Logarithmic mean

From the mean value theorem, there exists a value ξ in the interval between x and y where the derivative f ′ equals the slope of the secant line:

${\displaystyle \exists \xi \in (x,y):\ f'(\xi )={\frac {f(x)-f(y)}{x-y}}}$

The logarithmic mean is obtained as the value of ξ by substituting ln for f and similarly for its corresponding derivative:

${\displaystyle {\frac {1}{\xi }}={\frac {\ln x-\ln y}{x-y}}}$

and solving for χ:

${\displaystyle \xi ={\frac {x-y}{\ln x-\ln y}}}$

The logarithmic mean can also be interpreted as the area under an exponential curve.

$${\displaystyle {\begin{aligned}L(x,y)={}&\int _{0}^{1}x^{1-t}y^{t}\ \mathrm {d} t={}\int _{0}^{1}\left({\frac {y}{x}}\right)^{t}x\ \mathrm {d} t={}x\int _{0}^{1}\left({\frac {y}{x}}\right)^{t}\mathrm {d} t\\[3pt]={}&\left.{\frac {x}{\ln {\frac {y}{x}}}}\left({\frac {y}{x}}\right)^{t}\right|_{t=0}^{1}={}{\frac {x}{\ln {\frac {y}{x}}}}\left({\frac {y}{x}}-1\right)={}{\frac {y-x}{\ln {\frac {y}{x}}}}\\[3pt]={}&{\frac {y-x}{\ln y-\ln x}}\end{aligned}}}$$

The area interpretation allows the easy derivation of some basic properties of the logarithmic mean. Since the exponential function is monotonic, the integral over an interval of length 1 is bounded by x and y. The homogeneity of the integral operator is transferred to the mean operator, that is ${\displaystyle L(cx,cy)=cL(x,y)}$.

Two other useful integral representations are

$${\displaystyle {1 \over L(x,y)}=\int _{0}^{1}{\operatorname {d} \!t \over tx+(1-t)y}}$$

and

$${\displaystyle {1 \over L(x,y)}=\int _{0}^{\infty }{\operatorname {d} \!t \over (t+x)\,(t+y)}.}$$

One can generalize the mean to n + 1 variables by considering the mean value theorem for divided differences for the n-th derivative of the logarithm.

We obtain

${\displaystyle L_{\text{MV}}(x_{0},\,\dots ,\,x_{n})={\sqrt[{-n}]{(-1)^{n+1}n\ln \left(\left[x_{0},\,\dots ,\,x_{n}\right]\right)}}}$

where ${\displaystyle \ln \left(\left[x_{0},\,\dots ,\,x_{n}\right]\right)}$ denotes a divided difference of the logarithm.

For n = 2 this leads to

${\displaystyle L_{\text{MV}}(x,y,z)={\sqrt {\frac {(x-y)(y-z)(z-x)}{2{\bigl (}(y-z)\ln x+(z-x)\ln y+(x-y)\ln z{\bigr )}}}}.}$

The integral interpretation can also be generalized to more variables, but it leads to a different result. Given the simplex ${\textstyle S}$ with

$${\displaystyle S=\{\left(\alpha _{0},\,\dots ,\,\alpha _{n}\right):\left(\alpha _{0}+\dots +\alpha _{n}=1\right)\land \left(\alpha _{0}\geq 0\right)\land \dots \land \left(\alpha _{n}\geq 0\right)\}}$$

and an appropriate measure ${\textstyle \mathrm {d} \alpha }$ which assigns the simplex a volume of 1, we obtain

${\displaystyle L_{\text{I}}\left(x_{0},\,\dots ,\,x_{n}\right)=\int _{S}x_{0}^{\alpha _{0}}\cdot \,\cdots \,\cdot x_{n}^{\alpha _{n}}\ \mathrm {d} \alpha }$

This can be simplified using divided differences of the exponential function to

${\displaystyle L_{\text{I}}\left(x_{0},\,\dots ,\,x_{n}\right)=n!\exp \left[\ln \left(x_{0}\right),\,\dots ,\,\ln \left(x_{n}\right)\right]}$.

Example n = 2:

${\displaystyle L_{\text{I}}(x,y,z)=-2{\frac {x(\ln y-\ln z)+y(\ln z-\ln x)+z(\ln x-\ln y)}{(\ln x-\ln y)(\ln y-\ln z)(\ln z-\ln x)}}.}$