First input argument must be a function handle.

2017 8 10
1 个回答
回答已采纳
更新时间：2017 8 10
8 次查看（30 天）

0 个投票

Hi, I am trying to integrate using the following function:
if true
syms h g x pi
h = 2.0545
g = 5.3442
C = 1
y = C - (exp(-2.*g.*i.*x./h)).*(C - cos(x).*((h.^2)./2) + (g.*x)/2.*h.*i)
yx = integral(y,-pi,pi)
end
however, I get the error "First input argument must be a function handle. "
Can anyone suggest a solution?
Thanks!

请先登录，再回答此问题。

 采纳的回答

KSSV
KSSV 2017-8-10

1 个投票

h = 2.0545 ;
g = 5.3442 ;
C = 1 ;
y = @(x) C - (exp(-2.*g.*1i.*x./h)).*(C - cos(x).*((h.^2)./2) + (g.*x)/2.*h.*1i) ;
yx = integral(y,-pi,pi)

3 个评论
显示 1更早的评论 隐藏 1更早的评论

Sergio Manzetti
Sergio Manzetti 2017-8-10
编辑：Sergio Manzetti 2017-8-10
Hi KSSV, tried your suggestion on a full model:
if true
syms h g x C
h = 1.0545;
g = 5.34428;
C = 40;
a = (C - (exp(-2.*g.*i.*x./h)).*(C);
b = (C - (exp(-2.*g.*-i.*x./h)).*(C);
y = @(x) a*b;
q = integral(y,10^-9,0);
end
but it gives:
Error in integral (line 88) Q = integralCalc(fun,a,b,opstruct);
José-Luis
José-Luis 2017-8-10
y = @(x) ((C - (exp(-2.*g.*i.*x./h)).*C) * ((C - (exp(-2.*g.*-i.*x./h)).*C);
In your anonymous function the @(x) is not the symbolic variable x.
Sergio Manzetti
Sergio Manzetti 2017-8-10
Thanks. It must have function and not a result or call to another function in other words..?

请先登录，再进行评论。

更多回答（0 个）

类别

在 帮助中心 和 File Exchange 中查找有关 Calculus 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by