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From: Allamarein on 25 Feb 2010 07:44
I want to define
f[x_]=a x^2
and after
g[a_,f_]= 2a+f
so g is an "a" "x" function and where "f" is the previous function.
Giving to g a value for "a" and "x", I'd get a correct "g" result.
Which is it the correct listed?

From: Christoph Lhotka on 25 Feb 2010 17:37
hi, as far as I understand your post, f[x_]:=a x^2, then
g[a_,x_]:=2a*f[x] and g[a,x] should return, what you desire, chr!

Allamarein wrote:
> I want to define
> f[x_]=a x^2
> and after
> g[a_,f_]= 2a+f
> so g is an "a" "x" function and where "f" is the previous function.
> Giving to g a value for "a" and "x", I'd get a correct "g" result.
> Which is it the correct listed?
>
>
>
>


From: Bill Rowe on 26 Feb 2010 04:08
On 2/25/10 at 7:45 AM, matteo.diplomacy(a)gmail.com (Allamarein) wrote:

>I want to define f[x_]=a x^2 and after g[a_,f_]= 2a+f so g is an "a"
>"x" function and where "f" is the previous function. Giving to g a
>value for "a" and "x", I'd get a correct "g" result. Which is it the
>correct listed?

There are several ways to achieve what you want. The smallest
change to your code needed to obtain the desired result would be
to define the functions as:

f[x_]=a x^2;
g[a_,x_}= 2 a + f[x];

But, usually (not always) it is better to use SetDelayed (:=)
rather than Set (=) so that evaluate occurs when the function is
used rather than when it is defined. That would be done as follows:

f[x_]:= a x^2
g[a_, x_]= 2 a + f[x];

Note, I've used Set not SetDelayed in the definition of g. Since
f is defined in terms of a single argument x, it is necessary to
evaluate f[x] when defining g. Otherwise, f will not see the
value of a provided to g.

While the above works, I would do the following:

f[x_, a_]:= a x^2
g[x_, a_]:= 2 a f[x, a]

This causes evaluation to occur when g is evaluated and makes
both f and g depend on arguments passed to them. Making
functions depend on things not directly passed to them often
causes unexpected results when the function is used elsewhere in
a notebook. And this kind of problem can be difficult to find.


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