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# Program to find root of an equations using secant method

The secant method is used to find the root of an equation f(x) = 0. It is started from two distinct estimates x1 and x2 for the root. It is an iterative procedure involving linear interpolation to a root. The iteration stops if the difference between two intermediate values is less than the convergence factor.

Examples :

Input : equation = x3 + x – 1
x1 = 0, x2 = 1, E = 0.0001
Output : Root of the given equation = 0.682326
No. of iteration=5

Algorithm

```Initialize: x1, x2, E, n         // E = convergence indicator
calculate f(x1),f(x2)

if(f(x1) * f(x2) = E); //repeat the loop until the convergence
print 'x0' //value of the root
print 'n' //number of iteration
}
else

## C++

 `// C++ Program to find root of an ` `// equations using secant method` `#include ` `using` `namespace` `std;` `// function takes value of x and returns f(x)` `float` `f(``float` `x)` `{` `    ``// we are taking equation as x^3+x-1` `    ``float` `f = ``pow``(x, 3) + x - 1;` `    ``return` `f;` `}`   `void` `secant(``float` `x1, ``float` `x2, ``float` `E)` `{` `    ``float` `n = 0, xm, x0, c;` `    ``if` `(f(x1) * f(x2) < 0) {` `        ``do` `{` `            ``// calculate the intermediate value` `            ``x0 = (x1 * f(x2) - x2 * f(x1)) / (f(x2) - f(x1));`   `            ``// check if x0 is root of equation or not` `            ``c = f(x1) * f(x0);`   `            ``// update the value of interval` `            ``x1 = x2;` `            ``x2 = x0;`   `            ``// update number of iteration` `            ``n++;`   `            ``// if x0 is the root of equation then break the loop` `            ``if` `(c == 0)` `                ``break``;` `            ``xm = (x1 * f(x2) - x2 * f(x1)) / (f(x2) - f(x1));` `        ``} ``while` `(``fabs``(xm - x0) >= E); ``// repeat the loop` `                                ``// until the convergence`   `        ``cout << ``"Root of the given equation="` `<< x0 << endl;` `        ``cout << ``"No. of iterations = "` `<< n << endl;` `    ``} ``else` `        ``cout << ``"Can not find a root in the given interval"``;` `}`   `// Driver code` `int` `main()` `{` `    ``// initializing the values` `    ``float` `x1 = 0, x2 = 1, E = 0.0001;` `    ``secant(x1, x2, E);` `    ``return` `0;` `}`

## Java

 `// Java Program to find root of an ` `// equations using secant method` `class` `GFG {` `    `  `    ``// function takes value of x and ` `    ``// returns f(x)` `    ``static` `float` `f(``float` `x) {` `        `  `        ``// we are taking equation ` `        ``// as x^3+x-1` `        ``float` `f = (``float``)Math.pow(x, ``3``) ` `                               ``+ x - ``1``;` `                               `  `        ``return` `f;` `    ``}` `    `  `    ``static` `void` `secant(``float` `x1, ``float` `x2,` `                                ``float` `E) {` `        `  `        ``float` `n = ``0``, xm, x0, c;` `        ``if` `(f(x1) * f(x2) < ``0``) ` `        ``{` `            ``do` `{` `                `  `                ``// calculate the intermediate` `                ``// value` `                ``x0 = (x1 * f(x2) - x2 * f(x1))` `                            ``/ (f(x2) - f(x1));` `        `  `                ``// check if x0 is root of` `                ``// equation or not` `                ``c = f(x1) * f(x0);` `        `  `                ``// update the value of interval` `                ``x1 = x2;` `                ``x2 = x0;` `        `  `                ``// update number of iteration` `                ``n++;` `        `  `                ``// if x0 is the root of equation ` `                ``// then break the loop` `                ``if` `(c == ``0``)` `                    ``break``;` `                ``xm = (x1 * f(x2) - x2 * f(x1)) ` `                            ``/ (f(x2) - f(x1));` `                            `  `                ``// repeat the loop until the ` `                ``// convergence ` `            ``} ``while` `(Math.abs(xm - x0) >= E); ` `                                                `  `            ``System.out.println(``"Root of the"` `+` `                    ``" given equation="` `+ x0);` `                    `  `            ``System.out.println(``"No. of "` `                      ``+ ``"iterations = "` `+ n);` `        ``} ` `        `  `        ``else` `            ``System.out.print(``"Can not find a"` `              ``+ ``" root in the given interval"``);` `    ``}` `    `  `    ``// Driver code` `    ``public` `static` `void` `main(String[] args) {` `        `  `        ``// initializing the values` `        ``float` `x1 = ``0``, x2 = ``1``, E = ``0``.0001f;` `        ``secant(x1, x2, E);` `    ``}` `}`   `// This code is contributed by Anant Agarwal.`

## Python3

 `# Python3 Program to find root of an ` `# equations using secant method `   `# function takes value of x ` `# and returns f(x) ` `def` `f(x):` `    `  `    ``# we are taking equation ` `    ``# as x^3+x-1 ` `    ``f ``=` `pow``(x, ``3``) ``+` `x ``-` `1``; ` `    ``return` `f; `   `def` `secant(x1, x2, E):` `    ``n ``=` `0``; xm ``=` `0``; x0 ``=` `0``; c ``=` `0``; ` `    ``if` `(f(x1) ``*` `f(x2) < ``0``):` `        ``while` `True``: ` `            `  `            ``# calculate the intermediate value ` `            ``x0 ``=` `((x1 ``*` `f(x2) ``-` `x2 ``*` `f(x1)) ``/` `                            ``(f(x2) ``-` `f(x1))); `   `            ``# check if x0 is root of ` `            ``# equation or not ` `            ``c ``=` `f(x1) ``*` `f(x0); `   `            ``# update the value of interval ` `            ``x1 ``=` `x2; ` `            ``x2 ``=` `x0; `   `            ``# update number of iteration ` `            ``n ``+``=` `1``; `   `            ``# if x0 is the root of equation ` `            ``# then break the loop ` `            ``if` `(c ``=``=` `0``): ` `                ``break``; ` `            ``xm ``=` `((x1 ``*` `f(x2) ``-` `x2 ``*` `f(x1)) ``/` `                            ``(f(x2) ``-` `f(x1)));` `            `  `            ``if``(``abs``(xm ``-` `x0) < E):` `                ``break``;` `        `  `        ``print``(``"Root of the given equation ="``, ` `                               ``round``(x0, ``6``)); ` `        ``print``(``"No. of iterations = "``, n); ` `        `  `    ``else``:` `        ``print``(``"Can not find a root in "``,` `                   ``"the given interval"``); `   `# Driver code `   `# initializing the values ` `x1 ``=` `0``; ` `x2 ``=` `1``; ` `E ``=` `0.0001``; ` `secant(x1, x2, E); `   `# This code is contributed by mits`

## C#

 `// C# Program to find root of an ` `// equations using secant method` `using` `System;`   `class` `GFG {` `    `  `    ``// function takes value of ` `    ``// x and returns f(x)` `    ``static` `float` `f(``float` `x) ` `    ``{` `        `  `        ``// we are taking equation ` `        ``// as x^3+x-1` `        ``float` `f = (``float``)Math.Pow(x, 3) ` `                                ``+ x - 1;` `        ``return` `f;` `    ``}` `    `  `    ``static` `void` `secant(``float` `x1, ``float` `x2,` `                    ``float` `E)                 ` `                    `  `    ``{` `        `  `        ``float` `n = 0, xm, x0, c;` `        ``if` `(f(x1) * f(x2) < 0) ` `        ``{` `            ``do` `{` `                `  `                ``// calculate the intermediate` `                ``// value` `                ``x0 = (x1 * f(x2) - x2 * f(x1))` `                    ``/ (f(x2) - f(x1));` `        `  `                ``// check if x0 is root of` `                ``// equation or not` `                ``c = f(x1) * f(x0);` `        `  `                ``// update the value of interval` `                ``x1 = x2;` `                ``x2 = x0;` `        `  `                ``// update number of iteration` `                ``n++;` `        `  `                ``// if x0 is the root of equation ` `                ``// then break the loop` `                ``if` `(c == 0)` `                    ``break``;` `                ``xm = (x1 * f(x2) - x2 * f(x1)) ` `                    ``/ (f(x2) - f(x1));` `                            `  `                ``// repeat the loop until ` `                ``// the convergence ` `            ``} ``while` `(Math.Abs(xm - x0) >= E); ` `                                                `  `            ``Console.WriteLine(``"Root of the"` `+` `                    ``" given equation="` `+ x0);` `                    `  `            ``Console.WriteLine(``"No. of "` `+ ` `                              ``"iterations = "` `+ n);` `        ``} ` `        `  `        ``else` `            ``Console.WriteLine(``"Can not find a"` `+ ` `                              ``" root in the given interval"``);` `    ``}` `    `  `    ``// Driver code` `    ``public` `static` `void` `Main(String []args) ` `    ``{` `        `  `        ``// initializing the values` `        ``float` `x1 = 0, x2 = 1, E = 0.0001f;` `        ``secant(x1, x2, E);` `    ``}` `}`   `// This code is contributed by vt_m.`

## PHP

 `= ``\$E``); ` `        `  `        ``echo` `"Root of the given equation="``. ``\$x0``.``"\n"` `;` `        ``echo` `"No. of iterations = "``. ``\$n` `;` `        `  `    ``} ``else` `        ``echo` `"Can not find a root in the given interval"``;` `}`   `// Driver code` `{` `    `  `    ``// initializing the values` `    ``\$x1` `= 0; ``\$x2` `= 1; ` `    ``\$E` `= 0.0001;` `    ``secant(``\$x1``, ``\$x2``, ``\$E``);` `    ``return` `0;` `}`   `// This code is contributed by nitin mittal.` `?>`

## Javascript

 ``

Time Complexity: O(1)
Auxiliary Space: O(1)

Reference
https://en.wikipedia.org/wiki/Secant_method
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