User Manual

Section 2: Working with f 51
01.0/  acr
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
.1084018188070.8
))/()11ln((
)ln(/
6
2
1
2
1
2





qrrqp
quupqudupW
r
r
Then
.
)11(
1
/
)/1)/()1((
1
22
2
1
222
du
quu
u
r
pW
p
duququupWV
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:




 dtex
t 2/
2
2
1
)(P

(normal distribution function)




x
t
dtexPx
2/
2
2
1
)(1)(Q

(complementary normal distribution function)



x
t
dtex
0
2
2
)(erf

(error function)




x
t
dtexx
2
2
)(erf1)(erfc

(complementary error function)
The program calculates these functions using the transformation
2
t
eu


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
.