2 ms·
Even better : did you know (-1)^x draws the unit circle in the complex plane ? No need for complex exp and i*pi
by ttoinou 2mo ago
Even better : did you know (-1)^x draws the unit circle in the complex plane ? No need for complex exp and i*pi
- srean 2mo agoThat'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.
- fph 2mo agoIf you plot it over which domain?
- ttoinou 2mo agoComplex domain
- WCSTombs 2mo agoYou 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 2mo agoI 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