Symmetry of Polynomials

  • #1

Main Question or Discussion Point

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?
 
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Answers and Replies

  • #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?
 
  • #3
Ok the combination is clear:
[tex]-f(b-x)+a=f(b+x)-a[/tex]

Thank you! :-)
 
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  • #4
Gib Z
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you meant [tex]a + f(x) = - f(-x) - a[/tex], didn't you?
because it must be equivalent to [tex]a + f(x) + a - f(x) = 0 [/tex]

Ok the combination is clear:
[tex]-f(b-x)+a=f(b+x)-a[/tex]

Thank you! :-)
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.
 
  • #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!
 
  • #6
Gib Z
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No problem =]
 

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