Coinciding Mean of the Two Symmetries on the Set of Mean Functions

Abstract

On the set $\mathcal{M}$ of mean functions, the symmetric mean of M with respect to mean $M_0$ can be defined in several ways. The first one is related to the group structure on $\mathcal{M}$, and the second one is defined trough Gauss’ functional equation. In this paper, we provide an answer to the open question formulated by B. Farhi about the matching of these two different mappings called symmetries on the set of mean functions. Using techniques of asymptotic expansions developed by T. Burić, N. Elezović, and L. Mihoković (Vukšić), we discuss some properties of such symmetries trough connection with asymptotic expansions of means involved. As a result of coefficient comparison, a new class of means was discovered, which interpolates between harmonic, geometric, and arithmetic mean.

1. Introduction

Function $M:{\mathbf{R}}^{+}\times {\mathbf{R}}^{+}\to \mathbf{R}$ is called a mean if for all $s,t\in {\mathbf{R}}^{+}$

$$min(s,t)\le M(s,t)\le max(s,t).$$

(1)

Mean M is symmetric if for all $s,t\in {\mathbf{R}}^{+}$

$$M(s,t)=M(t,s)$$

and homogeneous (of degree 1) if for all $\lambda ,s,t\in {\mathbf{R}}^{+}$

$$M(\lambda s,\lambda t)=\lambda M(s,t).$$

This paper was motivated by the problem of matching two different mappings on the set of mean functions formulated in paper [1] in which author introduced algebraic and topological structures on the set ${\mathcal{M}}_{\mathcal{D}}$ of symmetric means on a symmetric domain $\mathcal{D}$ with additional property

$$M(s,t)=s\Rightarrow s=t,\forall (s,t)\in \mathcal{D}.$$

The first mapping is related to the group structure and the second one is defined trough Gauss’ functional equation. It was found that those mappings coincide for arithmetic, geometric, and harmonic mean, but the question of the existence of other solutions remained open. We shall take $\mathcal{D}={\mathbf{R}}^{+}\times {\mathbf{R}}^{+}$.

First, let ${\mathcal{A}}_{\mathcal{D}}$ be set of all functions $f:\mathcal{D}\to \mathbf{R}$ such that

$$(\forall (x,y)\in \mathcal{D})f(x,y)=-f(y,x).$$

$({\mathcal{A}}_{\mathcal{D}},+)$ is an abelian group with the neutral element 0. Function $\phi :{\mathcal{M}}_{\mathcal{D}}\to {\mathcal{A}}^{\mathcal{D}}$ defined by

$$\phi \left(M\right)(x,y):=\left\{\begin{array}{cc}log\left(-\frac{M(x,y)-x}{M(x,y)-y}\right),\hfill & x\ne y,\hfill \\ 0,\hfill & x=y,\hfill \end{array}\right.$$

is a bijection. The composition law $\ast :{\mathcal{M}}_{\mathcal{D}}\times {\mathcal{M}}_{\mathcal{D}}\to {\mathcal{M}}_{\mathcal{D}}$ is defined by

$$M_1\ast M_2={\phi }^{-1}(\phi \left(M_1\right)+\phi \left(M_2\right)).$$

Thus $({\mathcal{M}}_{\mathcal{D}},\ast )$ is an abelian group with the neutral element ${\phi }^{-1}\left(0\right)=A$. It can also easily be shown that the explicit formula for the composition law ∗ holds:

$$(M_1\ast M_2)(x,y)=\left\{\begin{array}{cc}\frac{x(M_1-y)(M_2-y)+y(M_1-x)(M_2-x)}{(M_1-x)(M_2-x)+(M_1-y)(M_2-y)},\hfill & x\ne y,\hfill \\ x,\hfill & x=y.\hfill \end{array}\right.$$

(2)

For the sake of simplicity, variables $(x,y)$ were omitted. By sum and difference of means, we assume usual pointwise addition and subtraction. More on the topological structures on set of bivariate means can also be found in [2].

Based on the operation ∗ defined in (2), the first type of the symmetry was defined.

Definition 1 

([1]). The symmetric mean $M_2$ to a mean $M_1$ with respect to mean $M_0$ via the group structure $({\mathcal{M}}_{\mathcal{D}},\ast )$ is defined with the expression

$$S_{M_0}\left(M_1\right)=M_2\iff M_1\ast M_2=M_0\ast M_0.$$

(3)

Combining (3) with (2), the explicit formula for symmetric mean of mean $M_1$ with respect to $M_0$ can easily be calculated:

$$S_{M_0}\left(M_1\right)=\frac{x(M_1-x){(M_0-y)}^2-y{(M_0-x)}^2(M_1-y)}{(M_1-x){(M_0-y)}^2-{(M_0-x)}^2(M_1-y)}.$$

(4)

We shall see the behavior of $S_{M_0}$ for some basic well known means $M_0$. For $(s,t)\in \mathcal{D}={\mathbf{R}}^{+}\times {\mathbf{R}}^{+}$ let

$$A(s,t)=\frac{s+t}{2},G(s,t)=\sqrt{st},H(s,t)=\frac{2st}{s+t},$$

be the arithmetic, geometric, and harmonic means, respectively.

Example 1 

([1]). For any mean $M\in {\mathcal{M}}_{\mathcal{D}}$, we have:

1.

$S_A\left(M\right)=2A-M,$

2.

$S_G\left(M\right)=\frac{G^2}{M}$,

3.

$S_H\left(M\right)=\frac{HM}{2M-H}$.

Notice that the denominator in $S_H\left(M\right)$ from Example 1 cannot be equal to 0, since $M=\frac{1}{2}H$ does not satisfy the left hand side inequality in (1) and, hence, it is not a mean.

Another type of symmetry, independent of the group structure $({\mathcal{M}}_{\mathcal{D}},\ast )$, can also be defined.

Definition 2 

([1]). Mean $M_2$ is said to be functional symmetric mean of $M_1$ with respect to $M_0$ if the following functional equation is satisfied:

$${\sigma }_{M_0}\left(M_1\right)=M_2\iff M_0(M_1,M_2)=M_0.$$

(5)

We can also say that mean $M_0$ is the functional middle of $M_1$ and $M_2$. Defining equation on the right side of the equivalence relation (5) is known as the Gauss functional equation. Some authors refer to means $M_1$ and $M_2$ as a pair of $M_0$-complementary means. Mean $M_0$ is also said to be $(M_1,M_2)$-invariant. For recent related results, see [3,4,5,6] and also survey article on invariance of means [7] and references therein. Furthermore, if functional symmetric mean exists, then it is unique.

With respect to the same means as in the latter exmple, we may calculate the symmetric means. For instance, when $M_0=H$, we have

$$H(M,{\sigma }_H\left(M\right))=H\iff \frac{2M{\sigma }_H\left(M\right)}{M+{\sigma }_H\left(M\right)}=H\iff 2M{\sigma }_H\left(M\right)=H(M+{\sigma }_H\left(M\right))\iff {\sigma }_H\left(M\right)=\frac{HM}{2M-H}.$$

Other symmetric pairs, with respect to A and G, are obtained in similar manner.

Example 2 

([1]). For any mean $M\in {\mathcal{M}}_{\mathcal{D}}$, we have:

1.

${\sigma }_A\left(M\right)=2A-M,$

2.

${\sigma }_G\left(M\right)=\frac{G^2}{M}$,

3.

${\sigma }_H\left(M\right)=\frac{HM}{2M-H}$.

Taking into account Examples 1 and 2, in which the same mappings appear with respect to arithmetic, geometric, and harmonic mean appear, the author in [1] states the following.

Open question. For which mean functions $M_0$ on $\mathcal{D}={\mathbf{R}}^{+}\times {\mathbf{R}}^{+}$ do the two symmetries, S and $\sigma$, with respect to $M_0$, coincide?

The goal of this paper is to analyze the open question and offer the answer in the setting of symmetric homogeneous means, which possess the asymptotic expansion. Techniques of asymptotic expansions were developed in [8,9,10] and appeared to be very useful in comparison and finding inequalities for bivariate means ([11,12]), comparison of bivariate parameter means ([10]), finding optimal parameters in convex combinations of means ([12,13]), and solving the functional equations of the form $B\left(A\right(x\left)\right)=C\left(x\right)$, where asymptotic expansions of B and C are known ([14]). In the latter example, A, B, and C are functions of a real variable, which possess asymptotic expansion as $x\to \infty$ with respect to asymptotic sequences ${\left(x^{w-n}\right)}_{n\in {\mathbf{N}}_0}$, ${\left(x^{u-n}\right)}_{n\in {\mathbf{N}}_0}$, and ${\left(x^{v-n}\right)}_{n\in {\mathbf{N}}_0}$, respectively, where w, u, and v are real numbers. When used with $B\left(x\right)=f\left(x\right)$ and $C\left(x\right)=\frac{1}{t-s}{\int }_s^{t}f(x+u)du,$ finding $A\left(x\right)$ is then equivalent to determining integral f-mean $I_f(x+s,x+t)=f^{-1}\left(\frac{1}{t-s}{\int }_{x+s}^{x+t}f\left(u\right)du\right)$ for a given function f as it was described in detail in above mentioned paper. We may perceive the significance of this approach when explicit formula for the inverse function is not known, which is case for the digamma function.

Techniques and results applyed in this paper were described in Section 2. In the next step, we obtained the algorithm for calculating the coefficients in the asymptotic expansions of means $M_2^S=S_{M_0}\left(M_1\right)$ and $M_2^{\sigma }={\sigma }_{M_0}\left(M_1\right)$. Comparing the first few obtained coefficents, we anticipated the general form of the coefficients in the asymptotic expansion of mean $M_0$ for which symmetries $S_{M_0}$ and ${\sigma }_{M_0}$ coincide, i.e., such that $M_2^S=M_2^{\sigma }$.

At the beginning of Section 3, we found closed formula and explored some properties, such as limit behavior and monotonocity with respect to the parameter. We proved that proposed function represents the well defined one parameter class of means. We have shown that it also covers, as the special cases, means from Examples 1 and 2.

Lastly, in Section 4, we have proved that this class of means answered the open question and stated the hypothesis that there were not any other solutions in the context of homogeneous symmetric means, which possess asymptotic power series expansions.

In addition, methods presented in this paper may be useful with similar problems regarding functional equations, especially in case when the explicit formula for included function was not known.

2. Asymptotic Expansions

Recall the definition of an asymptotic power series expansion as $x\to \infty$.

Definition 3. 

The series ${\sum }_{n=0}^{\infty }{c}_n{x}^{-n}$ is said to be an asymptotic expansion of a function $f\left(x\right)$ as $x\to \infty$ if for each $N\in \mathbf{N}$

$$f\left(x\right)=\sum _{n=0}^N{c}_n{x}^{-n}+o\left(x^{-N}\right).$$

Main properties of asymptotic series and asymptotic expansions can be found in [15]. Taylor series expansion can also be seen as an asymptotic expansion, but the converse is not generally true, and the asymptotic series may also be divergent. The main characteristic of asymptotic expansion is that it provides good approximation using a finite number of terms while letting $x\to \infty$.

Beacause of the intrinsity (1), mean M would possess the asymptotic power series as $x\to \infty$ of the form

$$M(x+s,x+t)=\sum _{n=0}^{\infty }{c}_n(s,t)x^{-n+1}$$

with $c_0(s,t)=1$. For a homogeneous symmetric mean, the coefficients $c_n(s,t)$ are also homogeneous symmetric polynomials of degree n in variables s and t, and for $s=-t$, they have a simpler form. Let the means included possess the asymptotic expansions as $x\to \infty$ of the form

$$\begin{array}{cc}\hfill M_0(x-t,x+t)& =\sum _{n=0}^{\infty }{c}_n{t}^{2n}{x}^{-2n+1},\hfill \\ \hfill M_1(x-t,x+t)& =\sum _{n=0}^{\infty }{a}_n{t}^{2n}{x}^{-2n+1},\hfill \\ \hfill M_2(x-t,x+t)& =\sum _{n=0}^{\infty }{b}_n{t}^{2n}{x}^{-2n+1}.\hfill \end{array}$$

(6)

Conversely, it can also be shown that the expansion in variables $(x-t,x+t)$ is sufficent to obtain the so-called two variable expansion, i.e., the expansion in variables $(x+s,x+t)$. Furthermore, note that

$$a_0=b_0=c_0=1.$$

(7)

In this section, we will find the asymptotic expansions of means $M_2^S=S_{M_0}\left(M_1\right)$ and $M_2^{\sigma }={\sigma }_{M_0}\left(M_1\right)$.

2.1. Symmetry $S_{M_0}$

Recall the recently developed results for tansformations of asymptotic series, i.e., the complete asymptotic expansions of the quotient and the power of asymptotic series.

Lemma 1 

([10], Lemma 1.1.). Let function $f\left(x\right)$ and $g\left(x\right)$ have the following asymptotic expansions ($a_0\ne 0,b_0\ne 0$) as $x\to \infty$:

$$f\left(x\right)\sim \sum _{n=0}^{\infty }{a}_n{x}^{-n},g\left(x\right)\sim \sum _{n=0}^{\infty }{b}_n{x}^{-n}.$$

Then, asymptotic expansion of their quotient $f\left(x\right)/g\left(x\right)$ reads as

$$\frac{f\left(x\right)}{g\left(x\right)}\sim \sum _{n=0}^{\infty }{c}_n{x}^{-n},$$

where coefficients $c_n$ are defined by

$$c_n=\frac{1}{b_0}\left(a_n-\sum _{k=0}^{n-1}{b}_{n-k}{c}_k\right).$$

Lemma 2 

([8,16]). Let $m\left(x\right)$ be a function with asymptotic expansion $(c_0\ne 0)$:

$$m\left(x\right)\sim \sum _{n=0}^{\infty }{c}_n{x}^{-n},(x\to \infty ).$$

Then, for all real r, it holds

$${\left[m\left(x\right)\right]}^r\sim \sum _{n=0}^{\infty }P[n,r,{\left(c_j\right)}_{j\in {\mathbf{N}}_0}]x^{-n}$$

where

$$\begin{array}{cc}\hfill P[0,r,{\left(c_j\right)}_{j\in {\mathbf{N}}_0}]& =c_0^r,\hfill \\ \hfill P[n,r,{\left(c_j\right)}_{j\in {\mathbf{N}}_0}]& =\frac{1}{n{c}_0}\sum _{k=1}^n[k(1+r)-n]c_{k}P[n-k,r,{\left(c_j\right)}_{j\in {\mathbf{N}}_0}].\hfill \end{array}$$

(8)

Symmetric mean with respect to mean $M_0$ of mean $M_1$ via the group structure $({\mathcal{M}}_{\mathcal{D}},\ast )$ as a consequence of (4) can be expressed as:

$$\begin{array}{cc}\hfill M_2^S& (x-t,x+t)=S_{M_0}\left(M_1\right)(x-t,x+t)\hfill \\ & =\frac{(x-t)(M_1-x+t){(M_0-x-t)}^2-(x+t){(M_0-x+t)}^2(M_1-x-t)}{(M_1-x+t){(M_0-x-t)}^2-{(M_0-x+t)}^2(M_1-x-t)}\hfill \\ & =\frac{(x-t)({\overline{M}}_1+t){({\overline{M}}_0-t)}^2-(x+t){({\overline{M}}_0+t)}^2({\overline{M}}_1-t)}{({\overline{M}}_1+t){({\overline{M}}_0-t)}^2-{({\overline{M}}_0+t)}^2({\overline{M}}_1-t)}\hfill \\ & =x+\frac{2t^2{\overline{M}}_0-t^2{\overline{M}}_1-{\overline{M}}_0^2{\overline{M}}_1}{t^2+{\overline{M}}_0^2-2{\overline{M}}_0{\overline{M}}_1},\hfill \end{array}$$

where ${\overline{M}}_i$, $i=1,2,3$, stands for $M_i-x$. The variables $(x-t,x+t)$ were omitted for the sake of symplicity. Further calculations reveal that:

$$\begin{array}{cc}& M_2^S(x-t,x+t)=x+t^2{x}^{-1}[(2c_1-a_1)\hfill \\ & +\sum _{n=0}^{\infty }\left(2c_{n+2}-a_{n+2}+\sum _{k=0}^n\left(\sum _{j=0}^k\left(c_{j+1}{c}_{k-j+1}\right)a_{n+1-k}\right)\right)t^{2n+2}{x}^{-2n-2}]\hfill \\ & \times {\left[1+\sum _{n=0}^{\infty }\sum _{k=0}^n{c}_{k+1}(c_{n-k+1}-2a_{n-k+1})t^{2n+2}{x}^{-2n-2}\right]}^{-1}.\hfill \end{array}$$

Coefficients $b_n^S$ for $n\ge 1$ are obtained using Lemma 1 for the division of asymptotic series. Hence, we have the following:

$$\begin{array}{cc}\hfill b_0^S& =1,\hfill \\ \hfill b_n^S& =nu{m}_n-\sum _{k=0}^{n-2}de{n}_{n-1-k}{b}_{k+1}^S,n\ge 1,\hfill \end{array}$$

where ${\left(nu{m}_n\right)}_{n\in {\mathbf{N}}_0}$ and ${\left(de{n}_n\right)}_{n\in {\mathbf{N}}_0}$ dentote auxiliary sequences, which appear in the numerator and the denominator:

$$\begin{array}{cc}\hfill nu{m}_0& =2c_1-a_1,\hfill \\ \hfill nu{m}_n& =2c_{n+1}-a_{n+1}+\sum _{k=0}^{n-1}\left(\sum _{j=0}^k\left(c_{j+1}{c}_{k-j+1}\right)a_{n-k}\right),n\ge 1,\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill de{n}_0& =1,\hfill \\ \hfill de{n}_n& =\sum _{k=0}^{n-1}{c}_{k+1}(c_{n-k}-2a_{n-k}),n\ge 1.\hfill \end{array}$$

We shall calculate the first few coefficients:

$$\begin{array}{cc}\hfill b_0^S& =1,\hfill \\ \hfill b_1^S& =2c_1-a_1,\hfill \\ \hfill b_2^S& =2c_2-a_2-2c_1{(a_1-c_1)}^2,\hfill \\ \hfill b_3^S& =2c_3-a_3-2(a_1-c_1)(2a_2{c}_1+c_1^2(2a_1^2-3a_1{c}_1+c_1^2)+(a_1-3c_1)c_2),\hfill \\ \hfill b_4^S& =2c_4-a_4-2(a_2^2{c}_1+4a_1^4{c}_1^3+4a_1^3{c}_1(-3c_1^3+c_2)\hfill \\ \hfill & +2a_2((3a_1-2c_1)(a_1-c_1)c_1^2+(a_1-2c_1)c_2)\hfill \\ \hfill & +a_1^2(13c_1^5-15c_1^2{c}_2+c_3)+2a_1(a_3{c}_1-3c_1^6+8c_1^3{c}_2-c_2^2-2c_1{c}_3)\hfill \\ \hfill & +c_1(-2a_3{c}_1+c_1^6-5c_1^3{c}_2+3c_2^2+3c_1{c}_3)),\hfill \\ \hfill b_5^S& =2c_5-a_5-2(-2a_4{c}_1^2+8a_1^5{c}_1^4+4a_3{c}_1^4-c_1^9-4a_3{c}_1{c}_2+7c_1^6{c}_2-10c_1^3{c}_2^2\hfill \\ \hfill & +c_2^3+a_2^2(6a_1{c}_1^2-5c_1^3+c_2)+4a_1^4{c}_1^2(-7c_1^3+3c_2)-5c_1^4{c}_3+6c_1{c}_2{c}_3\hfill \\ \hfill & +2a_1^3(19c_1^6-24c_1^3{c}_2+c_2^2+2c_1{c}_3)+2a_2(a_3{c}_1+8a_1^3{c}_1^3-3c_1^6+8c_1^3{c}_2\hfill \\ \hfill & -c_2^2+6a_1^2{c}_1(-3c_1^3+c_2)-2c_1{c}_3+a_1(13c_1^5-15c_1^2{c}_2+c_3))+3c_1^2{c}_4\hfill \\ \hfill & +a_1^2(6a_3{c}_1^2-5c_1(5c_1^6-13c_1^3{c}_2+3c_2^2+3c_1{c}_3)+c_4)+2a_1(a_4{c}_1\hfill \\ \hfill & +a_3(-5c_1^3+c_2)+2(2c_1^8-9c_1^5{c}_2+6c_1^2{c}_2^2+4c_1^3{c}_3-c_2{c}_3-c_1{c}_4)\left)\right).\hfill \end{array}$$

2.2. Symmetry ${\sigma }_{M_0}$

The problem of functional symmetic mean corresponds the functional equation

$$M_0(x-t,x+t)=M_0(M_1(x-t,x+t),M_2(x-t,x+t))$$

which we will solve in terms of asymptotic series. To this end, we shall use the following result from Burić and Elezović about the asymptotic expansion of the composition of means.

Theorem 1 

([17], Theorem 2.2.). Let M and N be given homogeneous symmetric means with asymptotic expansions

$$M(x-t,x+t)=\sum _{k=0}^{\infty }{a}_k{t}^{2k}{x}^{-2k+1},N(x-t,x+t)=\sum _{k=0}^{\infty }{b}_k{t}^{2k}{x}^{-2k+1},$$

and let F be homogeneous symmetric mean with expansion

$$F(x-t,x+t)=\sum _{k=0}^{\infty }{\gamma }_k{t}^{2k}{x}^{-2k+1}.$$

Then, the composition $H=F(M,N)$ has asymptotic expansion

$$H(x-t,x+t)=\sum _{k=0}^{\infty }{h}_n{t}^{2n}{x}^{-2n+1},$$

where coefficients $\left(h_n\right)$ are calculated by

$$h_n=\sum _{k=0}^{\lfloor \frac{n}{2z}\rfloor }{\gamma }_k\sum _{j=0}^{n-2zk}P[j,2k,{\left(d_m\right)}_{m\in {\mathbf{N}}_0}]P[n-2zk-j,-2k+1,{\left(c_m\right)}_{m\in {\mathbf{N}}_0}].$$

Sequences $\left(c_n\right)$ and $\left(d_n\right)$ are defined by

$$c_n=\frac{1}{2}(a_n+b_n),d_n=\frac{1}{2}(a_{n+z}-b_{n+z}),n\ge 0,$$

where z is the smallest number such that $d_n\ne 0$.

Applying Theorem 1 on $M=M_1$, $N=M_2$ (or equivalently $M=M_2$, $N=M_1$) and $F=M_0$, we obtain the asymptotic expansion of the composition $M_0(M_1,M_2)$. Since the equation $M_0=M_0(M_1,M_2)$ holds, on the other side, in Theorem 1, we also have $H=M_0$. The coeficients in the asymptotic expansion of the composition $M_0(M_1,M_2)$ equal the coefficients $c_n$ in the asymptotic expansion of mean $M_0$. In the end, we obtain the recursive algorithm for coefficients $c_n$:

$$\begin{array}{cc}\hfill c_0& =1;\hfill \\ \hfill c_n& =\sum _{k=0}^{\lfloor \frac{n}{2z}\rfloor }{c}_k\sum _{j=0}^{n-2zk}P[j,2k,{\left(\frac{1}{2}(a_m-b_m^{\sigma })\right)}_{m\ge z}]P[n-2zk-j,-2k+1,{\left(\frac{1}{2}(a_m+b_m^{\sigma })\right)}_{m\in {\mathbf{N}}_0}],\hfill \end{array}$$

(9)

where $P[n,r,{\left(c_m\right)}_{m\in {\mathbf{N}}_0}]$, $n\in {\mathbf{N}}_0$ denotes the n-th coefficient in the asymptotic expansion of r-th power of the asymptotic seires with coefficients ${\left(c_m\right)}_{m\in {\mathbf{N}}_0}$, as it was defined in (8). Because of (7), z is always greater or equal to 1.

For $z=1$ we calculate the first few coefficients:

$$\begin{array}{cc}\hfill c_0& =1,\hfill \\ \hfill c_1& =\frac{1}{2}(a_1+b_1^{\sigma }),\hfill \\ \hfill c_2& =\frac{1}{2}(a_2+b_2^{\sigma })+\frac{1}{4}{(a_1-b_1^{\sigma })}^2{c}_1,\hfill \\ \hfill c_3& =\frac{1}{2}(a_3+b_3^{\sigma })-\frac{1}{8}(a_1-b_1^{\sigma })(a_1^2-4a_2-{\left(b_1^{\sigma }\right)}^2+4b_2^{\sigma })c_1,\hfill \\ \hfill c_4& =\frac{1}{2}(a_4+b_4^{\sigma })+\frac{1}{16}\left(\right(a_1^4+4a_2^2-8a_3{b}_1^{\sigma }+{\left(b_1^{\sigma }\right)}^4+2a_2({\left(b_1^{\sigma }\right)}^2-4b_2^{\sigma })\hfill \\ \hfill & -2a_1^2(3a_2+{\left(b_1^{\sigma }\right)}^2-b_2^{\sigma })-6{\left(b_1^{\sigma }\right)}^2{b}_2^{\sigma }+4{\left(b_2^{\sigma }\right)}^2\hfill \\ \hfill & +4a_1(2a_3+b_1^{\sigma }(a_2+b_2^{\sigma })-2b_3^{\sigma })+8b_1^{\sigma }{b}_3^{\sigma })c_1+{(a_1-b_1^{\sigma })}^4{c}_2),\hfill \\ \hfill c_5& =\frac{1}{2}(a_5+b_5^{\sigma })-\frac{1}{32}\left(\right(a_1^5+a_1^4{b}_1^{\sigma }-4a_2^2{b}_1^{\sigma }+16a_4{b}_1^{\sigma }-4a_3{\left(b_1^{\sigma }\right)}^2+{\left(b_1^{\sigma }\right)}^5\hfill \\ \hfill & -2a_1^3(4a_2+{\left(b_1^{\sigma }\right)}^2)+16a_3{b}_2^{\sigma }-8{\left(b_1^{\sigma }\right)}^3{b}_2^{\sigma }+12b_1^{\sigma }{\left(b_2^{\sigma }\right)}^2\hfill \\ \hfill & -8a_2(2a_3+b_1^{\sigma }{b}_2^{\sigma }-2b_3^{\sigma })+2a_1^2(6a_3-{\left(b_1^{\sigma }\right)}^3+4b_1^{\sigma }{b}_2^{\sigma }-2b_3^{\sigma })+12{\left(b_1^{\sigma }\right)}^2{b}_3^{\sigma }\hfill \\ \hfill & -16b_2^{\sigma }{b}_3^{\sigma }-16b_1^{\sigma }{b}_4^{\sigma }+a_1(12a_2^2-16a_4-8a_3{b}_1^{\sigma }+{\left(b_1^{\sigma }\right)}^4+8a_2({\left(b_1^{\sigma }\right)}^2-b_2^{\sigma })\hfill \\ \hfill & -4{\left(b_2^{\sigma }\right)}^2-8b_1^{\sigma }{b}_3^{\sigma }+16b_4^{\sigma }\left)\right)c_1-{(a_1-b_1^{\sigma })}^3(3a_1^2-8a_2-3{\left(b_1^{\sigma }\right)}^2+8b_2^{\sigma })c_2).\hfill \end{array}$$

The connetcion between $b_n^{\sigma }$ and $c_n$ with the highest index n in each equation is linear. In the expression (9), $b_n^{\sigma }$ appears ony in the second part

$$P[n-2zk-j,-2k+1,{\left(\frac{1}{2}(a_m+b_m^{\sigma })\right)}_{m\in {\mathbf{N}}_0}],$$

(10)

when $k=j=0$. Then, (10) becomes $P[n,1,{\left(\frac{1}{2}(a_m+b_m^{\sigma })\right)}_{m\in {\mathbf{N}}_0}]$, which represents the n-th coefficient in the ${\sum }_{n=0}^{\infty }\frac{1}{2}(a_n+b_n^{\sigma })t^{2n}{x}^{-2n+1}$ to the power of 1, which equals $\frac{1}{2}(a_n+b_n^{\sigma })$. So, we can easily extract $b_n^{\sigma }$. The first few coefficients $b_n^{\sigma }$ are:

$$\begin{array}{cc}\hfill b_0^{\sigma }& =1,\hfill \\ \hfill b_1^{\sigma }& =2c_1-a_1,\hfill \\ \hfill b_2^{\sigma }& =2c_2-a_2-\frac{1}{2}{c}_1(a_1-b_1^{\sigma }),\hfill \\ \hfill b_3^{\sigma }& =2c_3-a_3+\frac{1}{4}(a_1-b_1^{\sigma })(a_1^2-4a_2-{\left(b_1^{\sigma }\right)}^2+4b_2^{\sigma })c_1,\hfill \\ \hfill b_4^{\sigma }& =2c_4-a_4-\frac{1}{8}\left(\right(a_1^4+4a_2^2-8a_3{b}_1^{\sigma }+{\left(b_1^{\sigma }\right)}^4+2a_2({\left(b_1^{\sigma }\right)}^2-4b_2^{\sigma })\hfill \\ \hfill & -2a_1^2(3a_2+{\left(b_1^{\sigma }\right)}^2-b_2^{\sigma })-6{\left(b_1^{\sigma }\right)}^2{b}_2^{\sigma }+4{\left(b_2^{\sigma }\right)}^2\hfill \\ \hfill & +4a_1(2a_3+b_1^{\sigma }(a_2+b_2^{\sigma })-2b_3^{\sigma })+8b_1^{\sigma }{b}_3^{\sigma })c_1+{(a_1-b_1^{\sigma })}^4{c}_2),\hfill \\ \hfill b_5^{\sigma }& =2c_5-a_5+\frac{1}{16}\left(\right(a_1^5+a_1^4{b}_1^{\sigma }-4a_2^2{b}_1^{\sigma }+16a_4{b}_1^{\sigma }-4a_3{\left(b_1^{\sigma }\right)}^2+{\left(b_1^{\sigma }\right)}^5\hfill \\ \hfill & -2a_1^3(4a_2+{\left(b_1^{\sigma }\right)}^2)+16a_3{b}_2^{\sigma }-8{\left(b_1^{\sigma }\right)}^3{b}_2^{\sigma }+12b_1^{\sigma }{\left(b_2^{\sigma }\right)}^2\hfill \\ \hfill & -8a_2(2a_3+b_1^{\sigma }{b}_2^{\sigma }-2b_3^{\sigma })+2a_1^2(6a_3-{\left(b_1^{\sigma }\right)}^3+4b_1^{\sigma }{b}_2^{\sigma }-2b_3^{\sigma })+12{\left(b_1^{\sigma }\right)}^2{b}_3^{\sigma }\hfill \\ \hfill & -16b_2^{\sigma }{b}_3^{\sigma }-16b_1^{\sigma }{b}_4^{\sigma }+a_1(12a_2^2-16a_4-8a_3{b}_1^{\sigma }+{\left(b_1^{\sigma }\right)}^4+8a_2({\left(b_1^{\sigma }\right)}^2-b_2^{\sigma })\hfill \\ \hfill & -4{\left(b_2^{\sigma }\right)}^2-8b_1^{\sigma }{b}_3^{\sigma }+16b_4^{\sigma }\left)\right)c_1-{(a_1-b_1^{\sigma })}^3(3a_1^2-8a_2-3{\left(b_1^{\sigma }\right)}^2+8b_2^{\sigma })c_2).\hfill \end{array}$$

For beter understanding the role of the parameter z, we shall recall the idea behind the proof of Theorem 1. The composition $F(M,N)$ has the asymptotic expansion

$$\begin{array}{cc}\hfill F\left(M\right(x-t,x+t)& ,N(x-t,x+t))\hfill \\ & =F\left(\frac{M+N}{2}-\frac{N-M}{2},\frac{M+N}{2}+\frac{N-M}{2}\right)\hfill \\ & =\sum _{k=0}^{\infty }{\gamma }_k{\left(\frac{N-M}{2}\right)}^{2k}{\left(\frac{M+N}{2}\right)}^{-2k+1}.\hfill \end{array}$$

Larger z corresponds with the equating $a_i$ and $b_i^{\sigma }$ and some parts of the coefficients $c_n$ reduce. Observation of the cases with $z>1$ in sequel did not provide any new information about the ceofficients $c_n$.

2.3. Comparison of Coefficients

Sequences ${\left(b_n^S\right)}_{n\in {\mathbf{N}}_0}$ and ${\left(b_n^{\sigma }\right)}_{n\in {\mathbf{N}}_0}$ represent the coefficients in asymptotic expansions of means, which are results of mappings $S_{M_0}\left(M_1\right)$ and ${\sigma }_{M_0}\left(M_1\right)$, respectively. Since we are looking for a mean $M_0$ such those mappings coincide, $b_n^S$ and $b_n^{\sigma }$ need to be equal. Since the equality must hold for any mean $M_1$, we may suppose that $z=1$, which is equivalent with $a_1\ne c_1$. Equating $b_0^S$ with $b_0^{\sigma }$ and $b_1^S$ with $b_1^{\sigma }$ does not provide any new information, except

$$b_0=b_0^S=b_0^{\sigma }=1\mathrm{and}{b}_1=b_1^S=b_1^{\sigma }=2c_1-a_1.$$

With such $b_1^{\sigma }$ we may express $b_2^{\sigma }$ as

$$b_2^{\sigma }=2c_2-a_2-2c_1{(a_1-c_1)}^2,$$

which is already equal to $b_2^S$. Now, we can substitute

$$b_2=b_2^S=b_2^{\sigma }=2c_2-a_2-2c_1{(a_1-c_1)}^2,$$

in $b_3^{\sigma }$ to obtain

$$b_3^{\sigma }=2c_3-a_3-2c_1(a_1-c_1)(2a_2+2c_1{(a_1-c_1)}^2+c_1^2-a_1{c}_1-2c_2),$$

which, after equating with $b_3^S$, gives the following condition

$${(a_1-c_1)}^2(c_1^2+c_1^3+c_2)=0.$$

Since we assumed that $a_1$ and $c_1$ are not equal, it is necessarily

$$c_2=-c_1^2(1+c_1).$$

Now, we have

$$b_3=b_3^S=b_3^{\sigma }=2c_3-a_3-2c_1(a_1-c_1)\left((3-4a_1)c_1^2+a_1(2a_1-1)c_1+2a_2+4c_1^3\right).$$

After substitutions, we observe the next coefficient

$$\begin{array}{cc}\hfill b_4^{\sigma }& =2c_4-a_4-2c_1(2a_2{c}_1\left(-a_1(6c_1+1)+3a_1^2+2c_1(2c_1+1)\right)\hfill \\ & +c_1\left(c_1\right(-4a_1^3(4c_1+1)+a_1^2(2c_1+1)(15c_1+2)-2a_1{c}_1(14c_1(c_1+1)+3)\hfill \\ & +4a_1^4+c_1^2(c_1(11c_1+15)+5))-2a_3)+2c_3(c_1-a_1)+a_2^2+2a_1{a}_3)\hfill \end{array}$$

which, after equating with $b_4^S$, gives the following condition:

$${(a_1-c_1)}^2\left(2c_1^3{(c_1+1)}^2-c_3\right)=0,$$

and we conclude that it must be

$$c_3=2c_1^3{(1+c_1)}^2.$$

We continue with this procedure as it was described above. Further calculations reveal that the first few coefficients $c_n$ have the following form:

$$\begin{array}{cc}\hfill c_0& =1,\hfill \\ \hfill c_1& =c,\hfill \\ \hfill c_2& =-c^2(1+c),\hfill \\ \hfill c_3& =2c^3{(1+c)}^2,\hfill \\ \hfill c_4& =-5c^4{(1+c)}^3,\hfill \\ \hfill c_5& =14c^5{(1+c)}^4,\hfill \\ \hfill c_6& =-42c^6{(1+c)}^5.\hfill \end{array}$$

After these first steps, it is natural to state the following hypothesis about the general formula for the coefficients in the asymptotic expansion of mean $M_0$:

$$\begin{array}{cc}\hfill c_0& =1,\hfill \\ \hfill c_n& ={(-1)}^{n-1}{C}_{n-1}{c}^n{(1+c)}^{n-1},n\ge 1,\hfill \end{array}$$

(11)

where $C_n$ denotes the n-th Catalan number. Catalan numbers appear in many occasions, and their behavior has been widely explored. Here, we mention only a few properties, which we will use in sequel. Catalan numbers are defined by

$$C_n=\frac{1}{n+1}\left(\genfrac{}{}{0pt}{}{2n}{n}\right),n\in {\mathbf{N}}_0$$

and they satisfy the recursive relation

$$C_{n+1}=\sum _{k=0}^n{C}_k{C}_{n-k},n\in {\mathbf{N}}_0.$$

Based on this recursive relation, the generating function for Catalan numbers can be obtained ([18]):

$$\sum _{n=0}^{\infty }{C}_n{y}^n=\frac{1-\sqrt{1-4y}}{2y},$$

(12)

which is convergent for $\left|y\right|<\frac{1}{4}$.

3. New Mean Function

In this section, we shall find closed a form for a mean whose coefficients are given in (11). We start from asymptotic expansion (6):

$$\begin{array}{cc}\hfill M_0(x-t,x+t)& =x+\sum _{n=1}^{\infty }{(-1)}^{n-1}{C}_{n-1}{c}^n{(1+c)}^{n-1}{t}^{2n}{x}^{-2n+1}\hfill \\ \hfill & =x+\sum _{n=0}^{\infty }{(-1)}^n{C}_n{c}^{n+1}{(1+c)}^n{t}^{2n+2}{x}^{-2n-1}\hfill \\ \hfill & =x+c{t}^2{x}^{-1}\sum _{n=0}^{\infty }{C}_n{\left[-\frac{c(1+c)t^2}{x^2}\right]}^n.\hfill \end{array}$$

(13)

Introducing the substitution $y=-\frac{c(1+c)t^2}{x^2}$, as $x\to \infty$ and thereby $y\to 0$, yields

$$\begin{array}{cc}\hfill M_0(x-t,x+t)& =x+c{t}^2{x}^{-1}\sum _{n=0}^{\infty }{C}_n{y}^n,\hfill \end{array}$$

and, then, according to the Formula (12), for $c+1\ne 0$, we obtain

$$\begin{array}{cc}\hfill M_0(x-t,x+t)& =x+c{t}^2{x}^{-1}\frac{1-\sqrt{1-4y}}{2y}\hfill \\ \hfill & =\frac{1+2c}{2(1+c)}x+\frac{1}{2(1+c)}\sqrt{x^2+4c(1+c)t^2}.\hfill \end{array}$$

(14)

Abandoning series expansion in this moment, from the Equation (14) with substitution

$$x=\frac{a+b}{2},t=\frac{b-a}{2},$$

we obtain the expression for $M_0$ in terms of variables a and b. For $c\in \mathbf{R}\setminus \{-1\}$ and $a,b>0$ we define function $L_c:{\mathbf{R}}^{+}\times {\mathbf{R}}^{+}\to {\mathbf{R}}^{+}$

$$L_c(a,b)=\frac{a+b}{2}\frac{1+2c}{2(1+c)}+\frac{1}{2(1+c)}\sqrt{{\left(\frac{a+b}{2}\right)}^2+4c(1+c){\left(\frac{b-a}{2}\right)}^2}.$$

(15)

Remark 1. 

Function $L_c$ is well defined for all $(a,b)\in {\mathbf{R}}^{+}\times {\mathbf{R}}^{+}$ as we can rearrange terms under the square root:

$$\begin{array}{cc}\hfill {\left(\frac{a+b}{2}\right)}^2+4c(1+c){\left(\frac{b-a}{2}\right)}^2& =\frac{1}{4}\left({\left(a+b\right)}^2+4c(1+c){\left(b-a\right)}^2\right)\hfill \\ \hfill & =\frac{1}{4}\left({(1+2c)}^2{(a-b)}^2+4ab\right)>0.\hfill \end{array}$$

Remark 2. 

For $c=-1$ function $L_c$ corresponds to the harmonic mean which will be proved in sequel. Therefore, definition (15) can be considered for all $c\in \mathbf{R}$.

Remark 3. 

Formula for $L_c$ can also be written in a following way:

$$L_c(a,b)=A(a,b)\frac{1+2c}{2(1+c)}+\frac{1}{2(1+c)}\sqrt{A{(a,b)}^2+4c(1+c)\left(A{(a,b)}^2-G{(a,b)}^2\right)}.$$

(16)

3.1. Limit Cases and Monotonicity

In this subsection, we study properties of $L_c$ with respect to parameter c. First, we state the following proposition, which can be proved using basic methods of mathematical analysis.

Proposition 1. 

For a fixed pair $(a,b)\in {\mathbf{R}}^{+}\times {\mathbf{R}}^{+}$, function $L_c$ holds

1.

${\displaystyle \underset{c\to -\infty }{lim}{L}_c(a,b)=min(a,b)}$,

2.

${\displaystyle \underset{c\to -1-}{lim}{L}_c(a,b)=\underset{c\to -1+}{lim}{L}_c(a,b)=\frac{2ab}{a+b}=H(a,b)}$

3.

${\displaystyle \underset{c\to +\infty }{lim}{L}_c(a,b)=max(a,b)}$,

It is well known that the following double inequality holds

$$H<A<G.$$

Also, $H=L_c$ for $c\to -1$, $G=L_c$ for $c=-\frac{1}{2}$, and $A=L_c$ for $c=0$. In the next Theorem, we explore the ordering of means $L_c$ with respect to parameter c.

Theorem 2. 

For a fixed pair $(a,b)\in {\mathbf{R}}^{+}\times {\mathbf{R}}^{+}$, $a\ne b$, function $f:\mathbf{R}\setminus \{-1\}\to \mathbf{R}$,

$$f\left(c\right)=L_c(a,b)$$

is strictly increasing.

Proof. 

Starting form the (16), with $A=A(a,b)$ and $G=G(a,b)$, we have

$$f\left(c\right)=A\frac{1+2c}{2(1+c)}+\frac{1}{2(1+c)}\sqrt{g\left(c\right)},$$

where

$$g\left(c\right)=A^2+4c(1+c)\left(A^2-G^2\right)>0$$

according to Remarks 1 and 3. The first derivative of function f equals

$$\begin{array}{cc}\hfill f^{\prime }\left(c\right)& =\frac{1}{2{(1+c)}^2}\left(A+2(1+2c)(1+c)(A^2-G^2)g{\left(c\right)}^{-\frac{1}{2}}-g{\left(c\right)}^{\frac{1}{2}}\right)\hfill \\ \hfill & =\frac{1}{2{(1+c)}^{2}g{\left(c\right)}^{\frac{1}{2}}}\left(Ag{\left(c\right)}^{\frac{1}{2}}+2(1+2c)(1+c)(A^2-G^2)-g\left(c\right)\right)\hfill \\ \hfill & =\frac{1}{2{(1+c)}^{2}g{\left(c\right)}^{\frac{1}{2}}}\left(Ag{\left(c\right)}^{\frac{1}{2}}+2(1+c)(A^2-G^2)-A^2\right).\hfill \end{array}$$

So, the condition $f^{\prime }\left(c\right)>0$ is equivalent to

$$Ag{\left(c\right)}^{\frac{1}{2}}>A^2-2(1+c)(A^2-G^2).$$

If the right-hand side is negative, than the inequality obviusly holds. If it is positive, then we may observe the squared inequality:

$$A^{2}g\left(c\right)>A^4-4(1+c)(A^2-G^2)A^2+4{(1+c)}^2(A^2-G^2)$$

which is equivalent to

$$4c(1+c)A^2(A^2-G^2)>-4(1+c)(A^2-G^2)A^2+4{(1+c)}^2(A^2-G^2)$$

and

$$(A^2-G^2){(1+c)}^2{G}^2>0$$

which is true for $a\ne b$ and $c\ne 1$. □

Since $L_c$ assumes values between minimum and maximum of a and b, we may conclude the following.

Corollary 1. 

For $c\in \mathbf{R}$ funcion $L_c$ is a mean.

Remark 4. 

Notice that we proved that $L_c$ is a strict mean, i.e., for $a\ne b$, strict inequalities hold:

$$min(a,b)<M(a,b)<max(a,b).$$

3.2. Special Cases

Before we continue further, let us see what happens with some of the special cases of parameter c. We shall also connect results form this paper with the previously obtained asymptotic expansions of classical means.

Example 3. 

(a) 

$c=-1$. Then mean has two non-zero coefficients:

$$c_0=1,c_1=c,c_n=0,n\ge 2.$$

Corresponding asymptotic expansion is finite. From (13), we obtain

$$L_c(x-t,x+t)=x-t^2{x}^{-1},$$

which, after substitution $x=\frac{a+b}{2},t=\frac{b-a}{2}$ becomes

$$L_c(a,b)=\frac{a+b}{2}-\frac{{(b-a)}^2}{4}·\frac{2}{a+b}=\frac{2ab}{a+b}=H(a,b).$$

(b) 

$c=0$. All coefficients except $c_0$ equal zero. Then, either from the (13) or (14), we obtain

$$L_c(x-t,x+t)=x,$$

and after the substitution

$$L_c(a,b)=\frac{a+b}{2}=A(a,b).$$

(c) 

$c=-\frac{1}{2}$. The coefficients are

$$c_0=1,c_n=-\frac{1}{2^{2n-1}}{C}_{n-1},n\ge 1.$$

(17)

Coefficients (17) correspond to the coefficients in asymptotic expansion of geometric mean obtained in [9] for $\alpha =0$ and $\beta =t$, and also to coefficients of power mean $M_p$ with $p=0$ obtained in [10]. On the other side, from the Formula (14), we obtain

$$L_c(x-t,x+t)=\sqrt{x^2-t^2},$$

and, after substitution

$$L_c(a,b)=\sqrt{ab}=G(a,b).$$

From the example above, we see that we covered the cases of means for which in [1] was stated that symmetries S and $\sigma$ coincide.

4. Answer to the Open Question

Theorem 3. 

For mean $L_c$, $c\in \mathbf{R}$, defined in (15), symmetries $S_{L_c}$ and ${\sigma }_{L_c}$ coincide.

Proof. 

Let us rewrite mean $L_c$ in the following manner:

$$L_c(a,b)=\frac{1}{4(1+c)}\left[(1+2c)(a+b)+\sqrt{{(a+b)}^2+4c(1+c){(b-a)}^2}\right].$$

For $M_0=L_c$ and variable mean $M_1=M$, there exists symmetric mean $\sigma ={\sigma }_{L_c}\left(M\right)$, i.e., the condition $L_c(M,\sigma )=L_c$ holds, which yields (for the sake of brevity, the variables will be ommited):

$$\frac{1}{4(1+c)}\left[(1+2c)(M+\sigma )+\sqrt{{(M+\sigma )}^2+4c(1+c){(M-\sigma )}^2}\right]=L_c,$$

or equivalently

$$\sqrt{{(M+\sigma )}^2+4c(1+c){(M-\sigma )}^2}=4(1+c)L_c-(1+2c)(M+\sigma ).$$

We rearrange the terms and, because of the existence of mean $\sigma ={\sigma }_{L_c}\left(M\right)$, we may square the latter expression:

$$\begin{array}{cc}\hfill M^2& {(1+2c)}^2+2M\sigma (1-4c-4c^2)+{\sigma }^2{(1+2c)}^2\hfill \\ \hfill & ={\left[4(1+c)L_c-(1+2c)M\right]}^2-2{\left[4(1+c)L_c-(1+2c)M\right]}^2+{\sigma }^2{(1-2c)}^2.\hfill \end{array}$$

The terms ${\sigma }^2{(1-2c)}^2$ cancel from both sides. Further calculation gives

$$\begin{array}{cc}\hfill 2M(1-4c-4c^2)& \sigma +2\left(4(1+c)L_c-(1+2c)M\right)(1+2c)\sigma \hfill \\ \hfill & =-M^2{(1+2c)}^2+{\left(4(1+c)L_c-(1+2c)M\right)}^2,\hfill \end{array}$$

and finally

$$\sigma =\frac{L_c\left((1+2c)M-2(1+c)L_c\right)}{2cM-(1+2c)L_c}.$$

(18)

Thus, we obtained the explicit expression for mean $\sigma ={\sigma }_{L_c}\left(M\right)$ in terms of M and $L_c$.

On the other side, from (4), we know that

$$S_{L_c}\left(M\right)=\frac{a(M-a){(L_c-b)}^2-b{(L_c-a)}^2(M-b)}{(M-a){(L_c-b)}^2-{(L_c-a)}^2(M-b)},$$

which may be written as

$$S_{L_c}\left(M\right)=\frac{K_{1}M-K_2}{K_{0}M-K_1},$$

(19)

where

$$\begin{array}{cc}\hfill K_0& ={(L_c-b)}^2-{(L_c-a)}^2,\hfill \\ \hfill K_1& =a{(L_c-b)}^2-b{(L_c-a)}^2,\hfill \\ \hfill K_2& =a^2{(L_c-b)}^2-b^2{(L_c-a)}^2.\hfill \end{array}$$

By equating the results of mappings $\sigma$ and S with respect to mean $L_c$ of a mean M and employing Formulas (18) and (19), we obtain

$$\frac{L_c\left((1+2c)M-2(1+c)L_c\right)}{2cM-(1+2c)L_c}=\frac{K_{1}M-K_2}{K_{0}M-K_1}$$

which needs to be proved. We calculate

$$L_c\left[2(1+c)L_c-(1+2c)M\right](K_{0}M-K_1)=\left[(1+2c)L_c-2cM\right](K_{1}M-K_2).$$

Grouping by the powers of M yields

$$\begin{array}{cc}\hfill \left[M_0(1+2c)K_0-2c{K}_1\right]M^2& +2\left[K_{2}c-(1+c)L_c^2{K}_0\right]M\hfill \\ \hfill & +L_c\left[2(1+c)L_c{K}_1-(1+2c)K_2\right]=0.\hfill \end{array}$$

(20)

Now, we simplify each coefficient by the powers of M. First,

$$\begin{array}{cc}\hfill M_0(1+2c)& K_0-2c{K}_1\hfill \\ \hfill & =M_0(1+2c)\left[{(L_c-b)}^2-{(L_c-a)}^2\right]-2c\left[a{(L_c-b)}^2-b{(L_c-a)}^2\right]\hfill \\ \hfill & =(a-b)\left[2(1+c)L_c^2-(a+b)(1+2c)L_c+2abc\right],\hfill \end{array}$$

second,

$$\begin{array}{cc}\hfill c{K}_2& -(1+c)L_c^2{K}_0\hfill \\ \hfill & =c\left[a^2{(L_c-b)}^2-b^2{(L_c-a)}^2\right]-(1+c)L_c^2\left[{(L_c-b)}^2-{(L_c-a)}^2\right]\hfill \\ \hfill & =-(a-b)L_c\left[2(1+c)L_c^2-(a+b)(1+2c)L_c+2abc\right],\hfill \end{array}$$

and third

$$\begin{array}{cc}\hfill 2& (1+c)L_c{K}_1-(1+2c)K_2\hfill \\ \hfill & =2(1+c)L_c\left[a{(L_c-b)}^2-b{(L_c-a)}^2\right]-(1+2c)\left[a^2{(L_c-b)}^2-b^2{(L_c-a)}^2\right]\hfill \\ \hfill & =(a-b)L_c\left[2(1+c)L_c^2-(a+b)(1+2c)L_c+2abc\right].\hfill \end{array}$$

Hence, the equation (20) factorizes as

$$(a-b)\left[2(1+c)L_c^2-(a+b)(1+2c)L_c+2abc\right]\left(M^2-2L_{c}M+L_c^2\right)=0.$$

(21)

Notice that the mean $L_c$ defined in (15) is one of the solutions of quadratic equation

$$2(1+c)L_c^2-(a+b)(1+2c)L_c+2abc,$$

and the condition (21) is fulfilled, which proves the theorem. □

We will close this section with a conjecture. Based on the analysis in this paper, we may conclude the following.

Hypothesis 1. 

Symmetric homogeneous mean, which has the asymptotic power series expansion and fulfills the requirements of the open question from [1] necessarily has the same coefficients as mean $L_c$, $c\in \mathbf{R}$.

5. Concluding Remarks

Using techniques of asymptotic expansions, we were able to compare two symmetries of different origins on the set of mean functions. Finding asymptotic series expansion for both of them, in terms of recursive algorithm for their coefficients, enabled us to carry out the coefficient comparison, which resulted in obtaining a class of means, which interpolates between harmonic, geometric, and arithmetic mean.

Methods presented in this paper may be useful with various problems regarding bivariate means and further. For example, in case of dual means, generalized inverses of means and similar problems where some functional connection is given, and especially when the explicit formula for some of the means involved, was not known.