05/29/2014, 01:31 PM
(This post was last modified: 05/29/2014, 02:47 PM by sheldonison.)
The op skipped an operation. What about \( x\uparrow x=-1 \)?
\( x^x=-1 \)
\( \ln(x)x=(2n-1)\pi i\;\;\; \) taking ln of both sides, for all ln(-1)
\( ye^y = W^{-1}(y) = (2n-1)\pi i\;\;\; \) where \( y=\ln(x)\;\;\; \) and W is the LambertW function
\( y=W((2n-1)\pi i)\;\;\; x=\exp(W((2n-1)\pi i))\;\;\; \) Solution with the LambertW function
for n=1, I get x=1.6903867571636 + 1.8699079640268i, which I solved numerically with Newton's method. Oh, and then x=-1 is the obvious solution as well! I missed the obvious solution the first time; where \( W(\pi i)=-\pi i \). Maybe that trivial solution of x=-1 is why the Op skipped asking about \( x\uparrow x=-1 \), but the complex solutions are analogous to the infinite number of complex solutions of \( x\uparrow \uparrow x=-1 \)
\( x^x=-1 \)
\( \ln(x)x=(2n-1)\pi i\;\;\; \) taking ln of both sides, for all ln(-1)
\( ye^y = W^{-1}(y) = (2n-1)\pi i\;\;\; \) where \( y=\ln(x)\;\;\; \) and W is the LambertW function
\( y=W((2n-1)\pi i)\;\;\; x=\exp(W((2n-1)\pi i))\;\;\; \) Solution with the LambertW function
for n=1, I get x=1.6903867571636 + 1.8699079640268i, which I solved numerically with Newton's method. Oh, and then x=-1 is the obvious solution as well! I missed the obvious solution the first time; where \( W(\pi i)=-\pi i \). Maybe that trivial solution of x=-1 is why the Op skipped asking about \( x\uparrow x=-1 \), but the complex solutions are analogous to the infinite number of complex solutions of \( x\uparrow \uparrow x=-1 \)
- Sheldon