# Symmetry of Polynomials

1. Oct 2, 2008

### littleHilbert

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. Oct 3, 2008

### Gib Z

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?

3. Oct 5, 2008

### littleHilbert

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

Thank you! :-)

Last edited: Oct 5, 2008
4. Oct 5, 2008

### Gib Z

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. Oct 5, 2008

### littleHilbert

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

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. Oct 5, 2008

### Gib Z

No problem =]