GeezerVison

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: $\exists \xi \in (x,y):f^{\prime }\left(\xi \right)=\frac{f\left(x\right)-f\left(y\right)}{x-y}$

The logarithmic mean is obtained as the value of ξ by substituting ln for f and similarly for its corresponding derivative: $\frac{1}{\xi }=\frac{\ln x-\ln y}{x-y}$

and solving for ξ: $\xi =\frac{x-y}{\ln x-\ln y}$

The logarithmic mean is also given by the integral $L(x,y)={\int }_0^1{x}^{1-t}{y}^{t}dt.$

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$\frac{1}{L(x,y)}={\int }_0^1\frac{dt}{tx+(1-t)y}$and$\frac{1}{L(x,y)}={\int }_0^{\infty }\frac{dt}{(t+x)(t+y)}.$

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 $L_{\text{MV}}(x_0,\dots ,x_n)=\sqrt[-n]{(-1{)}^{n+1}n\ln \left([x_0,\dots ,x_n]\right)}$ where $\ln \left([x_0,\dots ,x_n]\right)$ denotes a divided difference of the logarithm. For n = 2 this leads to $L_{\text{MV}}(x,y,z)=\sqrt{\frac{(x-y)(y-z)(z-x)}{2\left(\right(y-z)\ln x+(z-x)\ln y+(x-y)\ln z)}}.$

The integral interpretation can also be generalized to more variables, but it leads to a different result. Given the simplex $S$ with $S=\{({\alpha }_0,\dots ,{\alpha }_n)\in {ℝ}^{n+1};{\alpha }_0+\cdots +{\alpha }_n=1\text{ and }{\alpha }_j\ge 0,\text{ for }j=0,\dots ,n\}$ and an appropriate measure $d\alpha$ which assigns the simplex a volume of 1, we obtain $L_{\text{I}}(x_0,\dots ,x_n)={\int }_S{x}_0^{{\alpha }_0}\cdots x_n^{{\alpha }_n}d\alpha .$

This can be expressed as the divided differences of the exponential function by $L_{\text{I}}(x_0,\dots ,x_n)=n!\exp [\ln \left(x_0\right),\dots ,\ln \left(x_n\right)].$ In the case of n = 2, it is $L_{\text{I}}(x,y,z)=-2\frac{x(\ln y-\ln z)+y(\ln z-\ln x)+z(\ln x-\ln y)}{(\ln x-\ln y)(\ln y-\ln z)(\ln z-\ln x)}.$