First Semester in Numerical Analysis with Julia, 2020a

CHAPTER 1. INTRODUCTION 45
the average of two numbers, a+b . For a =2.954 and b = 100.9, the true average is 51.927.
2
However, four-digit arithmetic with rounding yields:
( ) ( ) ( )
100.9+2.954 fl(103.854) 103.9
fl
= fl
= fl =51.95,
2
2
2
which has a relative error of 4.43 × 10 −4 . If we rewrite the averaging formula as a + b−a,on
2
the other hand, we obtain 51.93, which has a much smaller relative error of 5.78 × 10 −5 .The
following table displays the exact and 4-digit computations, with the corresponding relative
error at each step.
a+b
b−a
a b a+ b b − a a + b−a
2 2 2
4-digit rounding 2.954 100.9 103.9 51.95 97.95 48.98 51.93
Exact 103.854 51.927 97.946 48.973 51.927
Relative error 4.43e-4 4.43e-4 4.08e-5 1.43e-4 5.78e-5
Example 21. There are two standard formulas given in textbooks to compute the sample
variance s 2 of the numbers x 1 , ..., x n :
[ ∑n
1. s 2 = 1
n−1 i=1 x2 i − 1 (∑ n
n i=1 x i) 2] ,
2. First compute ¯x = 1 n
∑ n
i=1 x i, and then s 2 = 1
n−1
∑ n
i=1 (x i − ¯x) 2 .
Both formulas can suffer from roundoff errors due to adding large lists of numbers if n is
large, as mentioned in the previous example. However, the first formula is also prone to error
due to cancellation of leading digits (see Chan et al [6] for details).
For an example, consider four-digit rounding arithmetic, and let the data be
1.253, 2.411, 3.174. The sample variance from formula 1 and formula 2 are, 0.93 and 0.9355,
respectively. The exact value, up to 6 digits, is 0.935562. Formula 2 is a numerically more
stable choice for computing the variance than the first one.
Example 22. We have a sum to compute:
e −7 =1+ −7
1 + (−7)2 + (−7)3
2! 3!
+ ... + (−7)n .
n!
The alternating signs make this a potentially error prone calculation.
Julia reports the "exact" value for e −7 as 0.0009118819655545162. If we use Julia to
compute this sum with n =20, the result is 0.009183673977218275. Here is the Julia code
for the calculation: