What dose #{x_n}# converges ? when #x_1=5/2,5x_(n+1)=x_n^2+6#

↳Redirected from "How to prove this ? #2|k^2# then #2|k# for some #k\inZZ#"
1 Answer
Cesareo R.
Aug 7, 2017

See below.

Explanation:

This is a nonlinear difference equation. Normally it is hard to analyze convergence in this case. So we will make some basic convergence considerations.

If #x_n # converges to #x^@# then

#5x^@=(x^@)^2+6# and solving for #x^@# we have

#x^@ = {2,3}#

Checking for #x = 2+delta# we conclude that #x^@=2# is an stable attraction point for #-oo < delta < 1# and #x^@ = 3# is an inestable attraction point. Resuming, for #-oo < x_1 < 3# the sequence converges to #2# otherwise is diverges.

Related questions
Impact of this question
1414 views around the world
You can reuse this answer
Creative Commons License