Two questions....
Travis Kidd
1) Does e^(e^x) have a Taylor expansion?
2) What function(s), if any, grows faster than any finite collection
of the following: {e^x, e^(e^x), e^(e^(e^x)), ...}?
-Travis
Philo D.
In article <tkidd.868462435@hubcap>, tk...@hubcap.clemson.edu (Travis Kidd)
wrote:
> 1) Does e^(e^x) have a Taylor expansion?
Yes. It is the composition of two analytic functions, so it is analytic.
>
> 2) What function(s), if any, grows faster than any finite collection
> of the following: {e^x, e^(e^x), e^(e^(e^x)), ...}?
Given any countable list of functions, you can construct a function
that grows faster than any of them using a diagonal argument.
For reference, try: G. H. Hardy, ORDERS OF INFINITY.
Zdislav V. Kovarik
In article <tkidd.868462435@hubcap>,
> of the following: {e^x, e^(e^x), e^(e^(e^x)), ...}?
1) Yes.
2) First, define recursively a sequence {A(n)} by
A(0) = 0, A(n+1) = e^A(n).
It does what you require at non-negative integers.
a) Then, since you did not require that the function be analytic, just
make it continuous and piecewise linear, with vertices at non-negative
integers.
b) If you did mean analytic, there is a theorem by Weierstrass which
gives a construction of an analytic function interpolating any function
defined on a sequence, as long as the (domain) sequence diverges to
infinity (and the non-negative integers do just that). The theorem does
not promise that the interpolating function will be monotonic.
Reference: most textbooks of complex analysis, chapters on entire
functions (that will supply a proof of your question 1)).
My favourite text is from Saks and Zygmund.
c) There is a monotonic solution, analytic on the open interval
(0, infinity), called "a" (in the reference, it is German
letter "a"), which is an "iteration counter" for the function
e(x) = e^x - 1. It means that it solves an instance of Abel's equation
a(e(x)) = a(x) + 1
and it is actually completely monotone ((-1)^(n-1) times n-th
derivative is positive, starting with n=1).
Its inverse grows faster than any iterate of e^x.
Reference: Marek Kuczma: Functional Equations in a Single Variable,
Polish Scientific Publishers, Warszawa 1968.
The function "a" is defined in Chapter VII, Sec. 5, and a general
construction of such functions is given in Ch. VII, Sec. 1.
Have fun, ZVK (Slavek).
Jose Fernando Camoes Mendonca Oliveira Silva
Travis Kidd wrote:
> 1) Does e^(e^x) have a Taylor expansion?
Yes.
--
Motto: Think it through! http://web.mit.edu/camoes/public/home.html
Bill Dubuque
Of course,
e^(e^x) = e e^(e^x-1) = e (1 + u + u^2/2 + u^3/6 + ...), u = e^x-1
= e (1 + x + x^2 + 5/6 x^3 + 5/8 x^4 + 13/30 x^5 + 203/720 x^6 + ...)
Analogous expansions exist at other nonzero finite points x=a.
| 2) What function(s), if any, grows faster than any finite collection
| of the following: {e^x, e^(e^x), e^(e^(e^x)), ...}?
More generally, let f << g denote lim(f(x)/g(x)) -> 0 as x -> oo.
Given any sequence of functions increasing in growth
f1 << f2 << f3 << ... << fn << ...
there exists a continuous function f that grows faster than
any function in the sequence fn: i.e. f >> fn for all n.
The proof is a simple "diagonal argument":
choose f(x) = 1 + Sum (x-n) f (x)
n <= x n
This proof goes back to du Bois-Reymond (1871), who was one
of the pioneers of work on so-called "orders of infinity"
(and the true inventor of the "diagonal argument" -- often
mistakenly attributed to Cantor).
See also G.H. Hardy's book "orders of Infinity" and the paper:
P. Erdos et. al., Scales of Functions, J. Austr. Math. Soc.
1 (1959/1960), 396-418.
For an extensive historical treatment of du Bois-Reymond's work
on orders of infinity see Gordon Fisher's "The Infinite and
Infinitesimal Quantities of do Bois-Reymond and their Reception",
Arch. Hist. Exact. Sci., 24, 1981, 101-163.
P.S. The Subject of your post (Two questions) is not very descriptive.
It would benefit all if you use more descriptive Subject fields
(imagine how useless the Subject field would become if all posts
used Subjects like "one qestion" or "two questions" ...). With
non-specific Subject fields you greatly reduce the chance of
attracting those with the sought knowledge.
-Bill Dubuque
Ilias Kastanas
In article <doozy-ya02408000...@mathserv.mps.ohio-state.edu>,
Philo D. <do...@earthling.net.nospam> wrote:
>In article <tkidd.868462435@hubcap>, tk...@hubcap.clemson.edu (Travis Kidd)
>wrote:
>
>
>
>> of the following: {e^x, e^(e^x), e^(e^(e^x)), ...}?
>
>that grows faster than any of them using a diagonal argument.
>
>For reference, try: G. H. Hardy, ORDERS OF INFINITY.
In fact, using e_2(x) for e^(e^x) etc., it is enough to define
f(x) = Sum (e_n(x)/e_n(n)) (n from 1 to inf)
Ilias