Elementary Symmetric Means as Quasi-Arithmetic Means. The elementary symmetric polynomial of degree k in n variables is e_{n,k}(x)=sum_{1 <= i_1 <= i_2 <= cdots <= i_k <= n}x_{i_1}x_{i_2} cdots x_{i_k}

Hunter Shah

Hunter Shah

Answered question

2022-10-27

Elementary Symmetric Means as Quasi-Arithmetic Means
The elementary symmetric polynomial of degree k in n variables is
e n , k ( x ) = 1 i 1 i 2 i k n x i 1 x i 2 x i k
and for positive data x, the corresponding elementary symmetric mean is
s n , k ( x ) := e n , k ( x ) ( n k ) k .
Quasi-arithmetic means are defined in terms of an invertible function f, and the arithmetic mean A
M f ( x ) := ( f 1 A f ) ( x ) = f 1 ( f ( x 1 ) + f ( x 2 ) + + f ( x n ) n ) .
For any n, the elementary symmetric mean in n variables of degree n is the geometric mean, which is a quasi-arithmetic mean:
s n , n ( x 1 , , x n ) = x 1 x 2 x n n = exp ( log ( x 1 ) + log ( x 2 ) + + log ( x n ) n ) .
My question is: are the other elementary symmetric means (with k < n) also quasi-arithmetic? If so, can the conjugating functions f n , k be described explicitly?
My attempt: Obviously s n , 1 = A is quasi-arithmetic, so the first nontrivial case is s 3 , 2 , with the functional equation
f 3 , 2 s 3 , 2 = A f 3 , 2 ,
or
f 3 , 2 ( x 1 x 2 + x 1 x 3 + x 2 x 3 3 ) = f 3 , 2 ( x 1 ) + f 3 , 2 ( x 2 ) + f 3 , 2 ( x 3 ) 3 .
I tried pre-composing f with a logarithm and differentiation, but it didn't seem to get me anywhere.
I would greatly appreciate any help on this.

Answer & Explanation

Davin Meyer

Davin Meyer

Beginner2022-10-28Added 13 answers

Step 1
If you require that f is, say, C1, then differentiating Mf with respect to any xi gives
x i M f = 1 f ( f ( x i ) n ) f ( x i ) n
which means that for any i j we have
x i M f x j M f = f ( x i ) f ( x j ) .
So we can check whether s 3 , 2 has this property. Actually it will be slightly more convenient for the purposes of calculating derivatives to check this property after conjugating s 3 , 2 by f ( x ) = x 2 to remove the outer square root, giving a modified mean
t 3 , 2 ( x 1 , x 2 , x 3 ) = x 1 x 2 + x 2 x 3 + x 3 x 1 3 .
Step 2
(Conjugating preserves the property of being quasi-arithmetic so this is fine.) We get
x 1 t 3 , 2 = x 2 + x 3 6 x 1
and similarly for x 2 , x 3 , which gives
x 1 t 3 , 2 x 2 t 3 , 2 = ( x 2 + x 3 ) x 2 ( x 1 + x 3 ) x 1 .
In particular, this quotient depends nontrivially on x 3 , so it's not of the form f ( x 1 ) f ( x 2 ) for any function f. A similar but more annoying calculation can be done for the other elementary symmetric means. Intuitively this is saying that the elementary symmetric means "mix" the x i too much to be quasi-arithmetic.

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