ramanujan and tetration
#8
Ivars Wrote:Yep, found it.

The function \( f(x) \) defined as series of n times iterated exponential \( \exp^{\circ n}_e(x) \) (with base e) terms divided by corresponding n-th hyperfactorials (2^3^4^...n....) is on p. 326-327, Berndt, Ramanujan Notebooks, Vol 1., in excerpts from Second Quaterly report. It would be interesting to see if the original contains something more.

Berndt states that it converges for every x and for every n>0 \( f(x)> \exp^{\circ n}_e(x) \).

He does not mention entire function there.

a function that has a power series representation
that converges for every x
is entire

Quote:P.S. by adhering to Berndt's endnote of giving way to formal math, we can define by simple analogy and make conjectures:

tetra e = 1+1/2^3+1/2^3^4+ .......... = 1+1/8 + rest = probably around 1,125...because of extreme slowness.

and use it as basis for taking tetra-logarithms of iterated exponential on iteration parameter ( if we have e[4] n = f(e,4,n) then basis tetra e would be used to take tetra-log f(e,4,n) = tetra-log (n). For other bases a the value would have to be adjusted. For other n-tations we need different base for penta log etc.

Further analogies may include tetra Taylor expansion, etc.
Code:
ramanujan was fascinated with the orders of growth
and played with many types of iteration

he has nested radical theorems
  continued fractions
  iterated trigonometric
and many much more complicated functional forms iterated

when combining iteration and summing
orders of growth are what dictate global convergence

there are quite a vast array of different forms one may sum

d'alembert's ratio test is the key algebraic tool for entire functions

if one were to formalise the theory to build natural generalisations
one needs to apply the classical hierarchies of orders
  to build a map of what interesting forms appear

it is clear that ramanujan does this in many places

these basic entire forms all look like

oo
---
\     f(x, j)
/     -------
---     g(j)
j=0

where forAll x,

| f(x, j+1) g(j) |
| -------------- |  --> 0  as j --> oo
| f(x, j) g(j+1) |

the exponential is the ur-entire function
and is very clever about the way it does this

for f(x, j)
  x^j is the taylor basis
and just multiplies x together j times

to get the right type of term growth
g(j) needs to grow like some j things multiplied together or faster

with the factorial
  the n things multiplied together grow as n->oo
  which compared to the "constant" x
always wins out
i.e. eventually for some n

n          n
---        ---
| |        | |
| |  x  <  | |  j
j=1        j=1


ramanujan took

/  n         \
|  O  exp(.) |(x)
\ j=1        /

as his f(x, n)
and

n+1
^
/|\
|   (j+1)
j=1

as his g(n)
(with n=0 creating the special term 1 instead of 2 expected)

although they are both exponential towers
the first has constants e for all but one entry
  which is the "constant" x
(in all this discussion - i mean constant as n->oo)
and the latter expression has terms that grow

possibly more symmetric with taylor may have been

n
^
/|\
|  x
j=1

as f(x, n)
and

n
^
/|\
|  (j+1)
j=1

as g(n)

e.g.

               x
         x    x
    x   x    x
1 + - + -- + --- + ...
    2    3     4
        2     3
             2

but there are trade-offs in properties

the e^e^..^x have pretty good derivative properties
e.g. d/dx (e^e^x) = e^x e^e^x

but

d/dx (x^x^x) = d/dx (e^(ln(x^x^x)) = d/dx (e^(x^x ln(x)))
= d/dx(x^x ln(x)) x^x^x = (d/dx (x ln(x)) x^x ln(x) + x^(x-1)) x^x^x
= ((ln(x) + 1) x^x ln(x) + x^(x-1)) x^x^x

(or something like that)
which is much more complicated to work with

..

in the formal setting
  jumping to iterating exponentiation misses a whole lot of other growth orders

in fact
  we can start with iterating the original functions
    the numerator over x
    and the denominator over factorial

this gives two different possible directions

for iterating the numerator
f(x,n) = x^n

                    2
                   n
so f(f(x,n), n) = x

this is multiplying n^2 copies of x
so an appropriate denominator might be

g(n) = (n^2)!
or (n!)^n

repeated iteration of f gives the various series of entire functions

     oo     k
(   ---    j     )oo
/   \     x      \
\   /    -----   /
(   ---    k     )k=1
    j=0  (j )!

     oo     k
(   ---    j     )oo
/   \     x      \
\   /    -----   /
(   ---      k   )k=1
    j=0  (j!)

and since k here plays the part of constant
there is the "limit" types with terms

   j             j
  j             j
x             x
-----   and   -----
  j               j
(j )!         (j!)

this can continue anew with iterations
  producing constants
    that can be eaten by some limit form
    (which play much the same role as limit ordinals
     in the order theory of function asymptotics)

[[notice that the second form is not just a projection of the exponential series]]

alternatively
  the factorial may be iterated
giving terms with j! numbers in the denominator
so a natural term would be

  j!
x
-----
(j!)!

this
  like the first term type above
is just a projection
but substituting j^j for j!
in any single place
  produces new forms that have more interesting structure

and this is key to the generalisation needed
because there are many ways to ensure the correct asymptotic order conditions
  using a variety of iterative techniques to build function orders

all of these lie between the realm of the exponential and ramanujan's beast
  and there is an infinite hierarchy even beyond
each waiting for a theory to develop and interesting relations to find

-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-
galathaea: prankster, fablist, magician, liar
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Messages In This Thread
ramanujan and tetration - by galathaea - 05/28/2008, 08:10 AM
RE: ramanujan and tetration - by Ivars - 05/28/2008, 10:47 AM
RE: ramanujan and tetration - by galathaea - 05/28/2008, 03:19 PM
RE: ramanujan and tetration - by Gottfried - 05/28/2008, 04:12 PM
RE: ramanujan and tetration - by Ivars - 05/28/2008, 05:57 PM
RE: ramanujan and tetration - by galathaea - 05/30/2008, 04:21 AM
RE: ramanujan and tetration - by bo198214 - 05/30/2008, 06:18 AM
RE: ramanujan and tetration - by Ivars - 05/30/2008, 12:38 PM
RE: ramanujan and tetration - by Ivars - 05/29/2008, 06:39 AM
RE: ramanujan and tetration - by Ivars - 05/29/2008, 05:04 PM
RE: ramanujan and tetration - by Ivars - 05/30/2008, 07:23 AM
Thats not tetration! - by bo198214 - 05/30/2008, 07:44 AM

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