Computer Oriented Numerical Methods (BCA) 3rd Sem Previous Year Solved Question Paper 2022

Step 1: Calculate the divided differences.

We'll calculate the divided differences using the given data:

1. ΔF[0] = F(1) - F(0) = 14 - 1 = 13

2. ΔF[1] = F(2) - F(1) = 2 - 14 = -12

3. ΔF[2] = F(3) - F(2) = 15 - 2 = 13

4. ΔF[3] = F(4) - F(3) = 4 - 15 = -11

5. ΔF[4] = F(5) - F(4) = 5 - 4 = 1

6. ΔF[5] = F(6) - F(5) = 6 - 5 = 1

7. ΔF[6] = F(7) - F(6) = 19 - 6 = 13

Step 2: Calculate the interpolating polynomial.

The Lagrange interpolating polynomial is given by:

P(x) = F(x0) + (x - x0)ΔF[0] + (x - x0)(x - x1)ΔF[0,1] + ...

Here, x0 = 2, x1 = 1, and ΔF[0] = 13. So, the interpolating polynomial is:

P(x) = 14 + (x - 2)ΔF[0] - (x - 2)(x - 1)ΔF[0,1]

Now, we need to calculate ΔF[0,1]:

ΔF[0,1] = (ΔF[1] - ΔF[0]) / (x1 - x0) = (-12 - 13) / (1 - 2) = 25

So, the interpolating polynomial becomes:

P(x) = 14 + (x - 2)13 - (x - 2)(x - 1)25

Step 3: Evaluate P(3) to find F(3):

P(3) = 14 + (3 - 2)13 - (3 - 2)(3 - 1)25 P(3) = 14 + 13 - 2 * 25 P(3) = 14 + 13 - 50 P(3) = 27 - 50 P(3) = -23

So, F(3) is approximately equal to -23.

Step 1: Calculate the divided differences.

We'll calculate the divided differences using the given data. The divided difference table is as follows:

Step 2: Calculate the interpolating polynomial.

The Lagrange interpolating polynomial is given by:

(x)=F(x0)+(x−x0)Δ0+(x−x0)(x−x1)Δ0,1+(x−x0)(x−x1)(x−x2)Δ0,1,2+...+(x−x0)(x−x1)...(x−xn−1)Δ0,1,...,n−1

Here, 0x0=0, 1x1=1, 2x2=2, 3x3=3, and we have already calculated the divided differences. So, the interpolating polynomial is:

(x)=1+xΔ0+x(x−1)Δ0,1+x(x−1)(x−2)Δ0,1,2+x(x−1)(x−2)(x−3)Δ0,1,2,3

Now, we need to substitute the values for Δ0, Δ0,1, Δ0,1,2, and Δ0,1,2,3:

Δ0 = 13 Δ0,1 = 25 Δ0,1,2 = 26 Δ0,1,2,3 = 4

So, the interpolating polynomial becomes:

P(x)=1+13x+25x(x−1)+26x(x−1)(x−2)+4x(x−1)(x−2)(x−3)

Step 3: Evaluate P(3) to find F(3):

p(3)=1+13(3)+25(3)(3−1)+26(3)(3−1)(3−2)+4(3)(3−1)(3−2)(3−3)P(3)=1+13(3)+25(3)(3−1)+26(3)(3−1)(3−2)+4(3)(3−1)(3−2)(3−3)

p(3)=1+39+150+78+0P(3)=1+39+150+78+0

p(3)=268

So F(3)=268.