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# MTH603 - Numerical Analysis Assignment No 1 Fall 2016 Due Date Nov 16, 2016

By using Iterative method, find the real roots of equation:, correct to four decimal places.

Evaluate  by using Newton Raphson method, correct to four decimal places.

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Please Discuss here about this assignment.Thanks

Our main purpose here discussion not just Solution

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sir yh theek hai

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# MTH603 - Numerical Analysis Assignment No. 1 Solution Fall 2016 Due Date: Nov 16, 2016

MTH603 - Numerical Analysis Assignment No. 1 Solution Fall 2016 Due Date: Nov 16, 2016

By using Iterative method, find the real roots of equation:, correct to four decimal places.

Evaluate  by using Newton Raphson method, correct to four decimal places.

Iteration Method
Let the given equation be f(x) = 0 and the value of x to be determined. By using the Iteration method you can find the roots of the equation. To find the root of the equation first we have to write equation like below
x = pi(x)
Let x=x0 be an initial approximation of the required root α then the first approximation x1 is given by x1 = pi(x0).

Similarly for second, thrid and so on. approximation
x2 = pi(x1
x3 = pi(x2
x4 = pi(x3
xn = pi(xn-1

Iteration Method Example:
Find the real root of the equation x3 + x 2 = 1 by iteration method.
Solution:
We can rewrite the above equation by
x3 + x 2 - 1 = 0;
Let f(x) = x3 + x 2 - 1
f(0) = -1 (positive)
f(1) = 1 (negative)
Hence the root value lie between 0 to 1

x3 + x 2 - 1 = 0
x2 (x + 1) = 1
x2 = 1/ (x + 1)
x = 1/ √(x + 1)
pi(x) = 1/ √(x + 1)

Let the initial approximation be x0 = 0.5

x1 = pi(x0) = 1/√1+ 0.5 = 0.81649

x2 = pi(x1) = 1/√1+ 0.81649 = 0.74196

x3 = pi(x2) = 1/√1+ 0.74196 = 0.75767

x4 = pi(x3) = 1/√1+ 0.75767 = 0.75427

x5 = pi(x4) = 1/√1+ 0.75427 = 0.75500

x6 = pi(x5) = 1/√1+ 0.75500 = 0.75485

x7 = pi(x6) = 1/√1+ 0.75485 = 0.75488

Since the difference between x6 and x7 are very small, so the root is 0.75488.

Aslamo aelkum

me is ka sulotion samjna b chahta hon or han kiya he above sulotion 100% ok he ?

plz tell me

Dear Students Don’t wait for solution post your problems here and discuss ... after discussion a perfect solution will come in a result. So, Start it now, replies here give your comments according to your knowledge and understandings....

Here Is The Solution For this Assignment... Please try to understand the solution rather than replicating it.. Thanks

After Rewriting the 1st Equation Use 3 and 4 in g(x) to get the initial approximations.

2nd Solution can also be found in Handout..

is this correct solution?

Aslam alikum... can u tell me one thing.... Q1 mein intial value x ki i.e;

X0 ap ne 3.6 kaise liya hai.... and kya koi b value ly sakty hain yahan for example 0.5

Wa Alaikum Salam Aese sis...

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2x=7
x=7/2
x=3.5 aai ga x0 =3.5 hojai ga

Sorry bhai smjhi nhi

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