A curious question

: 136.376 Intro to Analysis: A curious question
   By umwoodf0 on Saturday, September 08, 2001 - 01:23 pm: Edit Post

Let a>0. Does there exist an analytic function r(x) on (-2,+¥) that satisfies the following:
1) r(0)=1, and
2) r(x+1)=ar(x)
for all x in (-1,+¥)???

Note 1. An approximation to this function is
s(x)=loga(x+2) , for x in (-2,-1],
s(x)=x+1, for x in (-1,0],
s(x)=ax , for x in (0,1],
s(x)=a(a(x-1)), for x in (1,2],
s(x)=a(a(a(x-2))), for x in (2,3], etc.

Note 2. a=e(1/e) is the largest value of a for which limx®¥r(x)=L for some L in R. For this case (a=e(1/e) ), L=e.


   By sasho on Monday, September 10, 2001 - 12:51 pm: Edit Post

Partial answer (for partial marks :-)).

a>0. Consider the following function, for the time being acting only over the positive integers: f(n)=a||n||=a(a(a(a...(aa)))), n many. We extend f to the set of rationals in the "usual" way : x||1/m|| is the inverse of x||m|| over positive real numbers; that is, y= x||1/m|| if x= y||m||, where y||m|| is the invertible function defined by y||m||=yyy..yy} m many times (here y is the variable, m is fixed). So then, f(1/m)= a||1/m||; further, f(n/m)=(a||1/m||)||m||. This gives f over rationals. Then extend to positive real numbers in the "usual" way : take a sequence r1, r2,... of rationals converging to a real number r and define ar = limi®¥ri.

So we got f(x)= a||x||, where x is real positive. It appears ("appears" is always a gap) that everything required is fulfilled except for the fact that f is defined over (0, ¥) rather than over (-2, ¥). I do not know if f could be extended. But the question could certainly be changed to accommodate the answer :-).


   By umwoodf0 on Monday, September 10, 2001 - 09:11 pm: Edit Post

There are some problems with such an approach.
Whereas for exponents, we have, for example that
(x^4)^(1/2)=x^2, it would not hold in this case.
If we take a=1.5, according to your definition, a^^2 (meaning a^a in your notation) is approximately 1.837117307.
a^^4 is approximately 2.349005319
finally, (a^^4)^^(1/2) is roughly 1.66839753, which is different from a^^2

In general, the rules for exponents
(a^bc)^(1/bd)=(a^c)^(1/d)=(a^(1/d))^c
do not hold for our "hyper-exponents".

Here is a second example:
(2^^4)^^(1/3)=2.58611054
However, (2^^(1/3))^^4=2.180596631

Perhaps, and this is just a conjecture, we could take a^^(p/q)=lim(n->infinity)(a^^(pn))^^(1/(qn))
if such a limit exists and equals
lim(n->infinity)(a^^(1/(qn))^^(pn).


   By sasho on Tuesday, September 18, 2001 - 09:58 pm: Edit Post

You are right: the extenstion to rationals is not well defined unless a^^(p/q) is equal to a^^(np/nq). And that is not true in general (essentially indicated in your last post). There is in fact one more problem: the function x^^n (fixed positive integer n) is not (necessarily) invertible over the set of positive real numbers x. For example, x^^4 looks like this:
x^^4.jpg
We can surely bypass this by restricting to reals not less than 1. Under that restriction, your idea with limits seems reasonable - but, of course, needs to be worked out.


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