Evaluating Nested Function Handles

5 次查看（过去 30 天）

Hasan Abbas 2016-8-15

评论： Hasan Abbas 2016-8-16

I have a function that I would like to define in steps. e.g.,

y1 = @(x) 1./sqrt(1 - x.^2);
y2 = @(x) 1./sqrt(1 + x.^2);
T = (y2 -y1)./(y2 + y1);

How can I directly evaluate T in terms of the underlying variable x, i.e. T(x)?

per isakson 2016-8-15

Did you try

T = @(x) (y2(x) -y1(x))./(y2(x) + y1(x));

and

>> T(0.4)
ans =
   -0.0805
>> T(1)
ans =
   NaN
>> T(1-eps)
ans =
   -1.0000

Hasan Abbas 2016-8-16

Thanks Per!

It did the trick.

MATLAB Language Fundamentals Data Types Structures

在 Help Center 和 File Exchange 中查找有关 Structures 的更多信息

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

Translated by