Re: [Axiom-mail] Beginner problem with repeat

[Bill Page]

On 10/17/07, Robert Funnell <address@hidden> wrote:
> Bill -
>
> Thanks! I'll try that tomorrow. I had read that everything but the
> last thing in the loop would be thrown away, but since the solve is
> the last thing in the loop, I thought I'd see its results. Why do I
> need an output for the solve in the loop when I don't need it for the
> one outside the loop?

Contrary to what I implied in my previous email, apparently Axiom's
for loop is defined to return only a "Void" result. In contrast a
sequence of statements does behave as you say so that for example the
value of 'x' after the execution of:

  x:=(1;2;3)

is 3.

>
> Is there a simple explanation for why you need to do 'for i in plist'
> then p=i rather than just 'for p in plist'?
>

The reason has to do with the scope of the variable inside the
for-loop. It is a different 'p' then the one that you used when you
defined the equation. You could write however:

  for p in [1000,2000,3000] repeat output solve(p^2 + 2*P + 5 = 0, 1.e-6)

now with p occuring inside the loop.

Alternatively you might pass p to a function that defines the equation
you want to solve:

  eq1(p) == p^2 + 2*P + 5 = 0
  for p in [1000,2000,3000] repeat output solve(eq1(p), 1.e-6)

Regards,
Bill Page.