Re: Bringing back an old tetration curiosity?

On Sep 21, 12:25 pm, "I.N. Galidakis" <morph...@xxxxxxxxxxxx> wrote:
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 withtetrationare 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.


Look it up from where? You're assuming we've predefined it for
decimals
again. The point here is to *find* a definition. In step 3, we have no
definition yet for what it means to tetrate to a real tower. In step
3,
f() would be defined as some function on the set R of real numbers, or
the set Q of rational numbers -- not the set "D" of decimal expansions
or "QE" of quotient expressions. In that case f(0.49999999999...) and
f(0.5) would be exactly the same since 0.499999999999... is
mathematically the same as 0.5 -- both are the same real number, the
same element of R, which is the set that f acts upon, and hence f
would be acting upon exactly the same element in both cases.

exp(x) is not defined using decimals, it is defined using Real
Numbers,
the set R.

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).


And it is, since the function is defined using *real* arguments, not
"decimal" arguments. It's defined on the set R of real numbers, like
I said, not on the set D of decimal expansions.

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.


I'm not sure. What is the reliance on decimal expansions or other
*representations* of numbers all about? You are not defining it
on the set of decimal expansions, or the set of quotient expressions,
you are defining it on the sets of real and rational numbers.

[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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