Smu Solved Assignments: BC0043 Computer Oriented Numerical Methods

BC0043 Computer Oriented Numerical Methods

Question 1. Verify Rolle’s theorem for the function f(x) = |x| in (–1, 1).

f(x) = -x for -1 < x < 0.

= 0 for x = 0

= x for 0 < x < 1

f(-1)= 1 f(1) -1

f(-1)= f(1)

f¹(x) = -1 for -1 < x < 0

f¹ (x) = 1 for 0 < x < 1

Therefore f¹(x) does not exist at x =0 and hence f(x) is not differentiable in (-1,1) Rolle’s theorem is not applicable to the function f(x) = ǀxǀ in (-1,1)

Question 2. Find the multiplicity root of the equation x³ – x² – x + 1 = 0 near x0 = 0.9

F(x) = x³ – x² –x + 1

F’(x) = 3x² -2x -1 and f’’ (x) =6x – 2

Let x₀ =0.9 (taken randomly) be the initial approximated value.

X₀ – m f(x₀) = x₀ – (m -1) f’(x₀)

F’(x₀) f’(x₀)

M=2, x₀ = 0.9, f(x₀) = f(x₀) = f(0.9) = 0.019, f’ (x₀) = f’(0.9) = -0.37 and f’’ (x₀) = f’’ (0.9) = 3.4

Therefore

X₀ – 2 f (x₀) = x₀ – f’(x₀)

F’(x₀) f’’(x₀)

We have x₀ – 2f(x₀) =0.9 - 2×0.019 = 1.003, and

F’(x₀) -0.37

X₀ – f’(x₀) = 0.9 – (-0.37) = 1.009

F’’(x₀) 3.4

The closeness of these values indicates that there is a double root near x =1.

For the next approximation we choose x₁ = 1.01

We get x₁ – 2f(x₁) = 1.01 - 2×0.0002 = 1.0001

F’(x₁) 0.0403

and x₁ – f’(x₁) = 1.01 – 0.0403 = 1.0001

f’’(x₁) 4.06

The show that there is a root at x =1.0001 which is quite near the actual root x =1.

Observation : for f(x) = x³-x²-x+1=(x-1)² (x+1). Therefore x=1 is a double root with multiplicity m= 2.

Question 3. Use Bisection method to solve x³+3x-5 defined in the interval [1,2]

F(x) = X³+ 3x-5

F(1) = 1+3×1-5

= 4 – 5 = -1

F(2) = 2³ + 3×2-5

= 8+6-5

= 9

F (1) is negative & f(2) is positive

X₁ = 1+2 = 3

2 2

= 1.5

F(x) = F(1.5)³ + 3×1.5 – 5

= 3.375 + 4.5-5

= 2.875 > 0

Hence the root ties between 1& 1.5. Since f(1.5) > 0 & f(1) < 0

Question 4. Find AB and BA, when = A =

and B =

Given A is a 2×3 matrix and B is a 3×2 matrix. So AB is a 2×2 matrix.

AB=

Question 5. Find the solution of the following system of equations.

z =

x+ y-

w =

x+z -

w =

y -

z+w =

using Gauss-Seidel method and perform the first five iterations.

The given system of equations can be rewritten as

........(1)

Taking the initial approximation as y =z =0 on the right side of the equation (1) we get x⁽¹⁾ =0.5.

Taking z=0 and w=0 and the current value of x, we get

Y⁽¹⁾ =0.5 + (0.25)(0.5) + 0 = 0.625 from the second equation of (1).

Now taking w = 0 and the current value of x, we get

Z⁽³⁾ = 0.25 + (0.25) (0.5) + 0 = 0.375 from the third equation of (1).

Lastly, using the current values of y and z, the fourth equation of (1) gives

W⁽¹⁾ = 0.25 + (0.25) (0.625) + (0.25)( 0.375) = 0.5 .

The Gauss-Seidal iterations for the given set of equations can be written as

........(1)

Now by Gauss – Seidel procedure , the second and the subsequent approximations can be obtained and the sequence of the first five approximations is tabulated below.

Iteration Number (r)	Variables
Iteration Number (r)	x	y	z	w
1	0.5	0.625	0.375	0.5
2	0.75	0.8125	0.5625	0.59375
3	0.84375	0.85938	0.60938	0.61719
4	0.86719	0.87110	0.62110	0.62305
5	0.87305	0.87402	0.62402	0.62451