fsolve yields wrong answer
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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
采纳的回答
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 个评论
Pooneh Shah Malekpoor
2021-11-10
Walter Roberson
2021-11-10
For real-valued x, then (x-20.6667) is real-valued, and (x-20.6667).^2 is always non-negative (could be 0).
57.89 is always positive.
The sum of a non-negative number (x-20.6667).^2 and a positive number is always positive.
fsolve() tries to find values for which the given expression is 0. But we just demonstrated that for any real value x, the expression is always positive. Therefore no real roots exist for the function.
If you look at Star Strider's answer, you might notice that the real part is the same as the 20.6667 being subtracted, so you get a 0 real contribution from x-20.6667 . And the imaginary part + 7.6085i is sqrt(-57.89) so 7.6085i ^2 is -57.89 and so when you add +57.89 to that you get a total of 0. So 20.6667 + 7.6085i is a correct solution, and no real-only solutions exist.
Pooneh Shah Malekpoor
2021-11-10
format long g
F = @(x)((x-20.6667).^2+57.89);
opts = optimoptions('fsolve', 'Display', 'off');
[x, fval, errorflag, output] = fsolve(@(x) F(x), 0, opts);
x
fval
errorflag
output
Notice the errorflag is negative, indicating a failure. Notice the output message says that the last steps it tried did not do any good, but the function value was not near 0.
If the question is how to find out whether fsolve() solved the problem, then look at errorflag
If the question is how fsolve() figures out that it is not going to be able to solve the problem, then the answer is approximately that it did numeric gradient estimation to get to a place where it could see that the gradient was 0 but that the function was not 0.
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
2021-11-10
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:
更多回答（2 个）
Matt J
2021-11-10
1 个投票
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.
parts=@(z) [real(z);imag(z)];
complx=@(x) complex(x(1),x(2));
F = @(x)parts((complx(x)-20.6667)^2+57.89);
opts = optimoptions('fsolve', 'Display', 'off');
x = complx( fsolve(F, [0,0], opts))
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