The solution of the nonlinear equation obtained by ‘fzero’ is different from that of 'fsolve'.
1 visualizzazione (ultimi 30 giorni)
Mostra commenti meno recenti
dcydhb dcydhb
il 13 Mag 2019
Commentato: dcydhb dcydhb
il 13 Mag 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
0 Commenti
Risposta accettata
Walter Roberson
il 13 Mag 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.
Più risposte (0)
Vedere anche
Categorie
Scopri di più su Calculus in Help Center e File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!