Solutions to the practice problems

Problem P01: Compute the first N decimal digits after the decimal point of sin(sin(sin 1)), rounded toward zero. We have sin(sin(sin 1)) ≈ 0.678: the first N decimal digits after the decimal point match the first N mantissa digits. We use a target decimal precision N1 > N , and a binary precision p. We compute x = ◦(sin 1), y = ◦(sinx), z = ◦(sin y), with all roundings to nearest. It is easy to see that since p ≥ 3, we have 1/2 ≤ x, y, z < 1, thus all rounding errors are bounded by 2−p−1. We can thus write x = sin 1 + x with | x| ≤ 2−p−1. It follows y = sin(sin 1 + x) + y with | y| ≤ 2−p−1; we can write sin(sin 1 + x) = sin(sin 1) + x cos θ, thus the absolute error on y is bounded by | x| + | y| ≤ 2−p. Similarly, the error on z is bounded by 3 · 2−p−1. With p ≥ 2 +N1 log 10 log 2 , we have 3 · 2 −p−1 < 1/2 · 10−N1 . Finally, we output the binary value z in decimal to N1 digits, with rounding to nearest. Since 1/2 ≤ z < 1, the last digit has weight 10−N1 , thus the total error — including that on z and the output error — is bounded by 10−N1 . Thus, unless the last N1 −N digits of the output are all zero, we can decide the correct output to N digits, rounded toward zero. Note: if the function sin(sin(sinx)) was D-finite, i.e. if it would satisfy a linear differential equation with polynomial coefficients, then it would be possible to compute sin(sin(sin 1)) to precision n in O(M(n) log n) using the “binary splitting” algorithm. Unfortunately, it does not seem that sin(sin(sin x)) is D-finite.