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HP HP-15C Advanced Functions Handbook

HP HP-15C
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182
Appendix: Accuracy
of
Numerical Calculations
Display
009-
26
Places
1 in
X-register.
010-43,30,
6
Tests
M
5*1.
011-
30
Calculates
u
1
when
!)
or
012-
013-
014-
10
Calculates
x/(u-
1/1.
20
Calculates
A(x).
43 32
The
calculated value
of u,
correctly rounded
by the
HP-15C,
is
u
= (1 + e) (1 + x),
where|e|
< 5 X
lO'10.
Ifu
= 1,
then
too,
in
which
case
the
Taylor series
X(x)
x
(I
Vzx
+
Vzx2
...)
tells
us
that
the
correctly rounded value
of
\(x)
must
be
just
x.
Otherwise,
we
shall
calculate
*
A(w
!)/(«
1)
fairly accurately
instead
of
k(x).
But
A(x)/x
= 1
V-ix
+
Vox'2
...
varies very
slowly,
so
slowly
that
the
absolute error
k(x)/x
A(u
l)/(u
~ 1) is no
worse
than
the
absolute error
x
(«
!)
=
e(l
+ x), and if x
^
1,
this
error
is
negligible relative
to
\(x)/x.
When
x > 1,
then
u 1 is
so
nearly
x
that
the
error
is
negligible again;
A(;t)
is
correct
to
nine
significant digits.
As
usual
in
error analyses,
the
explanation
is far
longer
than
the
simple procedure being explained
and
obscures
an
important fact:
the
errors
in
ln(a)
and u
1
were ignored during
the
explanation
because
we
knew they
would
be
negligible. This knowledge,
and
hence
the
simple procedure,
is
invalid
on
some other calculators
and big
computers! Machines
do
exist which calculate
ln( u)
and/or
1
u
with small absolute error,
but
large
relative
error when
u is
near
1; on
those machines
the
foregoing calculations must
be
wrong
or
much more complicated,
often
both.
(Refer
to the
discussion under Level
2 for
more about this.)
Back
to
Susan's
sum.
By
using
the
foregoing simple procedure
to
calculate
\(i/n)
=
ln(l
+
i/n)
=
3.567351591
X
1CT9,
she
obtains
a
better value:
(1
+
i/n)n
=
enk(l/n)
=
1.119072257

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HP HP-15C Specifications

General IconGeneral
BrandHP
ModelHP-15C
CategoryCalculator
LanguageEnglish

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