Migration of inflection points in y = b # x, for e^(1/e) < b < +oo

[Ivars]

[quote=andydude]
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=I/p, y=p/I we get:

d/dt((I/p)^(p/I) =(I/p)^((p-I)/I)((p/I) dx/dt+I/plnI/p (d(p/I)/dt)

d/dt((I/p)^(p/I) =(I/p)^((p-I)/I)((p/I)(-I/p^2)dp/dt+(I/p)lnI/p*(1/I )dp/dt))

d/dt((I/p)^(p/I) =(I/p)^((p-I)/I)(-1/p+(1/p)ln(I/p))dp/dt

Again, apply h on both sides over time:

d/dt h(((I/p)^(p/I) =h((I/p)^((p-I)/I)(-1/p+(1/p)ln(I/p))) dp/dt

and as h((I/p)^(p/I))= I/p if p> 1 and removing dt

d(I/p) = h((I/p)^(p-2i)/i* ln(I/e*p) dp

Now sinse time is not present any more , and I is not dependent on time, we can calculate:

d(I/p) = (dI*p +Idp)/p^2

so (dI/p+ I dp/p^2) )/dp = h((I/p)^(p-2i)/i* ln(I/e*p) or

dI/dp= p*(h((I/p)^(p-2I)/I* ln(I/e*p) - I/(p^2)

There are so many brackets I lost count, but I/(p^2 is outside tetration.

Which probably is a differential equation linking x=p and I in complex plane to be used.

if p=const, dI=0.

We could have taken also p/I and I/q - but then things get more complicated as infinite tetration of ( (I/p)^(q/I)) I do not know, must be rational numbers if p,q integers, but they can be any>1.

That was so long there has to be some mistakes