Matlab 中匿名函数定义内的匿名函数调用
anonymous function calls within anonymous Function definitions in Matlab
我很想知道为什么这段代码不起作用。如果有人知道导致 matlab 发现这么多错误的原因，我们将不胜感激。
m = 1;
c = 1.5;
fun =@(x, epsilon) 1 .* (1 - (1 - cos(x))/(2.*epsilon)).^c .* cos(m.*x);
a = @(ep) acos(1-(2*ep));
lm =@(e) 1/(2.*pi) .* integral(@(x)fun(x, e), -1.*a(e), a(e));
fprintf('ball bearing at 0.6 is %4.4f', lm(0.6));
我要复制的函数是() =1/2∫[1 − (1 − cos())/2]^ cos()dx
据我所知，乘法不需要点修饰符，但是 Matlab 抱怨说这需要逐元素运算，即使不涉及矩阵。
根据 integral 的文档，要集成的函数必须矢量化：
For scalar-valued problems, the function y = fun(x) must accept a vector argument, x, and return a vector result, y. This generally means that fun must use array operators instead of matrix operators. For example, use .* (times) rather than * (mtimes).
我很想知道为什么这段代码不起作用。如果有人知道导致 matlab 发现这么多错误的原因，我们将不胜感激。
m = 1;
c = 1.5;
fun =@(x, epsilon) 1 .* (1 - (1 - cos(x))/(2.*epsilon)).^c .* cos(m.*x);
a = @(ep) acos(1-(2*ep));
lm =@(e) 1/(2.*pi) .* integral(@(x)fun(x, e), -1.*a(e), a(e));
fprintf('ball bearing at 0.6 is %4.4f', lm(0.6));
我要复制的函数是() =1/2∫[1 − (1 − cos())/2]^ cos()dx
据我所知，乘法不需要点修饰符，但是 Matlab 抱怨说这需要逐元素运算，即使不涉及矩阵。
根据 integral 的文档，要集成的函数必须矢量化：
For scalar-valued problems, the function
y = fun(x)must accept a vector argument,x, and return a vector result,y. This generally means that fun must use array operators instead of matrix operators. For example, use.*(times) rather than*(mtimes).