Function with 90 degree rotational symmetry about the origin?

[lackasexical]

I haven't done calculus level math in a long time, and this question has got me stumped

Comments

[lackasexical]

Well, we did get to this point:



But it wouldn't be defined for 0, when we need it to be defined for all real numbers - so could we add a component for f(x) = 0 when x = 0?

[phlebas]

Assuming what you have works (you'll need to tweak a couple of inequalities, but something along those lines sounds good) you can just define f(0)=0, yes. Then prove that it satisfies the conditions and you're done.

[lackasexical]

Ah, you're right, there were some opportunities for missing points with those inequalities!

We've adjusted it like so:



Is that what you were referring to by "tweaking", or is it something more glaring we're overlooking?

[phlebas]

That was what I had in mind, yes - otherwise you end up with some values of x having two values defined for f(x) and some having none.

[notmrgarrison]

With that you have f(1) = 1 and f(-1) = 1.

When you rotate that 90 degrees you'll either get f(-1) = -1 or f(1) = -1.