Group Abstract Group Abstract

Message Boards Message Boards

0
|
3.1K Views
|
3 Replies
|
1 Total Like
View groups...
Share
Share this post:

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

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

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

Thanks, Dear Gianluca - it was a typo.

Yes, I meant

Pi ~= Sqrt[4 E - 1]

Also there was a typo in the link to the OEIS sequence which I also corrected.

What are those two distinct limit points?

Thanks,

Best Regards,

Alex

Reply to this discussion
Community posts can be styled and formatted using the Markdown syntax.
Reply Preview
Attachments
Remove
or Discard