How can I pass a anonymous function as an output to another function?

function [a0,a1,a2]= fcn(s,v,a,T)
%a0,a1,a2 are coefficients of a polynomial which are calculated based on inputs s,v,a,T
end

The above function returns coefficients. The polynomial g(t) needs be calculated at time t =T , based on equation g(t) = a0 + a1* t^2+ a2* t^3.

for doing this I have added the following lines to the function above

function [a0,a1,a2]= fcn(s,v,a,T)
%a0,a1,a2 are coefficients of a polynomial which are calculated based on inputs s,v,a,T

g = @(T) 
a2*T^2 + a1*T + a0 
g(T)
end

But I am unable to pass g(t) as output to the function fcn as shown below:

function [a0,a1,a2,g(t)]= fcn(s,v,a,T)
%a0,a1,a2 are coefficients of a polynomial which are calculated based on inputs s,v,a,T

g = @(T) 
a2*T^2 + a1*T + a0 
g(T)
end

Looks likes I am not allowed to do that . I would like to know how this can be achieved

Note: If I just use g in the output array, it doesn't return the g(T) value instead it returns a2*T^2 + a1*T + a0 because g is a function handle.

Answer 1

[Stephen23]

"Looks likes I am not allowed to do that . I would like to know how this can be achieved"

Your attempted code has several bugs in it, e.g.:

• g(t) is not a valid output argument.
• g = @(T) is not a valid anonymous function.
• you do not define the output arguments a0,a1,a2.
• a2*T^2 + a1*T + a0 is not assigned to anything, so its result is discarded.
• a2*T^2 + a1*T + a0 is not "...based on equation g(t) = a0 + a1* t^2+ a2* t^3."
• probably others, I gave up checking at that point.

A function handle is a variable just like any other, it can be returned just like any other variable, e.g.

function out = myfun()
out = @sin;
end

and tested:

>> f = myfun();
>> f(pi/2)
ans =  1

"The polynomial g(t) needs be calculated at time t =T , based on equation g(t) = a0 + a1* t^2+ a2* t^3."

For that you do not need to return a function handle. you can just calculate the value directly and return that:

function g = fcn(s,v,a,t)
a0 = ..;
a1 = ..;
a2 = ..;
g = a0 + a1.*t.^2+ a2.*t.^3;
end

[Sree Veepuri]

Thank you so much for your answer.