.
Procedure
 1. Find the residual d
i
 = y
i
 - f (x
i
) (i = 1, 2, …, n) for 
every pair of observed value y
i
 and f(x
i
), the value of 
the functional relation f (x) at x = x
i
 
 2. Find the sum of the squares of residuals corresponding 
to all pairs of values of y
i
 and f(x
i
) and let it be S
  i.e.,Sy fx
i
i
n
=-
=
?
(( )) .
2
1
 3. Find the values of the parameters in f(x) such that S is 
minimum.
Fitting a Straight Line Let y = a + bx be a straight line to 
be fitted to the data (x
1
 y
1
), (x
2
 y
2
), …, (x
n
 y
n
).
\ Residual = d
i
 = y
i
 - (a + bx
i
), i = 1, 2, …, n
Sum of the squares of the residuals = S = ?(y
i
 - (a + bx
i
 ))
2 
Now we have to find the parameters a and b such that S is 
minimum
?
?
=- +-
?
S
a
ya bx
ii
21 [( )( )] and
?
?
=- +-
?
S
b
ya bx x
ii i
[( ( ))( )] 2
 
For S to be minimum,
?
?
=
S
a
0 and 
?
?
=
S
a
0
Chapter 06.indd   126 5/31/2017   10:59:33 AM
Page 3


    
  
 
 
               
 
 
   
 
 
 
      
 
     
     
 
 
Geometrical Interpretation of 
Newton–Raphson Formula
y
f(x
0
)
f(x
f(x
2
)
x
0
x
x
1
x
2
y = f(x)
O
1
)
The geometrical meaning of Newton–Raphson method is a 
tangent is drawn at the point [x
0
, f (x
0
)] to the curve y = f (x). 
It cuts the x-axis at x
1
 which will be a better approximation 
of the root. Now drawing another tangent at [x
1
, f (x
1
)] which 
cuts the x-axis at x
2
 which is a still better approximation and 
the process can be continued till the desired accuracy has 
been achieved.
Convergence of Newton–Raphson Method
The order of convergence of Newton–Raphson method is 2 
or the convergence is quadratic. It converges if |f (x). f ?(x)| < 
|f '(x) |
2
. Also this method fails if f '(x) = 0
Newton’s Iterative Formula to Find bth Root 
of a Positive Real Number a
The iterative formula is given by x
n+1
=- +
?
?
?
?
?
?
?
?
?
?
-
1
1
1
b
bx
a
x
n
n
b
()
Newton’s Iterative Formula to Find a 
Reciprocal of a Number N
The iterative formula is given by 
x
i+1
 = x
i
 (2 - x
i
N)
Example
Find a real root of the equation -4x + cos x + 2 = 0, by 
Newton–Raphson method upto four decimal places assum-
ing x
0
 = 0.5
Solution
Let f (x) = -4x + cos x + 2 and
 f '(x) = -4 - sin x
Chapter 06.indd   125 5/31/2017   10:59:31 AM
    
Also f (0) = 1 + 2 = 3 > 0 and 
f (1) = -4 + cos 1 + 2 = -1.4596 < 0
So, a root lies between 0 and 1. 
Given x
0
 = 0.5
\ The first approximation
xx
fx
fx
10
0
0
=-
'
()
()
 
=-
-+ +
--
05
40 50 52
40 5
.
[( .) cos( .)
sin( .)
 
=-
-+ +
--
(. )
[cos(. )
sin.
05
22 05
40 5
 
=-
-
05
0 8775
4 4794
.
.
.
  = 0.5 + 0.1958 = 0.6958.
Example 
Obtain the cube root of 14 using Newton–Raphson method, 
with the initial approximation as 2.5.
Solution
We know that, the iterative formula to find a
b
 is
x
b
bx
a
x
nn
n
b + -
=- +
?
?
?
?
?
?
?
?
?
?
1 1
1
1 ()
Here b = 3 and a = 14 and let x
0
 = 2.5
?= +
?
?
?
?
?
?
?
?
?
?
xx
x
10
0
2
1
3
2
14
x
1 2
1
3
22 5
14
25
=+
?
?
?
?
?
?
(. )
(. )
=+
?
?
?
?
?
?
=+ =
1
3
5
14
625
1
3
52 24 2 413
.
{. }.
x
2 2
1
3
22 413
14
2 413
=+
?
?
?
?
?
?
(. )
(. )
=+
?
?
?
?
?
?
1
3
4 826
14
5 822569
.
.
.
.
=+ =
1
3
4 826 2 4044 2 410 {. .} .
\ The approximate cube root of 14 is 2.41.
Example 
Find the reciprocal of 24 using Newton–Raphson method 
with the initial approximation as 0.041.
Solution
The iterative formula to find 
1
N
 is,
x
n+1
 = x
n
(2 - x
n
N)
Let x
0
 = 0.041. Then x
1
 = x
0
(2 - x
0
 (24))
? x
1
 = (0.041) (2 - (24) (0.041))
= 0.04165
x
2
 = (0.0416) {2 - (24) (0.04165)} = 0.04161, similarly 
proceeding we get x
3
 = 0.041666
\ The reciprocal of 24 is 0.04166.
Curve Fitting 
In engineering applications, many a times, we need to find 
a suitable relation or law that may exist between the vari-
ables x and y from a given set of observed values (x
1
, y
1
), 
(x
2
, y
2
), …, (x
n
, y
n
), The relation connecting x and y is called 
as empirical law. 
The process of finding the equation of the curve of best 
fit which may be most suitable for predicting the value of y 
for a given value of x is known as curve fitting.
Least Squares Approximation
Least squares approximation method is one of the best 
methods available for curve fitting. 
Let (x
1
, y
1
), (x
2
, y
2
), …, (x
n
, y
n
) be the pairs of observed 
set of values of x and y. Let y = f (x) be the functional rela-
tionship sought between the variables x and y where f (x) 
consists of some unknown parameters. We need to find the 
relationship y = f (x) by using the observed values.
Procedure
 1. Find the residual d
i
 = y
i
 - f (x
i
) (i = 1, 2, …, n) for 
every pair of observed value y
i
 and f(x
i
), the value of 
the functional relation f (x) at x = x
i
 
 2. Find the sum of the squares of residuals corresponding 
to all pairs of values of y
i
 and f(x
i
) and let it be S
  i.e.,Sy fx
i
i
n
=-
=
?
(( )) .
2
1
 3. Find the values of the parameters in f(x) such that S is 
minimum.
Fitting a Straight Line Let y = a + bx be a straight line to 
be fitted to the data (x
1
 y
1
), (x
2
 y
2
), …, (x
n
 y
n
).
\ Residual = d
i
 = y
i
 - (a + bx
i
), i = 1, 2, …, n
Sum of the squares of the residuals = S = ?(y
i
 - (a + bx
i
 ))
2 
Now we have to find the parameters a and b such that S is 
minimum
?
?
=- +-
?
S
a
ya bx
ii
21 [( )( )] and
?
?
=- +-
?
S
b
ya bx x
ii i
[( ( ))( )] 2
 
For S to be minimum,
?
?
=
S
a
0 and 
?
?
=
S
a
0
Chapter 06.indd   126 5/31/2017   10:59:33 AM
    
?- -+ =
?
[( ( ))] 20 ya bx
ii
 and 
     [( ( ))( )] 20 ya bx x
ii i
-+ -=
?
?= +
??
ynab x
ii
 (1)
and xy ax bx
ii ii
=+
? ? ?
2
  (2)
Eqs. (1) and (2) are known as normal equations.
By solving these equations, we get the values of ‘a’ and b
Fitting a Parabola (Quadratic Equation)
To fit a parabola of the type y = a + bx + cx
2
 to the set of data 
points (x
1
, y
1
), (x
2
, y
2
), …, (x
n
, y
n
), by proceeding as above, we 
get the normal equations as 
ynab xc x
ii i
=+ +
?? ?
2
 (1)
xy ax bx cx
ii ii i ?? ??
=+ +
23
  (2)
xy ax bx cx
ii ii i
22 34
?? ??
=+ +  (3)
By solving Eqs. (1), (2) and (3), we can get the values of 
the parameters a, b and c
Fitting of various exponential curves that can be brought 
into the form of a straight line: Exponential curves of the 
type y = ax
b
, y = ab
x
 and y = ae
bx
 can be fitted to the given 
data by transforming it into the form of a straight line by 
applying logarithm as follows.
Equation of 
the curve to 
be fitted
Equation obtained 
after applying  
log
e
(= ln)
T ransformed  equation 
into the form  
of a straight line
y = ax
b
 lny = lna + blnx Y = A + bX where Y = lny; 
A = lna and X = lnx
y = ab
x
lny = lna + xlnb Y = A + Bx where Y = lny; 
A = lna and B = lnb
y = ae
bx
lny = lna +bx Y = A + bx where Y = lny; 
A = lna
Example 
Using the method of least squares, fit a straight line y = a + 
bx to the following data.
x 1 2 3 4
y 4 11 35 100
Hence find the value of y at x = 5.
Solution
We have to fit the line y = a + bx to the given data.
The normal equations are
?= +? ynab x
ii
 (1)
?= ?+ ? xy ax bx
ii ii
2
 (2)
The required values in the normal equations can be found 
using the following table
x
i
y
i
x
i  
y
i
x
i
2
1 4 4 1
2 11 22 4
3 35 105 9
4 100 400 16
xy xy x
ii ii i ?? ??
== == 10 150 531 30
2
;; ;
Substituting these values in Eqn. (1) and (2), we get
150 = 4a +10b and 531 = 10a + 30b
? 4a +10b = 150
10a + 30b = 531
Solving these linear equations, we get
a = -40.5 and b = 31.2
\ The straight line that fits to the given data is y = a + bx
? y = -40.5 + 31.2x
The value of y at x = 5 is
y = -40.5 +31.2 × 5 ? y = 115.5.
Example 
Fit a quadratic equation y = a + bx + cx
2
 to the following 
data by the method of least squares.
x -2 -1 0 1 2
y 1 5 10 22 38
Solution
We have to fit the curve y = a + bx + cx
2
 to the given data.
Here the normal equations are
?= +? +? ynab xc x
ii i
2
 (1)
? x
i 
y
i
 = a? x
i
 + b? x
i
2 
+ c? x
i
3
 (2)
? x
i
2
y
i
 = a? x
i
2
 + b? x
i
3
 + c? x
i
4
 (3)
The values required in the normal equations can be obtained 
by the following table:
x
i
y
i
x
i 
y
i
x
i
2
x
3
i
x
i
4
x
i
2 
y
i
-2 1 -2 4 -8 16 4
-1 5 -5 1 -1 1 5
0 10 0 0 0 0 0
1 22 22 1 1 1 22
2 38 76 4 8 16 152
? x
i
 = 0; ? y
i
 
= 76; ? x
i 
y
i
 = 91; ? x
i
2
 = 10; ? x
i
3
 = 0; ? x
i
4
 = 34; 
? x
i
2 
y
i
 = 183
Substituting these values in the normal equation we have
 76 = 5a + b × 0 + c × 10
 91 = a × 0 + b × 10 + c × 0
Chapter 06.indd   127 5/31/2017   10:59:35 AM
Page 4


    
  
 
 
               
 
 
   
 
 
 
      
 
     
     
 
 
Geometrical Interpretation of 
Newton–Raphson Formula
y
f(x
0
)
f(x
f(x
2
)
x
0
x
x
1
x
2
y = f(x)
O
1
)
The geometrical meaning of Newton–Raphson method is a 
tangent is drawn at the point [x
0
, f (x
0
)] to the curve y = f (x). 
It cuts the x-axis at x
1
 which will be a better approximation 
of the root. Now drawing another tangent at [x
1
, f (x
1
)] which 
cuts the x-axis at x
2
 which is a still better approximation and 
the process can be continued till the desired accuracy has 
been achieved.
Convergence of Newton–Raphson Method
The order of convergence of Newton–Raphson method is 2 
or the convergence is quadratic. It converges if |f (x). f ?(x)| < 
|f '(x) |
2
. Also this method fails if f '(x) = 0
Newton’s Iterative Formula to Find bth Root 
of a Positive Real Number a
The iterative formula is given by x
n+1
=- +
?
?
?
?
?
?
?
?
?
?
-
1
1
1
b
bx
a
x
n
n
b
()
Newton’s Iterative Formula to Find a 
Reciprocal of a Number N
The iterative formula is given by 
x
i+1
 = x
i
 (2 - x
i
N)
Example
Find a real root of the equation -4x + cos x + 2 = 0, by 
Newton–Raphson method upto four decimal places assum-
ing x
0
 = 0.5
Solution
Let f (x) = -4x + cos x + 2 and
 f '(x) = -4 - sin x
Chapter 06.indd   125 5/31/2017   10:59:31 AM
    
Also f (0) = 1 + 2 = 3 > 0 and 
f (1) = -4 + cos 1 + 2 = -1.4596 < 0
So, a root lies between 0 and 1. 
Given x
0
 = 0.5
\ The first approximation
xx
fx
fx
10
0
0
=-
'
()
()
 
=-
-+ +
--
05
40 50 52
40 5
.
[( .) cos( .)
sin( .)
 
=-
-+ +
--
(. )
[cos(. )
sin.
05
22 05
40 5
 
=-
-
05
0 8775
4 4794
.
.
.
  = 0.5 + 0.1958 = 0.6958.
Example 
Obtain the cube root of 14 using Newton–Raphson method, 
with the initial approximation as 2.5.
Solution
We know that, the iterative formula to find a
b
 is
x
b
bx
a
x
nn
n
b + -
=- +
?
?
?
?
?
?
?
?
?
?
1 1
1
1 ()
Here b = 3 and a = 14 and let x
0
 = 2.5
?= +
?
?
?
?
?
?
?
?
?
?
xx
x
10
0
2
1
3
2
14
x
1 2
1
3
22 5
14
25
=+
?
?
?
?
?
?
(. )
(. )
=+
?
?
?
?
?
?
=+ =
1
3
5
14
625
1
3
52 24 2 413
.
{. }.
x
2 2
1
3
22 413
14
2 413
=+
?
?
?
?
?
?
(. )
(. )
=+
?
?
?
?
?
?
1
3
4 826
14
5 822569
.
.
.
.
=+ =
1
3
4 826 2 4044 2 410 {. .} .
\ The approximate cube root of 14 is 2.41.
Example 
Find the reciprocal of 24 using Newton–Raphson method 
with the initial approximation as 0.041.
Solution
The iterative formula to find 
1
N
 is,
x
n+1
 = x
n
(2 - x
n
N)
Let x
0
 = 0.041. Then x
1
 = x
0
(2 - x
0
 (24))
? x
1
 = (0.041) (2 - (24) (0.041))
= 0.04165
x
2
 = (0.0416) {2 - (24) (0.04165)} = 0.04161, similarly 
proceeding we get x
3
 = 0.041666
\ The reciprocal of 24 is 0.04166.
Curve Fitting 
In engineering applications, many a times, we need to find 
a suitable relation or law that may exist between the vari-
ables x and y from a given set of observed values (x
1
, y
1
), 
(x
2
, y
2
), …, (x
n
, y
n
), The relation connecting x and y is called 
as empirical law. 
The process of finding the equation of the curve of best 
fit which may be most suitable for predicting the value of y 
for a given value of x is known as curve fitting.
Least Squares Approximation
Least squares approximation method is one of the best 
methods available for curve fitting. 
Let (x
1
, y
1
), (x
2
, y
2
), …, (x
n
, y
n
) be the pairs of observed 
set of values of x and y. Let y = f (x) be the functional rela-
tionship sought between the variables x and y where f (x) 
consists of some unknown parameters. We need to find the 
relationship y = f (x) by using the observed values.
Procedure
 1. Find the residual d
i
 = y
i
 - f (x
i
) (i = 1, 2, …, n) for 
every pair of observed value y
i
 and f(x
i
), the value of 
the functional relation f (x) at x = x
i
 
 2. Find the sum of the squares of residuals corresponding 
to all pairs of values of y
i
 and f(x
i
) and let it be S
  i.e.,Sy fx
i
i
n
=-
=
?
(( )) .
2
1
 3. Find the values of the parameters in f(x) such that S is 
minimum.
Fitting a Straight Line Let y = a + bx be a straight line to 
be fitted to the data (x
1
 y
1
), (x
2
 y
2
), …, (x
n
 y
n
).
\ Residual = d
i
 = y
i
 - (a + bx
i
), i = 1, 2, …, n
Sum of the squares of the residuals = S = ?(y
i
 - (a + bx
i
 ))
2 
Now we have to find the parameters a and b such that S is 
minimum
?
?
=- +-
?
S
a
ya bx
ii
21 [( )( )] and
?
?
=- +-
?
S
b
ya bx x
ii i
[( ( ))( )] 2
 
For S to be minimum,
?
?
=
S
a
0 and 
?
?
=
S
a
0
Chapter 06.indd   126 5/31/2017   10:59:33 AM
    
?- -+ =
?
[( ( ))] 20 ya bx
ii
 and 
     [( ( ))( )] 20 ya bx x
ii i
-+ -=
?
?= +
??
ynab x
ii
 (1)
and xy ax bx
ii ii
=+
? ? ?
2
  (2)
Eqs. (1) and (2) are known as normal equations.
By solving these equations, we get the values of ‘a’ and b
Fitting a Parabola (Quadratic Equation)
To fit a parabola of the type y = a + bx + cx
2
 to the set of data 
points (x
1
, y
1
), (x
2
, y
2
), …, (x
n
, y
n
), by proceeding as above, we 
get the normal equations as 
ynab xc x
ii i
=+ +
?? ?
2
 (1)
xy ax bx cx
ii ii i ?? ??
=+ +
23
  (2)
xy ax bx cx
ii ii i
22 34
?? ??
=+ +  (3)
By solving Eqs. (1), (2) and (3), we can get the values of 
the parameters a, b and c
Fitting of various exponential curves that can be brought 
into the form of a straight line: Exponential curves of the 
type y = ax
b
, y = ab
x
 and y = ae
bx
 can be fitted to the given 
data by transforming it into the form of a straight line by 
applying logarithm as follows.
Equation of 
the curve to 
be fitted
Equation obtained 
after applying  
log
e
(= ln)
T ransformed  equation 
into the form  
of a straight line
y = ax
b
 lny = lna + blnx Y = A + bX where Y = lny; 
A = lna and X = lnx
y = ab
x
lny = lna + xlnb Y = A + Bx where Y = lny; 
A = lna and B = lnb
y = ae
bx
lny = lna +bx Y = A + bx where Y = lny; 
A = lna
Example 
Using the method of least squares, fit a straight line y = a + 
bx to the following data.
x 1 2 3 4
y 4 11 35 100
Hence find the value of y at x = 5.
Solution
We have to fit the line y = a + bx to the given data.
The normal equations are
?= +? ynab x
ii
 (1)
?= ?+ ? xy ax bx
ii ii
2
 (2)
The required values in the normal equations can be found 
using the following table
x
i
y
i
x
i  
y
i
x
i
2
1 4 4 1
2 11 22 4
3 35 105 9
4 100 400 16
xy xy x
ii ii i ?? ??
== == 10 150 531 30
2
;; ;
Substituting these values in Eqn. (1) and (2), we get
150 = 4a +10b and 531 = 10a + 30b
? 4a +10b = 150
10a + 30b = 531
Solving these linear equations, we get
a = -40.5 and b = 31.2
\ The straight line that fits to the given data is y = a + bx
? y = -40.5 + 31.2x
The value of y at x = 5 is
y = -40.5 +31.2 × 5 ? y = 115.5.
Example 
Fit a quadratic equation y = a + bx + cx
2
 to the following 
data by the method of least squares.
x -2 -1 0 1 2
y 1 5 10 22 38
Solution
We have to fit the curve y = a + bx + cx
2
 to the given data.
Here the normal equations are
?= +? +? ynab xc x
ii i
2
 (1)
? x
i 
y
i
 = a? x
i
 + b? x
i
2 
+ c? x
i
3
 (2)
? x
i
2
y
i
 = a? x
i
2
 + b? x
i
3
 + c? x
i
4
 (3)
The values required in the normal equations can be obtained 
by the following table:
x
i
y
i
x
i 
y
i
x
i
2
x
3
i
x
i
4
x
i
2 
y
i
-2 1 -2 4 -8 16 4
-1 5 -5 1 -1 1 5
0 10 0 0 0 0 0
1 22 22 1 1 1 22
2 38 76 4 8 16 152
? x
i
 = 0; ? y
i
 
= 76; ? x
i 
y
i
 = 91; ? x
i
2
 = 10; ? x
i
3
 = 0; ? x
i
4
 = 34; 
? x
i
2 
y
i
 = 183
Substituting these values in the normal equation we have
 76 = 5a + b × 0 + c × 10
 91 = a × 0 + b × 10 + c × 0
Chapter 06.indd   127 5/31/2017   10:59:35 AM
    
 183 = a × 10 + b × 0 + c × 34
 ? 5a + 10c = 76
 10b = 91
 10a + 34c = 183
Solving these equations for a, b and c we get
a = 10.77, b = 9.1 and c = 2.21 
Substituting these in y = a + bx + cx
2
, we get the required 
parabola as
y = 10.77 + 9.1x + 2.21x
2
.
Interpolation
The process of finding the most appropriate estimate for 
the unknown values of a function y = f (x) at some values 
of x by using the given pairs of values (x, f(x)) is called 
interpolation.
Assumptions in Interpolation
 1. The frequency distribution is normal and not marked 
by sudden ups and downs.
 2. The changes in the series are uniform within a period.
  Before looking into interpolation, let us get 
familiarity with the finite differences which we use in 
interpolation.
Finite Differences
 1. Forward differences: Consider a function y = f(x). 
Let we were given the following table representing 
the values of y = f(x) corresponding to the values x
i
 
, 
x
2
 
, …, x
n
 
of x that are equally spaced (i.e., x
i
 
= x
0
 
+ ih; 
i = 1, 2, …, n).
x = x
i
x
1
 
x
2
x
3
… x
n
y = f(x
i  
) y
1
y
2
y
3
y
n
  The forward difference of f(x) denoted by Dy = D[ f(x)] 
can be defined as
  D y = D[ f(x)] = f(x +h) - f(x)
  \ Dy
0
 = f(x
0
 + h) - f(x
0
) = y
1
 - y
0
  D y
1
 = y
2
 - y
1
  D y
2
 = y
3
 - y
2
  .
  .
  .
  D y
n-1
 = y
n
 - y
n-1
  where D is called the forward difference operator and 
D y
0
, D y
1
, …, D y
n-1
 are called the first order forward 
differences of y = f(x)
  Similarly, D 2
y = D[  f(x + h)] - D[  f(x)]
  \ D 2
y
0
 = Dy
1
 - Dy
0
  D 2
y
1
 = Dy
2
 - Dy
1
  D 2
y
n- 2
 = Dy
n- 1
 - Dy
n- 2
  where D 2
y
0
 , D 2
y
1
,…, D 2
y
n-2
 are called the second order 
forward differences.
  And in general, the nth order forward differences 
are given by
  D n
y = D n
[  f(x)] = D n-1
[  f(x + h)] - D n-1
[  f(x)]
  \ D n
y
0
 = D n-1
y
1
 - D n-1
y
0
  D n
y
1
 = D n-1
y
2
 - D n-1
y
1
  .
  .
  .
  These forward differences of various orders can be 
found and represented in a table called the forward 
difference table as shown below
x y = f(x) Dy D 2
y D 3
y D 4
y D 5
y
x
0
x
1
 = x
0
 + h
x
2
 = x
0
 + 2h
x
3
 = x
0
 + 3h
x
4
 = x
0
 + 4h
x
5
 = x
0
 + 5h
y
0
y
1
y
2
y
3
y
4
y
5
Dy
0
Dy
1
Dy
2
Dy
3
Dy
4
D 2
y
0
D 2
y
1
D 2
y
2
D 2
y
3
D 3
y
0
D 3
y
1
D 3
y
2
D 4
y
0
D 4
y
1
D 5
y
0
 2. Backward differences: Consider a function y = f (x). 
Let we were given the following table representing 
the values of y = f (x) corresponding to the values x
1
, 
x
2
 , …, x
n
 of x that are equally spaced (i.e; x
i
 = x
0
 + ih, 
i = 1, 2, … n)
x = x
i
x
1
x
2
x
3
… x
n
y
i
 = f(x
i
) y
1
y
2
y
3
… y
n
  The backward difference of f(x) denoted by ?y (or) 
?[  f(x)] can be defined as 
  ? y = ?[  f(x)] = f(x) - f(x - h)
  \ ?y
1
= y
1
 - y
0
  ? y
2
 = y
2
 - y
1
  ? y
3
 = y
3
 - y
2
  .
  .
  .
  ?y
n
 = y
n
 - y
n-1
  where ? is called the backward difference operator and 
?y
1
, ?y
2
, …, ?y
n
 are called the first order backward 
differences of y = f(x)
  Similarly, ?
2
y = ?
2
[    f(x)] = ?[f(x) - ?[  f(x - h)]
Chapter 06.indd   128 5/31/2017   10:59:35 AM
Page 5


    
  
 
 
               
 
 
   
 
 
 
      
 
     
     
 
 
Geometrical Interpretation of 
Newton–Raphson Formula
y
f(x
0
)
f(x
f(x
2
)
x
0
x
x
1
x
2
y = f(x)
O
1
)
The geometrical meaning of Newton–Raphson method is a 
tangent is drawn at the point [x
0
, f (x
0
)] to the curve y = f (x). 
It cuts the x-axis at x
1
 which will be a better approximation 
of the root. Now drawing another tangent at [x
1
, f (x
1
)] which 
cuts the x-axis at x
2
 which is a still better approximation and 
the process can be continued till the desired accuracy has 
been achieved.
Convergence of Newton–Raphson Method
The order of convergence of Newton–Raphson method is 2 
or the convergence is quadratic. It converges if |f (x). f ?(x)| < 
|f '(x) |
2
. Also this method fails if f '(x) = 0
Newton’s Iterative Formula to Find bth Root 
of a Positive Real Number a
The iterative formula is given by x
n+1
=- +
?
?
?
?
?
?
?
?
?
?
-
1
1
1
b
bx
a
x
n
n
b
()
Newton’s Iterative Formula to Find a 
Reciprocal of a Number N
The iterative formula is given by 
x
i+1
 = x
i
 (2 - x
i
N)
Example
Find a real root of the equation -4x + cos x + 2 = 0, by 
Newton–Raphson method upto four decimal places assum-
ing x
0
 = 0.5
Solution
Let f (x) = -4x + cos x + 2 and
 f '(x) = -4 - sin x
Chapter 06.indd   125 5/31/2017   10:59:31 AM
    
Also f (0) = 1 + 2 = 3 > 0 and 
f (1) = -4 + cos 1 + 2 = -1.4596 < 0
So, a root lies between 0 and 1. 
Given x
0
 = 0.5
\ The first approximation
xx
fx
fx
10
0
0
=-
'
()
()
 
=-
-+ +
--
05
40 50 52
40 5
.
[( .) cos( .)
sin( .)
 
=-
-+ +
--
(. )
[cos(. )
sin.
05
22 05
40 5
 
=-
-
05
0 8775
4 4794
.
.
.
  = 0.5 + 0.1958 = 0.6958.
Example 
Obtain the cube root of 14 using Newton–Raphson method, 
with the initial approximation as 2.5.
Solution
We know that, the iterative formula to find a
b
 is
x
b
bx
a
x
nn
n
b + -
=- +
?
?
?
?
?
?
?
?
?
?
1 1
1
1 ()
Here b = 3 and a = 14 and let x
0
 = 2.5
?= +
?
?
?
?
?
?
?
?
?
?
xx
x
10
0
2
1
3
2
14
x
1 2
1
3
22 5
14
25
=+
?
?
?
?
?
?
(. )
(. )
=+
?
?
?
?
?
?
=+ =
1
3
5
14
625
1
3
52 24 2 413
.
{. }.
x
2 2
1
3
22 413
14
2 413
=+
?
?
?
?
?
?
(. )
(. )
=+
?
?
?
?
?
?
1
3
4 826
14
5 822569
.
.
.
.
=+ =
1
3
4 826 2 4044 2 410 {. .} .
\ The approximate cube root of 14 is 2.41.
Example 
Find the reciprocal of 24 using Newton–Raphson method 
with the initial approximation as 0.041.
Solution
The iterative formula to find 
1
N
 is,
x
n+1
 = x
n
(2 - x
n
N)
Let x
0
 = 0.041. Then x
1
 = x
0
(2 - x
0
 (24))
? x
1
 = (0.041) (2 - (24) (0.041))
= 0.04165
x
2
 = (0.0416) {2 - (24) (0.04165)} = 0.04161, similarly 
proceeding we get x
3
 = 0.041666
\ The reciprocal of 24 is 0.04166.
Curve Fitting 
In engineering applications, many a times, we need to find 
a suitable relation or law that may exist between the vari-
ables x and y from a given set of observed values (x
1
, y
1
), 
(x
2
, y
2
), …, (x
n
, y
n
), The relation connecting x and y is called 
as empirical law. 
The process of finding the equation of the curve of best 
fit which may be most suitable for predicting the value of y 
for a given value of x is known as curve fitting.
Least Squares Approximation
Least squares approximation method is one of the best 
methods available for curve fitting. 
Let (x
1
, y
1
), (x
2
, y
2
), …, (x
n
, y
n
) be the pairs of observed 
set of values of x and y. Let y = f (x) be the functional rela-
tionship sought between the variables x and y where f (x) 
consists of some unknown parameters. We need to find the 
relationship y = f (x) by using the observed values.
Procedure
 1. Find the residual d
i
 = y
i
 - f (x
i
) (i = 1, 2, …, n) for 
every pair of observed value y
i
 and f(x
i
), the value of 
the functional relation f (x) at x = x
i
 
 2. Find the sum of the squares of residuals corresponding 
to all pairs of values of y
i
 and f(x
i
) and let it be S
  i.e.,Sy fx
i
i
n
=-
=
?
(( )) .
2
1
 3. Find the values of the parameters in f(x) such that S is 
minimum.
Fitting a Straight Line Let y = a + bx be a straight line to 
be fitted to the data (x
1
 y
1
), (x
2
 y
2
), …, (x
n
 y
n
).
\ Residual = d
i
 = y
i
 - (a + bx
i
), i = 1, 2, …, n
Sum of the squares of the residuals = S = ?(y
i
 - (a + bx
i
 ))
2 
Now we have to find the parameters a and b such that S is 
minimum
?
?
=- +-
?
S
a
ya bx
ii
21 [( )( )] and
?
?
=- +-
?
S
b
ya bx x
ii i
[( ( ))( )] 2
 
For S to be minimum,
?
?
=
S
a
0 and 
?
?
=
S
a
0
Chapter 06.indd   126 5/31/2017   10:59:33 AM
    
?- -+ =
?
[( ( ))] 20 ya bx
ii
 and 
     [( ( ))( )] 20 ya bx x
ii i
-+ -=
?
?= +
??
ynab x
ii
 (1)
and xy ax bx
ii ii
=+
? ? ?
2
  (2)
Eqs. (1) and (2) are known as normal equations.
By solving these equations, we get the values of ‘a’ and b
Fitting a Parabola (Quadratic Equation)
To fit a parabola of the type y = a + bx + cx
2
 to the set of data 
points (x
1
, y
1
), (x
2
, y
2
), …, (x
n
, y
n
), by proceeding as above, we 
get the normal equations as 
ynab xc x
ii i
=+ +
?? ?
2
 (1)
xy ax bx cx
ii ii i ?? ??
=+ +
23
  (2)
xy ax bx cx
ii ii i
22 34
?? ??
=+ +  (3)
By solving Eqs. (1), (2) and (3), we can get the values of 
the parameters a, b and c
Fitting of various exponential curves that can be brought 
into the form of a straight line: Exponential curves of the 
type y = ax
b
, y = ab
x
 and y = ae
bx
 can be fitted to the given 
data by transforming it into the form of a straight line by 
applying logarithm as follows.
Equation of 
the curve to 
be fitted
Equation obtained 
after applying  
log
e
(= ln)
T ransformed  equation 
into the form  
of a straight line
y = ax
b
 lny = lna + blnx Y = A + bX where Y = lny; 
A = lna and X = lnx
y = ab
x
lny = lna + xlnb Y = A + Bx where Y = lny; 
A = lna and B = lnb
y = ae
bx
lny = lna +bx Y = A + bx where Y = lny; 
A = lna
Example 
Using the method of least squares, fit a straight line y = a + 
bx to the following data.
x 1 2 3 4
y 4 11 35 100
Hence find the value of y at x = 5.
Solution
We have to fit the line y = a + bx to the given data.
The normal equations are
?= +? ynab x
ii
 (1)
?= ?+ ? xy ax bx
ii ii
2
 (2)
The required values in the normal equations can be found 
using the following table
x
i
y
i
x
i  
y
i
x
i
2
1 4 4 1
2 11 22 4
3 35 105 9
4 100 400 16
xy xy x
ii ii i ?? ??
== == 10 150 531 30
2
;; ;
Substituting these values in Eqn. (1) and (2), we get
150 = 4a +10b and 531 = 10a + 30b
? 4a +10b = 150
10a + 30b = 531
Solving these linear equations, we get
a = -40.5 and b = 31.2
\ The straight line that fits to the given data is y = a + bx
? y = -40.5 + 31.2x
The value of y at x = 5 is
y = -40.5 +31.2 × 5 ? y = 115.5.
Example 
Fit a quadratic equation y = a + bx + cx
2
 to the following 
data by the method of least squares.
x -2 -1 0 1 2
y 1 5 10 22 38
Solution
We have to fit the curve y = a + bx + cx
2
 to the given data.
Here the normal equations are
?= +? +? ynab xc x
ii i
2
 (1)
? x
i 
y
i
 = a? x
i
 + b? x
i
2 
+ c? x
i
3
 (2)
? x
i
2
y
i
 = a? x
i
2
 + b? x
i
3
 + c? x
i
4
 (3)
The values required in the normal equations can be obtained 
by the following table:
x
i
y
i
x
i 
y
i
x
i
2
x
3
i
x
i
4
x
i
2 
y
i
-2 1 -2 4 -8 16 4
-1 5 -5 1 -1 1 5
0 10 0 0 0 0 0
1 22 22 1 1 1 22
2 38 76 4 8 16 152
? x
i
 = 0; ? y
i
 
= 76; ? x
i 
y
i
 = 91; ? x
i
2
 = 10; ? x
i
3
 = 0; ? x
i
4
 = 34; 
? x
i
2 
y
i
 = 183
Substituting these values in the normal equation we have
 76 = 5a + b × 0 + c × 10
 91 = a × 0 + b × 10 + c × 0
Chapter 06.indd   127 5/31/2017   10:59:35 AM
    
 183 = a × 10 + b × 0 + c × 34
 ? 5a + 10c = 76
 10b = 91
 10a + 34c = 183
Solving these equations for a, b and c we get
a = 10.77, b = 9.1 and c = 2.21 
Substituting these in y = a + bx + cx
2
, we get the required 
parabola as
y = 10.77 + 9.1x + 2.21x
2
.
Interpolation
The process of finding the most appropriate estimate for 
the unknown values of a function y = f (x) at some values 
of x by using the given pairs of values (x, f(x)) is called 
interpolation.
Assumptions in Interpolation
 1. The frequency distribution is normal and not marked 
by sudden ups and downs.
 2. The changes in the series are uniform within a period.
  Before looking into interpolation, let us get 
familiarity with the finite differences which we use in 
interpolation.
Finite Differences
 1. Forward differences: Consider a function y = f(x). 
Let we were given the following table representing 
the values of y = f(x) corresponding to the values x
i
 
, 
x
2
 
, …, x
n
 
of x that are equally spaced (i.e., x
i
 
= x
0
 
+ ih; 
i = 1, 2, …, n).
x = x
i
x
1
 
x
2
x
3
… x
n
y = f(x
i  
) y
1
y
2
y
3
y
n
  The forward difference of f(x) denoted by Dy = D[ f(x)] 
can be defined as
  D y = D[ f(x)] = f(x +h) - f(x)
  \ Dy
0
 = f(x
0
 + h) - f(x
0
) = y
1
 - y
0
  D y
1
 = y
2
 - y
1
  D y
2
 = y
3
 - y
2
  .
  .
  .
  D y
n-1
 = y
n
 - y
n-1
  where D is called the forward difference operator and 
D y
0
, D y
1
, …, D y
n-1
 are called the first order forward 
differences of y = f(x)
  Similarly, D 2
y = D[  f(x + h)] - D[  f(x)]
  \ D 2
y
0
 = Dy
1
 - Dy
0
  D 2
y
1
 = Dy
2
 - Dy
1
  D 2
y
n- 2
 = Dy
n- 1
 - Dy
n- 2
  where D 2
y
0
 , D 2
y
1
,…, D 2
y
n-2
 are called the second order 
forward differences.
  And in general, the nth order forward differences 
are given by
  D n
y = D n
[  f(x)] = D n-1
[  f(x + h)] - D n-1
[  f(x)]
  \ D n
y
0
 = D n-1
y
1
 - D n-1
y
0
  D n
y
1
 = D n-1
y
2
 - D n-1
y
1
  .
  .
  .
  These forward differences of various orders can be 
found and represented in a table called the forward 
difference table as shown below
x y = f(x) Dy D 2
y D 3
y D 4
y D 5
y
x
0
x
1
 = x
0
 + h
x
2
 = x
0
 + 2h
x
3
 = x
0
 + 3h
x
4
 = x
0
 + 4h
x
5
 = x
0
 + 5h
y
0
y
1
y
2
y
3
y
4
y
5
Dy
0
Dy
1
Dy
2
Dy
3
Dy
4
D 2
y
0
D 2
y
1
D 2
y
2
D 2
y
3
D 3
y
0
D 3
y
1
D 3
y
2
D 4
y
0
D 4
y
1
D 5
y
0
 2. Backward differences: Consider a function y = f (x). 
Let we were given the following table representing 
the values of y = f (x) corresponding to the values x
1
, 
x
2
 , …, x
n
 of x that are equally spaced (i.e; x
i
 = x
0
 + ih, 
i = 1, 2, … n)
x = x
i
x
1
x
2
x
3
… x
n
y
i
 = f(x
i
) y
1
y
2
y
3
… y
n
  The backward difference of f(x) denoted by ?y (or) 
?[  f(x)] can be defined as 
  ? y = ?[  f(x)] = f(x) - f(x - h)
  \ ?y
1
= y
1
 - y
0
  ? y
2
 = y
2
 - y
1
  ? y
3
 = y
3
 - y
2
  .
  .
  .
  ?y
n
 = y
n
 - y
n-1
  where ? is called the backward difference operator and 
?y
1
, ?y
2
, …, ?y
n
 are called the first order backward 
differences of y = f(x)
  Similarly, ?
2
y = ?
2
[    f(x)] = ?[f(x) - ?[  f(x - h)]
Chapter 06.indd   128 5/31/2017   10:59:35 AM
    
  \ ?
2
y
2
 = ?y
2
 - ?y
1
   ?
2
y
3
 = ?y
3
 - ?y
2
  .
  .
  .
  ?
2
y
n
 = ?y
n
 - ?y
n-1
  where ?
2
y
2
, ?
2
y
3
, …, ?
2
y
n
 are called the second order 
backward differences.
  And in general, the nth order backward differences 
are given by
  ?
n
y = ?
n
[  f(x)] = ?
n-1
[  f(x)] - ?
n-1
[  f(x - h)]
  \ ?
n
y
n
 = ?
n-1
y
n
 - ?
n-1
y
n-1
  These backward differences of various orders can be 
found and represented in a table called the backward 
difference table as shown below
x y = f(x) ?y ?
2
y ?
3
y ?
4
y ?
5
y
x
0
x
1
 = x
0
 + h
x
2
 = x
0
 + 2h
x
3
 = x
0
 + 3h
x
4
 = x
0
 + 4h
x
5
 = x
0
 + 5h
y
0
y
1
y
2
y
3
y
4
Y
5
?y
1
?y
2
?y
3
?y
4
?y
5
?
2
y
2
?
2
y
3
?
2
y
4
?
2
y
5
?
3
y
3
?
3
y
4
?
3
y
5
?
4
y
4
?
4
y
5
?
5
y
5
Relation between forward and backward differences: 
First order: ?[  f(x + h)] =  D[  f(x)]
Second order: ?
2
[  f (x + 2h)] = D 2 
[  f (x)]
Third order: ?
3 
[  f(x + 3h)] = D 3 
[  f (x)]
In general, the nth order forward and backward differ -
ences are connected by the relation.
?
n
[f (x+ nh)] = D n 
[f (x)]
 3. Divided differences: Consider a function y = f(x). 
Let we were given a table of values of y = f (x) at x
1
, 
x
2
, x
3
,…, x
n
, (need not be equally spaced) as shown 
below.
x = x
i
x
1
x
2
x
3
… x
n
y
i
 = f( x
i 
) y
1
y
2
y
3
… y
n
  Then the first order divided differences are given by:
  [] xx
yy
xx
x 0
10
10
=
-
-
 
  [] xx
yy
xx
12
21
21
=
-
-
 
  .
  .
  .
  []
,
xx
yy
xx
nn
nn
nn
-
-
-
=
-
-
1
1
1
  
  The second order divided differences are given by:
 
[, ,]
|, || ,|
[, ,]
[, ][ ,]
xx x
xx xx
xx
xx x
xx xx
x
01 2
12 01
30
12 3
23 12
=
-
-
=
-
3 31
- x
 
  Similarly, the third order divided differences are given 
by 
  [, ,, ]
|, ,| |, ,|
xxx x
xx xx xx
xx
0 234
12 30 12
30
=
-
-
 
  
[, ,, ]
|, ,| |, ,|
xx xx
xx xx xx
xx
1 234
234 12 3
41
=
-
-
 
  Note that [x
0
, x
1
] = [x
1
, x
0
]
  And [x
0
, x
1
, x
2
] = [x
1
, x
2
, x
0
] = [x
2
, x
0
, x
1
]
Interpolation Formulae
 1. Newton’s forward interpolation formula: If the 
function y = f (x) takes the values y
0
, y
1
, y
2
, …, y
n
 
respectively at the equally spaced points x
0
, x
1
, x
2
, 
…, x
n
 (i.e., x
i+1
 = x
0
 + ih (OR) x
i+1
 = x
i
 + h, then the 
Newton’ s forward interpolation formula is given by:
  
yy py
pp
y
pp p
y
pp pp
p
=+ +
-
+
--
++
--
00
2
0
3
0
1
2
12
3
12
??
?
()
!
()()
!
()() (

 - -- ())
!
n
n
y
n
1
0
?
  where p
xx
h
=
-
0
 (OR) x = x
0
 + ph.
 2. Newton’s backward interpolation formula: If the 
function y = f(x) takes the values y
0
, y
1
, y
2
, …,  y
n
 
respectively at the equally spaced points x
0
, x
1
, x
2
, 
…., x
n
 (i.e., x
i+1
 = x
0
 + ih (OR) x
i+1
 = x
i
 + h), then the 
Newton’ s backward interpolation formula is given by 
  Y
p
= ?y
n
 + p?y
n
 + 
pp
y
pp p
n
()
!
()()
!
+
?+
++ 1
2
12
3
2
  ?+ +
++ …+ -
?
3
12 1
y
pp pp n
n
y
n
n
n

()() (( ))
!
 
  where p
xx
h
xx ph
n
n
=
-
=+ () . OR 
Chapter 06.indd   129 5/31/2017   10:59:37 AM