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Recurrence converging limit evaluation

Posted 2 years ago

Long time ago I came up with the primitive 2 decimal digits Pi approximation:

Pi ~= Sqrt[4 E - 1] 

see

https://oeis.org/A135821

and formula (9) in

https://mathworld.wolfram.com/PiApproximations.html

I was thinking how to improve it and in trying so lately came up with the following recurrence:

RecurrenceTable[{u[n + 1] == (1 + 1/u[n])^(Sqrt[4 E - 1] + 1),
  u[0] == Sqrt[4 E - 1] + 1}, u, {n, 0, 35}]

It appears that the even and odd indexed terms of the rational numbers sequence A(n), generated by the above recurrence, are converging to some limit value ? when

$n-->infinity$

A(n)={4.14215, 2.44921, 4.12963, 2.4552, 4.11755, 2.46102,
  4.10589, 2.46668, 4.09462, 2.4722, 4.08372, 2.47757,
  4.07317, 2.4828, 4.06296, 2.4879, 4.05306, 2.49287,
  4.04347, 2.49772, 4.03416, 2.50246, 4.02513, 2.50708,
  4.01636, 2.5116, 4.00783, 2.51601, 3.99954, 2.52033,
  3.99148, 2.52455, 3.98364, 2.52868, 3.976, 2.53271,...}

but I am using free version of Wolfram Alpha and it only works for the first 36 terms. How to evaluate the converging limit of above recurrence?

3 Replies

You get the three solutions of f[f[x]]==x from Wolfram|Alpha with the query

Solve (1 + (1 + 1/x)^(-1 - Sqrt[-1 + 4 E]))^(1 + Sqrt[-1 + 4 E])=x

Just throw away the middle one, which is the fixed point of f.

POSTED BY: Gianluca Gorni

Perhaps you meant

Pi ~= Sqrt[4 E - 1]

The function

f[x_] = (1 + 1/x)^(1 + Sqrt[-1 + 4 E])

has a fixed point, but it is repulsive, because the derivative is less than -1. Hence your sequence does not converge. However, it has two distinct limit points, that are fixed points for f[f[x]]:

POSTED BY: Gianluca Gorni
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