Problem solved in full
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Showing 355/113 is the closest fraction to π with a denominator under 1000 7 steps
Show that 355/113 is the closest fraction to π with a denominator under 1000, and find out how far its reign actually runs.
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Run the expansion first. π = [3; 7, 15, 1, 292, …], so the convergents are 3, then 22/7, then 333/106, then 355/113.
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Each convergent comes from the two before it: multiply by the new term and add the one further back. Nothing in that step needs π itself, only its terms.
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355/113 comes out at 3.1415929204 against π's 3.1415926536. The error is 2.7 × 10⁻⁷, which is seven significant figures from three digits over three digits.
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Now the search. Take every denominator from 1 to 112, pick the nearest numerator for each, and keep the closest result. The winner is 333/106, out by 8.3 × 10⁻⁵ — 312 times further than 355/113.
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Widen the search to 1000 and nothing new appears. 355/113 is still the closest fraction there is.
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So how far does it hold? Sweeping upward, the first fraction that beats it is 52163/16604. That means 355/113 is the closest fraction to π for every denominator up to 16,603.
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And look at what beating it buys. 52163/16604 has a denominator 147 times larger and reduces the error by 0.2%. That is the shape of the whole subject: the good fractions are enormously better than their neighbours, and the next real improvement is a long way up.
Answer
355/113, out by 2.7 × 10⁻⁷, and the closest fraction to π for every denominator up to 16,603. Where this stops is the part worth knowing. √2 runs 2, 2, 2, … for ever and e runs 1, 1, 2, 1, 1, 4, 1, 1, 6, …, so both patterns can be stated and proved. Nobody knows a pattern in π's terms. Nobody knows whether they stay bounded, or whether some vast term is waiting further out and with it a fraction far better than 355/113. The 292 is not explained by anything. It is simply what π does.
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