# Logarithmic mean

Three-dimensional plot showing the values of the logarithmic mean.

In mathematics, the logarithmic mean is a function of two non-negative numbers which is equal to their difference divided by the logarithm of their quotient. This calculation is applicable in engineering problems involving heat and mass transfer.

## Definition

The logarithmic mean is defined as:

\displaystyle{ \begin{align} M_\text{lm}(x, y) &= \lim_{(\xi, \eta) \to (x, y)} \frac{\eta - \xi}{\ln(\eta) - \ln(\xi)} \\[6pt] &= \begin{cases} x & \text{if }x = y ,\\ \frac{y - x}{\ln(y) - \ln(x)} & \text{otherwise,} \end{cases} \end{align} }

for the positive numbers $\displaystyle{ x, y }$.

## Inequalities

The logarithmic mean of two numbers is smaller than the arithmetic mean and the generalized mean with exponent one-third but larger than the geometric mean, unless the numbers are the same, in which case all three means are equal to the numbers.

$\displaystyle{ \sqrt{x y} \leq \frac{x - y}{\ln(x) - \ln(y)}\leq \left(\frac{x^{1/3}+y^{1/3}}2\right)^3 \leq \frac{x + y}{2} \qquad \text{ for all } x \gt 0 \text{ and } y \gt 0. }$[1][2][3]

Toyesh Prakash Sharma generalizes the arithmetic logarithmic geometric mean inequality for any $\displaystyle{ n }$ belongs to the whole number as

$\displaystyle{ \sqrt{x y} (\ln(\sqrt{x y}))^{(n-1)} (\ln(\sqrt{x y})+n)\leq \frac{x {\ln^n(x)} - y {\ln^n(y)}}{\ln(x) - \ln(y)}\leq \frac{x(\ln(x))^{(n-1)} (\ln(x)+n) + y(\ln(y))^{(n-1)} (\ln(y)+n)} {2} }$

Now, for $\displaystyle{ n=0 }$

$\displaystyle{ \sqrt{x y} (\ln(\sqrt{x y}))^{(-1)} (\ln(\sqrt{x y}))\leq \frac{x - y}{\ln(x) - \ln(y)}\leq \frac{x(\ln(x))^{(-1)} (\ln(x)) + y(\ln(y))^{(-1)} (\ln(y))} {2} }$
$\displaystyle{ \sqrt{x y} \leq \frac{x - y}{\ln(x) - \ln(y)}\leq \frac{x + y} {2} }$

This is the arithmetic logarithmic geometric mean inequality. similarly, one can also obtain results by putting different values of $\displaystyle{ n }$ as below

For $\displaystyle{ n=1 }$

$\displaystyle{ \sqrt{x y} (\ln(\sqrt{x y})+1)\leq \frac{x {\ln(x)} - y {\ln(y)}}{\ln(x) - \ln(y)}\leq \frac{x (\ln(x)+1) + y(\ln(y)+1)} {2} }$

for the proof go through the bibliography.

## Derivation

### Mean value theorem of differential calculus

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 $\displaystyle{ \xi }$:

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

### Integration

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

\displaystyle{ \begin{align} 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{align} }

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$\displaystyle{ {1 \over L(x,y)} = \int_0^1 {\operatorname{d}\!t \over t x + (1-t)y} }$and$\displaystyle{ {1 \over L(x,y)} = \int_0^\infty {\operatorname{d}\!t \over (t+x)\,(t+y)}. }$

## Generalization

### Mean value theorem of differential calculus

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

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

### Integral

The integral interpretation can also be generalized to more variables, but it leads to a different result. Given the simplex $\displaystyle{ S }$ with $\displaystyle{ S = \{\left(\alpha_0,\, \dots,\, \alpha_n\right) : \left(\alpha_0 + \dots + \alpha_n = 1\right) \land \left(\alpha_0 \ge 0\right) \land \dots \land \left(\alpha_n \ge 0\right)\} }$ and an appropriate measure $\displaystyle{ \mathrm{d}\alpha }$ 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 $\displaystyle{ n = 2 }$

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

## Connection to other means

• Arithmetic mean: $\displaystyle{ \frac{L\left(x^2, y^2\right)}{L(x, y)} = \frac{x + y}{2} }$
• Geometric mean: $\displaystyle{ \sqrt{\frac{L\left(x, y\right)}{L\left( \frac{1}{x}, \frac{1}{y} \right)}} = \sqrt{x y} }$
• Harmonic mean: $\displaystyle{ \frac{ L\left( \frac{1}{x}, \frac{1}{y} \right) }{L\left( \frac{1}{x^2}, \frac{1}{y^2} \right)} = \frac{2}{\frac{1}{x}+\frac{1}{y}} }$