1.10: Machine Epsilon
- Page ID
- 96036
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Machine epsilon \(\left(\epsilon_{\mathrm{mach}}\right)\) is the distance between 1 and the next largest number. If \(0 \leq \delta<\epsilon_{\mathrm{mach}} / 2\), then \(1+\delta=1\) in computer math. Also since
\[x+y=x(1+y / x) \nonumber \]
if \(0 \leq y / x<\epsilon_{\operatorname{mach}} / 2\), then \(x+y=x\) in computer math.
Find \(\epsilon_{\text {mach }}\)
The number 1 in the IEEE format is written as
\[1=2^{0} \times 1.000 \ldots 0, \nonumber \]
with \(230^{\prime}\) s following the binary point. The number just larger than 1 has a 1 in the 23 rd position after the decimal point. Therefore,
\[\epsilon_{\text {mach }}=2^{-23} \approx 1.192 \times 10^{-7} \nonumber \]
What is the distance between 1 and the number just smaller than 1? Here, the number just smaller than one can be written as
\[2^{-1} \times 1.111 \ldots 1=2^{-1}\left(1+\left(1-2^{-23}\right)\right)=1-2^{-24} \nonumber \]
Therefore, this distance is \(2^{-24}=\epsilon_{\operatorname{mach}} / 2\).
The spacing between numbers is uniform between powers of 2, with logarithmic spacing of the powers of 2 . That is, the spacing of numbers between 1 and 2 is \(2^{-23}\), between 2 and 4 is \(2^{-22}\), between 4 and 8 is \(2^{-21}\), etc. This spacing changes for denormal numbers, where the spacing is uniform all the way down to zero.
Find the machine number just greater than 5
A rough estimate would be \(5\left(1+\epsilon_{\text {mach }}\right)=5+5 \epsilon_{\text {mach }}\), but this is not exact. The exact answer can be found by writing
\[5=2^{2}\left(1+\frac{1}{4}\right) \nonumber \]
so that the next largest number is
\[2^{2}\left(1+\frac{1}{4}+2^{-23}\right)=5+2^{-21}=5+4 \epsilon_{\text {mach }} \nonumber \]