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# Assignment # 1 (Lecture# 01 - 08) Of MTH603 (Spring 2014)

Maximum Marks: 30

Due Date: 12 May, 2014

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Question: 1                                                                                                                               Marks: 10

The roots of the following equation are and. Using Muller method approximates the root upto four decimal places.

Question: 2                                                                                                                               Marks: 10

Using Newton-Raphson method approximates the root of the following equation in the interval  upto the three decimal places.

Note: All the calculations should be in radian.

Question: 3                                                                                                                               Marks: 10

Using Regula Falsi method approximates the root of the following equation upto four decimal places.

Note: The following graph of above equation can be very helpful in finding the interval of the root. So, I am not mentioning the interval there as mentioned above. So, try to find the appropriate interval then apply the method. Otherwise, the convergence would be very slow.

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muller method konsa ha

MULLER METHOD

Muller's method is a generalization of the secant method. Instead of starting with two initial values and then joining them with a straight line in secant method, Mullers method starts with three initial approximations to the root and then join them with a second degree polynomial (a parabola), then the quadratic formula is used to find a root of the quadratic for the next approximation. That is if x0, x1 and x 2 are the initial approximations then x3 is obtained by solving the quadratic which is obtained by means of x0, x 1 and x2. Then two values among x0, x1 and x2 which are close to x3are chosen for the next iteration

x3 = x2 + z           ( * )

 where z = -2c b±Ö( b2-4ac)

a = D1/D2 ,  b = D 2 /D     and   c = f(x2)

D = h0h1(h0-h1) ,      D1 = (f0-c)h1-(f1-c)h,        D2 = (f1 -c)h02 - (f0-c)h12

h0 = x0-x,   h= x1-x2

Algorithm - Muller's Scheme

Given an equation f(x) = 0
Let the initial guesses be x0,  xand x2
Let xi = x2,  xi-1 = xand xi-2 = x0
Compute f(xi-2), f(xi-1) and f(xi)

Do

Compute
h = x - xi,  hi = xi - xi-1  and  hi-1 = xi-1 - xi-2
li = hi / hi-1di = 1 + li
gi = li2 fi-2 - di2 fi-1 +  ( ldi fi
ci = li  (li fi-2 - di fi-1 +   fi )
li+1 = min (  -2di fi / (g± Ö(gi2- 4di fici)  )
(take minimum in absolute sense)

Then xi+1 =  xi + ( x -  xi-1 li+1 ,       i = 2, 3, 4, . . .

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koi es asgnmnt ka sol nhiiiiiiiiiiiiiiiiiiiiii day ga kya

Numerical methods for solving either algebraic or transcendental equation are classified

into two groups

1. Direct Methods
2. Iterative Methods

Direct Methods:

Those methods which do not require any information about the initial approximation of

root to start the solution are known as direct methods.

1.   Graefee root squaring method
2.   Gauss elimination method
3.   Gauss Jordan method

(All these methods do not require any type of initial approximation.)

Iteratative Methods:

These methods require an initial approximation to start.

1.   Bisection method
2.   Newton raphson method
3.   Secant method
4.   jacobie method
5.   Muller’s Method

What is root?

x + 7= 0   is a linear equation

x = -7  -7 is root of  equation  x+7= 0

x-3=2  is a linear equation

x=5  5 is root of  equation  x-3=2 3 and 4 are the roots of this quadratic equation

neton raphson method ka solution k ly b guid kr dain

Question: 1 Marks: 10 The roots of the following equation are and. Using Muller method approximates the root upto four decimal places. What is root?

x + 7= 0   is a linear equation

x = -7  -7 is root of  equation  x+7= 0

x-3=2  is a linear equation

x=5  5 is root of  equation  x-3=2

Regula-Falsi method (Method of false position)
Here we choose two points xn and n 1 x − such that 1 ( ) ( ) n n f x and f x− have opposite signs.
Intermediate value property suggests that the graph of the y=f(x) crosses the x-axis
between these two points and therefore, a root lies between these two points.

.