UVIR Wrote:bo198214 Wrote:Nobody denies that the tetraroot is the inverse of \( {}^nx \) (by definition) however to define \( {}^{1/n}x \) as being the tetraroot is quite arbitrary
Why? What is "arbitrary" about it? And if it's "arbitrary", in what "sense" is it "arbitrary"?
It is in the same sense arbitrary as say that I define
\( {}^{1/n}x=1 \). No rule justifies that \( {}^{1/n}x \) should be the inverse of \( {}^nx \).
If you not even demand that \( {}^xe \) is continuous, what will then remain?
Quote:Again, if the definition of the n-th order tetraroot as \( {^{1/n}}x \) is "arbitrary", surely the tetraroot can be defined better as \( {^{m/n}}x \) for some m,n. Do you care to define m and n in some other consistent way for the tetraroot?
The n-th tetraroot is already clearly defined as the inversion of \( {^nx} \), no need to redefine it. What we all are seeking for is a sound definition for the tetration for non-integer exponents. Natural demands are \( {^xb} \) being analytic, strict increase and \( {^{x+1}b}=b^{^xb} \).
Quote:So? Has a divine judge decided on the validity of any of the proposed methods so far?
The interesting thing is not about validity of each single method, but rather that 3 completely different methods yield identical values.
Quote:So, if you tell me for example, take Andrew's or Gottfried's or your method and "input" 0.666... or 0.6363..., and then see what the function outputs, this is already suffering badly as a definition.
Can not see how this suffers.
Take for example the method of continuous iteration at a fixed point.
There you move the fixed point of \( b^x \) to 0, say \( f(x)=b^{x-a}+a \) where a is the fixed point of \( b^x \).
Then you compute \( g=f^{\circ 1/n} \) by unique analytic solution of the equation \( g^{\circ n}=f \) and then you have the value
\( {^{1/n}b}=g(1+a)-a \).
