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I have a brief question related to an example in my textbook. In my book, the following theorem on Bisection Method is presented:

If $[a_0,b_0], [a_1,b_1],. . .,[a_n,b_n]. . .$ denote the intervals in the bisection method, then the limits $\lim_{n \to \infty} a_n$, and $\lim_{n \to \infty} b_n$ exist, are equal, and represent a zero of $f$. If $r=\lim_{n \to \infty} c_n$ and $c_n = \frac{1}{2}(a_n + b_n)$, then

$$|r-c_n| \leq 2^{-(n+1)}(b_0 - a_0)$$

Next, there is the following example:

Suppose that the bisection method is started with the interval $[50,63]$. How many steps should be taken to compute a root with relative accuracy of one part in $10^{-12}$?



and then solve this for $n$. However, the book example says:

The stated requirement on relative accuracy means that

$$|r-c_n|/|r| \leq 10^{-12}$$

We know that $r \geq 50$, and thus it suffices to secure the inequality

$$|r-c_n|/50 \leq 10^{-12}$$

By means of the theorem above, we infer that the following condition is sufficent:

$$2^{-(n+1)}\cdot (13/50) \leq 10^{-12}$$

Solving this for $n$, we conclude that $n \geq 37$.

OK, so what I don't understand here is why the example begins by writing $|r-c_n|/|r| \leq 10^{-12}$ instead of just $|r-c_n| \leq 10^{-12}$. What is the motivation for including the $|r|$ in the denominator on the left side of the inequality?


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Because of relative in "relative accuracy". The relative error is the absolute error divided by the magnitude of the exact value. See here.

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Ah! Thanks a lot. I wasn't aware of this definition of relative accuracy, as the section I am reading is the first section in the book that is part of the curriculum (I now see that the difference beteween the two types of error are explained in an earlier section not included in the curriculum). Cheers! Appreciate it a lot. –  Kristian May 12 '12 at 11:55

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