by PWrong » Fri Jun 18, 2004 3:44 am
Ok, I've figured out some more stuff for tetration.
the inverse of tetration can only be done with the Newton-rhapson algorithm.
Example, to find the square "tetroot" of 2:
2=x^x
0=x^x-2
By Newton-Rhapson's algorithm, if x is a good approximation for f(x)=0, then x-f(x)/f'(x) is a better approximation.
x^x is equivalent to e^(x*lnx)
By the product rule, the derivative is
(d/dx(x)*lnx + x*d/dx(lnx)) e^(x*lnx)
= (lnx + x/x) * e^(x*lnx)
= (lnx + 1)x^x
so f'(x)=(lnx + 1)x^x
Iterate the following:
xi+1=xi-(xi^xi-2)/(d/dx(xi^xi)
x(i+1)=xi-(xi^xi-2)/((lnx+1)xi^xi
If you do this only a few times, starting with x1 = 2, you'll get to
1.5596... in only a few iterations. The only way to give an exact value is to say tetroot(2).
It's much more difficult for higher powers, because the derivatives are more complicated.
That's the fastest way to find the tetroot of a number. If
You can work out x tetrated to a rational number p/q in two different ways, implying that there are 2 possible answers.
You can find the tetroot of (x^p), or you can find the tetroot of x, then tetrate this number to p. Several examples show that these are not always equal.
Tell me if any of this doesn't make sense. I'm pretty sure it's right, since I've checked extensively.
I'm working on some general laws for tetration and tetroots, so I'll keep you informed.[/i]