MaplePrimes - answers and comments on Question, Extracting information from a procedure?

bug

Joe Riel — Sat, 27 Feb 2010 22:55:14 Z

What are a and s doing? The assignment a[s] := a[s],a[0] will almost certainly give an error. That is, it is similar to

restart;
x := x+1;
Error, recursive assignment

sorry the above wasn't very

slipper — Sun, 28 Feb 2010 00:12:37 Z

sorry the above wasn't very clear, heres a new post.

iter := proc(f::mathfunc,x0::complex,x1::complex,N::integer,gs::realcons)         local a,k,n:
>                a[0] := x0:
>                a[1] := x1:
>                   for k to N do
>                        a[0] := evalf(a[0]-f(a[0])*((a[0]-a[1])/(f(a[0])-f(a[1])))):
>                        a[t] := a[t],a[0]:
>                          print(abs(a[t+1]-a[t])):
>                           if gs>abs(a[t+1]-a[t]) and k<=N then return
>                          print(abs(a[t+1]-a[t])):
>                       end if:
>                    end do:
> end proc:

i want to evaluate a[t] at the value for a[0], then if gs>abs(a[t+1]-a[t]) and k<=N . i want to return the first iterate a[t+1] that meets this criteria and print an out put to see if it converges.

many thanks for any help

regards

slip

Assign variables

Joe Riel — Sun, 28 Feb 2010 00:36:16 Z

What is t? What is a? With a unassigned (which it is here) you cannot do

a[t] := a[t],a[0]

Technically, you can make the assignment, in a procedure, however, when later you try to reference a[t] it you get an infinite recursion.

sorry, i will start from

slipper — Sun, 28 Feb 2010 03:20:05 Z

sorry,

i will start from the beginning. i have this procedure (below), which emulates the secant method for finding a solution to f(x)=0, with initial estimates x0 and x1.

iter := proc(f::mathfunc,x0::complex,x1::complex,N::integer)

local a,k:
>                 a[0] := x0:
>                 a[1]:= x1:
>                  for k to N do
>                        a[0] := evalf(a[0]-f(a[0])*((a[0]-a[1])/(f(a[0])-f(a[1])))):
>                        print(a[0]):
>                  end do:

> end proc:
>
>
>
>
> f := x->x^2-2;

                                      2
                           f := x -> x - 2

> iter(f,1.,2,10);

I am trying to modify to obtain procedure iter2(f,x0,x1,N,gs), where gs is real number. iter2 to return first iterate x[k] for which |x[k]-k[k-1]|<gs and k<=N, if |x[k]-k[k-1]|>gs, then an error message saying covergence not achieved.

The first two posts were my attempts to include a further statement, in the above procedure to obtain this.

regards

slip

careful

longrob — Sun, 28 Feb 2010 11:30:31 Z

This is assessed homework. I think you need to be careful with what you are asking.