solve vs. fsolve

I am trying to solve non linear equations with solve or fsolve.

But I found difficulties in using them.

for example,

i would like to solve the equation below.

a^2 = 4

then, i can get the answer with the script below.

syms a;
eqn1 = a^2 == 4;
[a1] = solve(eqn1);

then, it gives me a = -2, 2

but when I used fsolve, it gives me only one value according to initial point.

this is the script.

fun = @test;
x0 = [1];
[a2] = fsolve(fun,x0)

function F = test(x)
F(1) = -x(1)^2 + 4;
end

if I set x0 as [-1], then it gives me another value but still only one value.

How can I get the all values with fsolve?

Answer 1

[Star Strider]

Give fsolve different starting points, one positive and one negative:

eqn1 = @(x) x.^2 - 4;
for k = 1:2
    a2(k) = fsolve(eqn1, 5*(-1)^k);
end
a2

producing:

a2 =
           -2            2

[byungkeuk cho]

Thank you for your quick response.

But if I have no idea on "how many staring points needed" or "what points they should be" due to complicated equations, then how can I solve it?

[Star Strider]

My pleasure.

The easiest way would be to plot it, at least to find the real roots. If some or all of the roots are complex, this becomes more difficult, however fsolve will take complex initial estimates and will use them to return complex roots. In that situation, it will be necessary to experiment.

Answer 2

[Walter Roberson]

Sometimes plotting and finding intervals by eye is not productive.

In the general case, theory says that it is impossible to construct a function that will find all the zeros of every arbitrary function.

In practice, sometimes you can do good enough (your function might be well-enough behaved)

https://www.mathworks.com/matlabcentral/fileexchange/55206-findroots

https://www.mathworks.com/matlabcentral/fileexchange/15122-real-roots-on-interval

https://www.mathworks.com/matlabcentral/fileexchange/50223-bairstow-s-method-of-finding-all-roots-of-a-polynomial (for the case where you are dealing with a polynomial)