MTH3051 Introduction to Computational Mathematics - User Web ...

School of Mathematical Sciences
Monash University
approach (one that can be used for other more challenging computations). We will start
by writing both x and y in the standard form
Thus we have
then
x = (a + 10 −N b) × 10 m
y = (c + 10 −N d) × 10 n
˜x = a × 10 m , E R (x) = 10 −N × (b × 10 m )
ỹ = c × 10 m , E R (y) = 10 −N × (d × 10 m )
x
y = a + 10−N b
c + 10 −N d 10m−n
On most computers N ≈ 15. So 10 −N d is much smaller than c. Thus we can use a
Taylor series for 1/(1 + 10 −N (d/c)) in powers of 10 −N to produce
x
y = 1 c (a + 10−N b)10 m−n (1 − 10 −N d d2
+ 10−2N
c c − · · · )
2
If we retain just the first two terms in the series then we obtain (after a wee bit of
algebra)
( (
x a b
y = c + 10−N c − ad ))
10 m−n + O ( 10 −2N)
c 2
Our computer will compute the approximation ˜z to the exact value z = x/y with a
round off error which we denote by E R (z). We have z = ˜z + E R (z) and (ignoring any
carries)
˜z = a c × 10m−n
E R (z) = 10 −N ( b
c − ad
c 2 )
10 m−n
= 10 −N O (z)
This last line shows that our N− digit computer will return an estimate for z = x/y
accurate to N digits. This is good! We rejoice (in moderation).
This is a very general technique and it can be applied to any function to analyse its
sensitivity to round-off errors. It is well worth your time studying the above example in
detail (as you always do, n’est pas?).
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