Application – HP 15c User Manual
Page 51
Section 2: Working with
f
51
01
.
0
/
a
c
r
However, this integral is nearly improper because q and r are both so nearly zero. But by
using an integral in closed form that sufficiently resembles the troublesome part of V, the
difficulty can be avoided. Try
.
10
8
4018188070
.
8
))
/(
)
1
1
ln((
)
ln(
/
6
2
1
2
1
2
q
r
r
q
p
q
u
u
p
q
u
du
p
W
r
r
Then
.
)
1
1
(
1
/
)
/
1
)
/(
)
1
(
(
1
2
2
2
1
2
2
2
du
q
u
u
u
r
p
W
p
du
q
u
q
u
u
p
W
V
r
r
The HP-15C readily handles this integral. Don't worry about
2
1 u
as u approaches 1
because the figures lost to roundoff aren't needed.
Application
The following program calculates the values of four special functions for any argument x:
dt
e
x
t
2
/
2
2
1
)
(
P
(normal distribution function)
x
t
dt
e
x
P
x
2
/
2
2
1
)
(
1
)
(
Q
(complementary normal distribution function)
x
t
dt
e
x
0
2
2
)
(
erf
(error function)
x
t
dt
e
x
x
2
2
)
(
erf
1
)
(
erfc
(complementary error function)
The program calculates these functions using the transformation
2
t
e
u
whenever |x| > 1.6.
The function value is returned in the X-register, and the uncertainty of the integral is returned
in the Y-register. (The uncertainty of the function value is approximately the same order of
magnitude as the number in the Y-register.) The original argument is available in register R
0
.