set x=.000000000001 Iterate x-> x+sin(x) it'll look like powers of two until it gets larger and then it crushes/folds into pi
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Let y_n = pi - x_n. Then y_{n+1} = y_n - sin(y_n), so y_{n+1} ~= (y_n)^3 / 6. This implies that the number of correct digits is approximately tripled with each iteration.
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