Vectorization and plotting user-defined function

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John Ball
John Ball 2020-9-4
评论： Adam Danz 2020-9-5
I am new to MATLAB and am having trouble understanding how to plot userdefined functions.
I have defined a function f via
fun=@(t,r)(t.^2-t.^4).*exp(r.*t.^2);
fun1=@(t,r)exp(r.*t.^2);
f=@(r)integral(@(t)fun(t,r),0,1)./integral(@(t)fun1(t,r),0,1);
Now if I specify say
r=linspace(-10,10);
and type
x=f(r);
I get errors such as
Matrix dimensions must agree.
Error in @(t,r)(t.^2-t.^4).*exp(r.*t.^2)
whereas
x=sin(r);
is accepted. I seem to have been careful to use .* etc. What is the difference between my f and sin, and how do I plot a parametrized graph (f(r),g(r))?

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采纳的回答

Adam Danz
Adam Danz 2020-9-4
编辑：Adam Danz 2020-9-4
To evaluate the integrals at each value in r,
x = arrayfun(f,r);
  2 个评论
John Ball
John Ball 2020-9-4
Many thanks Adam for the superfast response. But if I do
>> x=arrayfun(T,r);
>> y=arrayfun(g,r);
>> fplot(T,g, [-100 100])
I get a plot but still get vectorization errors
Warning: Function behaves unexpectedly on array inputs. To improve performance,
properly vectorize your function to return an output with the same size and shape as
the input arguments.
So it seems I am not understanding something.
Adam Danz
Adam Danz 2020-9-4
I don't get that error when I apply this solution to the function in you question (I doubt you get that error with those functions, either). So, something is different between T and g compared to f.
What line is producing the error? Could you provide a new minimal working example that reproduces the error?

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更多回答（1 个）

Steven Lord
Steven Lord 2020-9-4
By default integral calls the integrand function with a vector of values at which the integrand should be evaluated. There's no guarantee that this vector will have a size compatible with the size of your r vector.
If you tell integral that your function is array-valued by passing the name-value pair 'ArrayValued', true then integral will call your integrand with a scalar and expect an array as output. Try running the "Vector-Valued Function" example on the documentation page for the integral function with and without the name-value pair and compare the results.
  2 个评论
John Ball
John Ball 2020-9-5
Thanks very much for this. Combining your and Adam's answer works fine, and I think I even understand why. I accepted his answer, but I would like to have accepted both, but don't see how. In any case, this being my first use of this forum I found the quick responses so helpful.
Adam Danz
Adam Danz 2020-9-5
@John, voting (thumbs up icon) is another way to show support/appreciation. Glad to hear that you found the forum useful!

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