Algorithm for Feigenbaum's Constant

[jim mcnamara]

4.6692.... is Feigenbaum's Constant. Any clue as to how to generate it?

If you look at it graphically you get:
Bifurc no.   Divides at    Length    This length/next length
------------------------------------------------------------
1              0.75        L1=1.0         L1/L2=2.0
2              1.25        L2=0.5         L2/L3=4.25
3              1.3677      L3=0.1147      L3/L4=4.492
4              1.3939      L4=0.0262      L4/L5=4.6208
5              1.39957     L5=0.00567     L5/L6=4.536
------------------------------------------------------------

Analytically it's:

      Lim N      L(n)
 F =  n -> inf   -------  = 4.669...
                 L(n+1)

I can't even fake it numerically. No Clue.

[Starman]

I don't know if this will be any help:

http://www.mathsoft.com/asolve/const...m/fgnbaum.html

[jim mcnamara]

Thanks kindly - I'll see what out there....