In two previous posts, we discussed the L’Hôpital Rule:

http://irthoughts.wordpress.com/2012/05/22/understanding-the-lhopital-rule/

http://irthoughts.wordpress.com/2012/01/23/lhopitals-rule-and-the-00-power-controversy/

This time we show how to use the L’Hôpital Rule to derive the geometric mean from the power mean.

**Deriving the Geometric Mean**

The power mean from a set of n numbers of x values is defined as

Power mean = [(sum of x^p)/n]^1/p

where p is any real number.

Taking natural logs,

ln(Power mean) = [ln(sum of x^p) – ln(n)]/p

With respect to p, the right side of this expression is of the form

f(p)/g(p)

where

f(p) = ln(sum of x^p) – ln(n)

g(p) = p

So, for p = 0,

f(p)/g(p) = 0/0

This indeterminate form asks for applying the L’Hôpital Rule, which reduces to evaluating

f(p)/g(p) = f’(p)/g’(p) in the limit where p vanishes (approaches zero).

Proceeding with the derivation,

f’(p) = (1/sum of x^p) [sum of (x^p)ln(x)]

g’(p) = 1

f’(p)/g’(p) = (1/sum of x^p) [sum of (x^p)ln(x)]

Now when p = 0,

f’(p)/g’(p) = (1/n) [sum of ln(x)]

However, since

sum of ln(x) = ln(product of x)

then

f’(p)/g’(p) = (1/n) [ln(product of x)] = ln(product of x)^1/n

Therefore, taking the antilog, we obtain the geometric mean,

Geometric mean = (product of x)^1/n

**Limitations of the geometric mean**

Evidently, to compute the geometric mean from n numbers of x values these must be positive, greater than zero, and not written as percentages.

In the next section, we describe workarounds for circumventing these limitations.

**Circumventing the limitations of the geometric mean**

For a dataset with a zero value, we may add a positive value, k, calculate the geometric mean and substract k to the result. We may use a similar approach for a dataset with a negative, but insuring that k is greater than the largest negative value of the set.

For example, to calculate the geometric mean of the values +12%, -8%, and +2%, we may add 1 to remove the negative value and calculate the geometric mean of their decimal multiplier equivalents of 1.12, 0.92, and 1.02, to compute a geometric mean of 1.0167. Subtracting 1 from this value gives the geometric mean of +1.67%.

Note that for a dataset consisting of percents, we divide each percent by 100, compute the geometric mean, and multiply the result by 100.

Additional workarounds are found in:

http://www.buzzardsbay.org/geomean.htm

http://www.waterboards.ca.gov/water_issues/programs/swamp/docs/cwt/guidance/3413.pdf

Have a mean day!

PS. Sorry. I fixed the last section to add an example describing the workaround.