First Semester in Numerical Analysis with Julia, 2020a

CHAPTER 1. INTRODUCTION 44
Next will use four-digit arithmetic with rounding to compute the roots:
fl( √ 117) = 10.82
(
fl(fl(11.0) + fl( √ ) ( ) ( )
117)) fl(11.0+10.82) 21.82
fl(r 1 )=fl
= fl
= fl =10.91
fl(2.0)
2.0
2.0
(
fl(fl(11.0) − fl( √ ) ( ) ( )
117)) fl(11.0 − 10.82) 0.18
fl(r 2 )=fl
= fl
= fl =0.09.
fl(2.0)
2.0
2.0
The relative errors are:
rel error in r 1 =
10.90832691 − 10.91
∣ 10.90832691 ∣ =1.5 × 10−4
rel error in r 2 =
0.09167308680 − 0.09
∣ 0.09167308680 ∣ =1.8 × 10−2 .
Notice the larger relative error in r 2 compared to that of r 1 , about a factor of 100, whichis
due to cancellation of leading digits when we compute 11.0 − 10.82.
One way to fix this problem is to rewrite the offending expression by rationalizing the
numerator:
r 2 = 11.0 − √ 117
2
( ) √
1 11.0 − 117
=
2 11.0+ √ 117
If we use this formula to compute r 2 we get:
(
)
2.0
fl(r 2 )=fl
fl(11.0+fl( √ 117))
(
11.0+ √ ) ( 1
117 =
2)
4
11.0+ √ 117 = 2
11.0+ √ 117 .
( ) 2.0
= fl =0.09166.
21.82
The new relative error in r 2 is:
( )
0.09167308680 − 0.09166
rel error in r 2 =
=1.4 × 10 −4 ,
0.09167308680
an improvement about a factor of 100, even though in the new way of computing r 2 there
are two operations where rounding error happens instead of one.
Example 20. The simple procedure of adding numbers, even if they do not have mixed signs,
can accumulate large errors due to rounding or chopping. Several sophisticated algorithms
to add large lists of numbers with accumulated error smaller than straightforward addition
exist in the literature (see, for example, Higham [11]).
For a simple example, consider using four-digit arithmetic with rounding, and computing