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Almost Pi

Problem 461

Published on Saturday, 1st March 2014, 04:00 pm; Solved by 401

Let fn(k) = ek/n - 1, for all non-negative integers k.

Remarkably, f200(6) + f200(75) + f200(89) + f200(226) = 3.141592644529… ≈ π.

In fact, it is the best approximation of π of the form fn(a) + fn(b) + fn(c) + fn(d) for n = 200.

Let g(n) = a2 + b2 + c2 + d 2 for a, b, c, d that minimize the error: | fn(a) + fn(b) + fn(c) + fn(d) - π|
(where |x| denotes the absolute value of x).

You are given g(200) = 62 + 752 + 892 + 2262 = 64658.

Find g(10000).