First Semester in Numerical Analysis with Julia, 2020a

CHAPTER 1. INTRODUCTION 38
Chopping & Rounding
Let x be a real number with more digits the computer can handle: x =0.d 1 d 2 ...d k d k+1 ...×
10 n . How will the computer represent x? Let’s use the notation fl(x) for the floating-point
representation of x. There are two choices, chopping and rounding:
• In chopping, we simply take the first k digits and ignore the rest: fl(x) =0.d 1 d 2 ...d k .
• In rounding, if d k+1 ≥ 5 we add 1 to d k to obtain fl(x). If d k+1 < 5, then we simply
do as in chopping.
Example 11. Find 5-digit (k =5) chopping and rounding values of the numbers below:
• π =0.314159265... × 10 1
Chopping gives fl(π) =0.31415 × 10 1 and rounding gives fl(π) =0.31416 × 10 1 .
• 0.0001234567
We need to write the number in the normalized representation first as 0.1234567×10 −3 .
Now chopping gives 0.12345 × 10 −3 and rounding gives 0.12346 × 10 −3 .
Absolute and relative error
Since computers only give approximations to real numbers, we need to be clear on how we
measure the error of an approximation.
Definition 12. Suppose x ∗ is an approximation to x.
• |x ∗ − x| is called the absolute error
• |x∗ −x|
|x|
is called the relative error (x ≠0)
Relative error usually is a better choice of measure, and we need to understand why.
Example 13. Find absolute and relative errors of
1. x =0.20 × 10 1 ,x ∗ =0.21 × 10 1
2. x =0.20 × 10 −2 ,x ∗ =0.21 × 10 −2
3. x =0.20 × 10 5 ,x ∗ =0.21 × 10 5
Notice how the only difference in the three cases is the exponent of the numbers. The
absolute errors are: 0.01 × 10 , 0.01 × 10 −2 , 0.01 × 10 5 . The absolute errors are different
since the exponents are different. However, the relative error in each case is the same: 0.05.