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jay_shark
02-08-2007, 02:26 PM
Show that the geometric mean of a set S of positive numbers equals the geometric mean of the geometric means of all non empty subsets of S .

Oh my , where to begin ....

arahant
02-08-2007, 02:42 PM
[ QUOTE ]
Show that the geometric mean of a set S of positive numbers equals the geometric mean of the geometric means of all non empty subsets of S .

Oh my , where to begin ....

[/ QUOTE ]

Well, since your question appears to be just where to begin, I would say 'logarithmic identities'... /images/graemlins/smile.gif

Enrique
02-08-2007, 03:09 PM
[ QUOTE ]
Show that the geometric mean of a set S of positive numbers equals the geometric mean of the geometric means of all non empty subsets of S .

Oh my , where to begin ....

[/ QUOTE ]

Focus on one element.
Notice that in the first geometric mean, that element will be raised to the 1/n (assuming there are n elements in S).

Now in the second, you'll have 2^n - 1 elements where that thing appears.
So the exponent will be:
[1 + (n-1)C1 * 1/2 + (n-1)C2 * 1/3 + ... + (n-1)C(n-1)* 1/n] / [2^n - 1].

So we need to show that SUM(i = 1 to n) of [{(n-1)C(i-1)}*{1/i}] = (2^n - 1)/n.

But now multiply by n on both sides:
The left side you get the sum from 1 to n of nCi. Since the sum from 0 to n is 2^n. The sum from 1 to n is 2^n -1.
Completing the proof.

Sorry, if it is hard to understand, but it is hard to write mathematics on a forum.

jay_shark
02-08-2007, 03:40 PM
ahh yes good job Enrique .

Take 3 elements {1,2,3} g(1,2,3)=(1*2*3)^1/3

There are 2^3-1 subsets excluding the null set .

Geometric mean of the geometric means is
[1*2*3*[(1*2)(1*3)(2*3)]^1/2*(1*2*3)^1/3]^1/7

[g(1,2,3)^3*g(1,2,3)^3*g(1,2,3)]^1/7 = g(1,2,3)

Enrique
02-08-2007, 06:19 PM
I thought the problem is also nice, because you can find a formula for the sum of (n-1)C(i-1) * 1/i. If someone asks that sum, you could think on this problem and come up with a solution. Pretty nice.

Since you like math problems, here is one I liked a lot that I solved recently:
"Let n>6 and 1 = a_1,...,a_(phi(n)) = n-1 are the numbers less than n relatively prime to n. Find all n such that the a_i's are in an arithmetic progression."