Question

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`?

Answer 1

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

Explanation:

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`