> The following is better. It is longer, but it will work even if n >=
> 1000, whereas the above will fail spectacularly due to Python's stack
> limit. Also, the following is a bit faster than the above:
>
> def compose(f, n, a):
> """
> Return f(f(...f(a)...)), where the composition occurs n times.
>
> INPUT:
> - `f` -- anything that is callable
> - `n` -- a nonnegative integer
> - `a` -- any input for `f`
>
> OUTPUT:
> result of composing `f` with itself `n` times and applying to `a`.
>
> EXAMPLES::
>
> sage: def f(x): return x^2 + 1
> sage: x = var('x')
> sage: compose(f, 3, x)
> ((x^2 + 1)^2 + 1)^2 + 1
> """
> n = Integer(n)
> if n <= 0: return a
> a = f(a)
> for i in range(n-1): a = f(a)
> return a
Two thoughts:
(1) For negative numbers, it should be the inverse function, if that
is feasible. Otherwise, an error.
(2) Secondly, could we get this to return a function, somehow? In the
present state, I imagine I could do something like:
sage: x=var('x')
sage: f = compose(sin, 10, x)
sage: f = lambda z : f(x=z)
#f= sin^10(x)
> I independently called mine "compose" too, so that must be the right name!
>
> I've made adding this to the library:
> http://trac.sagemath.org/sage_trac/ticket/7742
>
> I hope somebody will make a patch, etc. (since this would be good
> practice for somebody, and I have other things to do).
>
> William
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