Exact and Unique solution for base e^(1/e)

[jaydfox]

Also your linear construction on sci.math.research is not completely clear to me.
...

Specifically I dont know what you mean with the function "frac" in your post in sci.math.research.

Frac(x) just means the fractional portion of x, defined by frac(x)=x-floor(x). Floor(x) means the greatest integer less than x. So frac(3.76)=0.76., and frac(-0.63) = 0.37.

Let me gather what I got so far. If we have the conditions
\( {}^1 b = b \) and \( {}^{x+1}b=b^{{}^x b} \) (1)

then to define \( g(x)={}^{x}b \) it suffices to define an initial function f on the interval \( (0,1] \) and
then derive the value on an interval \( (n,n+1\] \) via \( g(x)=\exp_b^{\circ n}(f(x-n)) \) (where \( \exp_b^{\circ n} \) is the n-times iterated \( \exp_b(x)=b^x \)).
For intervals \( (-n,-n+1] \) left of \( (0,1] \) one has similarly \( g(x)=\log_b^{\circ n}(f(x+n)) \).
If \( g(x) \) is continuous at 1 then
\( g(1)=b^{g(0)} \), \( g(1)=b^{b^{g(-1)}} \) and \( g(1)=b^{b^{b^{g(-2)}}} \) from which \( {}^0b=1 \), \( {}^{-1}b=0 \), \( {}^{-2}b=-\infty \) follows.

So it makes sense to define g on \( (-2,\infty) \).

Correct so far.

Then somehow you wanted to define the initial function \( f : (0,1]\to(1,b] \) to be linear and from there stopped my understanding.
Especially I didnt see the problem you mentioned with \( b=e^e,10,1.5 \).

Actually, in most implementations I've seen that use linear interpolation, the initial function is typically defined as linear on (-1, 0], with endpoints of 0 and 1, respectively. Then you perform a logarithm to get (-2, -1], and iterated exponents to get intervals (0, 1], (1, 2], etc.

Edit: Now it seems to me that \( \text{frac}(x)=1+(b-1)*x \) is the initial linear function that maps \( (0,1]\to (1,b] \). Ok, but if I graph the resulting function g(x), I see angles at x=0 and x=1, for both \( b=e \) and \( b=e^{1/e} \).

Not sure I follow. Run the numbers again, with the frac(x) function I described above, and see if the angles you describe go away. But remember, for base e, you'll be using (-1, 0] as the interval, so linear interpolation will go through points like <-0.9, 0.1>, <-0.35, 0.65>, etc. For base 2, you'll be using an interval of (-0.5156131..., 0.4843868...], as shown in my sci.math.research post. (Note: I left out the minus sign on -0.5156..., but I hope it's obvious that it should be there).

On that interval, you'll be interpolating between log_2(e) and log_2(log_2(e)) as your y values, so your function should get the following points:

<-0.515613137, 0.528766373>
<-0.4, 0.634428533>
<-0.2, 0.817214266>
<0.3, 1.2741786>
<0.484386863, 1.442695041>

I'll probably be away until early next week. When I return, I plan to go back in and provide numerical results to demonstrate some of the formulae I used. I'll also provide graphs where appropriate.

By the way, in the seventh post in that sci.math.research thread, I describe my first attempt at an exact solution. It was a glorious failure, which is to say, it's infinitely differentiable, satisfies the iterated exponential property, and it's extremely "wrong". I posted a graph where you can see that the second derivative is very lumpy, whereas I would hope it to be far more smooth. In fact, the second derivative of the logarithm of the first derivative (the yellow curve in the graph) shows just how wavy the function can get.

I bring this up, because that solution is now attached to the discussion I started in the first few posts of that thread, which I still consider very insightful. I don't want to get the two ideas mixed up, since the one seems valid, the other quite "wrong", as I'm willing to admit.