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This is to be tried on your trusty scientific Voyager or Spice:
Let's say you want 1.415926536 on the display. Do it using four steps at most. Assume radians mode only.
Gerson.
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How's this:
3 0 COS ACOS
I'm assuming the appropriate FIX setting and that we're not currently entering a number. Prepending a CLx fixes this at the cost of another step.
- Pauli
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Congratulations!
The equation acos(cos(x)) = 10(pi - 3) has infinite solutions, of which only two are integers (if I am not wrong): 30 and -30.
Also interesting:
3 0 SIN ASIN
3 0 TAN ATAN
Gerson.
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Grson,
The solutions to most of these quizes are based on some equation one finds in a math handbook or web site.
Namir
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Namir,
I doubt the solution to this particular one is in the web or any math book because I created it myself. I was trying SIN(30 deg) on th 12CP with my program. Until I obtained -0.988031624 I was not sure what angle mode the program was in. When I computed ASIN the answer was -1.415926536, that is, -10 * FRAC(pi). Obviously these so many digits could not be a coincidence as a further investigation has confirmed.
I have yet to create a quiz Paul Dale is not able to solve, or at least he doesn't solve it instantly... :-)
Gerson.
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Another discovery that comes by coincidence. All good to me!!!
Namir
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Another interesting one, without going into radians, which alone takes two steps. On my 48SX: (-1 ln IM 3 -) = 0.14159265359... (I have IM on my CST menu. 10x will bring it up to 1.4159... if necessary.)
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Hi John,
This is interesting indeed. However on my 48S I'd rather do the following, if I didn't want to take the trouble to change the angle mode do RAD:
<< pi FP 10 * ->NUM >>
Actually, those quizzes make sense only on RPN calcs, on which every step counted. It was funny to program them, but it was funnier do it using as few steps as possible.
Regards,
Gerson.
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One more interesting note...
acos(cos(0))=0.00000
acos(cos(1))=1.00000
acos(cos(2))=2.00000
acos(cos(3))=3.00000
acos(cos(4))=2.28318
acos(cos(5))=1.28318
acos(cos(6))=0.28318
continuing this we get
acos(cos(29))=2.4159265359
acos(cos(30))=1.4159265359
acos(cos(31))=0.4159265359
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Really interesting! It seems I missed those... These shouldn't be hard to prove: start by graphing y = acos(cos(x)) (from x=28 to x=32).
Now, compute this on one of your calculators:
250 999 ___
--- * ----- * \/ 2
7 16061
I came up with this yesterday, when observing sqrt(8) / (0.1 * pi)= 9.003163161. Not so interesting, except perhaps the palindromic fraction...
Gerson.
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Quite similar to this is the (probably) well-known variant which asks to create the value Pi/10 in an HP-15C in just 3 program steps (the obvious sequence PI, 1, 0, /, is *four* steps long in the HP-15C).
Best regards from V.
Edited: 19 Dec 2006, 7:38 a.m.
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I thought that the idea was to NOT use the PI key. That would lead to too obvious solutions!
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The idea of whom ? Mine is a different variant, with different objetive and rules. Can you do it in less than four (program) steps in an HP-15C ?
It's fairly trivial, by the way.
Best regards from V.
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Hi Valentin,
I didn't know this one, but I do remember one of your articles on the HP-15C:
pi
STO RAN#
RCL RAN#
Best regards,
Gerson.
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Xacto.
Best regards from V.
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