Taylor expansion in elementary symmetric polynomials Assume that I have an analytic, symmetric function $F(x,y,z)$. Such a function should have an expansion of the form
$$  F(x,y,z)  = \sum_{k,m,n\geq 0} c_{k,m,n} e_1^ne_2^m e_3^k$$
where $e_1,e_2,e_3$ are the elementray symmetric polynomials. How would I go about calculating these $c_{m,n,k}$ in an efficient way (by efficient I mean something better than rewriting the polynomials $$ P_{n_1,n_2,n_3}(x_1,x_2,x_3) = \sum_{\sigma \in S_3} x^{n_{\sigma(1)}}y^{n_{\sigma(2)}}z^{n_{\sigma(3)}}$$
 in terms of elementary polynomials by hand...). 
More concretely I would be interested in knowing whether the function
$$ G(x_1,x_2,x_3)  =  \sum_{\sigma \in S_3} \operatorname{exp}\Big[ Ax_{\sigma(1)}+Bx_{\sigma(2)}+Cx_{\sigma(3)}\Big] $$ 
can be rewritten as a function in elementary symmetric polynomials in a nice form.
 A: Elementary polynomial for the elementary symmetric function
$$f\left( t \right) = {t^3} - {e_1}{t^2} + {e_2}t - {e_3} = \left( {t - x} \right)\left( {t - y} \right)\left( {t - z} \right)$$
Symmetric function
$$F\left( {x,\;y,\;z} \right) = \sum\limits_{{\alpha _1} + 2{\alpha _2} + 3{\alpha _3} = {\beta _x} + {\beta _y} + {\beta _z} = 0}^\infty  {K\left( {{\alpha _1},\;{\alpha _2},\;{\alpha _3},\;{\beta _x},\;{\beta _y},\;{\beta _z}} \right)\frac{{e_1^{{\alpha _1}}e_2^{{\alpha _2}}e_3^{{\alpha _3}}}}{{{\beta _x}!{\beta _y}!{\beta _z}!}}\frac{{{\partial ^{{\beta _x} + {\beta _y} + {\beta _z}}}F}}{{\partial {x^{{\beta _x}}}\partial {y^{{\beta _y}}}\partial {z^{{\beta _z}}}}}(0,0,0)} $$
Coefficient $K$ has a generating function $J$;
$$K\left( {{\alpha _1},\;{\alpha _2},\;{\alpha _3},\;{\beta _x},\;{\beta _y},\;{\beta _z}} \right) = \mathop {\lim }\limits_{\left( {p,\,q,\,x,\,y,\,z} \right) \to \left( {0,\;0,\;0,\;0,0} \right)} \frac{{{{\left( { - 1} \right)}^{{\beta _x} + {\beta _y} + {\beta _z}}}}}{{{\alpha _2}!{\alpha _3}!{\beta _x}!\left( {{\beta _x} + {\beta _y}} \right)!\left( {{\beta _x} + {\beta _y} + {\beta _z}} \right)!}}\frac{{{\partial ^{{\alpha _2} + {\alpha _3} + 3{\beta _x} + 2{\beta _y} + {\beta _z}}}}}{{\partial {p^{{\alpha _2}}}\partial {q^{{\alpha _3}}}\partial {x^{{\beta _x}}}\partial {y^{{\beta _x} + {\beta _y}}}\partial {z^{{\beta _x} + {\beta _y} + {\beta _z}}}}}J_{{\beta _x},{\beta _y}}\left( {p,\,q,\,x,\,y,\,z} \right)$$
$$J_{u,v}\left( {p,\,q,\,x,\,y,\,z} \right) = \frac{{{y^{u + 1}}{z^{u + v + 2}}\left( {y + z + yz} \right)\left( {1 + z + p{z^2}} \right)}}{{\left( {xy + xz + yz + xyz} \right)\left( {y\left( {y + z} \right)\left( {1 + z} \right) + {z^2} + p{y^2}{z^2}} \right)\left( {1 + z + p{z^2} + q{z^3}} \right)}}$$
It is possible to write down a closed-form expression for the coefficient $J$. We would recommend the use of a math formula processing software like Mathematica to get a concrete number. {I am very old and terminally ill, I do not have enough vigor to complete the calculation, sorry. I tested some examples using Mathematica though.}
Full proof  is lengthy for the generating function and not having enough space here, so we give a sketch of proof:
In this proof, we switch notation $x,y,z$. Now, ${x_1},{x_2},{x_3}$ are three roots of the elementary polynomial $f$ assuming, without loss of generality, they are all distinct. It is a standard assumption, especially when we deal with symmetric expressions.
Review that the Lagrangean interpolation formula is the same as polynomial remainder.
$$\sum\limits_{i = 1}^3 {g\left( {{x_i}} \right)\frac{{f\left( x \right)}}{{\left( {x - {x_i}} \right)f'\left( {{x_i}} \right)}}}  = g\left( x \right)\mathop {\bmod }\limits_x f\left( x \right)$$
where $f'$ is the derivative of $f$. Notice that the remainder of the cubic polynomial is a quadratic polynomial, where the constant term is a symmetric function. Apply this observation three times to our symmetric function of interest, if $F$ is a symmetric function, then
$$F\left( {{x_1},{x_2},{x_3}} \right) = \left\{ {\left\{ {F\left( {x,y,z} \right)\mathop {\bmod }\limits_{x \to 0} \frac{{f\left( x \right)}}{{\left( {x - y} \right)\left( {x - z} \right)}}} \right\}\mathop {\bmod }\limits_{y \to 0} \frac{{f\left( y \right)}}{{y - z}}} \right\}\mathop {\bmod }\limits_{z \to 0} f\left( z \right)$$
Now remember the Cauchy integration result on a symmetric sum
$$\sum\limits_{i = 1}^3 {g\left( {{x_i}} \right)}  = \frac{1}{{2\pi \sqrt { - 1} }}\oint {g\left( x \right){{\left\{ {\log f} \right\}}^\prime }\left( x \right)dx} $$
this contour contains all roots of the elementary polynomial $f$, we apply this Cauchy formula to the Lagrangean interpolation/polynomial remainder calculation. The equivalent sequential remainder expression for the symmetric function has a slight problem that divisors are rational functions, but this problem can be easily fixed replacing them by difference polynomials, also do not forget replacing similarly $g$‘s coefficients in the original Lagrange formula in the Cauchy integration form. Consequently, our symmetric function $F$ is equivalently expressed as three-dimensional multiple complex integration in ${\mathbb{C}^3}$. After we replace integration variables to their reciprocal and so does each contour reverse direction accordingly, we find an aesthetic candidate.
$$\hat J\left( {p,\,q,\,x,\,y,\,z} \right) = \left( {1 - \frac{{qx\left( {x - y} \right)\left( {x - z} \right)}}{{1 + x + p{x^2} + q{x^3}}}} \right)\left( {1 - \frac{{q{y^2}\left( {y - z} \right)}}{{1 + y + p{y^2} + q{y^3}}}} \right)\left( {1 - \frac{{q{z^3}}}{{1 + z + p{z^2} + q{z^3}}}} \right)$$
it is not yet a generating function unfortunately, so we prepare a re-modification of complex modulo integrations, we find the result. QED
The Symmetric sum of exponential functions
$$\begin{array}{c}
G\left( {{x_1},\,{x_2},\,{x_3}} \right) = \sum\limits_{\sigma  \in {S_3}} {\exp \left( {{C_1}{x_{\sigma \left( 1 \right)}} + {C_2}{x_{\sigma \left( 2 \right)}} + {C_3}{x_{\sigma \left( 3 \right)}}} \right)} \\
 = \sum\limits_{{\alpha _1} + 2{\alpha _2} + 3{\alpha _3} = {\beta _1} + {\beta _2} + {\beta _3} = 0}^\infty  {K\left( {{\alpha _1},\;{\alpha _2},\;{\alpha _3},\;{\beta _1},\;{\beta _2},\;{\beta _3}} \right)\frac{1}{{{\beta _1}!{\beta _2}!{\beta _3}!}}\left\{ {\sum\limits_{\sigma  \in {S_3}} {C_{\sigma \left( 1 \right)}^{{\beta _1}}C_{\sigma \left( 2 \right)}^{{\beta _2}}C_{\sigma \left( 3 \right)}^{{\beta _3}}} } \right\}e_1^{{\alpha _1}}e_2^{{\alpha _2}}e_3^{{\alpha _3}}} 
\end{array}$$
