fsolve yields wrong answer
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Pooneh Shah Malekpoor
el 9 de Nov. de 2021
Comentada: Pooneh Shah Malekpoor
el 10 de Nov. de 2021
Hello
It is an example of many equations in my lengthy code. I want to use fsolve to solve it as it is more efficient; however, it yields wrong answer:
F = @(x)((x-20.6667)^2+57.89);
opts = optimoptions('fsolve', 'Display', 'off');
x(1) = fsolve(@(x) F(x), 0, opts);
in this case it yields x(1)=20.6667 ; howver, it should not yield any solution in real domain!!!!I want to get real domain roots by fsolve.
What should i do? I dont want to use solve as it results in longer run times.
Bests
Pooneh
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Respuesta aceptada
Star Strider
el 9 de Nov. de 2021
Give fsolve a complex initial estimate to get a complex root —
F = @(x)((x-20.6667).^2+57.89);
opts = optimoptions('fsolve', 'Display', 'off');
x0 = 1 + 1i;
x(1) = fsolve(@(x) F(x), x0, opts)
.
6 comentarios
Star Strider
el 10 de Nov. de 2021
The fsolve function will find a complex root if given a complex initial estimate, or if a purely imaginary root is the only root.
One option to eliminate complex numbers could be to take their absolute values —
x = 20.6667 + 7.6085i;
xa = abs(x)
and the other option of course is just to use the real part.
However changing the code to accommodate complex numbers may be the best option, because it will likely provide the most accurate results.
.
Matt J
el 10 de Nov. de 2021
Give fsolve a complex initial estimate to get a complex root —
But note that this will only lead to a proper complex-domain solution under certain conditions:
Más respuestas (2)
Matt J
el 10 de Nov. de 2021
I want to use fsolve to solve it as it is more efficient; however, it yields wrong answer:
In your example, the function is a polynomial. Is this always the case for you? If so, it is more efficient to use roots(). Moreover, roots() will find all solutions, both real and complex, whereas fsolve can find only one solution.
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