Obtaining exact decimals in bisection method

Question

While studying the bisection for the approximation of roots of non-linear equations I was given the following bound for the error:

$|x_n-s| \leq \frac{(b-a)}{2^{n+1}}$

where $x_n$ is the n-th approximation of the root. Now consider the following problem:

Proof that the equation $x^3+4x^2=10$ has a unique solution on [1,2]. Using the bisection method give an approximation of the solution with at least 5 exact decimals.

The problem is I don't know what error should I use, in my class they worked out n from the following expression $\frac{1}{10^5}>\frac{2-1}{2^{n+1}}$.

Now what confuses me comes from the following hypothetical situation

Say the exact result is $1.00001$ and the error is $0.000005$ then between the posible approximations is $1.000005$ (substracting the error) and then I wouldn't have 5 exact decimals.

I need someone to point me out what am I missing here. Is it an error theory concept I'm missing? Should take another bound for the error?

Answer 1

This is known as the "table makers dilemma", where a result is so close to $\frac12$ after the last decimal place wanted that there is no way to decide what the last digit shown should be.

Look it up - Google returns a number of quite relevant links.

Answer 2

Formally, your concern about crossing a breakpoint is well founded. We could have a problem where the true answer is $1.0000050000000000000001$ As you approach it with bisection your intervals will cross $1.000005$ until they get very small. To be sure you have the correct rounded five decimal expression, you need a lot of steps. To get the minimum number of steps, you would need to check after each step that the uncertainty range remaining crosses $1.000005$.

Less formally, we are not so concerned about breakage and rounding. We would say we want an answer that is correct within some range, say $\pm 0.000005$ This, or even with a few more zeros, is not enough to guarantee that it will round correctly, but you can quote the result as $1.0000050 \pm 0.000005$. You have given a range that includes the correct result. Anyone who sees your result should be able to tell if it is precise enough for his purpose. Now you can compute exactly how many steps of bisection are required because you know the range gets cut in half each step.