View Question
Q: Simplifying a sum of powers ( No Answer,   3 Comments )
 Question
 Subject: Simplifying a sum of powers Category: Science > Math Asked by: wirelesswizard-ga List Price: \$20.00 Posted: 07 Jul 2006 11:54 PDT Expires: 06 Aug 2006 11:54 PDT Question ID: 744166
 ```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.```
 There is no answer at this time.

 ```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 i0. Note that f(0)= mu (the mean of the x data). So f maps [0,infinity) to [mu,x1). So if mu<=a
 ```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
 ```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<