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Nothing's perfect. (Posted on 2016-12-24) Difficulty: 3 of 5

The bisection of the Fibonacci series, Sloane A001906 {1, 3, 8, 21, 55, 144,...}, naturally produces approximations to phi^2, by the division of the nth term by its predecessor: a(n)/a(n-1). ; e.g 55/21, 144/55, etc.

WolframAlpha also lists these fractions as convergents to 5pi/6.

In fact, there will always be a small shortfall between the two: (phi)^2 - 5pi/6 is not zero.

For sufficiently large n, how is the shortfall best approximated, in terms of a rational fraction, say 1/x?

  Submitted by broll    
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Solution: (Hide)
Direct evaluation of (3+sqrt(5))/2 - 5pi/6 in WolframAlpha gives 0.000040110758400..., (4*10^-5), whose continued fraction expansion begins [0; 24930, 1, 29, 2, 3, 1, 2, 3, 1, 4, 2,....].

Hence the best rational fraction approximation, in terms of 1/x, is 1/24931. With sufficiently large n, this augments accuracy to 5*10^(-11).

Doubtless more precise approximations exist, but not in the required form.

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  Subject Author Date
Some Thoughtscomputer explorationCharlie2016-12-26 08:56:34
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