Logarithmic mean

Mean value theorem of differential calculus

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

Integration

The logarithmic mean is also given by the integral $${\displaystyle L(x,y)=\int _{0}^{1}x^{1-t}y^{t}\,\mathrm {d} t.}$$

This interpretation allows the derivation of some 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.

Two other useful integral representations are$${\displaystyle {1 \over L(x,y)}=\int _{0}^{1}{\operatorname {d} \!t \over tx+(1-t)y}}$$and$${\displaystyle {1 \over L(x,y)}=\int _{0}^{\infty }{\operatorname {d} \!t \over (t+x)\,(t+y)}.}$$

Mean value theorem of differential calculus

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

Integral

The integral interpretation can also be generalized to more variables, but it leads to a different result. Given the simplex ${\textstyle S}$ with $${\displaystyle S=\{\left(\alpha _{0},\dots ,\alpha _{n}\right)\in \mathbb {R} ^{n+1}\,;\,\alpha _{0}+\dots +\alpha _{n}=1{\text{ and }}\alpha _{j}\geq 0,{\text{ for }}j=0,\dots ,n\}}$$ and an appropriate measure ${\textstyle \mathrm {d} \alpha }$ which assigns the simplex a volume of 1, we obtain $${\displaystyle L_{\text{I}}\left(x_{0},\dots ,x_{n}\right)=\int _{S}x_{0}^{\alpha _{0}}\cdots x_{n}^{\alpha _{n}}\,\mathrm {d} \alpha .}$$

This can be expressed as the divided differences of the exponential function by $${\displaystyle L_{\text{I}}\left(x_{0},\dots ,x_{n}\right)=n!\exp \left[\ln \left(x_{0}\right),\dots ,\ln \left(x_{n}\right)\right].}$$ In the case of n = 2, it is $${\displaystyle L_{\text{I}}(x,y,z)=-2{\frac {x(\ln y-\ln z)+y(\ln z-\ln x)+z(\ln x-\ln y)}{(\ln x-\ln y)(\ln y-\ln z)(\ln z-\ln x)}}.}$$