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.
I hope you are deligent enough to do heavy duty algebraic computation. I use the Cauchy itegration form of remainder calculation,
$\mathop {\lim }\limits_{x\; \to \;0} \left\{ {f\;\bmod g} \right\} = \frac{1}{{2\pi \sqrt { - 1} }}\oint {\frac{{{x^{\deg f - 1}}}}{{f\left( x \right)}}g\left( x \right)dx} $
The contour includes all the roots of $f$.
Then answer to your question is obtainable in two steps
$\Phi \left( z \right) = \mathop {\lim }\limits_{x\; \to \;0} \left\{ {F\left( {x,{e_1} - z - x,z} \right)\;{{\bmod }_x}\left( {{x^2} - \left( {{e_1} - z} \right)x + {e_2} - z\left( {{e_1} - z} \right)} \right)} \right\}$
then
$F\left( {{e_1},{e_2},{e_2}} \right) = \mathop {\lim }\limits_{z\; \to \;0} \left\{ {\Phi \left( z \right){{\bmod }_z}\left( {{z^3} - {e_1}{z^2} + {e_2}z - {e_3}} \right)} \right\}$
I let Mathematica do the messy computation, for
$F\left( {x,y,z} \right) = \sum\limits_{{m_1} + 2{m_2} + 3{m_3} = {n_1} + {n_2} + {n_2}} {{C_{{n_1},{n_2},{n_3}}}\left({{m_1},{m_2},{m_3}} \right)\left\{{\frac{1}{{{n_1}!{n_2}!{n_3}!}}\frac{{{\partial^{{n_1}+{n_2}+{n_2}}}\;F}}{{\partial{x^{{n_1}}}\partial{y^{{n_2}}}\partial{z^{{n_2}}}}}\left({0,0,0}\right)} \right\}{e_1^{{m_1}}}{e_2^{{m_2}}}{e_3^{{m_3}}}}$
I got
${C_{{n_1},{n_2},{n_3}}}\left( {{m_1},{m_2},{m_3}} \right) = \sum\limits_{{k_1} + 2{k_2} = {n_1} + {n_2}} {\left( {\begin{array}{*{20}{c}}{{n_2} - {m_2}}\\
{{m_1}}\end{array}} \right)\frac{{{{\left( { - 1} \right)}^{{n_1} + {n_3} + {m_1} + {m_3} + {k_2}}}}}{{{m_3}!}}\sum\limits_{r = 0}^{{m_1} \wedge {k_2}} {\left( {\begin{array}{*{20}{c}}{{k_2}}\\r\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{m_1} + {m_2} + {m_3} - {k_1} - {k_2}}\\{{m_1} - r}\end{array}} \right)\frac{{\left( {r + {m_2} + {m_3} - {k_2}} \right)!}}{{\left( {r + {m_2} - {k_2}} \right)!}}} } $
I hope my coding is correct.
I hope you are deligent enough to do heavy duty algebraic computation. I use the Cauchy itegration form of remainder calculation,
$\mathop {\lim }\limits_{x\; \to \;0} \left\{ {f\;\bmod g} \right\} = \frac{1}{{2\pi \sqrt { - 1} }}\oint {\frac{{{x^{\deg f - 1}}}}{{f\left( x \right)}}g\left( x \right)dx} $
The contour includes all the roots of $f$.
Then answer to your question is obtainable in two steps
$\Phi \left( z \right) = \mathop {\lim }\limits_{x\; \to \;0} \left\{ {F\left( {x,{e_1} - z - x,z} \right)\;{{\bmod }_x}\left( {{x^2} - \left( {{e_1} - z} \right)x + {e_2} - z\left( {{e_1} - z} \right)} \right)} \right\}$
then
$F\left( {{e_1},{e_2},{e_2}} \right) = \mathop {\lim }\limits_{z\; \to \;0} \left\{ {\Phi \left( z \right){{\bmod }_z}\left( {{z^3} - {e_1}{z^2} + {e_2}z - {e_3}} \right)} \right\}$
I let Mathematica do the messy computation, for
$F\left( {x,y,z} \right) = \sum\limits_{{m_1} + 2{m_2} + 3{m_3} = {n_1} + {n_2} + {n_2}} {{C_{{n_1},{n_2},{n_3}}}\left({{m_1},{m_2},{m_3}} \right)\left\{{\frac{1}{{{n_1}!{n_2}!{n_3}!}}\frac{{{\partial^{{n_1}+{n_2}+{n_2}}}\;F}}{{\partial{x^{{n_1}}}\partial{y^{{n_2}}}\partial{z^{{n_2}}}}}\left({0,0,0}\right)} \right\}{e_1^{{m_1}}}{e_2^{{m_2}}}{e_3^{{m_3}}}}$
I got
${C_{{n_1},{n_2},{n_3}}}\left( {{m_1},{m_2},{m_3}} \right) = \sum\limits_{{k_1} + 2{k_2} = {n_1} + {n_2}} {\left( {\begin{array}{*{20}{c}}{{n_2} - {m_2}}\\
{{m_1}}\end{array}} \right)\frac{{{{\left( { - 1} \right)}^{{n_1} + {n_3} + {m_1} + {m_3} + {k_2}}}}}{{{m_3}!}}\sum\limits_{r = 0}^{{m_1} \wedge {k_2}} {\left( {\begin{array}{*{20}{c}}{{k_2}}\\r\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{m_1} + {m_2} + {m_3} - {k_1} - {k_2}}\\{{m_1} - r}\end{array}} \right)\frac{{\left( {r + {m_2} + {m_3} - {k_2}} \right)!}}{{\left( {r + {m_2} - {k_2}} \right)!}}} } $
I hope my coding is correct.