"circular" operators, "circular" derivatives, and "circular" tetration.
#14
Quote:Now that was all fairly simple to understand, but now we come to the part where we must define circular multiplication.  It's still fairly simple, but may seem a bit awkward.

\( x \odot (x^{\circ n}) = x^{\circ n+1} \)
cxp(3π/4)=0. If [Image: svg.image?x\odot(x%5E%7B\circ%20n%7D)=x%5E%7B\circ%20n+1%7D], then [Image: svg.image?\delta\odot0] would equal cxp(3π/4+1)=-√(2)sin(1)~-1.190.
cxp(-π/4)=0. If [Image: svg.image?x\odot(x%5E%7B\circ%20n%7D)=x%5E%7B\circ%20n+1%7D], then [Image: svg.image?\delta\odot0] would equal cxp(-π/4+1)=sin(1)~0.841.
Please remember to stay hydrated.
ฅ(ミ⚈ ﻌ ⚈ミ)ฅ Sincerely: Catullus /ᐠ_ ꞈ _ᐟ\
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RE: "circular" operators, "circular" derivatives, and "circular" tetration. - by Catullus - 06/30/2022, 06:20 AM

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