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General Category => Everything Else => Topic started by: vh on September 28, 2012, 06:37:54 PM

Title: Math help
Post by: vh on September 28, 2012, 06:37:54 PM: bla or matty or someone who knows what they're doing:

1 <= floor((L)/(2(x-floor(x/3))^2+2(x-floor(x/3))+1)) < infinity
for a given value of L, what values of x fit the constraints?

(an algorithmic-like process would be best)
Title: Re: Math help
Post by: FiahOwl on September 28, 2012, 06:41:06 PM: This message is only viewable with Universe Sandbox Galaxy Edition. Access it and much more with promo-code '82559'.
Title: Re: Math help
Post by: vh on September 28, 2012, 06:41:55 PM: no this isn't a joke. it has practical applications, believe it or not.
Title: Re: Math help
Post by: matty406 on September 28, 2012, 06:50:47 PM: Are you seriously asking me for help with math.
Title: Re: Math help
Post by: vh on September 28, 2012, 06:51:33 PM: [09:47.20] <+Darvince> for L = 70, x is between 13 and 0

1 >= floor((L)/(2(x-floor(x/3))^2+2(x-floor(x/3))+1)) > infinity
1 >= floor((70)/(2(13-floor(13/3))^2+2(13-floor(13/3))+1)) > infinity
1 >= floor((70)/(2(9)^2+2(9)+1)) > infinity
1 >= floor((70)/(2*81+2(9)+1)) > infinity
1 >= floor((70)/(181)) > infinity
1 >= floor((70)/(181) > infinity
1 >= floor(70/181) > infinity
1 >=0 > infinity
Title: Re: Math help
Post by: blotz on September 28, 2012, 06:53:47 PM: holy moly.
Title: Re: Math help
Post by: vh on September 28, 2012, 06:58:28 PM: i flipped the symbols by accident. fixed.
Title: Re: Math help
Post by: Darvince on September 28, 2012, 07:16:31 PM: 1 <= floor((L)/(2(x-floor(x/3))^2+2(x-floor(x/3))+1)) < infinity

x = 16 test

1 <= floor((70)/(2(16-floor(16/3))^2+2(16-floor(16/3))+1)) < infinity
1 <= floor((70)/(2(11)^2+2(12)) < infinity
1 <= floor((70)/2*121+24) < infinity
1 <= floor((70)/242+24) < infinity
1 <= floor((70)/264) < infinity
1 <= 0 < infinity
false.

x=2 test

1 <= floor((70)/(2(2-floor(2/3))^2+2(2-floor(2/3))+1)) < infinity
1 <= floor((70)/(2(2)^2+2(3)) < infinity
1 <= floor((70)/(8+6)) < infinity
1 <= floor(70/14) < infinity
1 <= 5 < infinity

true.
Title: Re: Math help
Post by: Darvince on September 28, 2012, 07:17:31 PM: ok that^ is as far as i get before brain kiplodes
Title: Re: Math help
Post by: vh on September 28, 2012, 07:26:26 PM: yes, but how do you get this? i need a formula or an algorithm

Quote from: Darvince on September 28, 2012, 07:16:31 PM
1 <= floor((L)/(2(x-floor(x/3))^2+2(x-floor(x/3))+1)) < infinity

x = 16 test

1 <= floor((70)/(2(16-floor(16/3))^2+2(16-floor(16/3))+1)) < infinity
1 <= floor((70)/(2(11)^2+2(12)) < infinity
1 <= floor((70)/2*121+24) < infinity
1 <= floor((70)/242+24) < infinity
1 <= floor((70)/264) < infinity
1 <= 0 < infinity
false.

x=2 test

1 <= floor((70)/(2(2-floor(2/3))^2+2(2-floor(2/3))+1)) < infinity
1 <= floor((70)/(2(2)^2+2(3)) < infinity
1 <= floor((70)/(8+6)) < infinity
1 <= floor(70/14) < infinity
1 <= 5 < infinity

true.
Title: Re: Math help
Post by: atomic7732 on September 28, 2012, 08:08:53 PM: what is floor
Title: Re: Math help
Post by: Darvince on September 28, 2012, 08:23:37 PM: truncation

floor(70/223) = 0
floor(21.974836475898942) = 21
Title: Re: Math help
Post by: Bla on September 29, 2012, 11:27:20 AM: Well I've never been taught anything about floors in Math, so I don't really know how to deal with them, but after stopping my eye bleeding from looking at the horrible mess of characters which apparently is an "inequation", I noticed that the denominator looked suspiciously like a quadratic equation if the floors were made of invisiglass. So I tried to solve it and got these results, maybe those numbers are somewhat magic because after all if you divide by 0 you get infinity but... Now I'm bored so see if you can use this first.

I'm not even entirely sure if I wrote it correctly to begin with, you better check everything thoroughly.
Title: Re: Math help
Post by: vh on September 29, 2012, 11:28:37 AM: KOLOK