**Government College of Engineering - Srirangam, Trichy - 12**
**Department of Mathematics - Second Semester**
**Internal Assessment Test - I**
**MA3251 - Statistics and Numerical Methods**
**CIVIL / CSE / EEE / MECH**
PART A - Answer All the Questions (6 x 2 = 12 Marks)
1. State the convergence condition for Newton-Raphson method.
  - The method converges if the initial guess is close to the actual root and if f(x), f'(x), and f''(x) are
continuous near the root. Also, f'(x) 0 in that interval.
2. Compare Gauss elimination method and Gauss-Jordan method.
  - Gauss Elimination: Converts matrix to upper triangular form and then uses back-substitution.
  - Gauss-Jordan: Reduces matrix to diagonal form for direct solution.
3. Why Gauss-Seidel is better than Gauss-Jacobis method?
  - Gauss-Seidel uses the most recently updated values in each iteration, hence converges faster.
4. Define interpolation.
  - Interpolation is the method of estimating unknown values within the range of a discrete set of
known data points.
5. State Newtons forward difference interpolation formula.
  - f(x) = f(x0) + f(x0) (x - x0)/h + f(x0)/2! (x - x0)(x - x1)/h +
6. What is the order of the error in Trapezoidal rule?
  - The order of error is O(h).
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PART B (3 x 16 = 48 Marks)
7(a)(i) Solve by Gauss-Seidel method:
Given system:
4x + 2y + z = 14
x + 5y - z = 10
x + y + 8z = 20
Iteration form:
x = (1/4)(14 - 2y - z)
y = (1/5)(10 - x + z)
z = (1/8)(20 - x - y)
Initial guess: x0 = 0, y0 = 0, z0 = 0
Iteration 1:
x1 = 3.5, y1 = 1.3, z1 = 1.525
Iteration 2:
x2 = 2.96875, y2 = 1.71125, z2 = 1.665
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7(a)(ii) Newton-Raphson Method:
Find root of x logx - 1.2 = 0
f(x) = x logx - 1.2
f'(x) = logx + 1/ln10
Initial guess x0 = 2
f(2) -0.5979, f'(2) 0.7353, x1 = 2.813
f(2.813) -0.026, x2 = 2.8424
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7(b)(i) Power Method:
Find the numerically largest Eigen value of:
A = [ [1, 4], [4, -1] ]
Initial vector: x(0) = [1, 0]
x(1) = Ax(0) = [1, 4] normalized = [0.25, 1]
Next [4.25, 0]
Converges to eigenvalue 4.123
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7(b)(ii) Solve 3x = (1 + sin x)
Iterative form: x = (1/3) (1 + sin x)
Initial guess x0 = 0.5
x1 = 0.4304
x2 = 0.4253
Converges to 0.4235
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8(a)(i) Jacobi Method:
A = [ [5, 0, 1], [-2, 1, 0], [0, 1, 5] ]
Jacobi iteration involves zeroing off-diagonal elements using rotations.
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8(a)(ii) Gauss-Jordan Method:
Equations:
2x + y + 4z = 12
8x - 3y + 2z = 20
4x + 11y - z = 33
Use row operations to get solution:
x = 1, y = 2, z = 2
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End of Solutions