how can i integrate a function?

Question

elham kreem 2018-2-16

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/383068-how-can-i-integrate-a-function

评论： elham kreem 2018-2-16

采纳的回答： Jan

i have this function fun4=@(y)(y.^2 - d1i'*y+d2i)* f1 * exp(-(y-bx).^2./(2*s*tau))

expectfun4=integral(fun4, -2,4,'ArrayValued',true)

but it is not running ; appear this :

| | * _Error using + Matrix dimensions must agree.

Error in @(y)(y.^2-d1i'*y+d2i)*f1*exp(-(y-bx).^2./(2*s*tau))

Error in integralCalc/iterateArrayValued (line 156) fxj = FUN(t(1)).*w(1);

Error in integralCalc/vadapt (line 130) [q,errbnd] = iterateArrayValued(u,tinterval,pathlen);

Error in integralCalc (line 75) [q,errbnd] = vadapt(@AtoBInvTransform,interval);

Error in integral (line 88) Q = integralCalc(fun,a,b,opstruct); ****

although i run as the same function but in another fuction

can you help me

thanks

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Torsten 2018-2-16

在 MATLAB Online 中打开

fun4=@(y)(y.^2 - d1i*y+d2i)* f1 .* exp(-(y-bx).^2./(2*s*tau))

if

(y.^2 - d1i*y+d2i)* f1 .* exp(-(y-bx).^2./(2*s*tau))

gives a senseful vector for y being a scalar value.

Best wishes

Torsten.

elham kreem 2018-2-16

在 MATLAB Online 中打开

i applied this function ; the result is :

 fun4=@(y)(y.^2 - d1i'*y + d2i)* f1. * exp(-(y-bx).^2./(2*s*tau))
                                     |
Error: Unexpected MATLAB operator.

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Answer 1

Jan 2018-2-16

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/383068-how-can-i-integrate-a-function#answer_305498

在 MATLAB Online 中打开

As soon as you write the formula a function instead of an anonymous function, you can debug it:

% Create anonymous function to provide parameters:
fun4 = @(y) fun4fcn(y, d1i, d2i, bx, s, tau);
...  Call the integrator
% The actual function:
function dy = fun4fcn(y, d1i, d2i, bx, s, tau)
dy = (y .^ 2 - d1i' * y + d2i) * f1 * exp(-(y-bx) .^ 2 ./ (2 * s * tau));
end

Now use the debugger:

dbstop if error

When Matlab stops at the error, you can check the dimensions of the variables. Because there is only one "+" operator, I guess that d2i has an unexpected shape.