Hi Andrew! On 24 Aug., 15:35, andrew ewart <aewartma...@googlemail.com> wrote: > suppose we define a function f(x)=x^3+1 > and define a_0=1 > and then had the iteration a_n=f(a_n-1)/(a_n-1) > how would one go about writing this in sage?
You can use the fact that Sage is built on Python. So, you can define a Python iterator, like that: sage: def F(f,a0): ....: an = a0 ....: while(1): ....: yield an ....: an = f(an)/an ....: Since the function uses "yield" rather than return, calling F with two arguments is an iterator. You want a function x->x^3+1 and a starting value, say, 1.0 (I use floating points; if you start with "1" instead of "1.0", all return values will be rational, not floating point). # The function: sage: f = lambda x: x^3+1 # the iterator: sage: L = F(f,1.0) sage: for i in range(10): # iterate over the first ten values ....: print L.next() ....: 1.00000000000000 2.00000000000000 4.50000000000000 20.4722222222222 419.160729391762 175695.719449953 3.08689858330367e10 9.52894286360222e20 9.08007520977956e41 8.24477658152533e83 You could also do "for x in L: print x". In theory, you'd get an infinite loop, but since the starting value was a float, it will soon yield an error as the numbers are too big. Cheers, Simon -- To post to this group, send email to sage-support@googlegroups.com To unsubscribe from this group, send email to sage-support+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/sage-support URL: http://www.sagemath.org
