Fractional powers of the derivative operator are still linear operators: the usual derivative sends e^kx to ke^kx; the fractional (d/dx)^a sends e^kx to (k^a)e^kx. Then you can build other functions out of Fourier modes if k is complex. #maths

In fact it's not that simple because k^a is ambiguous when k is complex and a is fractional, this leads to multiple definitions. See: https://sciencedirect.com/science/article/pii/S0377042714000065

SchrÃ¶der's equation is also interestingly related https://en.wikipedia.org/wiki/Schr%C3%B6der%27s_equation #maths

...and you can even have one on vector fields. This whole thread was caused by me finding âˆšâˆ’âˆ‡Â² in a formula#maths #physics?

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Stephen Brooks ðŸ¦†@[email protected]In fact it's not that simple because k^a is ambiguous when k is complex and a is fractional, this leads to multiple definitions. See: https://sciencedirect.com/science/article/pii/S0377042714000065

SchrÃ¶der's equation is also interestingly related https://en.wikipedia.org/wiki/Schr%C3%B6der%27s_equation

#maths