We have a set {X_1, X_2, ... , X_N} and three positive values

N = number of elements in the set
y = real number greater than 0
a = real number between 0 and 1.

I am looking for the value "y" that satisfies the following equality

 N
SUM (X_i)^(y+1)
i=1
--------------- = a
  N
 SUM (X_i)^y
 i=1

"_" symbolizes a subscript and "^" symbolizes power of.  The set of
X_i is *not* a series of integers!  It's any set of positive values.

Here are some thoughts on this nice problem. Let f(y) = [sum over i of
xi^(y+1)]/[sum over i of xi^y]. Suppose that they are arranged in
decreasing order with x1 larger than any of the others. Then f(y) =
x1^(y+1)/x1^y * [1 + (x2/x1)^(y+1) + ... ]/[1 + (x2/x1)^y = ...]. As y
goes to infinity this expression goes to x1*1/1 = x1.

Also I claim that f'(y)>0 for any y>=0. To see this, use the quotient
rule. The denomninator is a square and so is positive. The numerator
is the sum over all i and j of [xi^y*xj^(y+1)*lnxj -
xi^(y+1)*xj^y*lnxj]. So the term is 0 when i=j. Combine the i,j term
with the j,i term to get the sum over all i<j of
[xi^y*xj^y*(xi-xj)*(ln(xi)-ln(xj)). Because the terms decrease in
size, this is >0.

Note that f(0)= mu (the mean of the x data).

So f maps [0,infinity) to [mu,x1). So if mu<=a<x1 then there is a unique y.

Gasp!  BerkeleyChocolate, you have uncovered the whole rationale
behind this function!  I designed it *precisely* because it
corresponds to mean(xi) at y=0 and it converges towards max(xi) for
y->infinity.  The reason why I want to solve for "y" is because I want
to create a weighted average of the xi that is equal to a*max(xi)
(0<a<1).  Maybe I should have specified *why* I'm looking for an
analytical expression (if it exists)!  The only half-way interesting
related question that I found by googling is
http://mathworld.wolfram.com/Newton-GirardFormulas.html.  Note that I
am looking for S_(k+1)/S_(k)=a.

Because your reasoning is correct, we should simply assume that
0<xi<1, by normalizing all values to xi=xi'/max(xi').

To wirelesswizard:

Here's one more thought on your question. One can easily obtain a
first approximation to y by using the first terms in x1*[ 1 +
(x2/x1)^(y+1) + ...]/[ 1 + (x2/x1)^y+...] = a. Let z=(x2/x1)^y and
ignore the +... to obtain the simple equation for z:

x1*[ 1 + (x2/x1)*z ]/[ 1+ z] = a.

Solve for z, then y = log(z)/log(x2/x1). This would give you a region
to search. If y is very large or x3<<x2 then this will be very close.