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 = 
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.
x- 
y-
z = 
y- z = x+ y- w =
- x+z - w =
x+z - w =
- y - 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)
........(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.
|
||||||||||||||||||||||||||||||||||
Question 6. Fit parabola of second degree y =
a + bx + 
for the data
for the data |
X
|
0
|
1
|
2
|
3
|
4
|
|
y
|
1
|
1.8
|
1.3
|
2.5
|
2.3
|
The normal
equations for y=a +bx+c
are
are
=na + b
+ c
Xy = a + b 
+c
y
= a + b+c
Here
n = 5.
The
relevant table is given below:
|
x
|
y
|
X2
|
X3
|
X4
|
xy
|
X2y
|
|
0
|
1
|
0
|
0
|
0
|
0
|
0
|
|
1
|
1.8
|
1
|
1
|
1
|
1.8
|
1.8
|
|
2
|
1.3
|
4
|
8
|
16
|
2.6
|
5.2
|
|
3
|
2.5
|
9
|
27
|
81
|
7.5
|
22.5
|
|
4
|
2.3
|
16
|
64
|
256
|
9.2
|
36.8
|
|
=
10 |
 8.9 |
2 = 30 |
 3 = 100 |
 354 |
 21.1 |
y
= 66.3 |
The
normal equations becomes
5a
+ 10 b + 30 c = 8.9
10
a +30 b +100 c =21.1
30
a +100 b +354 c = 66.3
Solving
these simultaneous equations, we get
a
= 1.078
b
= 0.414
c
= -0.021
Hence
the required parabola is
Y
= 1.078 + 0. 414 x – 0.021 x2
