Lecture 4
Numerical
Analysis
Solution of
Non-Linear
Equations
Chapter 2
Introduction
Bisection Method
Regula-Falsi Method
Method of iteration
Newton - Raphson Method
Muller’s Method
Graeffe’s Ro...
Bisection
Method
(Bolzano)
ExampleExample
SolveSolve xx33
– 9– 9xx + 1 = 0+ 1 = 0
for the root betweenfor the root between
xx = 2 and= 2 and xx = 4 b...
Solution GivenSolution Given ff ((xx) =) = xx33
– 9– 9xx + 1.+ 1.
HereHere ff (2) = -9,(2) = -9, ff (4) = 29.(4) = 29.
The...
We further define,We further define,
and note thatand note that ff ((xx33) =) = ff (2.5) < 0, so that(2.5) < 0, so that
ff...
These results are presented in the table.These results are presented in the table.
nn xxnn f ( xf ( xnn ))
22 33 1.01.0
33...
Regula-Falsi
Method
Method of falseMethod of false
positionposition
Method of falseMethod of false
positionposition
Here, we choose two points xn
and xn -1 such that f (xn) and f
(xn-1) are of opposite signs.
Intermediate value property
s...
Thus, to find a real root of
f (x) = 0 using Regula-Falsi
method, we replace the part of
the curve between the points
A[xn...
Now, the equation of the chord
joining the points A and B is
(2.1)
Setting y = 0 in Eq. (2.1), we get
1 1
( )
( ) ( )
n n
...
Hence, the first approximation to the root
of f (x) = 0 is given by
(2.2)
we observe that f (xn-1) and f (xn+1) are of
opp...
Example
Use the Regula-Falsi method to
compute a real root of the equation
x3
– 9x + 1 = 0,
(i) if the root lies between 2...
Solution
Let f (x) = x3
– 9x + 1.
f (2) = – 9 and f (4) = 29.
Since f (2) and f (4) are of
opposite signs, the root of
f (...
Taking x1 = 2, x2 = 4 and
using Regula-Falsi method,
the first approximation is
given by
2 1
3 2 2
2 1
( )
( ) ( )
2 29
4 ...
Now f (x3) = –6.12644.
Since f (x2) and f (x3) are of
opposite signs, the root lies
between x2 and x3.
The second approxim...
Therefore f (x4) = – 3. 090707.
Now, since f (x2) and f (x4) are
of opposite signs, the third
approximation is obtained
fr...
This procedure can be continued till
we get the desired result. The first
three iterations are shown as in the
table.
n xn...
(ii) f (2) = – 9 and f (3) = 1. Since
f (2) and f (3) are of opposite signs,
the root of f (x) = 0 lies between 2
and 3. T...
Since f (x2) and f (x3) are of
opposite signs, the root lies
between x2 and x3.
The second approximation
to the root is gi...
Now, we observe that f (x2)
and f (x4) are of opposite
signs, the third approximation
is obtained from
4 2
5 4 4
4 2
( )
(...
This procedure can be
continued till we get the desired
result. The first three iterations
are shown as in the table.
n xn...
We observe that the value of the
root as a third approximation is
evidently different in both the
cases, while the value o...
Example
Use Regula-Falsi method to
find a real root of the equation
log x – cos x = 0
accurate to four decimal places
afte...
Solution
Given f (x) = log x – cos x.
We observe that
f (1) = 0-0.5403,and
f (2)=0.69315+0.41615=1.1093
Since f (1) and f ...
The first approximation is obtained
from
2 1
3 2 2
2 1
3
( )
( ) ( )
1.1093
2 1.3275
1.6496
( ) 02833 0.2409 0.0424
x x
x ...
Now, since f (x1) and f (x3) are
of opposite signs, the second
approximation is obtained as
4
3
4
(.3275)(.0424)
1.3275
0....
Similarly, we observe that f (x1)
and f (x4) are of opposite signs,
so, the third approximation is
given by
The required r...
Lecture 4
Numerical
Analysis
of 30

Numerical Analysis (Solution of Non-Linear Equations)

Bisection Method Regula-Falsi Method Method of iteration Newton - Raphson Method Muller’s Method Graeffe’s Root Squaring Method
Published on: Mar 3, 2016
Published in: Education      
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Transcripts - Numerical Analysis (Solution of Non-Linear Equations)

  • 1. Lecture 4 Numerical Analysis
  • 2. Solution of Non-Linear Equations Chapter 2
  • 3. Introduction Bisection Method Regula-Falsi Method Method of iteration Newton - Raphson Method Muller’s Method Graeffe’s Root Squaring Method
  • 4. Bisection Method (Bolzano)
  • 5. ExampleExample SolveSolve xx33 – 9– 9xx + 1 = 0+ 1 = 0 for the root betweenfor the root between xx = 2 and= 2 and xx = 4 by the= 4 by the bisection method.bisection method.
  • 6. Solution GivenSolution Given ff ((xx) =) = xx33 – 9– 9xx + 1.+ 1. HereHere ff (2) = -9,(2) = -9, ff (4) = 29.(4) = 29. Therefore,Therefore, ff (2)(2) ff (4) < 0 and hence(4) < 0 and hence the root lies between 2 and 4.the root lies between 2 and 4. LetLet xx00 = 2,= 2, xx11 = 4. Now, we define= 4. Now, we define as a first approximation to a root ofas a first approximation to a root of ff ((xx) = 0 and note that) = 0 and note that ff (3) = 1, so(3) = 1, so thatthat ff (2)(2) ff (3) < 0.(3) < 0. Thus the root liesThus the root lies between 2 and 3between 2 and 3 0 1 2 2 4 3 2 2 x x x + + = = =
  • 7. We further define,We further define, and note thatand note that ff ((xx33) =) = ff (2.5) < 0, so that(2.5) < 0, so that ff (2.5)(2.5) ff (3) < 0. Therefore, we define the(3) < 0. Therefore, we define the mid-point,mid-point, Similarly,Similarly, xx55 = 2 . 875 and= 2 . 875 and xx66 = 2.9375= 2.9375 and the process can be continued untiland the process can be continued until the root is obtained to the desiredthe root is obtained to the desired accuracy.accuracy. 0 2 3 2 3 2.5 2 2 x x x + + = = = 3 2 4 2.5 3 2.75, etc. 2 2 x x x + + = = =
  • 8. These results are presented in the table.These results are presented in the table. nn xxnn f ( xf ( xnn )) 22 33 1.01.0 33 2.52.5 -5.875-5.875 44 2.752.75 -2.9531-2.9531 55 2.8752.875 -1.1113-1.1113 66 2.93752.9375 -0.0901-0.0901
  • 9. Regula-Falsi Method Method of falseMethod of false positionposition Method of falseMethod of false positionposition
  • 10. Here, we choose two points xn and xn -1 such that f (xn) and f (xn-1) are of opposite signs. Intermediate value property suggests that the graph of y = f (x) crosses the x-axis between these two points and therefore, a root say lies between these two points.
  • 11. Thus, to find a real root of f (x) = 0 using Regula-Falsi method, we replace the part of the curve between the points A[xn, f(xn)] and B[xn-1, f (xn-1)] by a chord in that interval and we take the point of intersection of this chord with the x-axis as a first approximation to the root.
  • 12. Now, the equation of the chord joining the points A and B is (2.1) Setting y = 0 in Eq. (2.1), we get 1 1 ( ) ( ) ( ) n n n nn n y f x x x f x f x x x− − − − = − − 1 1 ( ) ( ) ( ) n n n n n n x x x x f x f x f x − − − = − −
  • 13. Hence, the first approximation to the root of f (x) = 0 is given by (2.2) we observe that f (xn-1) and f (xn+1) are of opposite sign. Thus, it is possible to apply to above procedure, to determine the line through B and A1 and so on. Hence, the successive approximations to the root of f (x) = 0 is given by Eq. (2.2). 1 1 1 ( ) ( ) ( ) n n n n n n n x x x x f x f x f x − + − − = − −
  • 14. Example Use the Regula-Falsi method to compute a real root of the equation x3 – 9x + 1 = 0, (i) if the root lies between 2 and 4 (ii) if the root lies between 2 and 3. Comment on the results.
  • 15. Solution Let f (x) = x3 – 9x + 1. f (2) = – 9 and f (4) = 29. Since f (2) and f (4) are of opposite signs, the root of f (x) = 0 lies between 2 and 4.
  • 16. Taking x1 = 2, x2 = 4 and using Regula-Falsi method, the first approximation is given by 2 1 3 2 2 2 1 ( ) ( ) ( ) 2 29 4 2.47368 38 x x x x f x f x f x − = − − × = − =
  • 17. Now f (x3) = –6.12644. Since f (x2) and f (x3) are of opposite signs, the root lies between x2 and x3. The second approximation to the root is given as 3 2 4 3 3 3 2 ( ) ( ) ( ) 2.73989 x x x x f x f x f x − = − − =
  • 18. Therefore f (x4) = – 3. 090707. Now, since f (x2) and f (x4) are of opposite signs, the third approximation is obtained from and f (x5) = – 1.32686. 4 2 5 4 4 4 2 ( ) 2.86125 ( ) ( ) x x x x f x f x f x − = − = −
  • 19. This procedure can be continued till we get the desired result. The first three iterations are shown as in the table. n xn+1 f (xn+1) 2 2.47368 -6.12644 3 2.73989 -3.090707 4 2.86125 -1.32686
  • 20. (ii) f (2) = – 9 and f (3) = 1. Since f (2) and f (3) are of opposite signs, the root of f (x) = 0 lies between 2 and 3. Taking x1 = 2, x2 = 3 and using Regula-Falsi method, the first approximation is given by 3 2 1 3 2 2 2 1 ( ) ( ) ( ) 1 3 2.9 10 ( ) 0.711 x x x x f x f x f x f x − = − − = − = =−
  • 21. Since f (x2) and f (x3) are of opposite signs, the root lies between x2 and x3. The second approximation to the root is given as 3 2 4 3 3 3 2 4 ( ) 2.94156 ( ) ( ) ( ) 0.0207 x x x x f x f x f x f x − = − = − = −
  • 22. Now, we observe that f (x2) and f (x4) are of opposite signs, the third approximation is obtained from 4 2 5 4 4 4 2 ( ) ( ) ( ) 2.94275 ( ) 0.0011896 x x x x f x f x f x f x − = − − = =−
  • 23. This procedure can be continued till we get the desired result. The first three iterations are shown as in the table. n xn+1 f (xn+1) 2 2.9 -0.711 3 2.94156 -0.0207 4 2.94275 -0.0011896
  • 24. We observe that the value of the root as a third approximation is evidently different in both the cases, while the value of x5, when the interval considered is (2, 3), is closer to the root. Important observation: The initial interval (x1, x2) in which the root of the equation lies should be sufficiently small.
  • 25. Example Use Regula-Falsi method to find a real root of the equation log x – cos x = 0 accurate to four decimal places after three successive approximations.
  • 26. Solution Given f (x) = log x – cos x. We observe that f (1) = 0-0.5403,and f (2)=0.69315+0.41615=1.1093 Since f (1) and f (2) are of opposite signs, the root lies between x1 = 1, x2 = 2.
  • 27. The first approximation is obtained from 2 1 3 2 2 2 1 3 ( ) ( ) ( ) 1.1093 2 1.3275 1.6496 ( ) 02833 0.2409 0.0424 x x x x f x f x f x f x − = − − = − = = − =
  • 28. Now, since f (x1) and f (x3) are of opposite signs, the second approximation is obtained as 4 3 4 (.3275)(.0424) 1.3275 0.0424 0.5403 1.3037 ( ) 1.24816 10 x f x − = − + = = ×
  • 29. Similarly, we observe that f (x1) and f (x4) are of opposite signs, so, the third approximation is given by The required real root is 1.3030. 5 4 5 (0.3037)(0.001248) 1.3037 0.001248 0.5403 1.3030 ( ) 0.6245 10 x f x − = − + = = ×
  • 30. Lecture 4 Numerical Analysis