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    SOLVED MTH603 Assignment 1 Solution and Discussion

    MTH603 - Numerical Analysis
    mth603 assignment 1 solution discussion fall 2019
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    • zareen
      zareen Cyberian's Gold last edited by

      Assignment NO. 1 MTH603 (Fall 2019)

      Maximum Marks: 20 Due Date: 24 -11-2019

      DON’T MISS THESE: Important instructions before attempting the solution of this assignment:
      • To solve this assignment, you should have good command over 01 - 12 lectures.
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      Question #1: Find the root of the equation, Perform three iteration of the equation,
      ln (x −1) + sinx = 0 by using Newton Raphson method.
      Question #2: Solve the system of linear equations with the help of Gaussian elimination method.
      x + y + z = 6; 2x − y + z = 3; x + z = 4

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      • zareen
        zareen Cyberian's Gold last edited by

        Assignment No: 01
        Question #1: Find the root of the equation, Perform three iteration of the equation,
        ln (x −1) + sinx =0 by using Newton Raphson method.

        Ans: Let f(x) = ln(x+1) + sinx = 0 and f(x) = 1/(x-1) + cosx

        F (1.5) = ln(0.5) + (1.5) = - 0.0667

        F(2) = ln(1) + sin(2) = 0.035

        Since f (1.5) f (2) < 0 so roots lies in interval [1.5, 2]

        Let x0 = 1.75 . x0 can be taken in the interval any real number [ 1.5 , 2 ], we let mid point

        of this interval .
        As we know Newton Raphson method is

        Xn+1 = xn – f ( xn ) / f(xn)
        First iteration
        X1 = x0 –f(x0) / f(x0) = 1.75 - f(1.75) / f(1.75)
        = 1.75 – (-0.2571 / 2.3329) = 1.8602
        Second iteration:
        X2 = x1 - f(x) / f(x) = 1.8602 –[ f(1.8602) / f(1.8602)]
        = 1.8602 - ( -0.1181 / 2.1620 ) = 1.9148
        Third iteration:
        X3 = x2- f(x2) / f(x2) = 1.9148 –f(1.9148) / f(1.9148)
        = 1.9148 – [-0.0556/2.0926]
        = 1.9414
        Question #2: Solve the system of linear equations with the help of Gaussian elimination method.
        x + y + z = 6;2x − y + z = 3;x + z = 4

        ANS: In Gaussian elimination method we convert the augmented matrix into reduce

        Echelon form therefore,

        Augmented matrix is

        R2- 2R1 , R3 – R1

        -1R2 , -1R3
        R23
        R3-3R2

        X + Y+ Z = 6 ;………………….(1)
        Y = 2,
        Z = 3
        Put into eq (1),
        we get X = 1 ,

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        • zareen
          zareen Cyberian's Gold last edited by

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