HP 15c - Manual (Page 161)

Appendix: Accuracy of Numerical Calculations 161
when we wish to calculate f(x), the best we could hope to get is f(x + Δx), but we actually get
F(x+Δx)=(f+δf)(x+Δx+δx), where |δf| ≤ ε|f| and |δx| ≤ η|x|.
What we get is scarcely worse than the best we could hope for provided the tolerances ε and
η are small enough, particularly if |Δx| is likely to be at least roughly as big as η|x|. Of course,
the best we could hope for may be very bad, especially if f possesses a singularity closer to x
than the tolerances upon x's perturbation Δx and δx.
Backward Error Analysis Versus Singularities
The word "singularity" refers to both a special value of the argument x and to the way f(x)
misbehaves as x approaches that special value. Most commonly, f(x) or its first derivative
f'(x) may become infinite or violently oscillatory as x approaches the singularity. Sometimes
the singularities of ln|f| are called singularities of f, thereby including the zeros of f among its
singularities; this makes sense when the relative accuracy of a computation of f is at issue, as
we shall see. For our purposes the meaning of "singularity" can be left a little vague.
What we usually want to do with singularities is avoid or neutralize them. For instance, the
function
 




 

otherwise 2 1
0 if
2
cos 1
) (
x x x
x c
has no singularity at x = 0 even though its constituents 1 − cos x and x
2
(actually, their
logarithms) do behave singularly as x approaches 0. The constituent singularities cause
trouble for the program that calculates c(x). Most of the trouble is neutralized by the choice
of a better formula
 













otherwise. 2 1
0 2 if
2
2
2 sin
2
1
) (
x
x
x
x c
Now the singularity can be avoided entirely by testing whether x/2 = 0 in the program that
calculates c(x).
Backward error analysis complicates singularities in a way that is easiest to illustrate with the
function λ(x) = ln(1 + x) that solved the savings problem in example 2. The procedure used
there calculated u = 1 + x (rounded) = 1 + x + Δx. Then






otherwise. ) 1 ( ) ln(
1 u if
) (
u x u
x
x

This procedure exploits the fact that λ(x)/x has a removable singularity at x = 0, which means
that λ(x)/x varies continuously and approaches 1 as x approaches 0. Therefore, λ(x)/x is
relatively closely approximated by λ(x+Δx) / (x+Δx) when |Δx| < 10
-9
, and hence
λ(x) = x(λ(x)/x) ≈ x(λ(x+Δx)/(x+Δx)) = x(ln(u)/(u−1)),