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An Identity Concerning Averages |
| A(a1, a2, ..., aN) = (a1 + a2 + ... + aN)/N |
If none of the numbers is 0, we also define their harmonic mean by
| H(a1,a2,...,aN) = N/(1/a1 + 1/a2 + ... + 1/aN) |
Let there be an integer M. We may consider all possible divisors of M, including 1 and M itself. We do not know how many there are. Nonetheless, for a given M, we may consider the arithmetic and harmonic means of the set of divisors. I denote them AM and HM, respectively. The task of computing either, especially the harmonic mean, may seem daunting. Thus it's all the more amusing to have the following
Theorem
M = AM · HM.
Examples
M = 6 has four divisors: 1, 2, 3,and 6.
A6 = (1 + 2 + 3 + 6)/4 = 3. On the other hand,H6 = 4/(1/1 + 1/2 + 1/3 + 1/6) = 4/(6/6 + 3/6 + 2/6 + 1/6) = 24/12 = 2. Therefore we indeed have 6 = 3·2.
M = 8 has also four divisors: 1,2,4, and 8.
A8 = (1 + 2 + 4 + 8)/4 = 15/4.
H8 = 4/(1/1 + 1/2 + 1/4 + 1/8) = 4/(8/8 + 4/8 + 2/8 + 1/8) = 32/15. Finally, 15/4·32/15 = 32/4 = 8.
Proof
The main point here is to realize that the divisors of a number come in pairs: d divides M iff there exists an integer c such that
| (1) | 1/d = c/M |
Now, let's enumerate all the divisors of M as d1, d2,... We know that
for every di there exists a ci such that
| HM = DM/(1/d1 + 1/d2 + ...) = DM/(c1/M + c2/M + ...) = M/AM. |
Which exactly means that M = AM · HM.
References
- O.Ore Number Theory and Its History, Dover, 1988
Copyright © 1996-2008 Alexander Bogomolny
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