andydude Wrote:Finding such a differential equation would be amazing! But maybe we should model it after the differential equation of exponentiation:
\( \frac{d}{dt}(x^y) = x^{y-1}\left(y \frac{dx}{dt} + x \ln(x) \frac{dy}{dt}\right) \)although I think your idea has a better chance of working.
Andrew Robbins
If we look at this equation and make substitution x=x, y=1/x we get:
d/dt(x^(1/x) = x^((1-x)/x) ( (1/x)dx/dt+xlnx (d(1/x)/dt)
d/dt(x^(1/x) = x^((1-x)/x) ( (1/x)dx/dt-(1/x)*lnx (dx/dt))
d/dt(x^1/x) = x^((1-x)/x)) (1/x)*(1-lnx)) dx/dt
let us apply infinite tetration operation to both sides and use the conjecture that if t is real time, infinite tetration does not depend on real time even if variable x does :
d/dt(h(x^1/x)) = h(x^((1-x)/x)(1/x)*(1-lnx)) dx/dt
then we can get rid of parameter t here:
d(h(x^(1/x))/dx = h((x^((1-x)/x)(1/x)*(1-lnx)
but h(x^(1/x)) = x at least for real x in convergence region, so
h((x^((1-x)/x)*(1/x)*(1-ln(x)))= 1
h((x^((1-2x)/x))*(1-ln(x)))=1
h((x^((1/x)-2)))*(1-ln(x)))=1[/b]
I guess I should have taken imaginary argument on the left side such that x=i/p, y=p/i so that left side will look (i/p)^(p/i) but that I will do later when I understand if it makes any sense- and it must, since by above equation, ( if it is correct) , we only cover small region of interest.
