I'm reading Sauers Numerical Analysis. There is an exercis where I have to use the bisect method to calculate a height within $\pm 1$mm. (1.1 computer problems nr 9.)
I use bisect method with a while loop that continues until the interval is within a specified tolerance. The obvious solution to me is to set the tolerance to 1 mm. However the answer in the book has calculate the answer with significantly higher tolerance. My bisect method for different tolerances(in m): $$\begin{matrix} TOL & Approx\\ 5e-9&0.635500807315111\\ 5e-6&0.635501861572266\\ 5e-3&0.6367187500000000\\ 1e-3&0.635742187500000\end{matrix}$$
From this I would say that the answer was $0.636\pm 0.001$. The answer in the book is $0.6355\pm 1mm$.
My thought was, to give an answer within a certain tolerance was to calculate with bisect method until the error was smaller than the set tolerance and then round the answer to the nearest number above. If it's not, is the correct method to iterate through different tolerances util the answer stabilises to within the given error bound?
Edit: For clarity I'll post the entire exercise.
Find the height reached by 1 cubic meter of water stored in a spherical tank of radius 1 meter. Give your answer $\pm 1mm$. (Hint: First note that the sphere will be less than half full. The volume of the bottom $H$ meters of a hemisphere of radius $R$ is $\pi H^2(R-1)/3H$.)