Little HP-15C challenge for the weekend


Hi, here's a new, little challenge for all HP-15C fans
out there. If you think you know your 15C inside out,
this is your chance to prove it:

The challenge

Assume a master-cleared HP-15C. We want to compute the value of
e^(-Pi/2) [=0.2078+] on a 15c. The direct way to do it would be:


That's 5 steps. The challenge is to do it in four steps. A simple
way would be using trigs, like this:

SIN-1 (arcsin)

but that doesn't work on a master-cleared HP-15C, because this
procedure assumes radians mode and a master-cleared HP-15C defaults
to degrees , so you would need an additional RAD step at the beginning,
making 5 total. Can you do it in four ?

By the way, should you be interested in the solution to my latest Short & Sweet Math Challenge #6,
I've posted two short, commented programs for the HP-71B,
see the post S&SMC#6: An HP-71B solution in its proper thread below.


By the way, a little remark is in order:

By "master-cleared" I don't mean an actually all-erased 15C (i.e: Pr Error), but just use the term to state that you cannot depend on any particular mode or register contents, i.e: you cannot assume that RAD mode is set, or even that the X-register initially contains a 0 value (which would be the case after a real "master clear").
If you need a particular mode you must set it, and if you need some definite value in X to begin with, you must put it in there yourself.


I don't even have a 15C, but that's easy, no?




Yes ! Give the man a cigar !! (just joking)

Now that you've found a 4-step solution so easily, let's
see if you would succeed in finding another 4-step solution that doesn't include any of the four operations you did use, namely without using neither Pi, nor CHS, nor e^x, nor SQRT !!


If you succeed under those conditions, a cigar won't suffice, you'll deserve the whole box ... (there is such
a solution, of course)



I don't have the 15C (longing for it for years...), only read the documentation at the hp-museum about entering complex numbers.

Assuming an empty stack (all levels are zero) it should work like that, unfortunately I can't prove it...

1 fI Enter ^

Best regards


HP-15C      HP-42S

1 1
y^x y^x

calculates j^j = exp(-pi/2) ~= 0.20788, where j = sqrt(-1)


Euler's Identity: exp(jw) = cos(w) + j*sin(w)

==> for w = pi/2 radians, exp(j*pi/2) = 0 + j*1 = j

j^j = exp(ln(j^j))

= exp(j*ln(j))

= exp(j*ln(exp(j*pi/2)))

= exp(j*j*pi/2)

= exp(-pi/2)


Glad to see that it worked that way. "I can't proof" meant I can't check it on a real 15C. However, thank you, Karl for explaining it so nicely.

BTW, what I always missed on my HP calcs (32sii, 48s, 40g) is that there is no direct way to enter i. My nephew's TI-83 e.g. has the i on the keyboard. May be HP decided so because, in everyday math, you don't it so often.



"I always missed on my HP calcs (32sii, 48s, 40g) is that there is no direct way to enter i."

Please, try this on your 48:




Raul wrote that


will enter i and calculate i^i on an HP-48.

It works! I never knew that. Also, the HP-49G has an "i" entry key. {blue left shift}, (TOOL}



Very happy for helping you: I'm enjoying your "New multiple Varaible 15C Solver"... ;-)



Taking advantage of the HP-15C's fine and complete implementation of functionality with domains of complex-valued input and output.

f I ( or, g SF 8)

g COS-1



Axel's post below is the cleverest, and quite likely the solution: So, perhaps the trick is to prove that

j^j = exp(-pi/2)

by Euler's identity: exp(j*w) = cos w + j*sin w

and the trig/hyberbolic identities for complex arguments.

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