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MTH603 Numerical Analysis Assignment 01 Spring 2021 Solution / Discussion

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MTH603 Numerical Analysis Assignment 01 Spring 2021 Solution Files

MTH603%20Assignment%201%20Solution%20Spring%202021.pdf

MTH603-Assignment-1-Solution-Spring-2021.docx

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MTH603-Assignment%201-IDEA-SOL-SPRING-2021.docx

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Mth603 – Numerical Analysis Assignment 1

Assignment 1 of session spring 2021has been uploaded. Its solution is given below, also downloadable file is attached at the end.

Question 1:

Find a root of the given equation using Newton-Raphson Method Keep values correct to four decimal places.

Solution:

x3 + 2x2 +10x – 20 = 0 Take initial value x =1.5 x3 + 2x2 +10x – 20 = 0

(x) = x3 + 2x2 +10x – 20 d/dx(x3 + 2x2 +10x – 20)

= 3x2 + 4+10

 (x) = 3x2 + 4+10

x= 1.3736

Iteration (1) 

(x= (1.5) = 2.875

0

() = (1.5) = 22.75

− () / ()

1 0 0 1

x1 = 1.3736

Iteration (2) 

(x= (1.3736) = 0.1018

1

 () = (1.3736) = 21.1551

− () / ()

2 1 1 1

x= 1.3688

Iteration (3)

(x=

2

 () =

(1.3688) = 0.0001

(1.3688) = 21.0962

− () / ()

3 2 2 2

x= 1.3688

Hence the calculated root of the given equation using Newton-Rophson Megthod Upto four decimal places is 1.3688

Question 2:

Identify the intitial bracket (x0 , x1 )

Solution:


x 0 1 2 3
F(x) -11 -11 -5 -13

Given

x3 − −11 = 0

(x) = x3 − −11

(2) = −5 < 0 and f (3) =13 > 0 Now root lie btween 2 and 3

(iteration 1) 

(2) = −5 < 0 and f (3) =13 > 0

x= 2 + 3 / 2 = 2.5

(x=

(2.5) = 2.215 > 0

(iteration 2) 

(2) = −5 < 0 and f (2.5) = 2.125 > 0

x= 2 + 2.5 / 2 = 2.25

(x= (2.5) = 1.8594 /sub> 0

Hence root lies between 2 and 2.25

(iteration 3) 

(2.25) = 1.8594 < 0

(2.25) = 2.125 > 0

Now root lie between 2.25 and 2.50

x2 = 2.25 +2.5 / 2 = 2.375

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