The solution of the nonlinear equation obtained by ‘fzero’ is different from that of 'fsolve'.
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dcydhb dcydhb
on 13 May 2019
Commented: dcydhb dcydhb
on 13 May 2019
i use the 'fzero' and 'fsove' to get the roots of the nonlinear equation,and the function figure is as this ,so we can see the first positive root is about 1.5 and the second positive root is about 5,
and in the 'fsolve' we get the first 2 positive roots of the equation
m1=1.786212;
m2=5.287127;
but use the 'fzero',it runs out
m1=1.570796;
m2=4.712389;
and it didn't have any hints ,how queer is it.
it means that we should use an interval in the 'fzero'?
and the most important is that is didn't have any warnings.
codes are as this
g=9.8;
h=20;
hgang=20;
omega=2;
syms m0;
syms m1;
nu=omega^2*hgang/g;
g=(i*m1)*tanh(i*m1)-nu;
misan=inline(g);
m=zeros(10,1);
format long
for n=1:2;
x0=[1.5+(n-1)*pi];
m(n)=fzero(misan,x0);
fprintf('m%d=%f;\n',n,m(n));
end
figure are as this
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/219150/image.png)
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Accepted Answer
Walter Roberson
on 13 May 2019
fzero() identified a location with a sign change. The sign happened to change because of a singularity there, but it is still a sign change.
So, if you are going to use fzero, either use an interval or use a better starting approximation.
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