How do you prove that the 4-sd approximation to the value of #log_2(2+1/log_2(2+1/log_2(2+...)))# is 1.428?

1 Answer
Jul 28, 2016

Answer:

#approx1.432#

Explanation:

Calling

#y = log_2(2+1/log_2(2+1/log_2(2+...)))# we have

#y = log_2(2+1/y)#

Now, calling

#y_{k+1} = log_2(2+1/y_k)#

substituting

#y_1 = 1.428# we obtain
#y_2=1.433# and sucessively
#y_3 =1.432# etc.

converging to

#( (y_1 = 1.428000000000000000000000000000), (y_2 = 1.433109072200861922041781326700), (y_3 = 1.431774625761248698196936857130), (y_4 = 1.432122371909281044466411003668), (y_5 = 1.432031697667491831893471498920), ( cdots), (y_21 = 1.432050448448111801269533316372), (y_22 = 1.432050448448122681455174642906), (y_23 = 1.432050448448119794875310617499), (y_24 = 1.432050448448120683053730317624), (y_25 = 1.432050448448120238964520467562), (y_26 = 1.432050448448120238964520467562) )#