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DejMar
(Possibly a robot)

 Posted: Mon Dec 24, 2012 2:15 pm    Post subject: 1 There are Mathematical Recreation Puzzles where one uses a given number of one digit and a limited set of mathematical symbols for operations and functions to represent a number. One of the more recognized forms of this puzzle is with five fives. For example, to represent the value 27 one might present 27=5×5+(5+5)/5. Use of any function that demands one or more letters (to include the Greek and Hebrew letters) in its notation is forbidden, this eliminates using many mathematical numerical constants like i, φ and π. The challenge here is to present a "Five Fives" expression under the rules for the values of each of these mathematical constants: i, φ and π.
groza528
No Place Like Home

 Posted: Mon Dec 24, 2012 4:27 pm    Post subject: 2 I'll tackle the easy one: i = sqrt(5+5-55/5)
Thok
Oh, foe, the cursed teeth!

 Posted: Mon Dec 24, 2012 4:31 pm    Post subject: 3 π = (5-5/5)*arctan(5/5)
Chuck
Daedalian Member

 Posted: Mon Dec 24, 2012 4:50 pm    Post subject: 4 φ = sqrt(5)×5/(5+5)+.5

 Posted: Mon Dec 24, 2012 4:59 pm    Post subject: 5 Doesn't sqrt contain 4 letters? And.. arctan is 6 letters. These are both more than one letter. I guess there could be a symbol for sqrt. But, I think arctan is still out given these rules.
Chuck
Daedalian Member

 Posted: Mon Dec 24, 2012 5:11 pm    Post subject: 6 I could edit in the symbol for square root if necessary.
Thok
Oh, foe, the cursed teeth!

Posted: Mon Dec 24, 2012 5:15 pm    Post subject: 7

 jadesmar wrote: Doesn't sqrt contain 4 letters? And.. arctan is 6 letters. These are both more than one letter.

You are going to have a very hard time making pi without using an inverse trig function. It might be provably impossible.

Edit: you might be able to use natural logs instead of inverse trig functions, but those will still require multiple letter functions. Or maybe do a funky definite integral.

Edit: it might be possible to use zeta(2)=pi^2/6 (where zeta is a single greek letter representing the Riemann zeta function). The best I can do with that line is pi=sqrt(zeta((5+5)/5)*(5+5/5)) which uses one too many fives.

Last edited by Thok on Mon Dec 24, 2012 5:56 pm; edited 2 times in total
Zag
Tired of his old title

 Posted: Mon Dec 24, 2012 5:27 pm    Post subject: 8 i without using sqrt: (- 55/55) ^ .5 or, if you are not even allowing a decimal point: (-5/5) ^ (5/(5+5))
DejMar
(Possibly a robot)

 Posted: Tue Dec 25, 2012 4:26 am    Post subject: 9 Great job, Zag in tackling i without using the decimal point. Both the decimal point and the radical sign are permitted under the normal rules - thus the notation SQRT( ) is acceptable as a proxy for the radical sign. Functions, operations and other mathematical notations that use letters - to include those of a non-English alphabet - are not otherwise permitted. Chuck's solution for φ (phi) and groza528's solution for i (the imaginary unit) are both valid. Thok's use of 'arctan' for the trigonometric function and 'zeta' for the Riemann zeta function are invalid solutions for π (pi) . But do not give up hope, there are still solutions for Archimedes constant that are still available.
Thok
Oh, foe, the cursed teeth!

Posted: Tue Dec 25, 2012 6:28 am    Post subject: 10

 DejMar wrote: But do not give up hope, there are still solutions for Archimedes constant that are still available.

If you are only allowing plus,minus,times, divide, radicals, and exponentials, you are going to run into the problem that pi is a transcendental number. In fact, an exponential must be used to make pi with that restriction.
DejMar
(Possibly a robot)

 Posted: Tue Dec 25, 2012 2:17 pm    Post subject: 11 A solution for Archimedes' constant: Pi, π: (Spoiler follows) π = (.5! / .5) ((5 + 5) / 5) Though it is not standard decimal notation, decimal notation without an integer part is an acceptable form of notation and is also used in the "Five Fives" puzzle. This permits the fractions 1/2 (.5) 11/20 (.55), etc. and even the recurring decimal 5/9 (.5...), etc. with using only one or two fives. The factorial is one of the principal unary functions that is also used. In some variations of the "Five Fives" puzzle it is permitted to use other factorial-like functions (and in some, to use the Greek letter functions, though I intentionally excluded these in this puzzle) The n!! (double factorial), !n (subfactorial), and n# (primorial) allow many integers to be represented without any additional decimal digits than the fives.
Thok
Oh, foe, the cursed teeth!

Posted: Tue Dec 25, 2012 4:25 pm    Post subject: 12

 DejMar wrote: The factorial is one of the principal unary functions that is also used.

My comment still stands. There isn't a good reason to allow ! but not arctan.

Posted: Tue Dec 25, 2012 5:43 pm    Post subject: 13

Thok wrote:
 DejMar wrote: The factorial is one of the principal unary functions that is also used.

My comment still stands. There isn't a good reason to allow ! but not arctan.

Except that the puzzle rules say not to use functions that use letters including Greek and Hebrew letters.

It's a puzzle, why do you need a better reason than that?
lostdummy
Daedalian Member

 Posted: Wed Dec 26, 2012 11:02 am    Post subject: 14 Actually, we can get pi with less than five 'fives' under those rules. Best I could get on short notice is 'pi' with only 3 fives. *EDIT* I found solution for pi with only two fives, which means that now I have solution with any number of fives between two and infinity ;p
DejMar
(Possibly a robot)

 Posted: Thu Dec 27, 2012 11:37 am    Post subject: 15 lostdummy, I am interested in your solution. Though it was not stated, multifactorials of higher than 5th degree should be disallowed. My personal position is that no distinct symbol can be used more than 5 times, i.e. ((5!)/5!!) - 5!! constitutes 5 distinct !, even though there is a separate double exclamation point character in Unicode. For clarity, not grouping requirements, more than five sets of brackets may be used. [These are personal rules of mine in the "Five Fives" puzzles that I have adopted. With these restrictions I have found solutions for the integers 0 to over 1000.]
lostdummy
Daedalian Member

 Posted: Thu Dec 27, 2012 12:17 pm    Post subject: 16 Well, for 'best' solution with only two fives, I dont use any symbol more than twice: pi(2*5)= (-.5)!*(-.5)! Any other combination with more fives than three can use solution with two fives and fact that (5-5)=0, so for example: pi(4*5) = (-.5)!*(-.5)!+ (5-5) pi(5*5) = (-.5)!*(-.5)!+ (5-5)*5 pi(6*5) = (-.5)!*(-.5)!+ (5-5)+ (5-5) etc It gets more tricky for solution with three fives. My initial idea was pi=(-.5)!^(5-5!!!) , but then I found out that !!! does not start from 1 (in which case 5!!! would be 3), but downward from 5 (so 5!!!=10). So, I found alternate way with three fives (which amount to alternate ways to get 2 with two fives): pi(3*5)= (-.5)!^(5!!!/5) pi(3*5)= (-.5)!^(5#/5!!) pi(3*5)= (-.5)!^sqrt(5!/5#)
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