f(n)=3^f(n-1)-2^f(n-2) - Printable Version +- Tetration Forum ( https://math.eretrandre.org/tetrationforum)+-- Forum: Tetration and Related Topics ( https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1)+--- Forum: Mathematical and General Discussion ( https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=3)+--- Thread: f(n)=3^f(n-1)-2^f(n-2) ( /showthread.php?tid=772) |

f(n)=3^f(n-1)-2^f(n-2) - tommy1729 - 02/18/2013
Consider f(n)=3^f(n-1)-2^f(n-2) with f(1)=0 and f(2)=a. For a>>1.51... the sequence goes to oo. For a<<1.51... the sequence goes to 0. Can we express the constant x=1.51... ? Is a sine type wave possible for f(2)= x ? What happens with complex numbers ? RE: f(n)=3^f(n-1)-2^f(n-2) - tommy1729 - 02/18/2013
Notice that 1.51... is NOT a real repelling fixpoint of x=3^x-2^x. The only fixpoints of x=3^x-2^x on the extended real line are 0,1 and +oo. The value 1 seems to suggest that f(2)=1.51... might make the sequence oscillate around 1. Assuming the above it might work to compute 1.51... by starting at value 1 and doing the iterations backwards until we reach approximately 0. But of course nicer ways should be available. I have not seen this sequence investigated before and it seems it might require new ideas, unless I overlooked something trivial. I must add that this thread also deserves some credit from my student mick who some of you have already met on MSE. Mick and I have given eachother permission to post eachothers (or common ) ideas concerning some math subjects including dynamical systems and tetration. RE: f(n)=3^f(n-1)-2^f(n-2) - tommy1729 - 02/26/2013
I must note that - although trivial - if the sequence goes above 1 the sequence explodes to oo. This implies that solving for 1 must indeed give my desired constant. The value of my constant is 0,516553200406323... This value reminds me of eulers constant ? tommy1729 |