BCA3010 – COMPUTER ORIENTED NUMERICAL METHODS

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ASSIGNMENT

 

PROGRAM BCA(REVISED FALL 2012)
Semester 3
SUBJECT CODE & NAME BCA3010 – COMPUTER ORIENTED NUMERICAL METHODS
CREDIT 4
BK ID B 1643
MAX.MARKS 60

 

Note: Answer all questions. Kindly note that answers for 10 marks questions should be approximately of 400 words. Each question is followed by evaluation scheme.

 

Question.1 Find the Taylors Series for (𝑥) = 𝑥3 − 10𝑥2 + 6 about 𝑥0 = 3

 

Solution: Consider the one dimensional initial value problem

y’ = f(x, y),   y(x0) = y0

where

f is a function of two variables x and y and (x0 , y0) is a known point on the solution curve.

If the existence of all higher order partial derivatives is assumed for y at x = x0, then by Taylor series the value of y at any neibhouring point x+h can be written as

 

y(x0+h) = y(x0) + h y'(x0) + h2 /2 y”(x0) + h3/3! y”'(x0) +  .  .  .  .  .  .

 

 

 

 

Question. 2. Find a real root of the transcendental equation cos x – 3x+1 = 0, correct to four decimal places using iteration method.

 

Solution: Iteration Method: Let the given equation be f(x) = 0 and the value of x to be determined. By using the Iteration method you can find the roots of the equation. To find the root of the equation first we have to write equation like below

x = pi(x)

Let x=x0 be an initial approximation of the required root α then the first approximation x1 is given by x1 = pi(x0).

 

Similarly for second, thrid and so on. approximation

X2 = pi(x1)

X3 = pi(x2)

 

 

 

 

Question.3.  Solve the equations

2x + 3y + z = 9

x + 2y + 3z = 6

3x + y + 2z = 8 by LU decomposition method.

 

Solution: We shall solve the system

2x + 3y + z = 9

x + 2y + 3z = 6

3x + y + 2z = 8

 

 

 

Question.4 Fit a second degree parabola y = a + bx + cx2 in the least square method for the following data and hence estimate y at x = 6.

X 1 2 3 4 5
Y 10 12 13 16 19

 

 

Solution: The given straight line fit be y = ax+b. The normal equations of least squre fit are

aSxi2 + bSxi = Sxiyi —————- (1)

and    aSxi + nb = Syi  ——————— (2)

 

 

 

 

 

Question.5 The population of a certain town is shown in the following table

 

Year X 1931 1941 1951 1961 1971
Population Y 40.62 60.80 79.95 103.56 132.65

 

Find the rate of growth of the population in 1961.

 

Solution:

Year X 1931 1941 1951 1961 1971
Population Y 40.62 60.80 79.95 103.56 132.65

 

 

 

 

Question. 6. Solve of 𝑦𝑛+2 − 2 𝐶𝑜𝑠𝛼 𝑦𝑛+1 + 𝑦𝑛 = 𝐶𝑜𝑠 𝛼𝑛.

 

Solution: The order of the difference equation is the difference between the largest and smallest arguments occur-ring in the difference equation divided by the unit of argument.Thus, the order of the difference equation=Largest argument Smallest argument Unit of argument.

 

We have yn = A2n + B(-2)n             => yn-A2n – B(-2)n

Yn+1 = 2A2n – 2B(-2)n                        =>yn+1 – 2A2n + 2B (-2)n

Yn+2 = 4A2n + 4B(-2)n                       =>yn+2-4A2n-4B(-2)n

 

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