MTH603 Final Term Solved Mega File by Ashfaq

Octal Number System (Base 8)=0,1,2,3,4,5,6,7

MTH603 FINAL TERM MEGA FILE BY ASHFAQ

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Number systems, Errors in computation, Methods of solving non-linear equations, Solution of linear system of equations and matrix inversion, Eigen value problems, power method, Jacobi’s method, Different techniques of interpolation, Numerical differentiation and integration, Numerical integration formulas, different methods of solving ordinary differential equations.

Representation of numbers Binary Number System (Base 2)=0,1 Used in Computer Internal Operations and Calculation

ERROR Types of error are,

•  Inherent errors,
•  Local round-off errors
•  Local truncation errors

Decimal Number System (Base 10)=0,1,2,3,4,5,6,7,8,9 World Wide used and the most common system Hexa Number System (Base 16)=0,1,2,3,4,5,6,78,9,A,B,C,D,E,F Used in Processor Register Addressing Errors in computations The initial approximation It may be found by two methods either by graphical method or Analytical method Graphical method Each Topic One Solved Question with Steps:Compiled from Past assignment Preliminary material:

Introduction To Numerical Analysis

Course Content:

x11  2.104492188 and x9  2.10546875 then x12  ( x11  x9 )

 ( 2.104492188  2.10546875)  2.104980469

Now f (2.104980469)  (2.104980469)3  (2.104980469)2  2.104980469  7  0.0011  0 As f (2.104492188) f (2.104980469)  0.0038(0.0011)  0.00000418  0

As the next root lies between 2.104492188 and three decimal places. 2.104980469 and these roots are equal up to

So, the required root up to three decimal places is 2.104 Example From Handout at page # 4 Solve x3 −9x+1=0 for the root between x=2 and x=4 by bisection method Solution: Here we are given the interval (2,4) so we need not to carry out intermediate value property to locate initial approximation. Here f(x) =x3  9x  1  0 Now f(2)=-9 and f(4)=29. Here f(2).f(4)<0 So the root lies between 2 and 4. So, xo  2, x1  4 and x2  xo  x 1 =3 (This formula predicts next iteration)

Now f(3)=1,Here f(2).f(3)<0 So the root lies between 2 and 3. Repeat above process to required number of iteration.

Regula Falsi Method

Formula for the Regula Falsi Method is x3  x2  x2  x1 f (x2)  f (x1) f (x2) or

x ·  x  · xn  xn1 · f  x 

( n1) n

f  x   f  x 

n

n n1

Steps: If the interval is given , check whether the root lies in between the interval or not by the principle that states if both number have opposite sign , then the root lies in between interval. Find the next approximation with the help of the formula.

Q2. Marks 10

Solution:

As f (x)  cos x  xex f (0)  cos 0  0e0  1 0 f (1)  cos11e1  0.5403  2.718  2.1779 So root of the eq. lie between 0 and 1.Let x0  0 and x1  1 Thus f (x0 )  1 and f (x1 )  2.1779

The formula for finding the root of the function f(x) by Regula-Fasli is given by. x  x  xn  xn1

f  x 

( n1) n f  x   f  x  n n n1 For n=1 we have, x  x  x1  xo f  x   1  1  0 -2.17797  1  2.17797  1  0.68533  0.31467 f  x   f  x  -2.17797 1 3.17797 o f  x   cos(0.31467)  0.31467 e0.31467  0.950899  0.314671.369806 =0.519863

Since f (x1) and f (x2) are of opposite signs, the root lies between x1 and x2. Therefore, for n=2 we have, x  x  x2  x1 f  x   0.31467  0.31467 1 0.519863 f  x   f  x  0.519863  -2.17797

 0.31467 

1. 68533 0.519863  0.44673
2. 697833

f  x   cos(0.44673)  0.44673 e0.44673  0.901864  0.446731.563191 =0.20353

Since f (x1) and f (x3) are of opposite signs, the root lies between x1 and x3. Therefore, for n=3 we have, x  x  x3  x1 f  x   0.44673 0.44673 1 0.20353

f  x3   f  x1  0.20353  -2.17797  0.44673  0.54327 0.20353  0.493159

2. 3815

The required root after 3rd iteration using Regula-Falsi method is 0.493159

Q2. (Note: accuracy up to four decimal places is required)

Marks: 10 SOLUTION

Use the Regula Falsi (method of false position) to solve the equation x3  4x  9  0 Use Regula-Fasli method to compute the root of the equation f (x)  cos x  xe x In the interval [0, 1] after third iteration.