aa r X i v : . [ m a t h . C A ] O c t ON HARMONIC REPRESENTATION OF MEANS
ALFRED WITKOWSKI
Abstract.
We characterize continuous, symmetric and homogeneous means M that can be represented in the form M ( x, y ) = Z dtN (cid:16) x + y − t x − y , x + y + t x − y (cid:17) . New inequalities for means are derived from such representation. Introduction, Definitions and notation
In paper [5] we investigated the representation of a symmetric, homogeneousmean M : R → R of the form(1) M ( x, y ) = | x − y | f (cid:16) | x − y | x + y (cid:17) The main observation was that every symmetric, homogeneous mean admits sucha representation. The mapping(2) M ( x, y ) ↔ f M ( z ) = zM (1 − z, z ) establishes one-to-one correspondence between the set of symmetric homogeneousmeans and the set of functions f : (0 , → R satisfying(3) z z ≤ f ( z ) ≤ z − z , called Seiffert functions , and the identity(4) M ( x, y ) = | x − y | f M (cid:16) | x − y | x + y (cid:17) holds. Moreover, the formula (1) transforms Seiffert function into a symmetric,homogeneous mean.Note that the outermost functions in (3) correspond to max and min means.In this note we discuss the representation of means in the form M ( x, y ) = Z dtN (cid:0) x + y + t x − y , x + y − t x − y (cid:1) , where N is also a homogeneous, symmetric mean.We shall be using two facts from [5] Date : October 10, 2013.2000
Mathematics Subject Classification.
Key words and phrases.
Seiffert mean, logarithmic mean, Seiffert , harmonic representation,AGM mean.
Property 1. [5, Section 7] If f is a Seiffert mean, then for arbitrary < t ≤ thefunction f { t } given by the formula f { t } ( z ) = f ( tz ) t is also a Seiffert mean. Lemma 1.1. If f is a Seiffert function corresponding to the mean M , then f { t } isa Seiffert function for M { t } ( x, y ) = M (cid:0) x + y + t x − y , x + y − t x − y (cid:1) . Proof.
Let z = | x − y | x + y . Then by (1) and (2) we have | x − y | f { t } ( z ) = t | x − y | f ( tz ) = t | x − y | M (1 − tz, tz )2 tz = x + y M (cid:18) − t | x − y | x + y , t | x − y | x + y (cid:19) = M { t } ( x, y ) . (cid:3) Following [5, Section 5], consider the integral operator on the set of continuousSeiffert functions, defined as(5) I ( f )( z ) = Z z f ( u ) u du. Property 2.
The operator I has the following properties: • is monotone - if f ≤ g , then I ( f ) ≤ I ( g ) , • preserves convexity - if f is convex, then so is I ( f ) and for all < z < the inequalities z ≤ I ( f )( z ) ≤ f ( z ) hold, ([5, Theorem 5.1]), • preserves concavity - if f is concave, then so is I ( f ) and for all < z < the inequalities z ≥ I ( f )( z ) ≥ f ( z ) hold, ([5, Theorem 5.1]), • I ( f ) is a Seiffert function, ([5, Corollary 5.1]).The next simple theorem characterizes the functions, which are of the form I ( f ) . Theorem 1.1.
Let g be a real function defined on the interval (0 , . The followingconditions are equivalent • lim z → g ( z ) = 0 , g is continuously differentiable, and for all < z < (6)
11 + z ≤ g ′ ( z ) ≤ − z , • there exist a continuous Seiffert function f such that g = I ( f ) .Proof. Multiplying (6) by z we see that f ( z ) = zg ′ ( z ) is a continuous Seiffertfunction and clearly I ( f ) = g .Conversely, if f is continuous, then g = I ( f ) is differentiable. Since lim z → f ( z ) /z =1 we claim lim z → g ( z ) = 0 . Differentiating g we obtain g ′ ( z ) = f ( z ) /z , which yields(6) because f fulfills (3). (cid:3) Now we are ready to formulate the main result of this note.2.
Harmonic representation of means
Definition 2.1.
We say that a continuous mean N is a harmonic representationof mean M if M ( x, y ) = Z dtN { t } ( x, y ) . N HARMONIC REPRESENTATION OF MEANS 3
Theorem 2.1.
A continuous mean M admits a harmonic representation if andonly if its Seiffert function m can be represented as I ( n ) , where n is a continuousSeiffert function.Proof. Let N be the harmonic representation of M and let z = | x − y | x + y . Denote by m and n the Seiffert functions of M and N respectively. Applying (1) and (2) wehave | x − y | I ( n )( z ) = 2 | x − y | Z z n ( u ) u du = 2 | x − y | Z n { t } ( z ) dt = Z dtN { t } ( x, y ) = 1 M ( x, y ) = 2 | x − y | m ( z ) , which yields m = I ( n ) . Conversely, if m = I ( n ) and N is a mean corresponding to n , then M ( x, y ) = 2 | x − y | m ( z ) = 2 | x − y | I ( n )( z ) = 2 | x − y | Z z n ( u ) u du = 2 | x − y | Z n { t } ( z ) dt = Z dtN { t } ( x, y ) . (cid:3) From (3) we obtain by integration the inequalities(7) log(1 + z ) ≤ I ( f )( z ) ≤ − log(1 − z ) , which shows, that every mean admitting harmonic representation satisfies the in-equalities | x − y | A ( x, y ) − log min( x, y )) ≤ M ( x, y ) ≤ | x − y | x, y ) − log A ( x, y )) . The inverse statement is not true. It is easy to construct a function satisfying (7)for which (6) fails. 3.
Examples I
Example 3.1.
The Seiffert function of the Seiffert mean P ( x, y ) = | x − y | z isobviously arcsin . Let g ( z ) = z √ − z . Then arcsin = I ( g ) and g is the Seiffertfunction of the geometric mean G ( x, y ) = √ xy . Thus we obtain the identity P ( x, y ) = (cid:18)Z dtG { t } ( x, y ) (cid:19) − . Example 3.2.
The second Seiffert mean is given by T ( x, y ) = | x − y | z . Let C ( x, y ) = x + y x + y be the contra-harmonic mean. Its Seiffert function is c ( z ) = z z and one can easily verify that I ( c ) = arctan , so T ( x, y ) = (cid:18)Z dtC { t } ( x, y ) (cid:19) − . ALFRED WITKOWSKI
Example 3.3.
For the logarithmic mean L ( x, y ) = x − y log x − log y = | x − y | z we get L ( x, y ) = (cid:18)Z dtH { t } ( x, y ) (cid:19) − , where H ( x, y ) = xyx + y denotes the harmonic mean. Example 3.4.
The Seiffert function of the root-mean square R = q x + y is thefunction r ( z ) = z √ z , thus I ( r )( z ) = arsinh z , which in turn is the Seiffert meanof the Neuman-Sándor mean M ( x, y ) = | x − y | z , so M ( x, y ) = (cid:18)Z dtR { t } ( x, y ) (cid:19) − , In [5] we have shown that sin , tan , sinh and tanh are also Seiffert function. Letus check if their corresponding means admit harmonic representations. To do it weshall use Theorems 1.1 and 2.1 Example 3.5.
For g ( z ) = sin z we want to show that g ′ satisfies (6). Obviously cos z < < / (1 − z ) . To prove the other part observe that (1 + z ) cos z > (1 + z )(1 − z / > z (1 − z/ > , thus (6) holds, and one easily verifies that z cos z is the Seiffert function of the mean M ( x, y ) = A ( x, y ) / cos | x − y | x + y , which implies x − y x − yx + y = (cid:18)Z dtM { t } ( x, y ) (cid:19) − . Example 3.6.
Now let g ( z ) = tan z . We have
11 + z < < z = 1(1 + sin z )(1 − sin z ) < − z , so z/ cos z is the Seiffert function. It corresponds to the mean M ( x, y ) = A ( x, y ) cos | x − y | x + y and x − y x − yx + y = (cid:18)Z dtM { t } ( x, y ) (cid:19) − . Example 3.7.
With the hyperbolic sine the situation is simple. We have < cosh z = ∞ X m =0 z m (2 m )! < ∞ X m =0 z m = 11 − z , thus z cosh z is the Seiffert function, and its mean M ( x, y ) = A ( x, y ) / cosh | x − y | x + y satisfies x − y x − yx + y = (cid:18)Z dtM { t } ( x, y ) (cid:19) − . Example 3.8.
The last function is the hyperbolic tangent. Its derivative is cosh − z and cosh − (1) ≈ . < , so the left inequality in (6) does not hold, and thisyields the mean x − y x − yx + y does not have a harmonic representation.We leave as a simple exercise the fact that there is no harmonic representationof the geometric mean. N HARMONIC REPRESENTATION OF MEANS 5 The arithmetic-geometric mean
This section is devoted to the arithmetic-geometric mean given by the formula
AGM ( x, y ) = π Z π/ dϕ p x cos ϕ + y sin ϕ ! − . To find its Seiffert mean let us recall the famous result of Gauss [3](8)
AGM (1 − z, z ) = π K ( z ) , where K is the complete elliptic integral of the first kind(9) K ( z ) = Z π/ dϕ p − z sin ϕ = Z dt √ − t √ − z t . Comparing (8) and (2) we see that f AGM ( z ) = π zK ( z ) . We shall show that AGM admits the harmonic representation. By Theorem 1.1 it is enough to show that f ′ AGM satisfies (6). To this end let us recall the power series expansion of K ([2,900.00])(10) K ( z ) = π ∞ X m =1 (cid:20) (2 m − m )!! (cid:21) z m ! . We have f ′ AGM ( z ) = 2 π (cid:18) K ( z ) + z dKdz (cid:19) = 1 + ∞ X m =1 (2 m + 1) (cid:20) (2 m − m )!! (cid:21) z m (11)Denoting the m th coefficient in (11) by c m we see that c m +1 c m = 2 m + 32 m + 1 (cid:20) (2 m + 1)!!((2 m )!!(2 m + 2)!!(2 m − (cid:21) = (2 m + 1)(2 m + 3)(2 m + 2) < , and since c = 3 / we conclude that c m < for all ≥ . Thus < f ′ AGM ( z ) < z + z + · · · = 1 / (1 − z ) .Theorem 1.1 implies that the arithmetic-geometric mean admits the harmonic rep-resentation. To derive its explicit form, recall that the derivative of K is given by K ′ ( z ) = E ( z ) z (1 − z ) − K ( z ) z (see. e.g. [2, 710.00]), thus zf ′ AGM ( z ) = 2 π (cid:0) zK ( z ) + z K ′ ( z ) (cid:1) = 2 π z − z E ( z ) , ( E ( z ) = R π/ p − z sin ϕdϕ is the complete elliptic integral of the second kind).As z − z is the Seiffert function of the harmonic mean we obtain the formula V ( x, y ) = πH ( x, y )2 E (cid:16) | x − y | x + y (cid:17) = πH ( x, y )2 E (cid:16)q − G ( x,y ) A ( x,y ) (cid:17) = πG ( x, y )2 R π/ q A ( x, y ) cos ϕ + G ( x, y ) sin ϕdϕ . This mean has a nice geometric interpretation: in the ellipsis with semi-axes G ( x, y ) and A ( x, y ) it represents the ratio of the area of inscribed disc to its semi-perimeter. ALFRED WITKOWSKI Hermite-Hadamard inequality for means
The Hermite-Hadamard inequality in its classic form says that if f is a convexfunction in an interval I , then for all a, b ∈ If (cid:18) a + b (cid:19) ≤ b − a Z ba f ( t ) dt ≤ f ( a ) + f ( b )2 . A stronger inequality also holds b − a Z ba f ( t ) dt ≤ (cid:20) f (cid:18) a + b (cid:19) + f ( a ) + f ( b )2 (cid:21) . Suppose now that the mean N is the harmonic representation of M and itsSeiffert function n is such that the function n ( u ) /u is convex. Then, applying theHermite-Hadamard inequality to (5) and taking into account that lim u → n ( u ) /u =1 we obtain(12) n ( z/ ≤ I ( n )( z ) ≤ z + n ( z )2 . This yields (with help of (2)) the inequalities for means(13) H ( A ( x, y ) , N ( x, y )) ≤ M ( x, y ) ≤ N (cid:18) x + y , x + 3 y (cid:19) . The stronger version of the Hermite-Hadamard reads in this case:(14) I ( n )( z ) ≤ (cid:20) n ( z/
2) + z + n ( z )2 (cid:21) , which yields(15) H ( A ( x, y ) , N { / } ( x, y ) , N { / } ( x, y ) , N ( x, y )) ≤ M ( x, y ) ≤ N (cid:18) x + y , x + 3 y (cid:19) . Obviously, if n ( u ) /u is concave, the inequalities in (12)–(15) are reversed.In the above we use the Hermite-Hadamard inequality with the left end fixed,so it may happen that (12) holds even if n ( u ) /u is not convex. Of course, in suchcase an individual treatment would be required.6. Examples II
Example 6.1.
Let N = G . By Example 3.1 we know that M = P is the firstSeiffert mean. Since n ( u ) /u = (1 − u ) − / is convex and G { / } = √ A + G / ,(13) and (14) yield AGA + G ≤ (cid:18) √ A + G + A + G (cid:19) − ≤ P ≤ √ A + G . Example 6.2.
The Seiffert function c from Example 3.2 does not satisfy the con-vexity condition, but the reversed inequalities in (12) hold anyway, by the followinglemma. Lemma 6.1.
The inequalities u u > arctan u > u u u hold for < u < N HARMONIC REPRESENTATION OF MEANS 7
Proof.
Let h ( u ) = u u − arctan u . As h (0) = 0 and h ′ ( u ) = u (4 − u )( u +1)( u +4) wesee that h has local maximum at u = 2 / √ and since h (1) > we conclude that h ( u ) > .Let now h ( u ) = arctan u − u u u . Then h (0) = 0 and h ′ ( u ) = u (1 − u )2( x +1) > ,and the proof is complete. (cid:3) Thus for the contraharmonic mean and the second Seiffert mean we have C { / } = 5 A − G A ≤ T ≤ H ( A, C ) Example 6.3.
The pair ( M, N ) = (
L, H ) (see Example 3.3) gives the inequalities G AA + G ≤ AG (3 A + G )3 A + 12 A G + G ≤ L ≤ A + G A Example 6.4.
For the root-mean square and Neuman-Sándor means (Example3.4) the convexity condition is not satisfied, but the following lemma shows thatthe reversed inequalities (12) are valid.
Lemma 6.2.
For < u < the inequalities u √ u + 4 ≥ arsinh u ≥ u u √ u + 1 hold.Proof. To prove the left inequality it suffices to show that the function h ( u ) =arsinh u − u √ u +4 decreases, because h (0) = 0 . Differentiating we obtain(16) h ′ ( u ) = ( u + 4) / − u + 1) / ( u + 4) / ( u + 1) / . Let p denote the numerator in (16). Then p ′ ( u ) = u (cid:16) p u + 4) − √ u +1 (cid:17) := uq ( u ) . The function q is a difference of an increasing and decreasing function, thusincreases from q (0) = − to q (1) = 3 √ − √ > , so we conclude that p has onelocal minimum in the interval (0 , . Since p (0) = 0 and p (1) = √ − √ < we see that p ( u ) < for all u , thus h ′ ( u ) < and we are done.For the right inequality the method is similar: h ( u ) = u u √ u + 1 − arsinh u, h ′ ( u ) = ( u + 1) / − (2 u + 1)2( u + 1) / p ( u ) = ( u + 1) / − (2 u + 1) , p ′ ( u ) = u (3 p u + 1 −
4) := uq ( u ) . As above, q increases from − to √ − , so p has one local minimum, and since p (0) = 0 and p (1) = √ − < we conclude h ′ < . (cid:3) Thus for the Neuman-Sándor mean M ( x, y ) = | x − y | | x − y | x + y the inequality (13)in this case reads R { / } = √ A − G ≤ M ≤ H ( A, R ) . ALFRED WITKOWSKI
Example 6.5.
In Example 3.5 we consider the Seiffert functions m ( z ) = sin z and n ( z ) = z cos z . Clearly n ( z ) /z is concave and thus x + y | x − y | x + y ≤ | x − y | | x − y | x + y ≤ x + y | x − y | x + y Example 6.6.
The function z is convex, thus we can apply (12) to the functionsfrom Example 3.6 to obtain ( x + y ) cos | x − y | x + y | x − y | x + y ≤ | x − y | | x − y | x + y ≤ A ( x, y )cos | x − y | x + y . Example 6.7.
In Example 3.7 the function cosh is convex, so we get x + y | x − y | x + y ≤ | x − y | | x − y | x + y ≤ x + y | x − y | x + y . Example 6.8.
In this example we deal with the
AGM mean and its harmonic rep-resentation V described in Section 4. The Seiffert mean of V is v ( z ) = π z − z E ( z ) ,so(17) v ( z ) z = 2 π Z π/ p − z sin ϕ − z dϕ. We shall show that this function is convex. For < a < let h a ( u ) = √ − au √ − u .Then h ′ a ( u ) = (1 − a ) u (1 − au ) / (1 − u ) / . Note the h ′ a is nonnegative and increasing, since its numerator increases whiledenominator decreases. Thus h a is positive, increasing and convex. The function g ( u ) = 1 / p (1 − u ) shares the same properties, so their product is convex [4,Theorem I.13C]. Since the integrands in (17) are convex, so is the left-hand side.Therefore by (13) AVA + V ≤ AGM ≤ V { / } . ==================================================================================================================================================================================================================================== References [1] J.M. Borwein, P.B. Borwein,
Pi and the AGM , John Wiley & Sons, New York 1987.[2] P.F. Byrd, M.D. Friedman,
Handbook of Elliptic Integrals for Engineers and Scientists ,Springer, New York, 1971.[3] C.F. Gauss,
Werke,
Bd. 3, Königlichen Gesell. Wiss., Göttingen, 1876, pp. 361–403.[4] A.W. Rogers, D.E.Varberg,
Convex Functions , Academic Press, New York and London, 1973[5] A. Witkowski. On Seiffert–like means. arXiv:1309.1244 [math.CA] , June 2013.
Institute of Mathematics and Physics, University of Technology and Life Sci-ences, Al. prof. Kaliskiego 7, 85-796 Bydgoszcz, Poland
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