# Theme and variations: the Newton-Girard identities

Time for another gemstone from symmetric function theory! (I am studying for my Ph.D. qualifying exam at the moment, and as a consequence, the next several posts will feature yet more gemstones from symmetric function theory. You can refer back to this post for the basic definitions.)

Start with a polynomial $p(x)$ that factors as
$$p(x)=(x-\alpha_1)(x-\alpha_2)\cdots(x-\alpha_n).$$ The coefficients of $p(x)$ are symmetric functions in $\alpha_1,\ldots,\alpha_n$ – in fact, they are, up to sign, the elementary symmetric functions in $\alpha_1,\ldots,\alpha_n$.

In particular, if $e_i$ denotes $\sum_{j_1<\cdots<j_i} \alpha_{j_1}\alpha_{j_2}\cdots\alpha_{j_i}$, then
$$p(x)=x^n-e_1x^{n-1}+e_{2}x^{n-2}-\cdots+(-1)^n e_n.$$ (These coefficients are sometimes referred to as Vieta’s Formulas.)

Since $p(\alpha_i)=0$ for all $i$, we can actually turn the equation above into a symmetric function identity by plugging in $\alpha_1,\ldots,\alpha_n$:
$$\begin{eqnarray*} \alpha_1^n-e_1\alpha_1^{n-1}+e_{2}\alpha_1^{n-2}-\cdots+(-1)^n e_n&=&0 \\ \alpha_2^n-e_1\alpha_2^{n-1}+e_{2}\alpha_2^{n-2}-\cdots+(-1)^n e_n&=&0 \\ \vdots\hspace{3cm}& & \end{eqnarray*}$$ and then summing these equations:
$$(\alpha_1^n+\cdots+\alpha_n^n)-e_1(\alpha_1^{n-1}+\cdots+\alpha_n^{n-1})+\cdots+(-1)^n\cdot ne_n=0$$ …and we have stumbled across another important basis of the symmetric functions, the power sum symmetric functions. Defining $p_i=\alpha_1^i+\cdots+\alpha_n^i$, every symmetric function in $\{\alpha_j\}$ can be uniquely expressed as a linear combination of products of these $p_i$s (we write $p_\lambda=p_{\lambda_1}p_{\lambda_2}\cdots p_{\lambda_n}$.)

So, we have
$$p_n-e_1p_{n-1}+e_2 p_{n-2}-\cdots+(-1)^n\cdot ne_n=0.$$ This is called the $n$th Newton-Girard identity, and it gives us a recursive way of expressing the $p$s in terms of the $e$s.

Well, almost. We have so far only shown that this identity holds when dealing with $n$ variables. However, we can plug in any number of zeros for the $\alpha_i$s to see that the identity also holds for any number of variables $k<n$. And, if we had more than $n$ variables, we can compare coefficients of each monomial individually, which only can involve at most $n$ of the variables at a time since the equation is homogeneous of degree $n$. Setting the rest of the variables equal to zero for each such monomial will do the trick.

So now we have it! The Newton-Girard identities allow us to recursively solve for $p_n$ in terms of the $e_i$s. Wikipedia does this nicely and explains the computation, and the result is:
$$p_n=\det\left(\begin{array}{cccccc} e_1 & 1 & 0 & 0 &\cdots & 0 \\ 2e_2 & e_1 & 1 & 0 & \cdots & 0 \\ 3e_3 & e_2 & e_1 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ (n-1)e_{n-1} & e_{n-2} & e_{n-3} & e_{n-4} & \ddots & 1 \\ ne_n & e_{n-1} & e_{n-2} & e_{n-3} & \cdots & e_1 \end{array}\right)$$ For instance, this gives us $p_2=e_1^2-2e_2$, which is true:
$$x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+zx).$$ The Wikipedia page derives a similar identity for expressing the $e$s in terms of the $p$s. It also does the same for expressing the complete homogeneous symmetric functions $h_n$ in terms of the $p$s and vice versa. However, it does not explicitly express the $e$s in terms of the $h$s or vice versa. In the name of completeness of the Internet, let’s treat these here.

Fix some number of variables $x_1,\ldots,x_n$. For any $d$, define $h_d$ to be the sum of all monomials of degree $d$ in $x_1,\ldots,x_n$. This is clearly symmetric, and we define
$$h_\lambda = h_{\lambda_1}h_{\lambda_2}\cdots h_{\lambda_k}$$ for any partition $\lambda$, as we did for the elementary symmetric functions last week. The $h_\lambda$s, called the complete homogeneous symmetric functions, form a basis for the space of symmetric functions.

It’s a fun exercise to derive the following generating function identities for $e_n$ and $h_n$:
$$H(t)=\sum_n h_n t^n = \prod_{i=1}^n \frac{1}{1-x_i t}$$ $$E(t)=\sum_n e_n t^n = \prod_{i=1}^n (1+x_i t)$$ (The first requires expanding out each factor as a geometric series, and then comparing coefficients. Try it!)

From these, we notice that $H(t)E(-t)=1$, and by multiplying the generating functions together and comparing coefficients, we find the identities
$$h_n=h_{n-1}e_1-h_{n-2}e_2+\cdots+(-1)^{n-1}e_n$$ Just as before, this gives us a recursion for $h_n$ in terms of the $e_i$s. With a bit of straightforward algebra, involving Cramer’s Rule, we can solve for $h_n$:

$$h_n=\det\left(\begin{array}{cccccc} e_1 & 1 & 0 & 0 &\cdots & 0 \\ e_2 & e_1 & 1 & 0 & \cdots & 0 \\ e_3 & e_2 & e_1 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ e_{n-1} & e_{n-2} & e_{n-3} & e_{n-4} & \ddots & 1 \\ e_n & e_{n-1} & e_{n-2} & e_{n-3} & \cdots & e_1 \end{array}\right)$$ We can also use the same equations to solve for $e_n$ in terms of $h_n$:

$$e_n=\det \left(\begin{array}{cccccc} h_1 & 1 & 0 & 0 &\cdots & 0 \\ h_2 & h_1 & 1 & 0 & \cdots & 0 \\ h_3 & h_2 & h_1 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ h_{n-1} & h_{n-2} & h_{n-3} & h_{n-4} & \ddots & 1 \\ h_n & h_{n-1} & h_{n-2} & h_{n-3} & \cdots & h_1 \end{array}\right)$$

I find these two formulas to be more aesthetically appealing than the standard Newton-Gerard formulas between the $p_n$s and $e_n$s, since they lack the pesky integer coefficients that appear in the first column of the matrix in the $p$-to-$e$ case. While perhaps not as mainstream, they are gemstones in their own right, and deserve a day to shine.

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