# Proof of Newton Girard formula symmetric polynomials

Newton Girard formula states that for $k>2$: $$p_k=p_{k-1}e_1-p_{k-2}e_2+\cdots +(-1)^{k}p_1e_{k-1}+(-1)^{k+1}ke_{k}$$ where $e_i$ are elementary symmetric functions and $p_0=n$ with $p_k=x_1^k+\cdots+x_n^k$.

I am using induction to prove this result. I am stuck at the inductive step, that is to show:

$$p_{k}e_1-p_{k-1}e_2+\cdots +(-1)^{k+1}p_1e_k+(-1)^{k+2}(k+1)e_{k+1}= x_1^{k+1}+\cdots+x_n^{k+1}$$ I am not able to see how I can use my inductive hypothesis in the left hand side of the above expression.

• I don't think induction on $k$ is that helpful here. The proof I remember involves expanding out the terms $p_i e_{k-i}$, and observing the subsequent cancellation. – Rolf Hoyer Apr 19 '15 at 23:45

## 1 Answer

By way of enrichment here is an alternate formulation using cycle indices.

Recall that the OGF of the cycle index $Z(P_n)$ of the unlabeled set operator $\mathfrak{P}_{=n}$ is given by $$G(w) = \sum_{n\ge 0} Z(P_n) w^n = \exp\left(\sum_{q\ge 1} (-1)^{q+1} a_q \frac{w^q}{q}\right).$$

Differentiating we obtain $$G'(w) = \sum_{n\ge 0} (n+1) Z(P_{n+1}) w^n = G(w) \left(\sum_{q\ge 1} (-1)^{q+1} a_q w^{q-1}\right).$$

Extracting coefficients we thus have $$[w^n] G'(w) = (n+1) Z(P_{n+1}) = \sum_{q=1}^{n+1} (-1)^{q+1} a_q Z(P_{n+1-q})$$ This is apparently due to Lovasz.

Substitute the cycle indices with the variables $X_1$ to $X_m$ to get $$(n+1) Z(P_{n+1})(X_1+\cdots+X_m) \\= \sum_{q=1}^{n+1} (-1)^{q+1} (X_1^q+\cdots+X_m^q) Z(P_{n+1-q})(X_1+\cdots+X_m)$$

This yields $$(n+1) e_{n+1}(X_1,\ldots,X_m) = \sum_{q=1}^{n+1} (-1)^{q+1} p_q(X_1,\ldots,X_m) e_{n+1-q}(X_1,\ldots,X_m).$$

Now a choice of variable names yields the result.

Remark. The identity for $G(w)$ follows from the EGF for the labeled species for permutations where all cycles are marked with a variable indicating length of the cycle.

This yields $$\mathfrak{P} \left(A_1 \mathfrak{C}_{=1}(\mathcal{W}) + A_2 \mathfrak{C}_{=2}(\mathcal{W}) + A_3 \mathfrak{C}_{=3}(\mathcal{W}) + \cdots \right).$$

Translating to generating functions we obtain $$G(w) = \exp\left(a_1 + a_2 \frac{w^2}{2} + a_3 \frac{w^3}{3} + \cdots\right).$$

The fact that $$Z(P_n) = \left.Z(S_n)\right|_{a_q := (-1)^{q+1} a_q}$$ then confirms the claim.