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MTH603 - Numerical Analysis

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  • MTH603 Quiz 1 Solution and Discussion

    Solved mth603 solution discussion fall 2019 quiz 1
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    cyberianC
  • MTH603 Quiz 3 Solution and Discussion

    mth603 quiz 3 solution discussion fall 2019
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    zaasmiZ
    Question # 3 of 10 ( Start time: 09:05:56 PM ) Total Marks: 1 To take the derivative of f(x) = 2x in the interval [-3,3], which of the following partition of subintervals will be suitable? Select correct option: Equally spaced Unequally spaced Union of equally spaced and unequally spaced intervals. Any arbitrary partition will work
  • MTH603 Quiz 2 Solution and Discussion

    mth603 quiz 2 solution discussion fall 2019
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    zaasmiZ
    @zainab-ayub said in MTH603 Quiz 2 Solution and Discussion: Mth603 ka koi student hai tu plz yeh question bta dy kis trha solve ho ga Given the following data x:1 2 5 y:1 4 10 Value of 1st order divided difference f[2 , 5] is [image: JgeUzkI.png]
  • MTH603 Assignment 1 Solution and Discussion

    Solved mth603 assignment 1 solution discussion fall 2019
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    zareenZ
    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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