Re: Bringing back an old tetration curiosity?

mike3 wrote:
[snip]

To calculate x^^(m/n), any m,n \in N, m/n \in [0,1], just do the
long division, call its decimal representation y=m/n and simply look
up the value of the function for this real y, i.e., x^^y?


Well, just like how you'd evaulate any other function, like exp for
example.

Well, it's not "exactly" the same, technically speaking. For the exp function
(to use an example) we have a _definition_ which allows us to figure out things
quickly.

The definition proceeds as follows for f(x)=exp(x).

1) define f(0)=1 [and optionally f(1)=e]
2) define f(n)=e*e*...*e (n times), n \in N
3) define f(m/n)=exp(m)^(1/n)=exp(1/n)^m, m,n \in N
4) define f(y)=lim_{n->oo}exp(c(n)), with lim_{n->oo}c(n)=y, c(n) a Cauchy
sequence.

The difference with tetration are steps 3) and 4). If f(y)=x^^y, then the steps
are (for [0,1]):

1) define f(0)=1 [and optionally f(1)=x]
2) define f(n)=x^x^...^x (n-times), n\in N
3) define f(m/n)=f(y), where y=m/n.
4) define f(y)=lim_{n->oo}x^^(c(n)), with lim_{n->oo}c(n), c(n) a Cauchy
sequence.

The difference is that whereas the exp function has an explicit definition with
step 3) which allows one to calculate directly using m and n (using
exp(m)^(1/n)=exp(1/n)^m), with tetration you have to resort to the decimal
representation of m/n. For example, to calculate x^^(2/15), I cannot use m=2,
n=15 to calculate directly (i.e. "fall back" onto the functions x^^m, x^^(1/n),
m,n \in N), rather I have to map to the decimal y=m/n *first* and THEN look up
values.

Step 4) suffers accordingly. In general whenever one defines a function with
decimal arguments, there can be problems, because the decimal representation
suffers from non-uniqueness.

For example, one should expect x^^(0.49999....) to be exactly x^^(0.5)=x^^(1/2).

This subtle point bothers me. Whereas m/n is an exact representation of a
rational number, its decimal representation may not have a unique
representation, so how the heck does the function decide what to "output" for
x^^(0.4999...) for example? Do we say, 0.4999...=0.5, hence the function should
output x^^(0.5)? Do we say something similar for all rationals which have
non-unique representations? Perhaps I see ghosts, but I see a problem here.

[snip]

I guess it might work.

I think so too.

Hehe! A little optimism never hoit anyone ;o)
--
I.N. Galidakis

.

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