Little HP-15C challenge for the weekend « Next Oldest | Next Newest »

 ▼ Ex-PPC member Unregistered Posts: 142 Threads: 24 Joined: Jan 1970 03-13-2003, 07:41 AM 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: ``` PI 2 / CHS e^x ``` That's 5 steps. The challenge is to do it in four steps. A simple way would be using trigs, like this: ``` 1 SIN-1 (arcsin) CHS e^x ``` 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. ▼ Ex-PPC member Unregistered Posts: 142 Threads: 24 Joined: Jan 1970 03-13-2003, 07:54 AM 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. Werner Huysegoms Unregistered Posts: 44 Threads: 0 Joined: Jan 1970 03-13-2003, 10:36 AM I don't even have a 15C, but that's easy, no? PI CHS e^x SQRT Werner ▼ Ex-PPC member Unregistered Posts: 142 Threads: 24 Joined: Jan 1970 03-13-2003, 10:47 AM 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) ▼ axel Unregistered Posts: 14 Threads: 3 Joined: Jan 1970 03-13-2003, 11:34 AM Hello, 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 Axel ▼ Karl Schneider Unregistered Posts: 1,792 Threads: 62 Joined: Jan 2005 03-14-2003, 02:33 AM ```HP-15C HP-42S 1 1 f I COMPLEX ENTER ENTER y^x y^x calculates j^j = exp(-pi/2) ~= 0.20788, where j = sqrt(-1) THE PROOF ========= 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) ``` ▼ axel Unregistered Posts: 14 Threads: 3 Joined: Jan 1970 03-14-2003, 04:23 AM 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. Regards Axel ▼ R Lion (Spain) Unregistered Posts: 294 Threads: 32 Joined: Jul 2005 03-14-2003, 11:41 AM "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: i ENTER ENTER y^x ->NUM Regards.Raul ▼ Karl Schneider Unregistered Posts: 1,792 Threads: 62 Joined: Jan 2005 03-15-2003, 12:12 AM Raul wrote that "i ENTER ENTER y^x ->NUM" 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} Thanks... ▼ R Lion (Spain) Unregistered Posts: 294 Threads: 32 Joined: Jul 2005 03-15-2003, 02:32 AM Very happy for helping you: I'm enjoying your "New multiple Varaible 15C Solver"... ;-) Raul Karl Schneider Unregistered Posts: 1,792 Threads: 62 Joined: Jan 2005 03-14-2003, 12:45 AM 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 CHS e^x 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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