Logarithmic mean

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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:

[math]\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} }[/math]

for the positive numbers [math]\displaystyle{ x, y }[/math].

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.

[math]\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. }[/math][1][2][3]

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

[math]\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} }[/math]

Now, for [math]\displaystyle{ n=0 }[/math]

[math]\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} }[/math]
[math]\displaystyle{ \sqrt{x y} \leq \frac{x - y}{\ln(x) - \ln(y)}\leq \frac{x + y} {2} }[/math]

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

For [math]\displaystyle{ n=1 }[/math]

[math]\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} }[/math]

for the proof go through the bibliography.

Derivation

Mean value theorem of differential calculus

From the mean value theorem, there exists a value [math]\displaystyle{ \xi }[/math] in the interval between x and y where the derivative [math]\displaystyle{ f' }[/math] equals the slope of the secant line:

[math]\displaystyle{ \exists \xi \in (x, y): \ f'(\xi) = \frac{f(x) - f(y)}{x - y} }[/math]

The logarithmic mean is obtained as the value of [math]\displaystyle{ \xi }[/math] by substituting [math]\displaystyle{ \ln }[/math] for [math]\displaystyle{ f }[/math] and similarly for its corresponding derivative:

[math]\displaystyle{ \frac{1}{\xi} = \frac{\ln(x) - \ln(y)}{x - y} }[/math]

and solving for [math]\displaystyle{ \xi }[/math]:

[math]\displaystyle{ \xi = \frac{x - y}{\ln(x) - \ln(y)} }[/math]

Integration

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

[math]\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} }[/math]

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 [math]\displaystyle{ x }[/math] and [math]\displaystyle{ y }[/math]. The homogeneity of the integral operator is transferred to the mean operator, that is [math]\displaystyle{ L(cx, cy) = cL(x, y) }[/math].

Two other useful integral representations are[math]\displaystyle{ {1 \over L(x,y)} = \int_0^1 {\operatorname{d}\!t \over t x + (1-t)y} }[/math]and[math]\displaystyle{ {1 \over L(x,y)} = \int_0^\infty {\operatorname{d}\!t \over (t+x)\,(t+y)}. }[/math]

Generalization

Mean value theorem of differential calculus

One can generalize the mean to [math]\displaystyle{ n + 1 }[/math] variables by considering the mean value theorem for divided differences for the [math]\displaystyle{ n }[/math]th derivative of the logarithm.

We obtain

[math]\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)} }[/math]

where [math]\displaystyle{ \ln\left(\left[x_0,\, \dots,\, x_n\right]\right) }[/math] denotes a divided difference of the logarithm.

For [math]\displaystyle{ n = 2 }[/math] this leads to

[math]\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)}} }[/math].

Integral

The integral interpretation can also be generalized to more variables, but it leads to a different result. Given the simplex [math]\displaystyle{ S }[/math] with [math]\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)\} }[/math] and an appropriate measure [math]\displaystyle{ \mathrm{d}\alpha }[/math] which assigns the simplex a volume of 1, we obtain

[math]\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 }[/math]

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

[math]\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] }[/math].

Example [math]\displaystyle{ n = 2 }[/math]

[math]\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)} }[/math].

Connection to other means

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

See also

References

Citations
  1. B. C. Carlson (1966). "Some inequalities for hypergeometric functions". Proc. Amer. Math. Soc. 17: 32–39. doi:10.1090/s0002-9939-1966-0188497-6. 
  2. B. Ostle; H. L. Terwilliger (1957). "A comparison of two means". Proc. Montana Acad. Sci. 17: 69–70. 
  3. Tung-Po Lin (1974). "The Power Mean and the Logarithmic Mean". The American Mathematical Monthly 81 (8): 879–883. doi:10.1080/00029890.1974.11993684. 
Bibliography