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problem with function handle

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Saneesh Ali
Saneesh Ali 2019-1-7
评论： madhan ravi 2019-1-7
alpha
alpha =
function_handle with value:
@(z)(z-b)/a
>> qa
qa =
function_handle with value:
@(z)sqrt((1./alpha.^2)-p.^2)
>> t = integral(qa,0,3)
Undefined operator '.^' for input arguments of type 'function_handle'.
Error in @(z)sqrt((1./alpha.^2)-p.^2)
Error in integralCalc/iterateScalarValued (line 314)
fx = FUN(t);
Error in integralCalc/vadapt (line 132)
[q,errbnd] = iterateScalarValued(u,tinterval,pathlen);
Error in integralCalc (line 75)
[q,errbnd] = vadapt(@AtoBInvTransform,interval);
Error in integral (line 88)
Q = integralCalc(fun,a,b,opstruct);

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采纳的回答

madhan ravi
madhan ravi 2019-1-7
编辑：madhan ravi 2019-1-7
EDITED
syms z
a=1; % an example datas
b=2;
p=3;
alpha=(z-b)/a;
qa=sqrt((1./(alpha.^2)-p.^2));
t = integral(matlabFunction(qa),0,3)
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Torsten
Torsten 2019-1-7
Alternatively:
qa=@(z)sqrt(1./alpha(z).^2-p.^2);

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