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harmonic mean (Definition)

If $ a_1,\,a_2,\,\ldots,\,a_n$ are positive numbers, we define their harmonic mean as the inverse number of the arithmetic mean of their inverse numbers:

$\displaystyle H.M.=\frac{n}{\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}}$


  • If you travel from city $ A$ to city $ B$ at $ x$ miles per hour, and then you travel back at $ y$ miles per hour. What was the average velocity for the whole trip?
    The harmonic mean of $ x$ and $ y$. That is, the average velocity is
    $\displaystyle \frac{2}{\frac{1}{x}+\frac{1}{y}}=\frac{2xy}{x+y}.$
  • If one draws through the intersecting point of the diagonals of a trapezoid a line parallel to the parallel sides of the trapezoid, then the segment of the line inside the trapezoid is equal to the harmonic mean of the parallel sides.
  • In the harmonic series
    $\displaystyle 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...$
    every term equals to the harmonic mean of the term preceding it and the term following it.



"harmonic mean" is owned by drini. [ full author list (3) | owner history (1) ]
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See Also: arithmetic mean, general means inequality, weighted power mean, power mean, arithmetic-geometric-harmonic means inequality, root-mean-square, proof of general means inequality, proof of arithmetic-geometric-harmonic means inequality, harmonic mean in trapezoid
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Cross-references: harmonic series, segment, sides, parallel, line, trapezoid, diagonals, point, arithmetic mean, inverse number, numbers, positive
There are 16 references to this entry.

This is version 7 of harmonic mean, born on 2001-10-20, modified 2007-06-18.
Object id is 408, canonical name is HarmonicMean.
Accessed 19541 times total.

Classification:
AMS MSC: 11-00 (Number theory :: General reference works )
Pending Errata and Addenda
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harmonic mean by pahio on 2004-06-06 06:13:40

Hi, drini! Can you add to your entry the mention, that every term in the harmonic series is the harmonic mean of preceding and subsequent term? I think that the name "harmonic series" is due to this fact.

Jussi
[ reply | up ]
working out the exercise by akrowne on 2002-01-16 07:12:12
Let the distance between the cities be d.
The velocity 1 way is x.
The velocity back is y.

so, with rate*time=distance, we have times:

x*t_1=d
y*t_2=d

hence

t_1=d/x
t_2=d/y

the total time is simply the sum of the time each way:

t=t_1+t_2

the overall velocity, v, will still conform to rate*time=distance, except the distance is twice the distance between the cities:

v*t=2d
v*(t_1+t_2)=2d
v=2d/(t_1+t_2)
 =2d/(d/x+d/y)
 =2/(1/x+1/y)

Hence we have derived the harmonic mean.

In general, when going between two cities n times, with {x_1,x_2,...,x_n} the velocities for each transit, we have

v*t=n*d
v*(t_1+t_2+...+t_n)=nd
v=nd/(t_1+t_2+...+t_n)
 =nd/(d/x_1+d/x_2+...+d/x_n)
 =n/(1/x_1 + 1/x_2 + ... + 1/x_n)

Which is the n-value harmonic mean.

-apk
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