Applied Mathematics for Electrical Engineering (3130908)

Question-2b

BE | Semester-3 Winter-2019 | 26-11-2019

Q2) (b)

If data are

(x_{0}, y_{0}), (x_{1}, y_{1}), \dots, (x_{n}, y_{n})

. Where,

x_{0}, x_{1}, x_{2}, \dots, x_{n}

are equally spaced,v

f (x) = y = y_{0} + p ∆ y_{0} + \frac{p (p - 1)}{2!} ∆^{2} y_{0} + \frac{p (p - 1) (p - 2)}{3!} ∆^{3} y_{0} + \dots

Where,

p = \frac{x - x_{0}}{h}

;

x_{0}

is the initial point.

Here,

p = \frac{x - x_{0}}{h} = \frac{1.6 - 1}{0.4} = \frac{3}{2}

The Newton’s forward difference formula is

f (x) = y = y_{0} + p ∆ y_{0} + \frac{p (p - 1)}{2!} ∆^{2} y_{0} + \frac{p (p - 1) (p - 2)}{3!} ∆^{3} y_{0} + \dots

f (x) = y = 3.49 + (\frac{3}{2}) (1.33) + \frac{1}{2!} (\frac{3}{2}) (\frac{3}{2} - 1) (- 0.19) + \frac{1}{3!} (\frac{3}{2}) (\frac{3}{2} - 1) (\frac{3}{2} - 2) (- 0.41)

f (x) = 3.49 + 1.9950 - 0.0713 + 0.0256

f (x) = 5.4393