Logarithmic mean

From the mean value theorem, there exists a value ${\displaystyle \xi }$ in the interval between x and y where the derivative ${\displaystyle f'}$ 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 ${\displaystyle \xi }$ by substituting ${\displaystyle \ln }$ for ${\displaystyle f}$ and similarly for its corresponding derivative:

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

and solving for ${\displaystyle \xi }$:

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

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

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

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 ${\displaystyle x}$ and ${\displaystyle y}$. 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 ${\displaystyle n+1}$ variables by considering the mean value theorem for divided differences for the ${\displaystyle n}$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 ${\displaystyle n=2}$ this leads to

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

The integral interpretation can also be generalized to more variables, but it leads to a different result. Given the simplex ${\textstyle S}$ with ${\textstyle S=\{\left(\alpha _{0},\,\dots ,\,\alpha _{n}\right):\left(\alpha _{0}+\dots +\alpha _{n}=1\right)\land \left(\alpha _{0}\geq 0\right)\land \dots \land \left(\alpha _{n}\geq 0\right)\}}$ 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 ${\textstyle n=2}$

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