points by ttoinou 15 hours ago

Even better : did you know (-1)^x draws the unit circle in the complex plane ? No need for complex exp and i*pi

WCSTombs 12 hours ago

You do in fact need the complex exponential to define this correctly because the function a^x for nonintegers x is only unambiguously defined when a is a positive real number. For example, your function could be either e^(pi i x) or e^(-pi i x), which trace the circle in opposite directions as x varies over the reals. (They happen to agree when x is an integer.)

  • ttoinou 11 hours ago

    I agree. I just meant the 2D function cos(pix),sin(pix) is quite natural to work with and it can be reflected easily in the formulation of (-1)^x

srean 15 hours ago

That's because

   a^b = exp (b ln  a)

That's equivalent to saying, no need for -1 because we have exp.

One can change based of the exponentiation operation. Exp happens to be a convenient base.

  • ttoinou 11 hours ago

    Yeah bad formulation on my part

    • srean 11 hours ago

      Not bad at all, just equivalent.

      • ttoinou 10 hours ago

        We do need complex exp to define my formula, you’re right

fph 14 hours ago

If you plot it over which domain?

  • ttoinou 11 hours ago

    Complex domain