# components of vector valued functions

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Wilhelm on 3 Feb 2014
Commented: Wilhelm on 3 Feb 2014
Given
F=@(x,y) [x+y,x-y.^2]
how do I refer to the second component of that function;
the command
syms x,y; F(x,y)(2)
gives an error, though I would expect to get x-y^2. My goal is to define a new function which only contains one particular component of a vector field like
f2=@(x,y) F(x,y)(2)
Probably simple but I did not find the correct syntax.
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### Accepted Answer

Walter Roberson on 3 Feb 2014
VSelect = @(V,IDX) V(IDX);
F = @(x,y) [x+y,x-y.^2]
f2 = @(x,y) VSelect(F(x,y),2);
There is no good direct way to do it. There is a way to do it all in one function call, but it is somewhat ugly and not easy to read.
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Wilhelm on 3 Feb 2014
Thx! Not obvious, though ...

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### More Answers (1)

Wilhelm on 3 Feb 2014
Edited: Walter Roberson on 3 Feb 2014
That gives
>> syms z
>> F(1,z)
ans =
[ z + 1, 1 - z^2]
in my example.
I want to define a new function handle however which yields the second component x-y^2 only (for my example).
Or did I misunderstood your answer?
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