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Symmetry of Polynomials

  1. Oct 2, 2008 #1
    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: Oct 2, 2008
  2. jcsd
  3. Oct 3, 2008 #2

    Gib Z

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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:

    [tex] a - f(x) = f(-x) - a[/tex]

    For symmetry around (b,0), [tex] f(b+x) = - f(b-x)[/tex].

    Perhaps you can combine these conditions?
     
  4. Oct 5, 2008 #3
    Ok the combination is clear:
    [tex]-f(b-x)+a=f(b+x)-a[/tex]

    Thank you! :-)
     
    Last edited: Oct 5, 2008
  5. Oct 5, 2008 #4

    Gib Z

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    Now that you bring it up, im not sure what I meant lol. Its something like that, though im sure its NOT equivalent to the condition a=0 lol.
     
  6. Oct 5, 2008 #5
    Ok the combination is clear:
    [tex]-f(b-x)+a=f(b+x)-a[/tex]

    Thank you! :-)

    Oh sorry I've realised that everything is ok at the very moment you've sent your post...and deleted the thing!
     
  7. Oct 5, 2008 #6

    Gib Z

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    No problem =]
     
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