03/04/2008, 11:17 AM
As I just read in Knoebel's Exponential Reiterated there is even a parametrization of the curve \( f \), already given by Goldbach:
Knoebel considers the equation \( x^y=y^x \) which is equivalent to \( x^{1/x}=y^{1/y} \) which means that the fixed points \( x \) and \( y \) have the same base. So \( y=f(x) \). The parametrization is:
\( x=s^{1/(s-1)} \) and \( y=s^{s/(s-1)} \).
We can easily verify that this indeed satisfy \( x^y=y^x \):
\( x^y=s^{s^{s/(s-1)}/(s-1)}=s^{s^{1+\frac{1}{s-1}}/(s-1)}=s^{s x/(s-1)}=y^x \)
For example for \( s=2 \) we get our famous fixed points \( x=2^{1/\left(2-1\right)}=2 \) and \( y=2^{2/\left(2-1\right)}=2^{2}=4 \)
There is lots of other interesting stuff in Knoebel's article, but read it yourself
For our consideration here let:
\( f_1(s)=s^{1/(s-1)} \) and \( f_2(s)=s^{s/(s-1)} \) so that \( f=f_2\circ f_1^{-1} \). So I wonder whether we can express \( f_1^{-1} \) (\( f_1 : (0,\infty)\to(1,\infty) \) and \( f_2 : (0,\infty)\to(1,\infty) \) are indeed bijective) with the Lambert W function. Any ideas?
Knoebel considers the equation \( x^y=y^x \) which is equivalent to \( x^{1/x}=y^{1/y} \) which means that the fixed points \( x \) and \( y \) have the same base. So \( y=f(x) \). The parametrization is:
\( x=s^{1/(s-1)} \) and \( y=s^{s/(s-1)} \).
We can easily verify that this indeed satisfy \( x^y=y^x \):
\( x^y=s^{s^{s/(s-1)}/(s-1)}=s^{s^{1+\frac{1}{s-1}}/(s-1)}=s^{s x/(s-1)}=y^x \)
For example for \( s=2 \) we get our famous fixed points \( x=2^{1/\left(2-1\right)}=2 \) and \( y=2^{2/\left(2-1\right)}=2^{2}=4 \)
There is lots of other interesting stuff in Knoebel's article, but read it yourself
For our consideration here let:
\( f_1(s)=s^{1/(s-1)} \) and \( f_2(s)=s^{s/(s-1)} \) so that \( f=f_2\circ f_1^{-1} \). So I wonder whether we can express \( f_1^{-1} \) (\( f_1 : (0,\infty)\to(1,\infty) \) and \( f_2 : (0,\infty)\to(1,\infty) \) are indeed bijective) with the Lambert W function. Any ideas?