How to get real root of a function using fminbnd?

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rezheen
rezheen 2025-4-30
评论： rezheen 2025-5-1
Hello, How can I force MATLAB to only give real solutions to a math function using fminbnd?
This is my code:
x=-1:0.01:8; y=@(x) (x).^(4/3); yinv=@(x) -((x).^(4/3)); x_1=-1; x_2=8;
[xmin, ymin]=fminbnd(y,x_1,x_2);
[xmax, ymax]=fminbnd(yinv,x_1,x_2);
fprintf('The local maximum is %.2f at x = %.2f\n', -ymax, xmax)
fprintf('The local minimum is %.2f at x = %.2f\n', ymin, xmin)
Gives this as output:
The local maximum is 16.00 at x = 8.00 % Good. This is what it's supposed to be based on the x domain
The local minimum is -0.50 at x = -1.00 % This is the problem. (-1)^(4/3)=1 (The real solution)
Also, when I test in the Command Window:
(-1)^(4/3)
I get
-0.5000-0.866i
I think my code is spitting out the real part of (-1)^(4/3) that I get in Command Window.
Thanks for your help
  1 个评论
John D'Errico
John D'Errico 2025-4-30
移动：John D'Errico 2025-4-30
@rezheen
I think you are confused. FMINBND is a MINIMIZER. It does not compute a root. If the minimum of your objective can be negative, you will get it.
You want to use fzero to compute a root. Even at that, you need to understand that raising a negative number to a fractional power has a complex result as the primary solution.

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John D'Errico
John D'Errico 2025-4-30
编辑：John D'Errico 2025-4-30
Ran in:
It sounds like you want to see the result:
(-1)^(4/3) == 1
To do that, you want to use nthroot, by splitting the fraction in that exponent into two parts.
x^(4/3) = (x^(1/3))^4
And therefore, we would have
f = @(x) nthroot(x,3)^4;
f(-1)
ans = 1
So now a nice real number.
This works as long as the denominator of the exponent is an integer, and thus we can use nthroot. Will it still work, if we tried that trick on x^0.63? Well, in theory, it migh seem so, since 0.63 = 63/100. But nthroot will fail then. Try it:
nthroot(-2,100)^63
Error using nthroot (line 20)
If X is negative, N must be an odd integer.
  1 个评论
rezheen
rezheen 2025-5-1
Perfect. the nthroot function does exactly what I'm asking. It works not only for this problem but 3 other similar problems of y functions of x with fractional exponents. Thank you

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更多回答（1 个）

Walter Roberson
Walter Roberson 2025-4-30
Ran in:
x=-1:0.01:8; y=@(x) (x).^(4/3); yinv=@(x) -((x).^(4/3)); x_1=-1; x_2=8;
[xmin, ymin]=fminbnd(y,x_1,x_2);
[xmax, ymax]=fminbnd(yinv,x_1,x_2);
fprintf('The local maximum is %.2f at x = %.2f\n', -ymax, xmax)
The local maximum is 16.00 at x = 8.00
fprintf('The local minimum is %.2f+%.2fi at x = %.2f\n', real(ymin), imag(ymin), xmin)
The local minimum is -0.50+0.87i at x = -1.00
fprintf() ignores the imaginary component of numbers.
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Steven Lord
Steven Lord 2025-4-30
Ran in:
The realpow and/or nthroot functions may also be of interest.
x1 = (-1)^(1/3)
x1 = 0.5000 + 0.8660i
x2 = nthroot(-1, 3)
x2 = -1
x = roots([1, 0, 0, 1]) % all three roots
x =
-1.0000 + 0.0000i 0.5000 + 0.8660i 0.5000 - 0.8660i

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