ENGINEERING MATHEMATICS III

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ENGINEERING MATHEMATICS - III
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Numerical Analysis
Introduction
Limitations of analytical methods led to the
evolution of Numerical methods. Numerical Methods
often are repetitive in nature i.e., these
consist of the repeated execution of the same
procedure where at each step the result of the
proceeding step is used. This process known as
iterative process is continued until a desired
degree of accuracy of the result is obtained.
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Solution of Algebraic and Transcendental
Equations The equation f(x) 0 said to be
purely algebraic if f(x) is purely a polynomial
in x. If f(x) contains some other functions like
Trigonometric, Logarithmic, exponential etc. then
f(x) 0 is called a Transcendental equation.
Ex (1) x4 - 7x3 3x 5 0 is algebraic
(2) ex - x tan x 0 is transcendental
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Bisection Method This method is used in
locating a root of the equation f(x) 0 between
a and b.   If f(x) is continuous
between a and b, f(a) and f(b) are opposite in
sign then exists a root between a and b.   For
simplicity, let f(a) lt 0 and f(b) gt 0 The first
approximation to the root is If f(x1) 0 then
x1 is the root of f(x) 0 If f(x1) is ve then
root lies between a and x1 and the second
approximation to the root is
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Now if f(x2) is - ve then the root lies between
x2 and x1 and the third approximation to the
root is This process is continued until the
root is found with desired accuracy
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Problem 01
Solve x3 - 9x 1 0 for the root lying between
2 and 3 using bisection method in six
stages. f(x) x3 - 9x 1 0 f(2) -9
-ve f(3) 55 ve ? Root lies between
2.5 and 3
Solution
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f(x2) f(2.75) -2.9 -ve ?
Root lies between 2.75 and 3
f(x3) 2.875 -1.111 -ve ? Root
lies between 2.875 and 3

f(x4) -0.09 -ve
? Root lies between 2.9375 and 3

f(x5) 0.451 ve ? Root lies
between 2.9375 and 2.969
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f(x5) 0.17 ve ?
Root lies between 2.9375 and 2.953
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Method of false position or Regula-Falsi
Method This is a method of finding a real root
of an equation f(x) 0 and is slightly an
improvisation of the bisection method. Let x0
and x1 be two points such that f(x0) and f(x1)
are opposite in sign.
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Let f(x0) gt 0 and f(x1) lt 0 \ The graph of y
f(x) crosses the x-axis between x0 and x1 ?
Root of f(x) 0 lies between x0 and x1 Now
equation of the Chord AB is
When y 0 we get x x2
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• Which is the first approximation
• If f(x0) and f(x2) are opposite in sign then
  second approximation
• This procedure is continued till the root is
  found with desired accuracy.

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Problem 02
Find a real root of x3 - 2x -5 0 by method of
false position correct to three decimal places
between 2 and 3.
Solution

• Let f(x) x3 - 2x - 5 0 f(2) -1
  f(3) 16
• a root lies between 2 and 3
• Take x0 2, x1 3 ? x0 2, x1 3

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2.0588 f(x2) f(2.0588)
-0.3908   ? Root lies between 2.0588 and 3
Taking x0 2.0588 and x1 3 f(x0) -0.3908,
f(x1) 16 Repeating the process the successive
approximations are x5 2.0915, x6 2.0934, x7
2.0941, x8 2.0943 Hence the root is 2.094
correct to 3 decimal places.
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2.0813 f(x3) f(2.0813)
-0.14680 ? Root lies between 2.0813 and
3 Taking x0 2.0813 and x1 3 f(x0)
0.14680, f(x1) 16
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Problem 03
Find the root of the equation xex cos x using
Regula falsi method correct to three decimal
places.
Solution
Let f(x) cosx - xex Observe f(0)
1 f(1) cos1 - e -2.17798 ? root lies
between 0 and 1 Taking x0 0, x1 1 f(x0)
1, f(x1) -2.17798
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f(x2) f(0.31467) 0.51987 ve ? Root lies
between 0.31467 and 1 x0 0.31467, x1
1 f(x0) 0.51987, f(x1) -2.17798
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f(x3) f(0.44673) 0.20356 ve ? Root lies
between 0.44673 and 1
Repeating this process x5 0.50995, x6
0.51520, x7 0.51692, x8 0.51748 x9
0.51767, etc Hence the root is 0.518 correct to
4 decimal places
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Newton Raphson Method This method is used to
find the isolated roots of an equation f(x) 0,
when the derivative of f(x) is a simple
expression. Let m be a root of f(x) 0 near
a. ? f(m) 0 We have by Taylor's series
Ignoring higher order terms f(m) f(a)
(m - a) f' (a) 0
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Let a x0, m x1
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Problem 04
Using Newton's Raphson Method find the real root
of x log10 x 1.2 correct to four decimal
places.
Solution
Let f(x) x log10 x - 1.2 f(1) -1.2, f(2)
-0.59794, f(3) 0.23136
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Let x0 2.5 (you may choose 2 or 3 also)
Repeating the procedure x3 2.7406
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Problem 05
Using Newton's Method, find the real root of xex
2. Correct to 3 decimal places.
Solution
Let f(x) xex - 2 f(0) -2 f(1) e
- 2 0.7182 Let x0 1 f' (x)
(x 1) ex We have
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Problem 06
Find by Newton's Method the real root of 3x
cosx 1 near 0.6, x is in radians. Correct for
four decimal places.
Solution
Let f(x) 3x - cosx - 1 f'(x) 3 sinx
Since x1 x2 . The desired root is 0.6071
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Problem 07
Obtain the iterative formula for finding the
square root of N and find
Solution
? f(x) x2 N f'(x) 2x Now
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Since x2 x3 6.4031
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Problem 08
Obtain an iterative formula for finding the p-th
root of N and hence find (10)1/3 correct to 3
decimal places.
Solution
Let xp N or xp - N 0 Let f(x)
xp N f (x) pxp-1
? Use x0 2, p 3, N 10
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Problem 09
Obtain an iterative formula for finding the
reciprocal of p-th root of N. Find (30)-1/5
correct to 3 decimal places.
Solution
Let x -p N or x -p - N 0 ?
f(x) x -p N f'(x) -px -p - 1 Now
We use x0 0.5, p 5, N 30
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Finite Differences Forward difference operator
(D) ?f(x) f(x h) - f(x) ?yr y r 1 -
yr, r 0, 1, 2, , n -1
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first forward differences
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Difference Table
?y0, ?2y0, ?3y0,. are called the leading
differences
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Ex The following table gives a set of values
of x and the corresponding values of y f(x)
Form the difference table and find ?f (10), ?2f
(10), ?3f (20) ?4f (15)
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Note The nth differences of a polynomial of n
the degree are constant.
If f(x) a0 xn a1 xn-1 a2 xn-2 an, a0 ?
0 then
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If y0 1, y1 11, y2 21, y3 28, y4 29.
Construct the difference table.
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Backward difference operator (?)
Let y f(x) We define ?f(x) f(x) - f(x -
h) i.e. ?y1 y1 - y0 ?y0 ?y2 y2 - y1
? y1 ' ' ?yn yn - yn-1 ? yn - 1
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Note ly ?n f(x nh) ??n f(x)
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Backward difference table
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Example 1. Form the difference table for
and find ?y (30), ?2y (70), ?5y (90)
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2. Given
Construct the difference table and write the
values of ?f (4), ?2f (4), ?3f (3)
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Central Difference Operator (?)
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Note 1
Note 2
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Central difference table
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Problem
1. Show that
Solution
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Problem
2. Show that
Solution
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Problem
3. Given f(-2) 12, f(-1) 16, f(0) 15, f(1)
18, f(2) 20 form the central difference
table and write down the values of ?y-3/2,
?2y0, ?3y1/2 by taking x0 0
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Solution
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Shift operator (E)
E f(x) f(x h) E2 f(x) f(x 2h) En f(x)
f(x nh)
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Note 1
Note 2
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Interpolation
The word interpolation denotes the method of
computing the value of the function y f(x) for
any given value of x when a set (x0, y0), (x1,
y1),(xn-1, yn-1) are given.
Note Since in most of the cases the exact form
of the function is not known. In such cases the
function f(x) is replaced by a simpler function
?(x) which has the same values as f(x) for x0,
x1, x2.,xn-1.
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is called the Newton Gregory forward difference
formula

• Note
• Newton forward interpolation is used to
  interpolate the values of y near the beginning of
  a set of tabular values.
• y0 may be taken as any point of the table but
  the formula contains those values of y which come
  after the value chosen as y0.

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Problem
The table gives the distances in nautical miles
of the visible horizon for the given heights in
feet above the earths surface.
Find the values of y when i) x 120, ii) y
218
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Solution
Choose x0 100
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i)
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ii) Let x 218, x0 200,
15.7
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2. Find the value of f(1.85)
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Problem
Given sin 45o 0.7071, sin 50o 0.7660, sin
55o 0.8192, sin 60o 0.8660. Find sin 48o.
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Solution
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Problem
From the following data find the number of
students who have obtained ? 45 marks. Also
find the number of students who have scored
between 41 and 45 marks.
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Solution
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f(45) - f(40) 70 Number of students who have
scored between 41 and 45.
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Problem
Find the interpolating polynomial for the
following data f(0) 1, f(1) 0, f(2) 1,
f(3) 10. Hence evaluate f(0.5)
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Solution
f(0.5) 0.625
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Problem
Find the interpolating polynomial for the
following data
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Newton Gregory Backward Interpolation formula
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Problem
The values of tan x are given for values of x in
the following table. Estimate tan (0.26)
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Solution
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Problem
The deflection d measured at various distances x
from one end of a cantilever is given by the
following table. Find d when x 0.95
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Solution
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Problem
The area y of circles for different diameters x
are given below
Calculate area when x 98
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Solution
y 7542
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Problem
Find the interpolating polynomial which
approximates the following data.
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Solution
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Central Difference Interpolation Formulae
Bessels and Stirlings are two of the central
difference interpolation formulae. These are used
to find y at a value of x that is in the
middle of x0 and xn
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Central difference interpolation formula
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a) Stirlings formula
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b) Bessel's formula
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Problem
Apply sterling's formula to find f(14.2) from
the following table
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Solution
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Problem
Apply stirling's formula to find a polynomial
f(x) of degree 4 that approximates the following
data and hence find f(2.5)
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Solution
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Problem
Using Bessels formula find f(12.5) for the data
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Solution
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Problem
Using Bessel's formula find 3rd degree
polynomial that approximates the following
data f(0) 2, f(1) 3, f(2) 8, f(3) 23
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Solution
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Interpolation with unequal intervals
Newton backward and forward interpolation is
applicable only when x0, x1,,xn-1 are equally
spaced.
Now we use two interpolation formulae for
unequally spaced values of x.
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i. Lagranges formula for unequal intervals If
y f(x) takes the values y0, y1, y2,.,yn
corresponding to x x0, x1, x2,,xn then
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is known as the lagrange's interpolation formula
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ii) Divided differences (?)
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second divided difference
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Newton's divided difference interpolation formula
Newton backward and forward interpolation is
applicable only when x0, x1,,xn-1 are equally
spaced.
is called the Newton's divided difference formula.
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Note Lagrange's formula has the drawback that
if another interpolation value were inserted,
then the interpolation coefficients need to be
recalculated.   Inverse interpolation Finding
the value of y given the value of x is called
interpolation where as finding the value of x for
a given y is called inverse interpolation.   Sinc
e Lagrange's formula is only a relation between x
and y we can obtain the inverse interpolation
formula just by interchanging x and y.
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is the Lagranges formula for inverse
interpolation
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Problem
The following table gives the values of x and y
Find x when y 12 using Lagranges inverse
interpolation formula.
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Solution
Using Langrages formula
3.55
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Problem
Given the values
Evaluate f(9) using (i) Lagrange's formula
(ii) Newton's divided difference formula
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Solution
i) Lagranges formula
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(ii) Newton's divided difference formula
f(9) 150 121 (9 - 5) 24 (9 - 5) (9 - 7)
1(9 - 5) (9 - 7) (9 - 11) 810
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Problem
Using i) Langranges interpolation and ii)
divided difference formula. Find the value of y
when x 10.
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Solution
Lagranges formula
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Divided difference
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Problem
If y(1) -3, y(3) 9, y(4) 30, y(6) 132 find
the lagranges interpolating polynomial that
takes the same values as y at the given
points. Given
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Solution
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Problem
Find the interpolating polynomial using Newton
divided difference formula for the following
data
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Solution
F(x) 2 (x - 0)(1) (x - 0) (x - 1) (4) (x
- 0) (x - 1) (x - 2) 1 x3 x2 - x 2
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Central Difference Interpolation Formulae
1.Find y(35) given y(20) 5/2, y(30) 439,
y(40) 346 and y(50) 243 We get the
following difference table
xp x0 ph ? 35 3010p ? p 0.5
?y(35) 439
394.375.
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Examples
2. Find the third degree polynomial that
approximates the function f(x) given by f(0)
2, f(1) 3, f(2)8 f(3) 23. Hence find
f(1.5) f(2.3). The difference table is given
by
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xp x0 ph ? x 1p ? p x - 1
f(x) x3 x2 x 2 is the polynomial.
? f(1.5) 4.625 and f(2.3) 11.177
Note Bessels formula gives better results when
there are an even number of
observations
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Example
.
Find the population of a town (in appropriate
units) for the year 1974 given that
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• The difference table for the data is given below

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• xp x0 ph ? 1974 1969 10p ? p 0.5

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y (1974) 32.461
Note Stirlings formula is used when an odd
number of observations is given.
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1. Find f(4) given f(0) -4, f(2) 2, f(3)
14 and f(6) 158 The divided difference
table is
? f(4) -4 (4 ) (3) (4) (2) (3) (4) (2)
(1) (1) 40
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2. Fit an interpolating polynomial for u4
48, u5 100, u6 180 , u8 448, u10 900 and
u11 1210
The table of divided differences is
? y 48 (x -4) 52 (x-4) (x-5) 14 (x-4)
(x-5) (x -6) x3- x2
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Examples
1. Find u5 if u0 1, u319, u449, u6 181
At x 5, u is given by
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Examples
2. Fit a polynomial x f(y) for the data

The polynomial, using Lagranges inverse
interpolation formula, is given by
? x y2 1 is the required polynomial
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Examples
3. Apply Lagranges formula and obtain the root
of f (x) 0 if f(30) -30,
f(34) -13, f(38) 3 and f(42) 18 The
data can be arranged in the form
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Numerical Differentiation
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NUMERICAL DIFFERENTIATION
Numerical differentiation is the determination of
approximate values of the derivatives of a
function f(x) that is given by a table of
values. When the values of y (a function of x)
at some equally spaced values of x are known,
approximate values of f (x), f(x) and so on
can be calculated by numerical methods using
interpolation formulas.
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If y0, y1, yn are the values of y
corresponding to x0, x1....xn (equally
spaced) then
at x x
0,
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Note 1. The above formulae are got by
differentiating Newtons forward
and backward difference interpolation formulae
w.r.t. p and using
2. If f (x) and f (x) are
to be found at a tabulated value
of x, the value of p is chosen to be 0 in the
formulae.
otherwise p is given by
or
according as x lies close to x0 or xn
respectively.
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Examples
1. Find y (0) and y (0) from the given
table
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Forming the difference table, we get
We have h 1 ? f (0) - 27.9
and f (0) 117.67
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Examples
2. The population of a certain town is shown
below . Estimate the rate of growth of
population in the year 1951
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• The above data, after rearranging, we get

Here h 10 . We get the rate of growth of
population as
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NUMERICAL INTEGRATION
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Numerical Integration
The process of evaluating a definite integral of
the form
where a b are given and f(x) is a function
given analytically by a formula or
empirically by a table of values.
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Geometrically, I is the area under the curve
f(x ) between x a and x b.
The method here is to approximate the integrand
f(x) by polynomials. The given interval (a ,
b) is divided into a number of subintervals of
equal width (say h). Replacing f(x) by
polynomials of different degrees lead to
different rules of integration.
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Numerical Integration
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Rules of integration
Trapezoidal rule
Approximating f(x) by a piecewise linear
function, we obtain the Trapezoidal rule given
by
y
where b a nh and n
is an integer
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y
a x0 x1 x2
xn-1
xn b Trapezoidal Rule
x
0
x
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Examples
1. Evaluate
, choosing 7 ordinates.
Hence find an approximation for ln 2.
7 ordinates implies 6 subintervals.
? h , width of a subinterval 1/6
we have the following table of values
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Examples
Trapezoidal rule gives

ln 2

But
?
ln 2 ? 0.2316 or ln 2 ? 0.6948
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Examples
choosing subintervals of equal width
2. Evaluate
Compare with the theoretical value.
Here y f(x) cosx, h
The table of values is
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Examples
By Trapezoidal rule,
But
? The error is ? 0.002
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Simpsons One Third Rule
Approximating f(x) by a piecewise quadratic
function gives Simpsons one third rule
Here the number of subintervals is a multiple of
2.
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Simpsons One Third Rule
y
0
x
a x0 x1 x2 x3 x4 - - - - -
- - - - x2n-2
x2n-1 x2nb Simpsons rule
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Examples
1. Evaluate
by dividing the interval (0,1) into 4
equal subintervals
and hence find the value of ?

correct to four decimal places.
. The table of values is
f(x)
h
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By Simpsons rule


But
? ? ? 4 (0.7854) 3.1416, correct to four
decimal places.
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Simpsons Three Eighth Rule
The integrand f(x), when approximated by a
piecewise cubic function, the required integral
reduces to
This is Simpsons three eighth rule. Here the
number of subintervals is a multiple of 3.
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Examples
Evaluate
taking 6 equal strips and hence
find an approximation for ? .
Here h 1/6, with this the following table
is formed
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Examples
By Simpsons 3/8th rule, the given integral is
equal to
But
?/6

• Using Simpsons 3/8th rule, an approximate value
  of
• ? is 3.1416.

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Weddles Rule
Weddles rule of integration is obtained when
f(x) is approximated by a piecewise sixth degree
function and is given by

Here the number of subintervals is a multiple
of 6.
Note Among all the four rules, Simpsons one
third rule is sufficiently accurate
for most of the problems.
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Examples
1. Evaluate
choosing 7 ordinates.
Compare with the theoretical value.
h 1/6 gives the following table of values
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Examples
By Weddles rule, the integral is
But

• The value of the integral is correct to 4
  decimal places
• by Weddles rule.

Note 1. Simpsons one third rule gives 0.5009
2. Simpsons three eighth rule
gives 0.50018
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Examples
2. Evaluate
using all the four rules and
compare with the exact value 4.05095.
Let the number of subintervals be 6. Then h
0.2
Letting y sin x ln x ex, the values
of xi and the corresponding yi are
tabulated as below
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Examples
we get the value of the given integral as
4.07146 by Trapezoidal,
4.05208 by Simpsons 1/3rd ,
4.05293 by Simpsons 3/8th
and 4.05139 by Weddles rule.
We observe that Weddles rule gives the best
approximation compared to the other three rules.
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Numerical Solution of first
Order ODE
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Numerical Solution of first Order ODE

• There are various numerical methods to solve the
  initial value
• problem

The methods yield solutions either as power
series in x or directly as a numeric at a given
x. Here we study four numerical methods.
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Methods of solution
Runge-Kutta Fourth order
Milnes predictor-corrector
Eulers Modified method
Taylors series
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Taylors Series Method
Given the problem
y(x) is expanded in the form of a Taylors series
about the point x x0 as
The value of y at any given x can be found by
substituting for y, y etc. using the given
equation. The above power series gives the
values of y for every x for which the
series converges.
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Examples
1. Evaluate y(0.1) correct to six places of
decimals if y(x) satisfies y xy 1
y(0) 1. y xy 1 ? (y)(0) 1.
Differentiating y with reapect to x we get
y xy y ? y(0) 1 y
xy 2y ? y(0) 2 y(IV)
xy 3y ? y(IV) (0) 3 y(V)
xyIV 4y ? yv(0) 8 yVI xyv
5yIV ? yvI(0) 15
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Examples
The Taylors series expansion for f(x) about
the point x x0 0 is

when x 0.1, y(0.1) 1.105347 correct to
6 decimal places.
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Example
2. Solve y y2 x , y(0) 1 using
Taylors series method and compute y(0.1)
and y(0.2) Given y y2 x, y(0) 1
y 2yy 1 ? y(0)
3 y 2y2 2yy ? y(0)
8 yIV 6yy 2yy ? yIV (0) 34
Taylors series expansion for y(x) about the
point x 0, upto fourth degree terms is
? y (0.1) 1.116475 and y (0.2) 1.272933
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Modified Eulers Method
This is a predictor - corrector method which
consists of two formulae predictor formula and
corrector formula. Given the problem
To find y at x x0 h, where h is a small
increment in x, follow the steps given below
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Step 1 Estimate y, using the formula y1 y0
h f (x0, y0), called the predictor
formula
Step 2 Find
, first refined value of the estimate y1 using
the corrector formula given by
y0
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Continue this refinement until we get
with y1(i) and y1(i-1) as equal (to the
desired accuracy) (the ith and the i-1th
corrections of y are equal).
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Numerical Solution of first order ODE
Modified Eulers Method
Note 1. Predictor corrector method is the
technique of refining an initially
crude estimate of yi by means of
a more accurate formula. 2..This
method is a modification of Eulers method.
Here the predictor formula or Eulers
formula yi yi-1 h
f(xi-1, yi-1) is used to find successive
values of y. 3. Geometrically,
the curve in this method is
approximated by a line through the point (x0,
y0) having the slope as the
average of the slopes at (x0, y0)
and (x0h, y1) which is manifested in the
corrector formula.

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Examples

• Apply 5 corrections of Eulers modified method to
  
• find an approximate value of y (0.1) for the
  problem

Let h 0.1. Using the predictor formula, we get
y1 1.1 By corrector formula we get the
corrections y1(1) 1.09167 y1(2)
1.09161 The 3rd, 4th and 5th corrections give
1.09161
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Examples
2. Given
y(1) 1, find an approximate value of
y at x 2 in steps of 0.2
First Step
y1 1 0.221 1.6
y1(1) 1 0.1 3 2
1.639
y1(2) 1 0.1 3 2
1.640
1.640
y1(3) 1 0.1 3 2
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Second Step
x0 1.2, y0 1.640 ? f(x0, y0)
3.403 y1 1.64 0.2 (3.403) 2.3206
y1(1) 1.64 0.1 3.403 2
2.361
y1(2) 1.64 0.1 3.403 2
2.362
y1(3) 2.362
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Third Step

• x0 1.4, y0 2.362 ? f(x0, y0)
  3.818
• y1 2.362 0.2 (3.818) 3.126

3.167
y1(1) 2.362 0.1 3.818 2
y1(2) 3.169 y1(3)
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Fourth Step

• x0 1.6, y0 3.169 ? f(x0, y0)
  4.252
• y1 3.169 0.2 (4.252) 4.02

y1(1) 3.169 0.1 4.252 2
4.063
y1(2) 4.065 y1(3)
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Modified Eulers Method
Examples
Fifth Step

• x0 1.8, y0 4.065 ? f(x0, y0)
  4.705
• y1 4.065 0.2 (4.705) 5.006

y1 (1) 4.065 0.1 4.705 2
5.052
y1(2) 5.053 y1(3) ?y(2)
5.053
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Runge Kutta Fourth Order Method
This numerical method, also referred to as
Runge-Kutta method, is a most commonly used
method to solve a linear as well as a non-linear
differential equation. This method does not
require higher order derivatives of the function.
To solve an initial value problem
in finding y at a given x.
Let x x0 h and the corresponding y
y0 k.
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For a known h, k is found as the weighted
mean of the quantities k1, k2, k3 and k4
which are given by the formulae
k1 h f(x0, y0),
k2 h f(x0 h/2, y0 k1/2),
k3 h f(x0 h/2, y0 k2/2)
and k4 h f(x0 h, y0 k3)
k is computed using k (k1 2 k2 2
k3 k4 )/6.
176
Examples
1. Find y(0.2) given
choosing h 0.1.
Step1
x0 0, y0 1, h 0.1, f(x0, y0) 0.5 k1
(0.1)(0.5) 0.05
k2
k3
k4
? k 0.06653 ? y(0.1) 1 0.06653
1.06653
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Step2
x0 0.1, y0 1.06653, h 0.1, f(x0, y0)
0.83327. k1 (0.1)(0.83327) 0.083327
k2
k3
k4
? k 0.10069 ? y (0.2) 1.1672.
178
Examples
2. Evaluate y(1.1) given
Let h 0.1, f(x,y)
x0 1, y0 1, f(x0, y0) 0
179

• k1 0.1 (0) 0

- 0.004535
k2
k3
- 0.004319
k4
- 0.007871

• k - 0.004263
• ? y(1.1) 0.9957

180
Milnes Method
This is a predictor- corrector method of solving
a differential equation that can be used to find
y at a given x, provided the previous four
consecutive values of y are known. The method is
explained below
Given
To find an approximate value of y at x xn
x0 nh.
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Using y(x0) , compute y1 y(x0 h), y2
y(x02 h) and y3 y(x03 h), by any method (
Taylors series method can be used). Then
calculate f0 f (x0, y0), f1 f(x1,y1),

f2 f (x2, y2) and f3 f (x3, y3)
182
To find y4 y (x0 4h), use the predictor
formula
Now, let f4 f(x0 4h, y4)
The value of y4 is corrected using the
corrector formula y4 (c) y2 h/3
(f2 4 f3 f4)
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Then an improved value of f4 is computed and
again the corrector is applied to improve y4.
This process is repeated until two consecutive
corrected values of y4 remain unchanged (to the
desired accuracy). The above procedure can be
continued to find further values of y.
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Examples
1. Given
Find y(0.4) correct to 3 decimal places given
x 0.1 0.2 0.3 y
1.105 1.223 1.355
Form the table of values
185
Examples
Predict y4 by y0 4h/3 (2f1- f2 2f3)
1.579 ? f4 1.998
Correct y4 by y2 h/3 (f2 4 f3 f4)
First correction
Second correction
Third correction
? y(0.4) 1.538 correct to 3 decimal places
186
Examples
2. Use Milnes method to find y(0.3) from
y x2 y2 y(0)1
Find the initial values y(-0.1), y(0.1) and
y(0.2) using Taylors series method.
Here h0.1, x0, y1, f(x,y)x2 y2
y x2 y2 ? y(0)
1 y2x2yy ?
y(0) 2 y22y22yy ?
y(0) 8 y6yy 2yy ?
y(0) 28
187
Examples
Using Taylors series expansion,
y(0) . . . .
y(x) y(0) x.y(0)
y(0)
y(0)
x4 upto 4th degree terms.
x3
1 x x 2
?y(-0.1) 0.9088, y(0.1) 1.1114, y(0.2)
1.2525
188
Examples
Form the table of values
y4(p) y0 4h/3(2f1- f2 2f3)
1.4384 ? f4 2.1591, using x4 0.3
? first correction

1.43937 ? f4 2.1618
189
Examples
second correction
third correction
? y(0.3) 1.4395 correct to 4 decimal places.