Symmetry of Odd-Degree Polynomials: Conditions for Symmetry in the Plane

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    Polynomials Symmetry
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Discussion Overview

The discussion revolves around the conditions that an odd-degree polynomial function must satisfy to exhibit symmetry about a point in the plane. Participants explore the mathematical relationships and conditions necessary for such symmetry, including translations from the origin.

Discussion Character

  • Technical explanation, Mathematical reasoning, Debate/contested

Main Points Raised

  • One participant inquires about the conditions for odd-degree polynomials to be symmetric about a point in the plane.
  • Another participant suggests that the conditions involve horizontal and vertical translations of the symmetry condition for odd functions, specifically around the points (0,a) and (b,0).
  • For symmetry around (0,a), the proposed condition is a - f(x) = f(-x) - a.
  • For symmetry around (b,0), the proposed condition is f(b+x) = -f(b-x).
  • A later reply attempts to clarify the combination of these conditions, suggesting a relationship: -f(b-x) + a = f(b+x) - a.
  • There is a correction regarding the earlier proposed condition, with a participant questioning the equivalence and expressing uncertainty about the formulation.
  • One participant acknowledges a realization that the conditions discussed may be correct, but expresses confusion about their earlier statements.

Areas of Agreement / Disagreement

The discussion contains multiple viewpoints and some uncertainty regarding the correct formulation of the conditions for symmetry. Participants do not reach a consensus on the exact conditions, and some statements are corrected or questioned without resolution.

Contextual Notes

There are unresolved aspects regarding the equivalence of the proposed conditions and the implications of the corrections made during the discussion.

littleHilbert
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Hi! Brief question:

I wonder which conditions should a polynomial function of odd degree fulfill in order to be symmetric to some point in the plane.
Are there such conditions?
 
Last edited:
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Of course there are! It's merely horizontal and vertical translations of the ordinary condition for an odd function, which is symmetric about (0,0).

For symmetry around (0,a) a function must satisfy:

a - f(x) = f(-x) - a

For symmetry around (b,0), f(b+x) = - f(b-x).

Perhaps you can combine these conditions?
 
Ok the combination is clear:
-f(b-x)+a=f(b+x)-a

Thank you! :-)
 
Last edited:
littleHilbert said:
you meant a + f(x) = - f(-x) - a, didn't you?
because it must be equivalent to a + f(x) + a - f(x) = 0

Ok the combination is clear:
-f(b-x)+a=f(b+x)-a

Thank you! :-)

Now that you bring it up, I am not sure what I meant lol. Its something like that, though I am sure its NOT equivalent to the condition a=0 lol.
 
Ok the combination is clear:
-f(b-x)+a=f(b+x)-a

Thank you! :-)

Oh sorry I've realized that everything is ok at the very moment you've sent your post...and deleted the thing!
 
No problem =]
 

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