Author 
Comment/Response 
Andrew DuBuisson

10/21/04 2:09pm
I am having trouble using conditions with functions. If this has been answered before, please just refer me. I'll do the work, I just haven't found an answer yet.
The text below is all in the notebook too.

This is an example notebook. In my actual notebook, z is a huge function that I would not like to type in.
Independent varables: y, and x of form {x1,x2}
Arbitrary funciton: z, its the part of the problem
Arbitrary function: func[{y_,x_}]
I define an arbitrary value z:
z = x[[1]]^2;
I understand and accept the warning messages regarding the depth of x. I use immeadiate assignment because I do not want the delayed assignment.
Here's an arbitrary function that uses a condition. Notice arbitrary variable z in the third line.
In[20]:=
Clear[func]
func[{y_, x_}] := Evaluate[x + y] /; x[[1]] > y
func[{y_, x_}] := Evaluate[x + 4] /; x[[1]] < z
func[{2, {3, 4}}]
func[{9, {0, 8}}]
func[{9, {9, 7}}]
Out[23]={5,6}
Out[24]=func[{9,{0,8}}]
Out[25]=func[{9,{9,7}}]
Unfortunately the condition doesn't work with the variable z in there.
I checked to see if it would work with the value of z instead of z itself.
In[32]:=
Clear[func]
func[{y_,x_}]:=Evaluate[x+y]/;x[[1]]>y
func[{y_,x_}]:=Evaluate[x+4]/;x[[1]]<x[[1]]^2
func[{2,{3,4}}]
func[{9,{0,8}}]
func[{9,{9,7}}]
Out[35]={5,6}
Out[36]=func[{9,{0,8}}]
Out[37]={5,11}
Hey it works! Unfortunately the functions I am dealing with are much longer than z = x[[1]]^2. So I need a way to get them into the condition, without hand typing them in.
I even tried reformatting the condition like one of the examples in the help notebook.
In[81]:=
Clear[func]
func[{y_,x_}]/;x[[1]]>y:=Evaluate[x+y]/;x[[1]]>y
func[{y_,x_}]/;x[[1]]<z:=Evaluate[x+4]
func[{2,{3,4}}]
func[{9,{0,8}}]
func[{9,{9,7}}]
Out[84]={5,6}
Out[85]=func[{9,{0,8}}]
Out[86]=func[{9,{9,7}}]

Thanks,
Andrew
Attachment: conditions.nb, URL: , 
