Subject says it all.
Thanks,
-P
Hi,
do you mean something like
root [0] TF1 f("f","x",-3,3)
(TF1 &) Name: f Title: x
root [1] TF1 g("g","x*x",-3,3)
(TF1 &) Name: g Title: x*x
root [2] TF1 h("h","f+g",-3,3)
Cheers,
Danilo
I mean the TF1 equivalent of
y=f(x)
z=g(y)=g(f(x))
-P
HI,
This is possible for TF1 functions which are created using a string expression (a TFormula, see 1-3 in
root.cern.ch/doc/master/classTF1.html.
It is not possible for TF1 created using a C++ function (types 4 to 6).
For example, this works:
root [0] TF1 f("f","x*x");
root [1] TF1 g("g","sin(f)");
Best Regards
Lorenzo