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

2020 7 9
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更新时间：2020 7 10
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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),

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I replaced the range for x=[0 5] as you mentioned with x0=5 but I get the following error" Exiting fzero: aborting search for an interval containing a sign change because no sign change is detected during search. Function may not have a root.
x =
NaN
fval =
NaN
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 个）

madhan ravi
madhan ravi 2020-7-9

0 个投票

fh = @(x) matlabFunction(eqn)

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Pooneh Shah Malekpoor
编辑：Pooneh Shah Malekpoor 2020-7-9
I changed the code as you mentioned:
fh = @(x) matlabFunction(eqn);
if c==1
x0=5;
else
x0=k
end
[x,fval,exitflag] =fzero([fh],{x0}),
Then I came across this error:"FZERO accepts inputs only of data type double."
fh = @(x) matlabFunction(eqn);
if c==1
x0=5;
else
x0=k
end
[x,fval,exitflag] = fzero(fh,x0);
Ok, then this error!
"Undefined operator '==' for input arguments of type 'function_handle'."
I prefered to use vpasolve as I got errors each time.
Thanks

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