Secant method Algorithm & Example-1 f(x)=x^3-x-1

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Home > Numerical methods calculators > Secant method example

5. Secant method example ( Enter your problem )

( Enter your problem )

Algorithm & Example-1 $f(x)=x^3-x-1$
Example-2 $f(x)=2x^3-2x-5$
Example-3 $x=sqrt(12)$
Example-4 $x=root(3)(48)$
Example-5 $f(x)=x^3+2x^2+x-1$

Other related methods

4. Fixed Point Iteration method
(Previous method)

$f(x)=2x^3-2x-5$
(Next example)

1. Algorithm & Example-1 $f(x)=x^3-x-1$

Algorithm

Secant method Steps (Rule)
Step-1:	Find points $x_0$ and $x_1$ such that $x_0 < x_1$ and $f(x_0) * f(x_1) < 0$ .
Step-2:	find next value using Formula-1 : $x_2=x_0-f(x_0)(x_1-x_0)/(f(x_1)-f(x_0))$ or Formula-2 : $x_2=(x_0f(x_1)-x_1f(x_0))/(f(x_1)-f(x_0))$ or Formula-3 : $x_2=x_1-f(x_1)(x_1-x_0)/(f(x_1)-f(x_0))$ (Using any of the formula, you will get same x2 value)
Step-3:	If $f(x_2) = 0$ then $x_2$ is an exact root, else $x_0 = x_1$ and $x_1 = x_2$
Step-4:	Repeat steps 2 & 3 until $f(x_i) = 0$ or $\|f(x_i)\| <= "Accuracy"$

Example-1
1. Find a root of an equation $f(x)=x^3-x-1$ using Secant method

Solution:
Here

$x^3-x-1=0$

Let

$f(x) = x^3-x-1$

Here

$x$	0	1	2
$f(x)$	-1	-1	5

$1^(st)$ iteration :

$x_0 = 1$ and

$x_1 = 2$

$f(x_0) = f(1) = -1$ and

$f(x_1) = f(2) = 5$

$:. x_2 = x_0 - f(x_0) * (x_1 - x_0) / (f(x_1) - f(x_0))$

$x_2 = 1 - (-1) xx (2 - 1)/(5 - (-1))$

$x_2 = 1.16667$

$:. f(x_2) = f(1.16667) = -0.5787$

$2^(nd)$ iteration :

$x_1 = 2$ and

$x_2 = 1.16667$

$f(x_1) = f(2) = 5$ and

$f(x_2) = f(1.16667) = -0.5787$

$:. x_3 = x_1 - f(x_1) * (x_2 - x_1) / (f(x_2) - f(x_1))$

$x_3 = 2 - 5 xx (1.16667 - 2)/(-0.5787 - 5)$

$x_3 = 1.25311$

$:. f(x_3) = f(1.25311) = -0.28536$

$3^(rd)$ iteration :

$x_2 = 1.16667$ and

$x_3 = 1.25311$

$f(x_2) = f(1.16667) = -0.5787$ and

$f(x_3) = f(1.25311) = -0.28536$

$:. x_4 = x_2 - f(x_2) * (x_3 - x_2) / (f(x_3) - f(x_2))$

$x_4 = 1.16667 - (-0.5787) xx (1.25311 - 1.16667)/(-0.28536 - (-0.5787))$

$x_4 = 1.33721$

$:. f(x_4) = f(1.33721) = 0.05388$

$4^(th)$ iteration :

$x_3 = 1.25311$ and

$x_4 = 1.33721$

$f(x_3) = f(1.25311) = -0.28536$ and

$f(x_4) = f(1.33721) = 0.05388$

$:. x_5 = x_3 - f(x_3) * (x_4 - x_3) / (f(x_4) - f(x_3))$

$x_5 = 1.25311 - (-0.28536) xx (1.33721 - 1.25311)/(0.05388 - (-0.28536))$

$x_5 = 1.32385$

$:. f(x_5) = f(1.32385) = -0.0037$

$5^(th)$ iteration :

$x_4 = 1.33721$ and

$x_5 = 1.32385$

$f(x_4) = f(1.33721) = 0.05388$ and

$f(x_5) = f(1.32385) = -0.0037$

$:. x_6 = x_4 - f(x_4) * (x_5 - x_4) / (f(x_5) - f(x_4))$

$x_6 = 1.33721 - 0.05388 xx (1.32385 - 1.33721)/(-0.0037 - 0.05388)$

$x_6 = 1.32471$

$:. f(x_6) = f(1.32471) = -0.00004$

Approximate root of the equation

$x^3-x-1=0$ using Secant method is

$1.32471$

$n$	$x_0$	$f(x_0)$	$x_1$	$f(x_1)$	$x_2$	$f(x_2)$	Update
1	1	-1	2	5	1.16667	-0.5787	$x_0 = x_1$ $x_1 = x_2$
2	2	5	1.16667	-0.5787	1.25311	-0.28536	$x_0 = x_1$ $x_1 = x_2$
3	1.16667	-0.5787	1.25311	-0.28536	1.33721	0.05388	$x_0 = x_1$ $x_1 = x_2$
4	1.25311	-0.28536	1.33721	0.05388	1.32385	-0.0037	$x_0 = x_1$ $x_1 = x_2$
5	1.33721	0.05388	1.32385	-0.0037	1.32471	-0.00004	$x_0 = x_1$ $x_1 = x_2$

This material is intended as a summary. Use your textbook for detail explanation.
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4. Fixed Point Iteration method
(Previous method)

$f(x)=2x^3-2x-5$
(Next example)

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