CODEWITHSUNDEEP
matlabroot
Eric Kim
Eric Kim

Reputation: 41

MATLAB finding roots, and @ operator

x=1;
f=@(x) x^3 - (5/x^2)-4*sin(x)-2;
fzero(f,x)
ans =

    1.9227

I am supposed to find the root of the equation, x^3 - (5/x^2)-4*sin(x)-2, and the above code is the solution for it.

I don't understand the general mechanism of this code.

(1) What does @ operator do?

I know its something like function handle, but I don't understand what function handle is.

(2) How does it work when it includes x in the parenthesis?

(3) How can there be a function fzero(), when I haven't made a script for fzero()?

(4) why are there two variables inside fzero()? I don't understand that the variable 'f' does there

(5) Why did it declare x=1 in the beginning?

Please consider that I am pretty much new to MATLAB, and don't know much.

Upvotes: 0

Views: 695

Answers (1)

kdrtkl
kdrtkl

Reputation: 101

f = @(x) ... is the way to declare an anonymous function in MATLAB, actually not very different than creating a function normally in MATLAB such as function output = f(input) .... It is just the pratical way especially when you are working with mathematical functions.

@(x) defines that x is the variable of which is the same as f(x) in mathematics. fzero() is MATLAB's existing function to calculate the x value for f(x) = 0 which means calculating roots of defined funtion. Giving your x a real value at the beginning does mean the starting point to find the root. It will find the roots greater than 1 in your case. It will be very clear for you when you read existing documentation of MATLAB.

Edit:

If you give an interval such as x = [0 1] instead of x = 1, fzero(f,x) would try to calculate roots of f function in given interval, if there is no roots exist in that interval it woud return a NaN value.

Upvotes: 1

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