fsolve yields wrong answer
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Pooneh Shah Malekpoor
2021 年 11 月 9 日
コメント済み: Pooneh Shah Malekpoor
2021 年 11 月 10 日
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
0 件のコメント
採用された回答
Star Strider
2021 年 11 月 9 日
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 件のコメント
Star Strider
2021 年 11 月 10 日
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
2021 年 11 月 10 日
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:
その他の回答 (2 件)
Matt J
2021 年 11 月 10 日
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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