Applied Mathematics for Electrical Engineering (3130908)

Question-1b

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

Q1) (b)

Let,

x = N^{\frac{1}{q}}

, where

q = 1, 2, 3, \dots \dots, n

and

N

is an natural number.

\Rightarrow x^{q} - N = 0

\Rightarrow f (x) = x^{q} - N

\Rightarrow f^{'} (x) = q x^{q - 1}

By the general formula we get,

x_{n + 1} = \frac{1}{q} [(q - 1) x_{n} + \frac{N}{x_{n}^{q - 1}}]

Where,

n = 0, 1, 2, 3, \dots

Let,

x = \sqrt{8}

\Rightarrow x^{2} - 8 = 0

\Rightarrow f (x) = x^{2} - 8

\Rightarrow f^{'} (x) = 2 x

Let,

x_{0} = 3

By N-R Method:

x_{n + 1} = \frac{1}{2} [x_{n} + \frac{N}{x_{n}}], n = 0, 1, 2, 3, . . .

x_{1} = \frac{1}{2} (x_{0} + \frac{8}{x_{0}})

x_{1} = \frac{1}{2} (3 + \frac{8}{3})

x_{1} = 2.83

Therefore,

x = 2.83

is the square root of

8