Using the secant method for a different function

Hi. I have this code for using the secant root for finding the the root of an equation. I am quite new to Matlab with little experience. I am just wondering rather than inputting my function(func = x^2 + x -1;) into the script how can I use a function from another script here? As I want to play around with the equation in another script and then not have to change this script when using different functions. I have tried removing my function from the script and inputting it into another script and then I tried f =inline(@func) but this does not work. Can someone please help me as I have hit a stump with this and I know it is probably something very simple. Thanks!

% Find the roots of using Secant Method

% func --> function is taken by user

% like sin(x) or x^2 + x - 1 or any other but use the same format

% i.e. use paranthesis as used above with 'x' for sin, cos,... etc

% and spacing between '+' or '-' operation

% a --> Xn-1

% b --> Xn

% c --> Xn+1

% maxerr --> Maximum Error in Root

syms x;

func = x^2 + x -1;

a = input('Ener Lower Limit: ');

b = input('Ener Upper Limit: ');

maxerr = input('Enter Maximum Error: ');

f = inline(func);

c = (a*f(b) - b*f(a))/(f(b) - f(a));

disp(' Xn-1 f(Xn-1) Xn f(Xn) Xn+1 f(Xn+1)'); disp([a f(a) b f(b) c f(c)]);

while abs(f(c)) > maxerr

    b = a;

    a = c;

    c = (a*f(b) - b*f(a))/(f(b) - f(a));

    disp([a f(a) b f(b) c f(c)]);

end

display(['Root is x =' num2str(c)]);

Answer 1

[Jan]

Do you really want a symbolical solution? If not:

Either define your function in a function:

function y = func(x)
y = x^2 + x -1;

Then provide @func as input to your secant method.

Or use an anonymous function:

func = @(x) x^2 + x -1;

[Ben]

Thanks for the help Jan. I am looking for the numerical solution to the root of the function. I have tried your approach however I am still getting the following error:

Error using inline (line 47) Input must be a string.

Error in secant2 (line 18)

f = inline(fun1);

And I still can't understand why this does not work?

[Jan]

Omit the inline, because it is not useful when you have a function handle already.