Is There Symmetry in the Signs of Newton's Identities?

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The discussion centers on the symmetry in the signs of Newton's identities, specifically examining the equations for p_1, p_2, and p_3. The correct formulation is established as p_1 = 1h_1 and p_2 = 2h_2 - h_1p_1. The proposed alternative equations, p_2 = 2h_2 + h_1p_1 and p_3 = 3h_3 + h_2p_1 + h_1p_2, are definitively incorrect due to arbitrary sign changes. The discussion emphasizes the importance of maintaining the integrity of the original equations.

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Jhenrique
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Look this relationship:

7a818659d257a5542f5121bd88429784.png


http://en.wikipedia.org/wiki/Newton's_identities#Related_identities

If I isolate the variable p, I'll have:

##p_1 = 1h_1##
##p_2 = 2h_2-h_1p_1##
##p_3 = 3h_3-h_2p_1-h_1p_2##

So, my question is: BTW, would be true that:

##p_1 = 1h_1##
##p_2 = 2h_2+h_1p_1##
##p_3 = 3h_3+h_2p_1+h_1p_2##

?

EDIT: I'm asking because with base in other formulas seems that there is some symmetry between the signals...
imagem.png
 
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Jhenrique said:
If I isolate the variable p, I'll have:

##p_1 = 1h_1##
##p_2 = 2h_2-h_1p_1##
##p_3 = 3h_3-h_2p_1-h_1p_2##

So, my question is: BTW, would be true that:

##p_1 = 1h_1##
##p_2 = 2h_2+h_1p_1##
##p_3 = 3h_3+h_2p_1+h_1p_2##
No.
The first equation just above is correct, but the next two aren't. You can' t just change the sign arbitrarily.
If ##p_2 = 2h_2-h_1p_1##, you can replace ##p_1## by ##h_1## to get
##p_2 = 2h_2 - h_1^2##
 

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