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Assignment
PROGRAM BCA(REVISED FALL 2012)
SEMESTER 3
SUBJECT CODE & NAME BCA3010-COMPUTER ORIENTED NUMERICAL
METHODS
CREDIT 4
BK ID B1643
MAX.MARKS 60
Q.1 Determine the relative error for the function (푥,,)=3푥2푦2+5푦2푧2−7푥2푧2+38
Where x = y = z = 1 and Δ푥=−0.05, Δ푦=0.001, Δ푧=0.02
Answer:- Relative error:- Let the true value of a quantity be x and the measured or inferred value x_0.
Then the relative error is defined by:
where Delta x is the absolute error. The relative error of the quotient or product of a number of
quantities is less than or equal to the sum of their relative errors. The percentage error is 100% times
the relative error.

2. Q.2 Solve by Gauss elimination method.
2x + y + 4z = 12
4x + 11y – z = 33
8x – 3y + 2z = 20
Answer: - Given equation is: -
2x + y + 4z = 12 -------------------------------------------- (1)
4x + 11y – z = 33 --------------------------------- (2)
8x – 3y + 2z = 20----------------------------------- (3)
Multiplying equation1 with 11 and subtract it from equation 2 we get : -
18x + 45z = 99 ---------------------------------------
Q.3 Apply Gauss – Seidal iteration method to solve the equations
3x1 + 20x2 –x3 = –18
2x1 – 3x2 + 20x3 = 25
20x1 + x2 – 2x3 = 17
Answer: - In Gauss seidal method the latest values of unknowns at each stage of iteration are used in
proceeding to the next stage of iteration.
Let the rearranged form of a given set of equation be

3. Q.4 Using the method of least squares, find the straight line y = ax + b that fits the following data:
X 0.5 1.0 1.5 2.0 2.5 3.0
Y 15 17 19 14 10 7
Answer: - The given straight line fit be y = ax+b. The normal equations of least squre fit are
2 + bxi = xiyi ---------------- (1)
axi
and axi + nb = yi --------------------- (2)
From the given data, we have
x y xy X2
0.5 15 7.5 0.25
1.0 17 17.0 1.00
Q.5 Using Lagrange’s interpolation formula, find the value of y corresponding to x = 10 from the
following data:
X 5 6 9 11
F(x) 380 2 196 508
Answer:- Formula for Lagrange’s interpolation :-
Let Y = f(x) be a function which assumes the values f(x0), f(x1) ….. f(xn) corresponding to the
values x: x1, x1 …..x¬n.

4. (  )(  )...( 
)
1 2
n
 ( )  ( )

0
(  )(  )...( 
)
0 1 0 2 0
(  )(  )...( 
)
0 2
1
1 0 1 2 1
( ) .........
n
n
( )( )...( )
n
x x x x x x
Y f x f x
x x x x x x
x x x x x x
f x
x x x x x x

  
We have x0 = 5, x1 = 6, x2 = 9, x3 = 11
Y0 = 380, y1 = 2, y2 = 196, y3 = 508
Using Lagrange’s
Q.6 Find 푓′ (3), 푓′′ (7) and 푓′′′(12) from the following data.
X 2 4 5 6 8 10
Y 10 96 196 350 868 1746
Answer:-Given data is: -
X 2 4 5 6 8 10
Y 10 96 196 350 868 1746
F(X) 2 4 5 6 8 10
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