><number> = d(n)*pi^n + d(n-1)*pi^(n-1) + . . . + d(2)*pi^2 + d(1)*pi + >d(0) + d(-1)*pi^-1 + d(-2)* pi^(-2) + . . . > >where n is the larger of zero or the integer such that pi^n .le. <number> >.lt. pi^(n-1), and the series formally does not terminate on the right, >though it would be possible that all the d(i) = 0 after a certain point. > >The d(i) are what we call in base ten "digits," but since that term is >sometimes considered only applicable in base ten, what shall we call >them? "pigits", perhaps? > >(And similarly with e instead of pi.)
I managed to follow that, actually. I'm impressed with myself. >>Then again, I'd probably better stop trying to figure this out before my >>head melts. > >Oh, come on and join the fun! It's a lot easier after the meltdown. > >;-) You certainly have a point there. Which might be supported by something I experienced while a philosophy undergraduate at UC Davis: I found that proving theorems in intentional modal logic was a hell of a lot easier after a fifth of Jameson's. Sliante, Richard S. Crawford http://www.mossroot.com AIM: Buffalo2K ICQ: 11646404 Y!: rscrawford MSN: [EMAIL PROTECTED] "It is only with the heart that we see rightly; what is essential is invisible to the eye." --Antoine de Saint Exup�ry "Push the button, Max!"
