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?
Gib Z
Oct4-08, 12:19 AM
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?
littleHilbert
Oct5-08, 07:18 AM
Ok the combination is clear:
-f(b-x)+a=f(b+x)-a
Thank you! :-)
Gib Z
Oct5-08, 07:25 AM
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, im not sure what I meant lol. Its something like that, though im sure its NOT equivalent to the condition a=0 lol.
littleHilbert
Oct5-08, 07:31 AM
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!