Roots of function - Fzero error

I'm trying to find the root of this function: f(x) = (1 - 3/4x)^(1/3) to then apply newton's method, but I don't know the initial guess. This is the code I'm trying:

>> fun = @(x) (1 - (3/4*x)).^(1/3);
>> x0 = 1;
>> x = fzero(fun, x0)

The error I'm receiving is: "Exiting fzero: aborting search for an interval containing a sign change because complex function value encountered during search. (Function value at 1.45255 is 0.22358+0.38725i.) Check function or try again with a different starting value."

I've tried other starting guesses, but then I get the error: "Function value at starting guess must be finite and real."

Any tips of how to overcome these errors/find the initial guess? Thanks!

Answer 1

[Andrei Bobrov]

>>fun0 = @(x) 1 - 3/4*x;
>> fzero(fun0,1)
ans =  1.3333
>>

[Britt]

hey, thanks! Are you able to tell me why the .^(1/3) isn't necessary in the code please?

[Andrei Bobrov]

our equation:

(1 - 3/4*x).^(1/3) = 0  

((1 - 3/4*x).^(1/3))^3 = 0^3

1 - 3/4*x = 0

Answer 2

[Walter Roberson]

That function is not suitable for newton's method.

The .^(1/3) is not going to add any roots, but it will cause problems when the expression before it is negative. But you need it to go negative in order to find a sign change.

The problem is that the MATLAB A.^B operator is defined as

exp(B * ln(A))

and if A is negative then ln(A) is complex and the overall result is likely to be complex.

I suggest you use

fun = @(x) nthroot((1 - (3/4*x)), 3);

[Andrei Bobrov]

+1

[Britt]

I know it's not suitable, I have to still write the code for it (which I've done) and plot it to 50 iterations (still trying to figure that out) and then explain why it isn't suitable.