HP HP-15C - Summary of Matrix Functions

Appendix: Accuracy of Numerical Calculations 175
Example 1 Explained. Heref(x) = s(r(x)), where
50
roots
and
The exponents are Vz50 = 8.8818 X KT16 and 250= 1.1259 X 1015.
Now, x must lie between 10~" and 9.999 ... X 10" since no positive
numbers outside that range can be keyed into the calculator. Since
r is an increasing function, r(x) lies between
r(l(r99) = 0.9999999999997975...
and
r(10100) = 1.0000000000002045 ....
This suggests that R(x), the calculated value of r(x), would be 1 for
all valid calculator arguments x. In fact, because of roundoff,
fo
\1.
9999999999 for 0 < x< 1
000000000 for 1 < x ^ 9.999999999 X 10".
If 0<x<l, then x ^ 0.9999999999 in a 10-digit calculator. We
would then rightly expect that -^x^ V0.9999999999, which is
0.999999999949999999998... , which rounds to 0.9999999999 again.
Therefore, if | Jx |is pressed arbitrarily often starting with x<l, the
result cannot exceed 0.9999999999 . This explains why we obtain
R(x) = 0.9999999999 for 0 < x < I above. When R(x) is squared 50
times to produce F(x) = S(R(x)), the result is clearly 1 for x ^ 1, but
why is F(x) = 0 for 0 < x < 1? When x < 1,
s(R(x)) < s(0.9999999999) = (1 - lO'10)250 « 6.14 X IQ'48898.