j^j and the HP33S Message #1 Posted by Karl Schneider on 27 Nov 2004, 11:57 p.m.
An unintuitive mathematical fact:
j^{j} = e^{pi/2} = 0.207879576
where j = sqrt(1) and j^{2} = 1
The proof (which is left as an exercise for the reader) requires the use of Euler's identity and trigonometric/hyperbolic identities derived therefrom:
e^{jx} = (cos x) + j*(sin x)
sin jx = j * sinh x
cos jx = cosh x
On any RPN or RPL calculator designed and engineered by HewlettPackard with complexnumber capabilities, j^{j} can always be evaluated correctly. For example:
HP15C HP42S
1 0
f Re<>Im ENTER
ENTER 1
y^x f CMPLX
ENTER
y^x
For the 32S, 32SII, 33S, and 41C* with Math or Advantage ROM, a different approach is used:
1
ENTER
0
ENTER
1
ENTER
0
f CMPLXy^x XEQ Z^W
(32/33) (41/Math/Adv)
This procedure works for all four calculators, but the 33S gives the incorrect answer e^{3*pi/2} = 0.008983291 when the zero in the zregister is negated (0).
The root cause of this problem is the same one that causes the incorrect rectangulartopolar conversions with 0 or 0 in the xregister, which I and others documented fully, several months ago in this forum.
Calculation of j^{j} requires the calculation of ln(j).
Re [ln(a+jb)] = ln (sqrt(a^{2}+b^{2}))
Im [ln(a+jb)] = atan2(a, b) [in radians]  primary solution
The antilogarithm of the j residing in the "upper" part of the stack is taken. Since the 33S calculates the phase angle of (0 + j1) incorrectly, the answer given for j^j is incorrect.
Norris (or anyone else), has HP addressed the polarangle and other bugs?
 KS
Edited: 28 Nov 2004, 12:01 a.m.
