# Lehmer mean

Let $p$ be a real number.  Lehmer mean of the positive numbers $a_{1},\,\ldots,\,a_{n}$ is defined as

 $\displaystyle L_{p}(a_{1},\,\ldots,\,a_{n})\;:=\;\frac{a_{1}^{p}+\ldots+a_{n}^% {p}}{a_{1}^{p-1}+\ldots+a_{n}^{p-1}}.$ (1)

This definition fulfils both requirements set for means (http://planetmath.org/Mean3).  In the case of Lehmer mean of two positive numbers $a$ and $b$ we see for  $a\leqq b$  that

 $a\;=\;\frac{a^{p}\!+\!ab^{p-1}}{a^{p-1}\!+\!b^{p-1}}\;\leqq\;\frac{a^{p}\!+\!b% ^{p}}{a^{p-1}\!+\!b^{p-1}}\;\leqq\;\frac{a^{p-1}b\!+\!b^{p}}{a^{p-1}\!+\!b^{p-% 1}}\;=\;b.$

The Lehmer mean of certain numbers is the greater the greater is the parametre $p$, i.e.

 $L_{p}(a_{1},\,\ldots,\,a_{n})\;\geqq\;L_{q}(a_{1},\,\ldots,\,a_{n})\quad% \forall\;p\;>\;q.$

This turns out from the nonnegativeness of the partial derivative of $L_{p}$ with respect to $p$; in the case  $n=2$  it writes

 $\frac{\partial L_{p}}{\partial p}\;=\;\frac{a^{p-1}b^{p-1}(a\!-\!b)(\ln{a}-\ln% {b})}{(a^{p-1}\!+\!b^{p-1})^{2}}\;\geqq\;0.$

Thus in the below list containing special cases of Lehmer mean, the is the least and the contraharmonic the greatest (cf. the comparison of Pythagorean means).

E.g. for two arguments $a$ and $b$, one has

• $\displaystyle L_{0}(a,\,b)\,=\,\frac{2ab}{a\!+\!b}$,  harmonic mean,

• $\displaystyle L_{1/2}(a,\,b)\,=\,\sqrt{ab}$,  geometric mean,

• $\displaystyle L_{1}(a,\,b)\,=\,\frac{a\!+\!b}{2}$,  arithmetic mean,

• $\displaystyle L_{2}(a,\,b)\,=\,\frac{a^{2}\!+\!b^{2}}{a\!+\!b}$,  contraharmonic mean.

Note.  The least (http://planetmath.org/LeastNumber) and the greatest of the numbers (http://planetmath.org/GreatestNumber) $a_{1},\,\ldots,\,a_{n}$ may be regarded as borderline cases of the Lehmer mean, since

 $\lim_{p\to-\infty}L_{p}(a_{1},\,\ldots,\,a_{n})\;=\;\min\{a_{1},\,\ldots,\,a_{% n}\},\quad\lim_{p\to+\infty}L_{p}(a_{1},\,\ldots,\,a_{n})\;=\;\max\{a_{1},\,% \ldots,\,a_{n}\}.$

For proving these equations, suppose that there are exactly $k$ greatest (resp. least) ones among the numbers and that those are  $a_{1}=\ldots=a_{k}$.  Then we can write

 $L_{p}(a_{1},\,\ldots,\,a_{n})\;=\;\frac{a_{1}^{p}\left[k+\!\left(\frac{a_{k+1}% }{a_{1}}\right)^{p}\!+\ldots+\!\left(\frac{a_{n}}{a_{1}}\right)^{p}\right]}{a_% {1}^{p-1}\left[k+\!\left(\frac{a_{k+1}}{a_{1}}\right)^{p-1}\!+\ldots+\!\left(% \frac{a_{n}}{a_{1}}\right)^{p-1}\right]}.$

Letting  $p\to+\infty$  (resp. $p\to-\infty$),  this equation yields

 $L_{p}(a_{1},\,\ldots,\,a_{n})\;\longrightarrow\;a_{1}.$
Title Lehmer mean LehmerMean 2013-03-22 19:02:06 2013-03-22 19:02:06 pahio (2872) pahio (2872) 11 pahio (2872) Definition msc 62-07 msc 11-00 OrderOfSixMeans LeastAndGreatestNumber MinimalAndMaximalNumber