Comment on Mathematicians trace source of Rogers-Ramanujan identities, find algebraic goldparentComments−TheLoneWolfling12yWhy not just directly get an approximation of the golden ratio by taking fib(n) / fib(n-1)?If you use an iterative / tail-recursive version you can actually do this directly, without even the overhead of having to calculate it twice.(I.e. define fib to return a tuple / struct / etc consisting of (fib(n), fib(n-1)).)−NAFV_P12yWhy not just directly get an approximation of the golden ratio by taking fib(n) / fib(n-1)?That has been done millions of times over the past 800 odd years. My method has no practical uses, but plenty of educational ones.
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Why not just directly get an approximation of the golden ratio by taking fib(n) / fib(n-1)?
If you use an iterative / tail-recursive version you can actually do this directly, without even the overhead of having to calculate it twice.
(I.e. define fib to return a tuple / struct / etc consisting of (fib(n), fib(n-1)).)
That has been done millions of times over the past 800 odd years. My method has no practical uses, but plenty of educational ones.