4.2.2. Accuracy of Floating Point Arithmetic

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+#LyX 2.4 created this file. For more info see https://www.lyx.org/
+\lyxformat 620
+\begin_document
+\begin_header
+\save_transient_properties true
+\origin unavailable
+\textclass book
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+\input defs
+\end_preamble
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+\index Index
+\shortcut idx
+\color #008000
+\end_index
+\secnumdepth 3
+\tocdepth 3
+\paragraph_separation indent
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+\end_header
+
+\begin_body
+
+\begin_layout Standard
+
+\emph on
+Note:
+
+\emph default
+ Normalized floating point arithmetic is assumed unless the contrary is specified.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+exerc4[10]
+\end_layout
+
+\end_inset
+
+Is it possible to have floating point numbers 
+\begin_inset Formula $u$
+\end_inset
+
+,
+ 
+\begin_inset Formula $v$
+\end_inset
+
+,
+ and 
+\begin_inset Formula $w$
+\end_inset
+
+ for which exponent overflow occurs during the calculation of 
+\begin_inset Formula $u\otimes(v\otimes w)$
+\end_inset
+
+ but not during the calculation of 
+\begin_inset Formula $(u\otimes v)\otimes w$
+\end_inset
+
+?
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+Yes.
+ If,
+ say,
+ 
+\begin_inset Formula $b=10$
+\end_inset
+
+,
+ 
+\begin_inset Formula $q=8$
+\end_inset
+
+,
+ and overflow occurs when the exponent reaches 16,
+ let 
+\begin_inset Formula $u=(15,.10000001)$
+\end_inset
+
+,
+ 
+\begin_inset Formula $v=(9,.33333330)$
+\end_inset
+
+,
+ and 
+\begin_inset Formula $w=(9,.30000000)$
+\end_inset
+
+.
+ Then 
+\begin_inset Formula $v\otimes w=(9,.99999990)$
+\end_inset
+
+ and 
+\begin_inset Formula $u\otimes(v\otimes w)=(16,.10000000)$
+\end_inset
+
+,
+ which raises an overflow,
+ but 
+\begin_inset Formula $u\otimes v=(15,.33333333)$
+\end_inset
+
+ and 
+\begin_inset Formula $(u\otimes v)\otimes w=(15,.99999999)$
+\end_inset
+
+,
+ which does not raise an overflow.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc8[20]
+\end_layout
+
+\end_inset
+
+Let 
+\begin_inset Formula $\epsilon=0.0001$
+\end_inset
+
+;
+ which of the relations
+\begin_inset Formula 
+\begin{align*}
+u & \prec v\quad(\epsilon), & u & \sim v\quad(\epsilon), & u & \succ v\quad(\epsilon), & u & \cong v\quad(\epsilon)
+\end{align*}
+
+\end_inset
+
+hold for the following pairs of base 10,
+ excess 0,
+ eight-digit floating point numbers?
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $u=(1,+.31415927)$
+\end_inset
+
+,
+ 
+\begin_inset Formula $v=(1,+.31416000)$
+\end_inset
+
+;
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $u=(0,+.99997000)$
+\end_inset
+
+,
+ 
+\begin_inset Formula $v=(1,+.10000039)$
+\end_inset
+
+;
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $u=(24,+.60221400)$
+\end_inset
+
+,
+ 
+\begin_inset Formula $v=(27,+.00060221)$
+\end_inset
+
+;
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $u=(24,+.60221400)$
+\end_inset
+
+,
+ 
+\begin_inset Formula $v=(31,+.00000006)$
+\end_inset
+
+;
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $u=(24,+.60221400)$
+\end_inset
+
+,
+ 
+\begin_inset Formula $v=(28,+.00000000)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\sim$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\approx$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\sim$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\approx$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\sim$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\approx$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\sim$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\sim$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc10[M25]
+\end_layout
+
+\end_inset
+
+(W.
+ M.
+ Kahan.) A certain computer performs floating point arithmetic without proper rounding,
+ and,
+ in fact,
+ its floating point multiplication routine ignores all but the first 
+\begin_inset Formula $p$
+\end_inset
+
+ most significant digits of the 
+\begin_inset Formula $2p$
+\end_inset
+
+-digit product 
+\begin_inset Formula $f_{u}f_{v}$
+\end_inset
+
+.
+ (Thus when 
+\begin_inset Formula $f_{u}f_{v}<1/b$
+\end_inset
+
+,
+ the least-significant digit of 
+\begin_inset Formula $u\otimes v$
+\end_inset
+
+ always comes out to be zero,
+ due to subsequent normalization.) Show that this causes the monotonicity of multiplication to fail;
+ in other words,
+ exhibit positive normalized floating point numbers 
+\begin_inset Formula $u$
+\end_inset
+
+,
+ 
+\begin_inset Formula $v$
+\end_inset
+
+,
+ and 
+\begin_inset Formula $w$
+\end_inset
+
+ such that 
+\begin_inset Formula $u<v$
+\end_inset
+
+ but 
+\begin_inset Formula $u\otimes w>v\otimes w$
+\end_inset
+
+ on this machine.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+Assume 
+\begin_inset Formula $p=4$
+\end_inset
+
+,
+ and let 
+\begin_inset Formula $u=.9999<1.000=v$
+\end_inset
+
+ and let 
+\begin_inset Formula $w=.2222$
+\end_inset
+
+.
+ Then 
+\begin_inset Formula $u\otimes w=.2221>.2220=v\otimes w$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc13[M25]
+\end_layout
+
+\end_inset
+
+Some programming languages (and even some computers) make use of floating point arithmetic only,
+ with no provision for exact calculations with integers.
+ If operations on integers are desired,
+ we can,
+ of course,
+ represent an integer as a floating point number;
+ and when the floating point operations satisfy the basic definitions in (9),
+ we know that all floating point operations will be exact,
+ provided that the operands and the answer can each be represented exactly with 
+\begin_inset Formula $p$
+\end_inset
+
+ significant digits.
+ Therefore—
+so long as we know that the numbers aren't too large—
+we can add,
+ subtract,
+ or multiply integers with no inaccuracy due to rounding errors.
+\end_layout
+
+\begin_layout Standard
+But suppose that a programmer wants to determine if 
+\begin_inset Formula $m$
+\end_inset
+
+ is an exact multiple of 
+\begin_inset Formula $n$
+\end_inset
+
+,
+ when 
+\begin_inset Formula $m$
+\end_inset
+
+ and 
+\begin_inset Formula $n\neq0$
+\end_inset
+
+ are integers.
+ Suppose further that a subroutine is available to calculate the quantity 
+\begin_inset Formula $\text{round}(u\bmod1)=u\mathring{\bmod}1$
+\end_inset
+
+ for any given floating point number 
+\begin_inset Formula $u$
+\end_inset
+
+,
+ as in exercise 4.2.1–15.
+ One good way to determine whether or not 
+\begin_inset Formula $m$
+\end_inset
+
+ is a multiple of 
+\begin_inset Formula $n$
+\end_inset
+
+ might be to test whether or not 
+\begin_inset Formula $(m\oslash n)\mathring{\bmod}1=0$
+\end_inset
+
+,
+ using the assumed subroutine;
+ but perhaps rounding errors in the floating point calculations will invalidate this test in certain cases.
+\end_layout
+
+\begin_layout Standard
+Find suitable conditions on the range of integer values 
+\begin_inset Formula $n\neq0$
+\end_inset
+
+ and 
+\begin_inset Formula $m$
+\end_inset
+
+,
+ such that 
+\begin_inset Formula $m$
+\end_inset
+
+ is a multiple of 
+\begin_inset Formula $n$
+\end_inset
+
+ if and only if 
+\begin_inset Formula $(m\oslash n)\mathring{\bmod}1=0$
+\end_inset
+
+.
+ In other words,
+ show that if 
+\begin_inset Formula $m$
+\end_inset
+
+ and 
+\begin_inset Formula $n$
+\end_inset
+
+ are not too large,
+ this test is valid.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+A suitable condition would be 
+\begin_inset Formula $|m|<2b^{p-1}$
+\end_inset
+
+;
+ for the proof we may assume 
+\begin_inset Formula $m,n\geq0$
+\end_inset
+
+ as the signs of the operands do not affect the check.
+ In every case,
+ if 
+\begin_inset Formula $n\mid m$
+\end_inset
+
+,
+ then 
+\begin_inset Formula $m\oslash n=\text{round}(\frac{m}{n})$
+\end_inset
+
+ will necessarily be an integer and 
+\begin_inset Formula $(m\oslash n)\mathring{\bmod}1=0$
+\end_inset
+
+.
+ For the reciprocal,
+ if 
+\begin_inset Formula $n\nmid m$
+\end_inset
+
+,
+ then 
+\begin_inset Formula $m\oslash n=\frac{m}{n}+\delta$
+\end_inset
+
+,
+ where 
+\begin_inset Formula $\delta\leq\frac{1}{2}b^{e_{m\oslash n}-p-q}$
+\end_inset
+
+.
+ Note that the exponent is not increased after rounding in the division;
+ if it were,
+ that would mean that 
+\begin_inset Formula $b^{e}(1-\frac{1}{2}b^{-p})\leq\frac{m}{n}<b^{e}$
+\end_inset
+
+ for some integer 
+\begin_inset Formula $e$
+\end_inset
+
+,
+ but then 
+\begin_inset Formula $nb^{e}(1-\frac{1}{2}b^{-p})\leq m<nb^{e}$
+\end_inset
+
+ and,
+ because both 
+\begin_inset Formula $m$
+\end_inset
+
+ and 
+\begin_inset Formula $nb^{e}$
+\end_inset
+
+ are integers,
+ 
+\begin_inset Formula $\frac{1}{2}nb^{e-p}\geq1$
+\end_inset
+
+,
+ so 
+\begin_inset Formula $nb^{e}\geq2b^{p}$
+\end_inset
+
+ and 
+\begin_inset Formula $m\geq2b^{p}-1>2b^{p-1}\#$
+\end_inset
+
+.
+ This means that 
+\begin_inset Formula $e_{m\oslash n}=\lfloor\log_{b}\frac{m}{n}\rfloor+1+q$
+\end_inset
+
+.
+ Now,
+ since 
+\begin_inset Formula $n\nmid m$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\log_{b}\frac{m}{n}\notin\mathbb{Z}$
+\end_inset
+
+,
+ so 
+\begin_inset Formula $\lfloor\log_{b}\frac{m}{n}\rfloor+1=\lceil\log_{b}\frac{m}{n}\rceil=\lceil\log_{b}\frac{m}{2}-\log_{b}\frac{n}{2}\rceil\leq\lceil\log_{b}\frac{m}{2}\rceil-\lfloor\log_{b}\frac{n}{2}\rfloor$
+\end_inset
+
+.
+ With this,
+\begin_inset Formula 
+\[
+\delta\leq\frac{1}{2}b^{\lceil\log_{b}\frac{m}{2}\rceil-\lfloor\log_{b}\frac{n}{2}\rfloor-p}\leq\frac{1}{2}b^{-1}b^{-\lfloor\log_{b}\frac{n}{2}\rfloor}<\frac{1}{2}b^{\cancel{-1}}b^{-(\log_{b}\frac{n}{2}\cancel{-1})}=\frac{1}{n},
+\]
+
+\end_inset
+
+but 
+\begin_inset Formula $\frac{m}{n}$
+\end_inset
+
+ differs from the nearest integer by 
+\begin_inset Formula $\frac{1}{n}$
+\end_inset
+
+ at most,
+ so 
+\begin_inset Formula $m\oslash n\notin\mathbb{Z}$
+\end_inset
+
+ and 
+\begin_inset Formula $(m\oslash n)\mathring{\bmod}1\neq0$
+\end_inset
+
+.
+ This is assuming that there's no exponent underflow,
+ which would be rare because it would mean that 
+\begin_inset Formula $q<p-1$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc15[M24]
+\end_layout
+
+\end_inset
+
+(H.
+ Björk.) Does the computed midpoint of an interval always lie between the endpoints?
+ (In other words,
+ does 
+\begin_inset Formula $u\leq v$
+\end_inset
+
+ imply that 
+\begin_inset Formula $u\leq(u\oplus v)\oslash2\leq v$
+\end_inset
+
+?
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+No.
+ For example,
+ if 
+\begin_inset Formula $b=10$
+\end_inset
+
+,
+ 
+\begin_inset Formula $p=5$
+\end_inset
+
+,
+ 
+\begin_inset Formula $u=5.9998$
+\end_inset
+
+,
+ and 
+\begin_inset Formula $v=5.9999$
+\end_inset
+
+,
+ then 
+\begin_inset Formula $u\oplus v=12.000$
+\end_inset
+
+ and 
+\begin_inset Formula $(u\oplus v)\oslash2=6.0000>v$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc25[15]
+\end_layout
+
+\end_inset
+
+When people speak about inaccuracy in floating point arithmetic they often ascribe errors to 
+\begin_inset Quotes eld
+\end_inset
+
+cancellation
+\begin_inset Quotes erd
+\end_inset
+
+ that occurs during the subtraction of nearly equal quantities.
+ But when 
+\begin_inset Formula $u$
+\end_inset
+
+ and 
+\begin_inset Formula $v$
+\end_inset
+
+ are approximately equal,
+ the difference 
+\begin_inset Formula $u\ominus v$
+\end_inset
+
+ is obtained exactly,
+ with no error.
+ What do these people really mean?
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+It really means that,
+ if the inputs carry a relative error due to rounding,
+ the relative error of the output is potentially much bigger.
+ Let 
+\begin_inset Formula $u_{0}$
+\end_inset
+
+ be the 
+\begin_inset Quotes eld
+\end_inset
+
+correct
+\begin_inset Quotes erd
+\end_inset
+
+ value of 
+\begin_inset Formula $u$
+\end_inset
+
+,
+ that is,
+ the value it would have if the operations so far had been carried out with infinite precision,
+ and let 
+\begin_inset Formula $v_{0}$
+\end_inset
+
+ be the 
+\begin_inset Quotes eld
+\end_inset
+
+correct
+\begin_inset Quotes erd
+\end_inset
+
+ value of 
+\begin_inset Formula $v$
+\end_inset
+
+,
+ similarly defined.
+ Let 
+\begin_inset Formula $u\eqqcolon u_{0}(1+\delta)$
+\end_inset
+
+ and 
+\begin_inset Formula $v\eqqcolon v_{0}(1+\delta')$
+\end_inset
+
+.
+ Then,
+ if 
+\begin_inset Formula $u$
+\end_inset
+
+ and 
+\begin_inset Formula $v$
+\end_inset
+
+ are nearly equal and 
+\begin_inset Formula $u\ominus v$
+\end_inset
+
+ is obtained exactly,
+ then 
+\begin_inset Formula 
+\[
+\frac{u\ominus v}{u_{0}-v_{0}}=\frac{u_{0}(1+\delta)-v_{0}(1+\delta')}{u_{0}-v_{0}}=1+\frac{u_{0}\delta-v_{0}\delta'}{u_{0}-v_{0}}.
+\]
+
+\end_inset
+
+Even if 
+\begin_inset Formula $\delta$
+\end_inset
+
+ and 
+\begin_inset Formula $\delta'$
+\end_inset
+
+ are small,
+ the new relative error 
+\begin_inset Formula $\left|\frac{u_{0}\delta-v_{0}\delta'}{u_{0}-v_{0}}\right|$
+\end_inset
+
+ can be quite big,
+ as there's no reason for 
+\begin_inset Formula $\delta$
+\end_inset
+
+ and 
+\begin_inset Formula $\delta'$
+\end_inset
+
+ to be similar.
+ In the worst case where 
+\begin_inset Formula $\delta=-\delta'$
+\end_inset
+
+,
+ the relative error of the inputs is multiplied by 
+\begin_inset Formula $\left|\frac{u_{0}+v_{0}}{u_{0}-v_{0}}\right|$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc29[M25]
+\end_layout
+
+\end_inset
+
+Give an example to show that the condition 
+\begin_inset Formula $b^{p}\geq3$
+\end_inset
+
+ is necessary in the previous exercise.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+
+\begin_inset Note Greyedout
+status open
+
+\begin_layout Plain Layout
+(I had to look up the solution.) 
+\end_layout
+
+\end_inset
+
+Here 
+\begin_inset Formula $\text{round}(x)=2^{e}$
+\end_inset
+
+ for some integer 
+\begin_inset Formula $e$
+\end_inset
+
+ such that 
+\begin_inset Formula $|x-2^{e}|$
+\end_inset
+
+ is lowest.
+ If 
+\begin_inset Formula $f(x)\coloneqq x^{99/100}$
+\end_inset
+
+,
+ then 
+\begin_inset Formula $g(y)\coloneqq y^{100/99}$
+\end_inset
+
+.
+ Now,
+ for integer 
+\begin_inset Formula $e$
+\end_inset
+
+,
+\begin_inset Formula 
+\begin{multline*}
+\text{round}(f(2^{e}))=\text{round}(2^{e\cdot99/100})<2^{e}\iff(2^{99/100})^{e}<\frac{3}{4}2^{e}\iff\\
+\iff(2^{-1/100})^{e}<\frac{3}{4}\iff e>41.
+\end{multline*}
+
+\end_inset
+
+Conversely,
+\begin_inset Formula 
+\begin{multline*}
+\text{round}(g(2^{e}))=\text{round}(2^{e\cdot100/99})\leq2^{e}\iff(2^{100/99})^{e}<\frac{3}{2}2^{e}\iff\\
+\iff(2^{1/99})^{e}<\frac{3}{2}\iff e<58.
+\end{multline*}
+
+\end_inset
+
+Thus,
+ if 
+\begin_inset Formula $e\in\{42,\dots,58\}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\hat{h}(2^{e})<2^{e}$
+\end_inset
+
+,
+ and it's easy to see that in fact 
+\begin_inset Formula $\hat{h}(2^{e})=2^{e-1}$
+\end_inset
+
+.
+ Thus 
+\begin_inset Formula $\hat{h}^{2}(2^{53})=2^{51}\neq2^{50}=\hat{h}^{3}(2^{53})$
+\end_inset
+
+.
+\end_layout
+
+\end_body
+\end_document
diff --git a/vol2/index.lyx b/vol2/index.lyx
index fba036c..22d5987 100644
--- a/vol2/index.lyx
+++ b/vol2/index.lyx
@@ -987,6 +987,18 @@ literal "false"
 \end_inset
 
 
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+
+\family typewriter
+A10+R25
+\end_layout
+
+\end_inset
+
+
 \end_layout
 
 \begin_layout Section
@@ -1006,12 +1018,47 @@ literal "false"
 \end_inset
 
 
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+
+\family typewriter
+A10+R25
+\end_layout
+
+\end_inset
+
+
 \end_layout
 
 \begin_layout Subsection
 Accuracy of Floating Point Arithmetic
 \end_layout
 
+\begin_layout Standard
+\begin_inset CommandInset include
+LatexCommand input
+filename "4.2.2.lyx"
+literal "false"
+
+\end_inset
+
+
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+
+\family typewriter
+A10+R25
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
 \begin_layout Subsection
 Double-Precision Calculations
 \end_layout