# Talk:Harmonic mean

## stats comment

added the definition when frequency is not unary. Nova67 16:16, 12 June 2005 (UTC)

I don't understand. Are you saying that you added such a definition? If so, there's really no need to mention it here; the article history will show that. Are you saying that such a definition needs to be added? Please explain what you mean. Ruakh 14:27, 15 Jun 2005 (UTC)

## ”Draining Pool” example

I don’t believe the pool-draining example highlights a special case, and I suspect the example might warrant clarification or removal. As this is not my area of expertise, I submit the following for your review: The rates provided (4 and 6) are “hours per pool”. The harmonic mean, then, is the average number of hours it would take one pump to empty the pool (4.8 hours). With twice as many pumps working at that average rate, it would of course take only half the time, or 2.4 hours. —Preceding unsigned comment added by Jenburton (talkcontribs) 00:31, 25 August 2009 (UTC)

## Electrical Resistance?

Is this true?: "in an electrical circuit you have two resistors connected in parallel, one with 40 ohms and the other with 60 ohms, then the average resistance is 48 ohms" 210.224.218.13 15:51, Nov 20, 2003 (UTC)

The total resistance is given by the harmonic mean divided by the number of resistors, not just the harmonic mean. The resistance should then be around 24 ohms. 128.84.45.4 20:55, Jul 06, 2004 (UTC)
Exactly. The average resistance is 48 ohms, while the equivalent resistance is 24 ohms. (I don't think "average resistance" is a term any physicist/electrician/electrical engineer would use, since it would have different meanings for resistors in parallel as for resistors in series, but that's how this article is using the term, and I think that's fine.) Ruakh 02:32, 8 Mar 2005 (UTC)
Shouldn't that read arithmetic mean resistance rather than average?. Also resistance should strictly be impedance Smidoid 12:18, 23 Feb 2008 (UTC)

## Negative Numbers?

What if the two numbers are negative? The harmonic mean of -1 and -2 is 2/(-1 + -(1/2)), which equals 2/(-1.5) or 2 * -(2/3) or -4/3, which is larger than the arithmetic mean, which is -1.5 (as compared to the geometric mean being -1.3......). However, the geometric mean would actually be the square root of 2... messed up for negative numbers, I suppose. ugen64 03:26, Dec 16, 2003 (UTC)

The article explicitly restricts the notion of harmonic mean to the positive reals. If it didn't, then we'd have the bigger issue of finding the harmonic mean of -1 and 1 (since 2/(1/-1 + 1/1) = 2/(-1 + 1) = 2/0, which is not defined). Ruakh 02:32, 8 Mar 2005 (UTC)

## Removed example.

I removed the following text that a non-registered user inserted into the article, as its placement and internal organization were confusing and ill-suited to the article:

===An example===
* An experiment yields the following data: 34,27,45,55,22,34 To get the harmonic mean
# How many items? There are 6. Therefore n=6
# What is the sum on the bottom of the fraction? It is 0.181719152307
# Get the reciprocal of that sum. It is 5.50299727522
# To get the the harmonic mean multiply that by n to get <U>33.0179836513</U>
--

If anyone wants to rescue it, feel free. It needs to be rewritten, given lead text, and placed appropriately (perhaps in a sidebar; the "ExampleSidebar" template could be used). Ruakh 19:48, 8 September 2005 (UTC)

## Average speed

This statement warrants a little qualification:

if for half the distance of a trip you travel at 40 miles per hour and for the other half of the distance you travel at 60 miles per hour, then your average speed for the trip is given by the harmonic mean of 40 and 60, which is 48; that is, the total amount of time for the trip is the same as if you traveled the entire trip at 48 miles per hour.

it may be true, of course, but not independent of the premises of the problem which are rather subtely expressed only. It is equally true that:

if for half the time of a trip you travel at 40 miles per hour and for the other half of the time you travel at 60 miles per hour, then your average speed for the trip is given by the arithmetic mean of 40 and 60, which is 50; that is, the total distance you covered is the same as if you traveled the entire trip at 50 miles per hour.

While this isn't central to a discussion of harmonic means, as I discovered recently people are easily lulled, by reading such concise statements, into the mistaken belief that the harmonic mean is THE correct manner of averaging independent speed measures. And this, it is not. It is A correct measure given certain premises concerning the problem. Bwechner 05:00, 25 November 2005 (UTC)

The statement as it stands isn't even strictly correct. The harmonic mean of 40 and 60 is hardly an integer -- in fact, it is closer to 49 than it is to 48! 142.162.10.169 (talk) 01:33, 20 March 2008 (UTC)

## Deriving arithmetic and geometric means when n>2

OK, so in the article, there is a derived method of finding arithmetic mean and geometric mean when n=2. What about when n=3 or 4? I;m sorry I don't know this- I'm just a high school student at the Illinois Mathematics and Science Academy The preceding unsigned comment was added by 143.195.150.93 (talk • contribs) .

The article gives the relationship that for just two numbers, H = G2/A. This relationship doesn't hold for more than two numbers, and there's no corresponding relationship for more than two numbers: for any n > 2 and any A and G, there are infinitely many possibilities for H, unless A = G, in which case A = G = H. Ruakh 01:58, 8 March 2006 (UTC)
In the example area, it's shown that for pumps p1 p2 p3 ... working to empty the same pool, it takes ${\frac {{p_{1}}\cdot {p_{2}}\cdot {p_{3}}}{p_{1}+p_{2}+p_{3}}}$ hours to empty it, I believe this is the part you're mentioning? I think this only works for n==2. Assume you have n pumps which take 2 hours to empty it. A) for n==1, this gives ${\frac {2}{2}}=1hour$ . B) for n==2 this gives ${\frac {{2}\cdot {2}}{2+2}}=1hour$ (correct) C) For n==1000 we get ${\frac {2^{1000}}{2000}}=5.36*10^{297}hours$ . For this problem, it makes more sense to have it be 2/n for this problem. If the pool is 2m gallons, they drain it in 2 hours, giving a rate of m gallons per hour, then the rate at which n pumps can drain a pool of 2m gallons is 2m/mn=2/n. The general version seems to be exactly the harmonic mean over n (as intended), I.E.
${\frac {H}{n}}={\frac {1}{{\frac {1}{a_{1}}}+{\frac {1}{a_{2}}}+\cdots +{\frac {1}{a_{n}}}}}={\frac {1}{\sum _{i=1}^{n}{\frac {1}{a_{i}}}}}$ Cassiline (talk) 19:59, 13 April 2008 (UTC)

## Trigonometric example

I added this part and I want to thank the person who did the editing. It really flows well.
This example only work with the double angle tangent identity. I tried $tanA=tan(3x)$ and the formula failed to work out. It will fail to work out for any $tan(nx)$ where $n/=2$ .

Opinionhead (talk) 20:09, 10 May 2011 (UTC)

Harmonic Mean

Harmonic mean is another measure of central tendency and also based on mathematic footing like arithmetic mean and geometric mean. Like arithmetic mean and geometric mean, harmonic mean is also useful for quantitative data. Harmonic mean is defined as: Harmonic mean is quotient of “number of the given values” and “sum of the reciprocals of the given values”.

## Omission - Relationship with other means.

I just deleted a statement that cannot be correct, or is at least not properlyy explained/defined. DrMicro wrote that

"If all the real variables in the set are > 0 then H ≥ M2 / m where m is the minimum in the set."

If this were the case (and M is the maximum as defined in the main article above) then the H would be greater than the maximum, meaning that it no longer qualified as a mean. If someone knows the correct statement of this inequality as stated in the paper (it's too late for me to be reading a maths paper right now) feel free to re-add and correct.

I'll message DrMicro about it too.

Cheers,

JPBrod (talk) 23:28, 21 April 2012 (UTC)

Thank you. It must have been a typo. I have read the reference and I cannot find this inequality in this paper. Im afraid I do not have an explanation for its inclusion. Well spotted. DrMicro (talk) 11:14, 22 April 2012 (UTC)

## "Theoretical values" section is confused

Currently the subsection "Theoretical values" of the section "Statistics" reads

Theoretical value
The variance of the harmonic mean is
$\operatorname {Var} \left({\frac {1}{x}}\right)={\frac {m\left[\operatorname {E} (1/x-1)\right]}{nm^{2}}}$ where m is the arithmetic mean of the reciprocals, x are the variates, n is the population size and E is the expectation operator. Asymptotically E(1 / x) is distributed normally.
The mean of the sample m is also distributed normally with variance s2.
$s^{2}={\frac {m[\operatorname {E} (1/x-1)]}{m^{2}n}}$ Unfortunately the citation for this is a paper at a 1972 conference, so there's no way to check the source to confirm that it's been copied correctly (or even that it was right in the first place since it appears not to have been refereed or published). Moreover, it's been changed in the present article from what it said earlier.

(1) This looks suspicious to me because the variance of the harmonic mean is given as the same expression as the variance of the arithmetic mean of the reciprocals.

(2) It looks suspicious because surely the original source would not have left something with m not cancelled from the numerator and denominator.

(3) It says the variance of the harmonic mean is denoted as var(1/x) where "x are the variates". That doesn't make sense to me -- if 1/x is the harmonic mean, then x is the mean of the reciprocals of the variates, not "the variates". So this expression, even if correct, may refer to the variance of the reciprocals of the variates rather than the variance of the sample harmonic mean.

Does anyone have a source for the true mean and variance of the sample harmonic mean? Duoduoduo (talk) 15:08, 5 June 2013 (UTC)

1. Jean WH (1984) The harmonic mean and other necessary conditions for stochastic dominance. J Finance 39 (2) 527
2. Zelen M (1972) Length-biased sampling and biomedical problems. In Biometric Society Meeting, Dallas, Texas