Is there an easier/better way (default arguments)?
- To: mathgroup at smc.vnet.net
- Subject: [mg27793] Is there an easier/better way (default arguments)?
- From: johntodd at fake.com (John Todd)
- Date: Sat, 17 Mar 2001 05:41:17 -0500 (EST)
- Sender: owner-wri-mathgroup at wolfram.com
Hello,
I recently was wrangling with the problem of taking the
derivative with respect to a particular variable of each element of a
list. As an example (I am using Mathematica 4.0):
(* First, create the list *)
myt = Table[Cos[\[Theta] + (n - 1) \[Alpha]], {n, 10}]
(* The method outlined next didn't work for me, and I feel it has to
do with the fact that the D function requires two arguments. If
someone could point out another way other than my forthcoming solution
to using a multi-parameter function with Map, I'd be obliged to see
it. *)
Map[D, myt]
(* I then decided to define my own function and supply and default
argument of alpha.... *)
mymapfunc[arg1_, arg2_:\[Alpha]] := D[arg1, arg2]
(* Now, when I call *)
Map[mymapfunc, myt]
(* It works as I had hoped it would, in that it takes the derivative
with respect to alpha of each element in myt. *)
(*I am quite new to Mathematica, and I hope that this question is not
too mundane and simple, but, since I got it to work, I figured I'd run
it by the group and see what the consensus is.
One other quick question, whenever I would 're-execute' my function
definition after changing the default value of arg2, it would retain
the old default value unless I called Clear[mymapfunc] and then
're-executed' the function definition.*)
(* For example, if this is my original mymapfunc .. *)
mymapfunc[arg1_, arg2_:\[Alpha]] := D[arg1, arg2]
(* And then I change it to *)
mymapfunc[arg1_, arg2_:\[Theta]] := D[arg1, arg2]
(* And press the 'Return' key, it retains the default of alpha. But
if I call Clear[mymapfunc] and then 're-execute' it has the new value
of theta as a default. Whenever I change any other part of my
function, the change sticks without first having to clear. Anyone
else come across this problem? *)
Thanks very much,
JT