Robert Shaw wrote:
>
>> For example, if you just have x -> x! and x -> sqrt(x),
>> can you get as close as you want to any number? This is
>> an extreme case, because you can't ()! non-integer numbers
>> [calling x! = Gamma(x + 1) is cheating!!!]
>
> Well, if you start with 0 or 1, you can only reach 0 or 1
> respectively. Start with 2 and you can a single sequence of
> numbers between 1 and 2
>
Of course you have to start non-trivially :-)
> If you start with 3 you can only generate a countable infinity
> of numbers. Take the factorial n times, then the square root
> m times. Because of the sparsity of the factorials I'd suspect
> this set isn't dense.
>
My intuition says that this set *is* dense - of course, it's
hard to prove stuff with numbers in the range of 3!!!!!!!
> Generally, composing just two functions will give an uncountable
>
Countable :-)
> infinity of numbers which I think means the set of numbers
> must be dense over a nonzero interval.
>
This is not true in general; if you take off an interval
from any dense set, the resulting set isn't dense.
> Sometimes.
> Galois proved that from x+1, x^(1/2), x^(1/3) and x^(1/5)
> you can't get the solution of x^5+x=1.
>
It was stronger than that - and x^5+x = 1 *has*
a simple solution, because x^5 + x - 1 = x^5 + x^2 - x^2 + x - 1 =
x^2 (x + 1) (x^2 - x + 1) - (x^2 - x + 1) etc.
>
> We can add one because of the trigometric identities
> but multiplying would seem to require an infinite
> set of such identities. I'd suspect there isn't any finite
> set of functions whose compositions include n*x for
> all integer n.
>
Hmmm.... I'm not so sure. Those trigonometric functions
always find funny ways of cancelling each other :-)
>> Can you claim that some number is *not* constructable?
>
> I'd claim 5^(1/5) isn't constructable, but I don't know
> how to prove it.
>
It's easily constructable if your calculator has
10^x and log10(x) :-)
Ok, non-cheating. Let's try to construct 7^(1/7)
Alberto Monteiro
