For #n > 1#, I have designated the value of #log_n(pi+1/log_n(pi+1/log_n(pi+...)))# as the function #pin(n)#. Inversely, given #pin (8)=0.72544666#, how do you approximate # pi#?

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Answer 1 · 2016-08-03T21:09:17

#pi = 8^(0.72544666)-1/(0.72544666) ~~ 3.1415928#

Suppose:

#t = log_n(pi+1/(log_n(pi+1/(log_n(pi+...)))))=log_n(pi+1/t)#

Then:

#n^t = pi+1/t#

So:

#pi = n^t-1/t#

In our example, we are told that #n = 8# and #t = 0.72544666#

So:

#pi = 8^(0.72544666)-1/(0.72544666)#

#~~ 4.5200539-1.3784611 = 3.1415928#