gjerawlins@watdaisy.UUCP (Gregory J.E. Rawlins) (10/20/86)
Hello,
Does anyone know if the following summation has a closed
form solution?
n
---------
\ i
\ b
/ a
/
---------
i=0
Where a and b are constants and the exponents are to be parsed as
a^(b^i) and not (a^b)^i.
We all know the first two "exponential iterates":
n
---------
\
\ ( n+1 )
/ i = ( 2 )
/
---------
i=0
(arithmetic progression)
and
n
---------
\ n+1
\ i a - 1
/ a = ---------- (|a| <> 1)
/ a - 1
---------
i=0
(geometric progression)
But i don't know the "next" one up. Does it have a closed form?
:::::::--------------------------:::::::
I became interested in this when, in analysing an algorithm, i
required a good bound on the following series:
----------
------ / ------
--- / --- / / ---
S = \/ n + \/ \/ n + \/ \/ \/ n + etc.
n
Since the inverse series is the sum of lglgn terms each of which
is the square of the previous one (so a^(2^i)).
If you can't help me with the first summation can you give me
better bounds on S than the following?
n
--- ---
\/ n < S < 2 \/ n
n
(sorry to take up so much space with an ascii version of the
symbols but i much prefer it to having to parse linear versions
of same.)
greg.
--
gjerawlins%watdaisy@waterloo.csnet 1-519-884-3852 Gregory J. E. Rawlins
gjerawlins%watdaisy%waterloo.csnet@csnet-relay.arpa CS Dept., U. Waterloo
{allegra,decvax,inhp4,utzoo}!watmath!watdaisy!gjerawlins Waterloo, Ont. N2L3G1