1.12: Roundoff Error Example
- Page ID
- 96038
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)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.