# A generalization of Newton's identities

Consider the following equations: $$A_1^1=\sum_iy_i=y_1+y_2+\ldots+y_m=a_1$$ $$A_2^1=\sum_{i_1,i_2}y_{i_1}y_{i_2}=a_2\,\,,i_1< i_2$$ $$A_3^1=\sum_{i_1,i_2,i_3}y_{i_1}y_{i_2}y_{i_3}=a_3\,\,,i_1< i_2< i_3$$ $$\vdots$$ $$A_{m-1}^1=\sum_{i_1,\ldots,i_{m-1}}y_{i_1}\ldots y_{i_{m-1}}=a_{m-1}\,\,,i_1< \ldots< i_{m-1}$$ $$A_m^1=y_{1}\ldots y_{{m}}=a_m$$ How to compute following expressions without computing exact $$y_i$$'s, i.e. in terms of $$a_i$$s? $$A_1^n=\sum_iy_i^n=y_1^n+y_2^n+\ldots+y_m^n=?$$ $$A_2^n=\sum_{i_1,i_2}y_{i_1}^ny_{i_2}^n=?\,\,,i_1 $$A_3^n=\sum_{i_1,i_2,i_3}y_{i_1}^ny_{i_2}^ny_{i_3}^n=?\,\,,i_1< i_2< i_3$$ $$\vdots$$ $$A_{m-1}^n=\sum_{i_1,\ldots,i_{m-1}}y_{i_1}^n\ldots y_{i_{m-1}}^n=?\,\,,i_1< \ldots< i_{m-1}$$ $$A_m^n=y_{1}^n\ldots y_{{m}}^n=a_m^n$$ Does anyone know a reference containing the results?

As an example, $$m=3$$,$$n=3$$:

$$(\sum_iy_i)^3=y_1^3+y_2^3+y_3^3+3y_1^2y_2+3y_1^2y_3+3y_2^2y_1+3y_2^2y_3+3y_3^2y_1+3y_3^2y_1+6y_1y_2y_3$$ $$(A_1^1)^3=A_1^3+3(y_1+y_2+y_3)(y_1y_2+y_2y_3+y_3y_1)-3y_1y_2y_3$$ $$A_1^3=(A_1^1)^3-3A_1^1A_2^1+3A_3^1=a_1^3-3a_1a_2+3a_3$$.

• Not an answer: given the $y_i$s, the LHS is given by the multinomial theorem. See en.wikipedia.org/wiki/Multinomial_theorem. Unfortunately, I don't know how to rearrange this into the form you want. Commented Sep 16, 2015 at 10:26
• I think some terms should be added and subtracted in that form, However it seems difficult to do it for all of the expressions or even one of them in general! Commented Sep 16, 2015 at 10:38

Partial result - for $A_1^n$, I stumbled upon the following Wikipedia page: 'Newton's Identities'!
In your notation with $A_0^1:=1$, this says that
$$kA_k^1 = \sum_{m=1}^k (-1)^{m-1} A_{k-m}^1\ A^m_1$$
which allows you to compute $A^k_1$ recursively, so long as you know $A^1_k$. A proof of this is given in the Wikipedia page.
• Thank you for answering. According to Wikipedia it should be $(-1)^{m-1}$ Commented Sep 16, 2015 at 16:03