Two fuzzy ideas for new sequence transformations.
Antti Karttunen
karttu at megabaud.fi
Fri Sep 7 15:08:27 CEST 2001
Cheers,
In currently unavailable Weisstein's Encyclopedia of Mathematics,
there was listed as last three equations
at page http://mathworld.wolfram.com/ContinuedFraction.html
some very ramanujanesque looking identities from Ramanujan, the last one being:
4* int((x*exp(-x*sqrt(5)))/cosh(x), y=0..infinity )
= (1/2)*(Zeta(2,((1/4)*(1+sqrt(5)))) - Zeta(2,((1/4)*(3+sqrt(5)))))
= 1/(1+(1/1+(1/(1+4/(1+(4/(1+(9/(1+(9/(1+(16/(1+(16/1 ... ))) )))))))))))
(I don't know what exactly means the above two-argument calling form
of Zeta-function, but at least the first and the last expression seem to give (about)
the same value 0.568..., forgiving the limited digit precision and other quirks of Maple)
Now, the ordinary simple infinite continued fractions are of the
form
a0 + (1/(a1 + (1/(a2 + (1/(a3 + (1/....)))))))
while all the three continued fractions given by Ramanujan seem to
be of the general form:
1 + (b1/(1 + (b2/(1 + (b3/(1 +(b4/...)))))))
(of which "Rogers-Ramanujan continued fractions" are special case?)
Question: Is there any practical way of converting between these
two forms of continued fractions, even for some limited subset of
integer sequences [a0, a1, a2, a3, a4, a5, ...] or [b1, b2, b3, b4, b5, ...] ???
(If such conversion were possible, then I guess we would have there yet another
integer sequence transformation).
ANOTHER IDEA:
Suppose we have an infinite group, with an infinite, but enumerable generating set
(e.g. the Braid group). Let's fix some standard order for the generators, and
call them say, g1, g2, g3, g4, g5, etc...
Let's call an element x of the group "realizable with one pass", if it can be written
as (an infinite or finite) product of the said generators, such those generators occur in the
same
order, with any integral exponents (if exponent is 0, then that generator is absent):
x = g1^e1 * g2^e2 * g3^e3 * g4^e4 * ...
Now, would it be possible to find such infinitely generated group, that such
elements produced with "one pass" would form a subgroup of their own?
Or even, if we could guarantee that two such one pass elements x and y,
when multiplied together with proper "correction constants" c1,c2 and c3:
would yield a third element y:
c1 * x * c2 * y * c3 = y
which would also certainly be expressible in the same format:
y = h1^f1 * h2^f2 * h3^f3 * h4^f4 * ...
Thus we would have a transformation which would give from two integer
sequences A and B, the third one C, which would reflect the multiplication
table of the said group. (If any such group exists).
Of course, if the generators of the group were such that no other exponents
than 0 and 1 would make much sense (like the transpositions in the symmetric group),
then it would be better to use characteristic sequences as the exponent sequences.
Terveisin,
Antti Karttunen
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