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- - **Convergence of a recurrence relation**
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Convergence of a recurrence relationGiven . Use a calculator to determine the behaviour of the sequence for each of the cases I had no problems doing and . For , the series increases and converges to 0.619 while for , the series decreases and converges to 0.619. However, when solving for , all i got was error on the calculator screen. |

Re: Convergence of a recurrence relation1 Attachment(s) I observed the same behavior you did, and for the series diverged. For it increased and converged to 0.619061286736 For it decreased and converged to 0.619061286736 I tried finding the closed form via Newton's method to create a first order IVP, but wound up with an integral without an anti-derivative expressible in elementary terms. :( I found by setting that the value the series converged to in the first two cases is one of the solutions to: The other solution is about 1.5121345516578424739. Here is a plot: [attachment=0:5ucqk92t]convergence.jpg[/attachment:5ucqk92t] The larger root seems to be the division between convergence and divergence for the given recurrence relation when used as the starting value. |

Re: Convergence of a recurrence relationSolving for fixed points, where W(z) is the Lambert W function, which can be assigned two values there, 0.619061 and 1.51213. The first is stable and the second is unstable; any starting value greater than 1.51213 will result in divergence. |

Re: Convergence of a recurrence relationIn fairness to [color=#0040FF]aswoods[/color], I was editing my post to include essentially the same information (albeit not as succinctly) in his post at the same time he was posting. :wink: |

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