The Chemistry Maths Book, Second Edition

20.2 Errors 559

1.2335 and 1.2345. The number ais said to have a maximum absolute error or

(absolute) error boundε

a

1 = 1 0.0005, and the bound can be indicated by writing

a 1 ± 1 ε

a

1 = 1 1.2340 1 ± 1 0.0005

0 Exercises 1, 2

In fixed-point arithmetic all numbers are rounded to the same number of decimal

places, and therefore have the same error bound, εsay. The sum or difference of two

numbers then has error bound 2 ε. More generally, the sum or difference of numbers

aand b, with error boundsε

a

andε

b

, has error boundε

a

1 + 1 ε

b

.

EXAMPLE 20.1The numbersa 1 = 1 1.234andb 1 = 1 3.468have been rounded to 4

significant figures. Determine the error bound fora 1 − 1 b.

The numbers aand bboth have error boundε

a

1 = 1 0.0005. The largest and smallest

possible values ofa 1 − 1 bare therefore

(a 1 + 1 ε) 1 − 1 (b 1 − 1 ε) 1 = 1 (a 1 − 1 b) 1 + 12 ε or 1.2345 1 − 1 3.4675 1 = 1 −2.2330 1 = 1 −2.234 1 + 1 0.001

(a 1 − 1 ε) 1 − 1 (b 1 + 1 ε) 1 = 1 (a 1 − 1 b) 1 − 12 ε or 1.2335 1 − 1 3.4685 1 = 1 −2.2350 1 = 1 −2.234 1 − 1 0.001

and the error bound of the difference is 2 ε 1 = 1 0.001. Thereforea 1 − 1 b 1 = 1 −2.234 1 ± 1 0.001

0 Exercise 3

For multiplication and division the error propagation is best described in terms of

the relative error. A number whose value is given by aand whose true value isa

t

has

relative error

(20.1)

When the error is small enough,a

t

in the denominator can be replaced by aand (20.1)

byr

a

1 = 1 ε

a

2 |a|.

0 Exercise 4

The relative error of the product or quotient of aand b, with relative errorsr

a

andr

b

, isr

a

1 + 1 r

b

. Thus, for multiplication, the largest and smallest possible values of

p 1 = 1 a 1 × 1 bare

(a 1 + 1 ε

a

) (b 1 + 1 ε

b

) 1 = 1 ab 1 + 1 aε

b

1 + 1 bε

a

1 + 1 ε

a

ε

b

(a 1 − 1 ε

a

) (b 1 − 1 ε

b

) 1 = 1 ab 1 − 1 aε

b

1 − 1 bε

a

1 + 1 ε

a

ε

b

r

aa

aa

a

t

t

a

t

=

−

=

ε

||