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From: Steen Schmidt on 5 Sep 2006 14:35 Yao Konan wrote: > *(seq(seq(1/(sqrt(2*i+1)+I*sqrt(2*j+1)),i,1,4),j,1,4))^-1 On a TI92/89 there's no difference between i and I. Regards Steen
From: Steen Schmidt on 5 Sep 2006 14:39 Veli-Pekka Nousiainen wrote: > > > But in UserRPL the recursive Fibonacci formula traverses to > > > FIBNUM(1000) in 4 seconds on a HP50G. > > Recursive? > You mean a loop... Sure, sorry. > for the recursion you need a 50G in USB... > oe an emulator on a PC Why is that? Regards Steen
From: Veli-Pekka Nousiainen on 5 Sep 2006 15:02 Steen Schmidt wrote: > Veli-Pekka Nousiainen wrote: > >>>> But in UserRPL the recursive Fibonacci formula traverses to >>>> FIBNUM(1000) in 4 seconds on a HP50G. >> >> Recursive? >> You mean a loop... > > Sure, sorry. > >> for the recursion you need a 50G in USB... >> oe an emulator on a PC > > Why is that? batteries will run out before the answer gets there ;-)
From: Yao Konan on 5 Sep 2006 15:42 I told in one of my previous post that I stands for the imaginary number. Regards, Konan Yao Steen Schmidt a écrit : > Yao Konan wrote: > > > *(seq(seq(1/(sqrt(2*i+1)+I*sqrt(2*j+1)),i,1,4),j,1,4))^-1 > > On a TI92/89 there's no difference between i and I. > > Regards > Steen
From: Steen Schmidt on 6 Sep 2006 01:00
Yao Konan wrote: > I told in one of my previous post that I stands for the imaginary > number. Ok, sorry - couldn't find that, so I didn't look thoroughly enough :-) Cheers, Steen |