the error"The function values at the interval endpoints must differ in sign" + fzero

29 ビュー (過去 30 日間)
Pooneh Shah Malekpoor
Pooneh Shah Malekpoor 2020 年 7 月 9 日
Hello. I want to solve the beneath nonlinear eqn(there is only one variable) as the code below. When using fzero, I get the error as"The function values at the interval endpoints must differ in sign.". I know there is a correct root at x=3.47 (which i am looking for).... Is my coding for fzero correct?
c=0
while Error>tol
c=c+1
syms x positive
for i=1:n+1
if i==1
EE(1,1)=0;
else
EE(i,1)=(((C*l(i-1)*x)+(((tan(phi*pi/180))*x)*((rho*9.81*A(i-1)*(cos(beta(i-1))))+(EE(i-1)*(sin(beta(i-1)-(te))))))-(rho*9.81*A(i-1)*(sin(beta(i-1))))+(EE(i-1)*(cos(beta(i-1)-(te)))))/((cos(beta(i-1)-(te)))+((sin(beta(i-1)-(te)))*(tan(phi*pi/180))*x)));
end
end
eqn = (simplify((EE(n+1,1))));
fh = @(x) (eqn);% you should use matlabFunction's 'vars' option to convert the variables into a vector.
if c==1
x0=[0 5];
else
x0=k
end
[x,fval,exitflag] =fzero(fh,x0),
  4 件のコメント
Pooneh Shah Malekpoor
Pooneh Shah Malekpoor 2020 年 7 月 9 日
Again, the same error.
As I could get the result from
[k] = vpasolve(eqn, x ,[0 5]);
Dont you think the function handle that i have created is wrong(i.e. fh = @(x) (eqn) )?

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回答 (1 件)

madhan ravi
madhan ravi 2020 年 7 月 9 日
fh = @(x) matlabFunction(eqn)
  5 件のコメント
madhan ravi
madhan ravi 2020 年 7 月 9 日
What’s value of c?
Pooneh Shah Malekpoor
Pooneh Shah Malekpoor 2020 年 7 月 10 日
I prefered to use vpasolve as I got errors each time.
Thanks

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