Pick up a number, say 12. Write down a list of all its divisors, including the number itself.
In our case: 1,2,3,4,6,12. Now evaluate AA and HA the arithmetic and the harmonic averages.
AA=(1+2+3+4+6+12)/6=28/6
HA=6/(1/1+1/2+1/3+1/4+1/6+1/12)=6*12/28
Now evaluate the product AA*HA: (28/6)*(6*12/28)
Surprise, surprise!  the answer is 12, a number picked by us.
Please explain why it works with any number.
The list of divisors, in this case, 1, 2, 3, 4, 6, 12, consists of 3 pairs of numbers that multiply to 12, 1*12, 2*6, and 3*4.
AA = (sum_divisors) / (num_divisors)
HA = (num_divisors) / (sum_inverse_divisors)
The sum of the inverses is the secret to it all. As every inverse will have a common denominator of our value,
X, adding the terms simply reverses the order in which the factors show up. That is, 1/1 * 12/12 = 12/12, 1 * 12 = 12 ... 1/2 * 12 / 12 = 6 / 12, 2 * 6 = 12 ... 1/3 * 12/12 = 4 / 12, 3*4 = 12 ... etc.
So, the final sum of these inverses winds up being:
12/12 + 6/12 + 4/12 + 3/12 + 2/12 + 1/12 = 28 / 12 ... or (sum_divisors) /
XThat leaves us with: HA = (num_divisors) / ((sum_divisors) /
X)
Multiplying HA by AA gives the following:
((sum_divisors) / (num_divisors)) * ((num_divisors) / ((sum_divisors) /
X))
As you can see, the sum_divisors cancel out, as do the num_divisors, leaving:
HA * AA = 1 / (1 /
X), which reduces to
X, the number with which we started.
I'm sorry about the weird formatting, but I didn't want to mess around with symbols and all that to show it correctly.
Edited on March 12, 2010, 3:40 pm

Posted by Justin
on 20100312 15:38:05 