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We do have elementary symmetric functions, SymmetricPolynomial[k, {x_1, ..., x_n}] .

But I didn't find complete homogeneous symmetric functions.

The induction method to compute $h_n$ from $e_i$ and $h_j$ ($j\leq n-1$) is not that efficient.

Is there any easier way to do this?

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3 Answers 3

up vote 7 down vote accepted

For example:

completeSymmetricPolynomial[i_?IntegerQ, vars_?ListQ] :=
      Total@Union@Tuples[Times @@ vars, {i}];

completeSymmetricPolynomial[2, {a, b, c, d}]

(* a^2 + a b + b^2 + a c + b c + c^2 + a d + b d + c d + d^2 *)


You can verify the fundamental relationship between complete and incomplete symmetric polynomials:

$$\sum_{i=0}^m (-1)^i e_i(X_1,\dots,X_n)h_{m-i}(X_1,\dots,X_n)=0$$

 Table[Sum[(-1)^i completeSymmetricPolynomial[i, {a, b, c, d}]     
                          SymmetricPolynomial[m - i, {a, b, c, d}], 
      {i, 0, m}], {m, 1, 4}]

(* -> {0, 0, 0, 0} *)
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I think you can replace Times[Sequence @@ vars] with Times @@ vars. –  Mr.Wizard Jun 9 '12 at 15:21
...and the Plus @@ stuff is more compactly written as Total[stuff]. –  J. M. Jun 9 '12 at 15:33
Thanks, edits done –  belisarius Jun 9 '12 at 16:08
The fundamental relationship belisarius is using to check his implementation is essentially the second method in my answer. –  J. M. Jun 9 '12 at 17:11
The only little issue is 0-th order is not well defined, i.e. completeSymmetricPolynomial[0, {a, b, c, d}] –  Osiris Xu Jun 12 '12 at 21:44

What I'd do, based on the generating function identity in your Wikipedia link:

completeSymmetricPolynomial[k_Integer, vars_List] := 
   Apply[Times, 1/(1 - vars \[FormalT])], {\[FormalT], 0, k}] /; 
  0 <= k <= Length[vars]

This is somewhat slower, but I want to demonstrate that the induction approach can be made to work as well (and is easily modified if you want all the $n$-variable polynomials all at once, as opposed to just one):

completeSymmetricPolynomial[k_Integer, vars_List] := 
     Table[(-1)^\[FormalK] SymmetricPolynomial[\[FormalK], vars],
           {\[FormalK], 0, Length[vars] - 1}], 
     UnitVector[Length[vars], 1]],
     -Table[(-1)^\[FormalK] SymmetricPolynomial[\[FormalK], vars],
            {\[FormalK], Length[vars]}]][[k]]]
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:) Yes, this works too. –  Osiris Xu Jun 9 '12 at 21:22

This variant seems competitive in terms of speed.

completeSymmetricPolynomial2[i_?IntegerQ, vars_?ListQ] :=
  Expand[(Total@vars)^i]/. aa_Integer*bb_ :> bb
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cool. Thanks. :) –  Osiris Xu Jun 9 '12 at 21:22

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