Blog: Mth603 Assignment No. 1 Fall 2011 Solution

Assignment #  1

MTH603 (Fall 2011)

          

Total marks: 10

Lecture # 01-08  

Due date: 26-10-2011                                                                

  

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Question#1                                                                                            Marks 10

Find the real root of the equation  by Bisection method.

Perform three iterations only.

Note: Take any interval in which roots of the equation lie.

Question#2                                                                                             Marks 10

Use Regula-Fasli method to find the real root of the equation

Correct to four decimal places after three successive approximations in (-2,-1).

Note: All the calculation should be done in the radian mode only.

Question#3                                                                                             Marks 10

Apply Newton-Raphson method to determine a root of the equation.

Correct to three decimal places using the initial approximation.

Only three iterations needed.

Note: All the calculation should be done in the radian mode only.

Question#4                                                                                             Marks 10

Find the root of the equation by Second Method taking

as two starting values. Do three iterations only.

SOLUTION:

:::Q1:::

f (x) = x3 – 3x – 5 = 0

f (1) = (1)3 – 3(1) – 5= -7

f (3) = (3)3 – 3(3) – 5 = 13

As f(1) f(3)< 1 therefore roots lies between 1 and 3

We have x0 = 1 , x1 = 3

Now x2 = x0 + x1 /2 = 1+3/2 = x2 = 2

f(2) = (2)3 – 3 (2) – 5 = -3

As f(3) f(2) <0 therefore roots lies between 3 and 2

X1 = 3 , x2 = 2,

X3 = x1 + x2 /2 = 3+2/2 = 2.5

F(2.5) = (2.5)3 -3(2.5) -5 = 3.125

As f(2) f(2.5) < 0 therefore roots lies between 2 and 2.5

X2 = 2 , x3 = 2.5

X4 = x2 + x3 =2+2.5 / 2 = 2.25

f(2.25) = (2.25)3 – 3(2.25) – 5 = -0.36

Now the Real root of the equation after three iterations is 2.25

:::Q2:::

x1 = -2 , x2 = -1

f(x) = x3 – sinx + 1 = 0

f(x1) = (-2)3 – sin(-2) + 1

= -8-(-0.9092)+1

= -6.090

f(x2) = (-1)3 – sin(-1) + 1

= -1 – (0.8414)+1

=0.8414

As f(x1) and f(x2) <0 so the root lies between x1 and x2 the first approximation is

obtained form

X3 = x2 – x2 – x1 / f(x2) – f(x1) f(x2)

x3 = (-1) – (-1) – (-2)/ 0.8414 – (-6.0908) 0.8414

= -1.12137

f(x3) = (-1.12137)3 – sin(-1.12137) + 1

= 0.49060

Now , since f(x1) and f (x3) are of opposite signs , the second approximation is

obtained as

Now we will find x4

X4 = x3 - x3 – x1 / f(x3) – f (x1) f(x3)

= -1.12137 - -1.12137 - (-2) / 0.4906 – (-6.0908) (0.4906)

= -1.18686

f(x4) = (-1.18686)3 –sin(-1.18686) + 1 = 0.25534

Now since f(x1) and f(x4) are of opposite signs , the second approximation is

obtained as now we will find x5

X5 = x4 - x4 – x1 / f(x4) – f(x1) f(x4)

= -1.18686 - -1.18686 - (-2) 0.25534

0.25534 – (-6.0908)

= -1.21957

f(x5) = (-1.21957)3 – sin (-1.21957) + 1 = 0.12502

:::Q3:::

f(x) = cos x – xex

f ’ (x) = -sin x –ex –xe x

now we have initial approximation x0 =1

f (x0) =cos (1) – (1)e1 = -2.1779

f ‘ (x0) = -sin(1) – e1 – 1.e1 = -6.2780

x1 = x0 – f (x0) /f ‘ (x0)

= 1 – (-2.1779) / (-6.2780) = 0.6530

f(x1) = cos(0.6530) – (0.6530) .e 0.6530 = -0.4603

f ‘ (x1) = - sin (0.6530)-e 0.6530 - 0.6530 .e 0.6530 =-3.7834

x2 = x1 – f(x1) / f ‘ (x1)

= 0.6530 – (-0.4603) / (-3.7834) = 0.5313

f (x2) = cos(0.5313) – (0.5313).e 0.5313 = -0.0416

f ‘ (x2) = -sin(0.5313) – e 0.5313 – 0.5313 .e 0.5313 = -3.1116

x3 = x2 - f (x2)

f ‘ (x2)

= 0.5313 – (-0.0416) / (-3.1116) = 0.5179

f(x3) = cos(0.5179) – 0.5179.e 0.5179 = -0.0004

f ‘ (x3) = -sin(0.5179) – e 0.5179 – 0.5179.e 0.5179 = -3.0428

therefore approximate roots after three iterations is 0.5179

:::Q4:::

f(x) = x3 – 3x +1

x0 = 1 , x1 = 0.4

f(x0) = (1)3 -3(1) +1 = -1

f(x1) = (0.4)3 – 3(0.4) +1 = -0.136

x2 = x0.f(x1) – x1.f(x0) /f(x1) – (x0)

x2 = 1.(-0.136) – 0.4(-1) / -0.136 – (-1) =0.264 / 0.864 = 0.3055

f(x2) = (0.3055)3 – 3(0.3055) + 1 = 0.1120

x3 = x1.f(x2) – x2.f(x1) / f(x2) – f(x1)

= 0.4(0.1120) – 0.3055(-0.136) / 0.1120 – (-0.136) = 0.0863 / 0.248 =0.3479

f(x3) = (0.3479)3 – 3(0.3479) + 1 = -0.0015

x4 = x2.f(x3) – x3.f(x2) / f(x3) – f(x2)

= 0.3055(-0.0015) – 0.3479(0.1120) / (-0.0015) – 0.1120 = -0.0394 / -0.1135

=0.3471

f(x4) = (0.3471)3 – 3(0.3471) + 1 = 0.0418 – 1.0413 + 1 = 0.0005