User blog:ArtismScrub/Generalizing the fast-growing hierarchy towards positive real heights

amazing how we've come up with a way to generalize THIS before the hyper-operators

Rules
For integer a,b, computation proceeds as normal.

fa(b) =  fa-1b(b) if a > 0

f 0 (b) = b+1

Now for the generalized version:

f a (b) =  f c ( f a-1 int(b) (b) )  if a > 1, where c is (a-1) ×frac(b)

if c < 1, then  f a (b) =  f c ( f a-1 int(b) (b) ) -  log 2 (1+(1-c)), to prevent  f 0 s from causing a major difference for integer b

f a (b) = b(a+1)  + log2(1+(1-a)) if b is between 0 and 1

Think this is unnecessarily complicated? I agree completely.

The last part took me quite a while to devise. It was easy to find a sequence between b and 2b, but I somehow had to fit that +1 at the beginning so  f 0 (b) wouldn't just equal b, but  f 1 (b) would still be 2b, not 2b+1. So, for lack of a better sequence approaching 0, I used the binary logarithm. Arbitrary? Somewhat, but at least it's a viable sequence that didn't involve infinite sums or circular references.

And even after that, I realized that if I didn't add the logarithm part to the first rule, then some normal function like  f 2 (3) would evaluate to  f 0 (

f <sub style="font-weight:400;">1 (

f <sub style="font-weight:400;">1 (

f <sub style="font-weight:400;">1 (3) ) ) ), which would yield a different result than normal, which is not acceptable.

But still, this allows us to define numbers like:

biroot-balum =  f <sub style="font-weight:400;">√2 (10)

goldalum =  f <sub style="font-weight:400;">φ (10) (or bifrac-unadd-biroot-quintalum)

Eulalum =  f <sub style="font-weight:400;">e (10)

pialum =  f <sub style="font-weight:400;">π (10)

all of which I can't be bothered to calculate, since the recursion is very awkward.

Now, this still comes with a few issues. Is there any way this can be improved to be less forced and perhaps support the equalities mentioned?
 * 1) The logarithm part is still kind of forced and arbitrary.
 * 2) I'm almost certain that this does not satisfy the equality f<sub style="font-weight:400;">2 (n) = n×2n  for non-integer n.
 * 3) Or the equality f<sub style="font-weight:400;">n (1) = 2.