john_jrambo2000 Posted January 15, 2005 Report Posted January 15, 2005 Hi, new guy here. Some friends and I work on puzzles and we developed an interesting formula relating to squares. Is there any way I can find out if this formula is useful or already published? Where a=b+1 and b=c+1 and c=d+1 [{(a3-a2)-(b3-b2)}-{(b3-b2)-(c3-c2)}]-[{(b3-b2)-(c3-c2)}-{(c3-c2)-(d3-d2)}]=6 All the twos and threes refer to squares and cubes.Like I said, we just kinda thought it was interesting, and didn't know how to search for it, since it may be under a different format. Don't really know if there's anything useful here. Thanks for any input.
maddog Posted January 15, 2005 Report Posted January 15, 2005 Hi, new guy here. Some friends and I work on puzzles and we developed an interesting formula relating to squares. Is there any way I can find out if this formula is useful or already published? Depends. :hihi: There are number of functions with test for convergence (sums, ..., etc). What's the formula ? Maddog
Bo Posted January 15, 2005 Report Posted January 15, 2005 hi and welcome! it is in principle not easy to find if a discovery is already published. Best bet is to search the internet, and look in some books on the subject. but please enlighten us: what did you find? Bo
hefner Posted January 15, 2005 Report Posted January 15, 2005 Hi, new guy here. Where a=b+1 and b=c+1 and c=d+1 [{(a3-a2)-(b3-b2)}-{(b3-b2)-(c3-c2)}]-[{(b3-b2)-(c3-c2)}-{(c3-c2)-(d3-d2)}]=6 All the twos and threes refer to squares and cubes.Hi John. I get -156=6 :hihi: when d=1 :rant:
john_jrambo2000 Posted January 15, 2005 Author Report Posted January 15, 2005 Then you didn't follow the formula. When d=1, c=2, b=3, and a=4[{(64-16)-(27-9)}-{(27-9)-(8-4)}]-[{(27-9)-(8-4)}-{(8-4)-(1-1)}]{(48-18)-(18-4)}-{(18-4)-(4-0)}(30-14)-(14-4)16-10=6
hefner Posted January 15, 2005 Report Posted January 15, 2005 I'm not laughing! that mathematical oddity is exquisite. Have you tested it pretty thoroughly with software? I'd guess you have. I commend you guys heartily! As for how to check where it has been derived heretofore, I've posted an inquiry on the Australia Mathematics newsgroup (I don't find one for USA). No replies as yet; monitorfor responses here. I hope that's a sensible approach ! :hihi: Maybe you should just call CBS or something :rant:
john_jrambo2000 Posted January 16, 2005 Author Report Posted January 16, 2005 I'm having a lot of trouble deciding if you're being sarcastic or not. lolYou didn't post it correctly. I'm surprised you didn't just cut and paste it.
hefner Posted January 16, 2005 Report Posted January 16, 2005 I'm terribly sorry! It works right for me, but maybe the link is dependent on my local cookies or something. Still no replies. Here is a link to the newgroup: http://groups-beta.google.com/group/aus.mathematics ..look for replies to the post by Ben Bean.
hefner Posted January 16, 2005 Report Posted January 16, 2005 Here is what appears to be a really excellent place to make your inquiry: http://groups.yahoo.com/group/numbertheory/ That's a very busy group with lots and lots of members and activity. Be sure to come back here later and tell us how you fared!
Bo Posted January 16, 2005 Report Posted January 16, 2005 interesting equation this is! however i am convinced that you can extend this equation to not just 4 numbers, but n numbers. I looked a while on this (well 5 minutes maybe :hihi:) and it looks quite complicated to achieve that... Bo
Tim_Lou Posted January 16, 2005 Report Posted January 16, 2005 hmm, interesting...after all the simplifying on my own, the equation did in fact come out as 6, all the other terms cancel with each other. hmm, where did this idea come from? just curious.
Tim_Lou Posted January 17, 2005 Report Posted January 17, 2005 if you change all variables to a,the equation comes out as:a2(a-1) - 3 (a-1)2(a-2) + 3(a-2)2(a-3) - (a-3)2(a-4)where (a-2)2= (a-2)^2 hmm, seems familiar, something to do with binomial? but doesnt quite fit it. :hihi:
john_jrambo2000 Posted January 17, 2005 Author Report Posted January 17, 2005 I don't know how you got that. But I know it doesn't work for just any amount of variables.
Bo Posted January 17, 2005 Report Posted January 17, 2005 my guess was based on the form that tim also gave of the formula; that looks to me as an alternating series that can be expanded beyond 4 terms. (but once again: i havent explicitly checked this! if i have time i will try to). btw this also looks interesting: http://mathworld.wolfram.com/CatalansConjecture.html Bo
Tim_Lou Posted January 17, 2005 Report Posted January 17, 2005 there seems to be some cool pattern, if you use c instead of aa2(a-1) - 3 (a-1)2(a-2) + 3(a-2)2(a-3) - (a-3)2(a-4)=6 (c+2)2 (c+1) - 3(c+1)2 ©+ 3©2 (c-1) - (c-1)2 (c-2)=6since a can be anything, c can be anything as well, both equations work. these equations do work for any variable, you can test it yourself.(i simplified it by factoring, ex: a3 - a2 = a2 (a-1)if you expand all the square and things like that, you get 3(a-3)(-3a+5)+ (9a2 -33a +36)simplified it further, i got 6, all other terms canceled with each other.... :hihi:
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