1.12: Roundoff Error Example

Consider solving the quadratic equation

$$x^{2}+2 b x-1=0 \nonumber$$

where $b$ is a parameter. The quadratic formula yields the two solutions

$$x_{\pm}=-b \pm \sqrt{b^{2}+1} \nonumber$$

Consider the solution with $b>0$ and $x>0$ (the $x_{+}$solution) given by

$$x=-b+\sqrt{b^{2}+1} \nonumber$$

As $b \rightarrow \infty$

$$\begin{aligned} x &=-b+\sqrt{b^{2}+1} \\ &=-b+b \sqrt{1+1 / b^{2}} \\ &=b\left(\sqrt{1+1 / b^{2}}-1\right) \\ & \approx b\left(1+\frac{1}{2 b^{2}}-1\right) \\ &=\frac{1}{2 b} . \end{aligned} \nonumber$$

Now in double precision, realmin $\approx 2.2 \times 10^{-308}$ and we would like $x$ to be accurate to this value before it goes to 0 via denormal numbers. Therefore, $x$ should be computed accurately to $b \approx 1 /(2 \times$ realmin $) \approx 2 \times 10^{307}$. What happens if we compute (1.1) directly? Then $x=0$ when $b^{2}+1=b^{2}$, or $1+1 / b^{2}=1 .$ That is $1 / b^{2}=\epsilon_{\text {mach }} / 2$, or $b=\sqrt{2} / \sqrt{\epsilon_{\text {mach }}} \approx 10^{8} .$ For a subroutine written to compute the solution of a quadratic for a general user, this is not good enough. The way for a software designer to solve this problem is to compute the solution for $x$ as

$$x=\frac{1}{b\left(1+\sqrt{1+1 / b^{2}}\right)} \nonumber$$

In this form, if $1+1 / b^{2}=1$, then $x=1 / 2 b$ which is the correct asymptotic form.