Applied Mathematics for Electrical Engineering (3130908)

Question-1c ( i )

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

Q1) (c ( i ))

(i) By Secant Method

Let,

f (x) = e^{- x} - x

f (x_{0}) = f (0) = 1 > 0

f (x_{1}) = f (1) = - 0.6321 < 0

Therefore, root lies between

[0, 1]

By Secant Method:

x_{n + 1} = x_{n} - \frac{x_{n} - x_{n - 1}}{f (x_{n}) - f (x_{n - 1})} f (x_{n}), n \geq 1 .

x_{2} = x_{1} - \frac{x_{1} - x_{0}}{f (x_{1}) - f (x_{0})} f (x_{1})

x_{2} = 1 - \frac{1 - 0}{- 0.6321 - 1} (- 0.6321)

x_{2} = 0.6127

$x_{0} > 0$	$x_{1} < 0$	$f (x_{0})$	$f (x_{1})$	$x_{n + 1} = x_{n} - \frac{x_{n} - x_{n - 1}}{f (x_{n}) - f (x_{n - 1})} f (x_{n})$	$f (x_{n + 1})$
$0$	$1$	$1$	$- 0.6321$	$0.6127$	$- 0.0708$
$1$	$0.6127$	$- 0.6321$	$- 0.0708$	$0.5638$	$0.0052$
$0.6172$	$0.5638$	$- 0.0708$	$0.0052$	$0.5671$	$0.0000$

Therefore,

x = 0.5671

is the required root.