The solution of the nonlinear equation obtained by ‘fzero’ is different from that of 'fsolve'.

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

Answer 1

[Walter Roberson]

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.

[dcydhb dcydhb]

but when i use a single value rather the interval,it still give the roots and didn't have any warnings,so it means that the avalibility of 'fzero' is less than the 'fsolve'?

x0=1.5+(n-1)*pi;