A NEW ANALOGUE OF GAUSS ’ FUNCTIONAL EQUATION

Gauss established a theory on the functional equation (Gauss' functional equation) 
 0} \right) \]" id="E1" xmlns:mml="http://www.w3.org/1998/Math/MathML"> f ( a + b 2 , a b ) = f ( a , b )   ( a , b > 0 ) , 
where f : R + × R + → R is an unknown function of the above equation.In this paper we treat the functional equation 
 0} \right) \]" id="E3" xmlns:mml="http://www.w3.org/1998/Math/MathML"> f ( a + b 2 , 2 a b a + b ) = f ( a , b )   ( a , b > 0 ) 
where f : R + × R + → R is an unknown function of this equation.The purpose of this paper is to prove new results on this functional equation by following 
the theory of Gauss' functional equation.


INTRODUCTION.
Gauss established a theory on the functional equation where f: R + R + R is an unknown function of the above equation. (Cf. [21, [31, [41, [5]) In this paper we treat the functional equation .(.+ 2o 2 'db] f(a,b) (a,b> 0), (1.2) where f R + x R + R is an unknown function of the above equation.
The purpose of this paper is to obtain new results for the solution of (1.2) by following the theory on Gauss' functional equation.The main result is to solve (1.2) under suitable natural condition of f.In the last section we conclude with an open problem for the functional equation (.).REMARK 1.For the arithmetic mean, geometric mean, harmonic mean cf.[1, pp.287-297].
2. A RELATION BETWEEN (1.2) AND THE ARITHMETIC-HARMONIC MEAN.
Then the following theorem holds: THEOREM A. (i) The two sequences {a,}=0 and {b,}=0 are convergent and the limits are equal.
(ii) y(a, b) G(a, b) is a solution of the functional equation (1.1).For G(a, b) see below.
PROOF.See [4], [51 REMARK 2. The limit in Theorem A is said to be the arithmetic-geometric mean of Gauss of a, b denoted by G(a, b) in this paper.
Then the following theorem holds: THEOREM B. The two sequences {a,}=o and (b,.,}=o converge to the same limit vZ'.PROOF.See for example (3 I. REMARK 8.The limit in Theorem B is said to be the arithmetic-harmonic mean of a,b, denoted by M(a,b)in this paper.Hence, by Theorem B we obtain M(a,b)= v/.PROOF.The proof goes along the same lines as that of Theorem A (ii), or by the fact that M(a, b) x/ (see Remark 3) the proof is clear.Gauss (cf.[2], [3], [4], [5]) considered the following definite integral l(a, b) which is closely
In [7], the following converse of Theorem C(i) is proved.(Cf.[8]) THEOREM D. Let f R + R + R be a function.If f can be represented by f(,,) p()dO (,, > 0), where r v/a cos 0 + b sin O,p R + R is a function such that p"(x) is continuous in R+, then the only solution of (1.1) is given by f(a, b) Al(a, b)+ B A G(a, b) + B, where A, B are arbitrary real constants.
The following provides the converse of Theorem 2(i).This is the main theorem of the paper.
THEOREM 3. Let f R + x R + R be a function.If f can be represented by f(a, b) "r q(s)dO (a, b > 0), where s asin: O+ bcos O,q R + --, R is a function such that q"(x) is continuous in R+, then the only solution of (1.2) is given by mX___ f(a, b) AJ(a, b)+ B ,/ + B, where A, B are arbitrary real constants.
Solving the above differential equation in R + yields 1 q(z) A= + B and thus q(s) A I-+B s where A, B are real constants.Substituting (6.2) into (5.1) and using (3.(6.3) Direct substitution of (ft.3) into (1.2) shows that (6.3) is a solution of our original functional equation (1.2).
In this last section we shall give an open problem for the functional equation (1.2).
OPEN PROBLEM: Let f" R + x R + -R be a continuous function in R + x R+.Is the only continuous solution of the functional equation (1.2) given by f(a, b) F(ab), where F" R + R is an arbitrary continuous function of a real variable ACKNOWLEDGMENT.The authors with to thank Professor J. Aczdl for suggesting the paper [3] in References.

THEOREM 1 .
f(a, b) M(a, b) is a solution of the functional equation(1.2).
3. INTEGRAL REPRESENTATIONS FOR THE ARITHMETIC-HARMONIC MEAN M(a, b).