Definition
edit
The logarithmic mean is defined as:
M
lm
(
x
,
y
)
=
lim
(
ξ
,
η
)
→
(
x
,
y
)
η
−
ξ
ln
(
η
)
−
ln
(
ξ
)
=
{
x
if
x
=
y
,
y
−
x
ln
y
−
ln
x
otherwise,
{\displaystyle {\begin{aligned}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,\\[2pt]{\dfrac {y-x}{\ln y-\ln x}}&{\text{otherwise,}}\end{cases}}\end{aligned}}}
for the positive numbers x, y .
Inequalities
edit
The logarithmic mean of two numbers is smaller than the arithmetic mean and the generalized mean with exponent greater than 1. However, it is larger than the geometric mean and the harmonic mean , respectively. The inequalities are strict unless both numbers are equal.
2
1
x
+
1
y
≤
x
y
≤
x
−
y
ln
x
−
ln
y
≤
x
+
y
2
≤
(
x
2
+
y
2
2
)
1
/
2
for all
x
>
0
and
y
>
0.
{\displaystyle {\frac {2}{\displaystyle {\frac {1}{x}}+{\frac {1}{y}}}}\leq {\sqrt {xy}}\leq {\frac {x-y}{\ln x-\ln y}}\leq {\frac {x+y}{2}}\leq \left({\frac {x^{2}+y^{2}}{2}}\right)^{1/2}\qquad {\text{ for all }}x>0{\text{ and }}y>0.}
[1] [2] [3] [4]
Toyesh Prakash Sharma generalizes the arithmetic logarithmic geometric mean inequality for any n belongs to the whole number as
x
y
(
ln
x
y
)
n
−
1
(
n
+
ln
x
y
)
≤
x
(
ln
x
)
n
−
y
(
ln
y
)
n
ln
x
−
ln
y
≤
x
(
ln
x
)
n
−
1
(
n
+
ln
x
)
+
y
(
ln
y
)
n
−
1
(
n
+
ln
y
)
2
{\displaystyle {\sqrt {xy}}\ \left(\ln {\sqrt {xy}}\right)^{n-1}\left(n+\ln {\sqrt {xy}}\right)\leq {\frac {x(\ln x)^{n}-y(\ln y)^{n}}{\ln x-\ln y}}\leq {\frac {x(\ln x)^{n-1}(n+\ln x)+y(\ln y)^{n-1}(n+\ln y)}{2}}}
Now, for n = 0 :
x
y
(
ln
x
y
)
−
1
ln
x
y
≤
x
−
y
ln
x
−
ln
y
≤
x
(
ln
x
)
−
1
ln
x
+
y
(
ln
y
)
−
1
ln
y
2
x
y
≤
x
−
y
ln
x
−
ln
y
≤
x
+
y
2
{\displaystyle {\begin{array}{ccccc}{\sqrt {xy}}\left(\ln {\sqrt {xy}}\right)^{-1}\ln {\sqrt {xy}}&\leq &{\dfrac {x-y}{\ln x-\ln y}}&\leq &{\dfrac {x(\ln x)^{-1}\ln x+y(\ln y)^{-1}\ln y}{2}}\\[4pt]{\sqrt {xy}}&\leq &{\dfrac {x-y}{\ln x-\ln y}}&\leq &{\dfrac {x+y}{2}}\end{array}}}
This is the arithmetic logarithmic geometric mean inequality. similarly, one can also obtain results by putting different values of n as below
For n = 1 :
x
y
(
1
+
ln
x
y
)
≤
x
ln
x
−
y
ln
y
ln
x
−
ln
y
≤
x
(
1
+
ln
x
)
+
y
(
1
+
ln
y
)
2
{\displaystyle {\sqrt {xy}}\left(1+\ln {\sqrt {xy}}\right)\leq {\frac {x\ln x-y\ln y}{\ln x-\ln y}}\leq {\frac {x(1+\ln x)+y(1+\ln y)}{2}}}
for the proof go through the bibliography.
Derivation
edit
Mean value theorem of differential calculus
edit
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 :
∃
ξ
∈
(
x
,
y
)
:
f
′
(
ξ
)
=
f
(
x
)
−
f
(
y
)
x
−
y
{\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 :
1
ξ
=
ln
x
−
ln
y
x
−
y
{\displaystyle {\frac {1}{\xi }}={\frac {\ln x-\ln y}{x-y}}}
and solving for ξ :
ξ
=
x
−
y
ln
x
−
ln
y
{\displaystyle \xi ={\frac {x-y}{\ln x-\ln y}}}
Integration
edit
The logarithmic mean can also be interpreted as the area under an exponential curve .
L
(
x
,
y
)
=
∫
0
1
x
1
−
t
y
t
d
t
=
∫
0
1
(
y
x
)
t
x
d
t
=
x
∫
0
1
(
y
x
)
t
d
t
=
x
ln
y
x
(
y
x
)
t
|
t
=
0
1
=
x
ln
y
x
(
y
x
−
1
)
=
y
−
x
ln
y
x
=
y
−
x
ln
y
−
ln
x
{\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 {\frac {y}{x}}}}\left({\frac {y}{x}}\right)^{t}\right|_{t=0}^{1}={}{\frac {x}{\ln {\frac {y}{x}}}}\left({\frac {y}{x}}-1\right)={}{\frac {y-x}{\ln {\frac {y}{x}}}}\\[3pt]={}&{\frac {y-x}{\ln y-\ln x}}\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 x and y . The homogeneity of the integral operator is transferred to the mean operator, that is
L
(
c
x
,
c
y
)
=
c
L
(
x
,
y
)
{\displaystyle L(cx,cy)=cL(x,y)}
.
Two other useful integral representations are
1
L
(
x
,
y
)
=
∫
0
1
d
t
t
x
+
(
1
−
t
)
y
{\displaystyle {1 \over L(x,y)}=\int _{0}^{1}{\operatorname {d} \!t \over tx+(1-t)y}}
and
1
L
(
x
,
y
)
=
∫
0
∞
d
t
(
t
+
x
)
(
t
+
y
)
.
{\displaystyle {1 \over L(x,y)}=\int _{0}^{\infty }{\operatorname {d} \!t \over (t+x)\,(t+y)}.}
Generalization
edit
Mean value theorem of differential calculus
edit
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
MV
(
x
0
,
…
,
x
n
)
=
(
−
1
)
n
+
1
n
ln
(
[
x
0
,
…
,
x
n
]
)
−
n
{\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
ln
(
[
x
0
,
…
,
x
n
]
)
{\displaystyle \ln \left(\left[x_{0},\,\dots ,\,x_{n}\right]\right)}
denotes a divided difference of the logarithm.
For n = 2 this leads to
L
MV
(
x
,
y
,
z
)
=
(
x
−
y
)
(
y
−
z
)
(
z
−
x
)
2
(
(
y
−
z
)
ln
x
+
(
z
−
x
)
ln
y
+
(
x
−
y
)
ln
z
)
.
{\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
edit
The integral interpretation can also be generalized to more variables, but it leads to a different result. Given the simplex
S
{\textstyle S}
with
S
=
{
(
α
0
,
…
,
α
n
)
:
(
α
0
+
⋯
+
α
n
=
1
)
∧
(
α
0
≥
0
)
∧
⋯
∧
(
α
n
≥
0
)
}
{\displaystyle 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
d
α
{\textstyle \mathrm {d} \alpha }
which assigns the simplex a volume of 1, we obtain
L
I
(
x
0
,
…
,
x
n
)
=
∫
S
x
0
α
0
⋅
⋯
⋅
x
n
α
n
d
α
{\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
L
I
(
x
0
,
…
,
x
n
)
=
n
!
exp
[
ln
(
x
0
)
,
…
,
ln
(
x
n
)
]
{\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 n = 2 :
L
I
(
x
,
y
,
z
)
=
−
2
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
)
.
{\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)}}.}
Connection to other means
edit
Arithmetic mean :
L
(
x
2
,
y
2
)
L
(
x
,
y
)
=
x
+
y
2
{\displaystyle {\frac {L\left(x^{2},y^{2}\right)}{L(x,y)}}={\frac {x+y}{2}}}
Geometric mean :
L
(
x
,
y
)
L
(
1
x
,
1
y
)
=
x
y
{\displaystyle {\sqrt {\frac {L\left(x,y\right)}{L\left({\frac {1}{x}},{\frac {1}{y}}\right)}}}={\sqrt {xy}}}
Harmonic mean :
L
(
1
x
,
1
y
)
L
(
1
x
2
,
1
y
2
)
=
2
1
x
+
1
y
{\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}}}}}
See also
edit
References
edit
Citations
Bibliography
Oilfield Glossary: Term 'logarithmic mean'
Weisstein, Eric W. "Arithmetic-Logarithmic-Geometric-Mean Inequality" . MathWorld .
Stolarsky, Kenneth B.: Generalizations of the logarithmic mean , Mathematics Magazine, Vol. 48, No. 2, Mar., 1975, pp 87–92
Toyesh Prakash Sharma.: https://www.parabola.unsw.edu.au/files/articles/2020-2029/volume-58-2022/issue-2/vol58_no2_3.pdf "A generalisation of the Arithmetic-Logarithmic-Geometric Mean Inequality , Parabola Magazine, Vol. 58, No. 2, 2022, pp 1–5