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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? $$A_1^n=\sum_iy_i^n=y_1^n+y_2^n+\ldots+y_m^n=?$$ $$A_2^n=\sum_{i_1,i_2}y_{i_1}^ny_{i_2}^n=?\,\,,i_1<i_2$$ $$A_3^n=\sum_{i_1,i_2,i_3}y_{i_1}^ny_{i_2}^ny_{i_3}^n=?\,\,,i_1< i_2< i_3$$ $$\vdots$$ $$A_{m-1}^n=\sum_{i_1,\ldots,i_{m-1}}y_{i_1}^n\ldots y_{i_{m-1}}^n=?\,\,,i_1< \ldots< i_{m-1}$$ $$A_m^n=y_{1}^n\ldots y_{{m}}^n=a_m^n$$ Does anyone know a reference containing the results?


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

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

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  • $\begingroup$ 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. $\endgroup$ Commented Sep 16, 2015 at 10:26
  • $\begingroup$ 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! $\endgroup$
    – SMA.D
    Commented Sep 16, 2015 at 10:38

1 Answer 1

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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.

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  • $\begingroup$ Thank you for answering. According to Wikipedia it should be $(-1)^{m-1}$ $\endgroup$
    – SMA.D
    Commented Sep 16, 2015 at 16:03
  • $\begingroup$ ah yes. :) thanks $\endgroup$ Commented Sep 16, 2015 at 16:03

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