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#1 Re: Help Me ! » Powers of 3 and integers » 2011-06-21 23:33:34
Just one last thing, has anyone converted soroban's algorithm to Mathematica language.
I'm not very good in programming, I'm trying to find a way but without success.
So please share if you have it. Or if you have it in another program, I will download it
and use it.
Paste the lines here thank you.
#2 Re: Help Me ! » Powers of 3 and integers » 2011-06-21 01:11:32
well i don't really know what you mean with the p-adic representation but the reason why i need both negative and positive
is because if you use powers only once, you have only one solution for each integer and it includes negative and positive terms, or just positive, or just negative. it's all in the table
#3 Re: Help Me ! » Powers of 3 and integers » 2011-06-20 21:56:02
Thank you for your reply soroban and dragonshade, although these methods don't seem to be what i'm looking for, they provide some clues.
It seems more of an iterative process using base 3 (with 'if' twice), sort of more like an algoritm/program than a function(perhaps im wrong, algorithms and functions can be very similar, if so plz explain to me so ill learn stuff ).
So to me a function would clearly be a more 'elegant' solution. Though your method will allow me to experiment as though i already had the function and for that i appreciate!
By the way, if you generate a base 3 array plot of integers in mathematica, you will see the similarity to my table immediately with slight deviation. Thank you for making that connection clearer.
I consider every post as clues! (and maybe one will have the answer!)
Thank you again! 
and don't stop here, there's always more to discover!,
#4 Re: Help Me ! » Powers of 3 and integers » 2011-06-20 06:28:55
here is the table i made
Dark X = -
X = +
blank = 0
#5 Re: Help Me ! » Powers of 3 and integers » 2011-06-20 05:50:43
im sure the formula exists somewhere, i can't be the first person who asks himself this question...
its just a matter of knowing what it's called, so someone with a doctors or something would probably be able to name it... i think
that's why im here, im using mathematica if anyone wants to know... or paste me something they came up with.
#6 Re: Help Me ! » Powers of 3 and integers » 2011-06-20 05:09:16
exactly, i guess i should have mentioned it
i made a little table up to like 42 and the pattern is easy to see, just hard to tranform into math terms
i also notices that for the 3^0, its goes [+,-,0] periodical
3^1 [0,+,+,+,-,-,-,0,0] periodical
and so forth
the so basically for the 3^0 its 2*Sqrt[3]*Cos[(2*\[Pi]*x/3 + (\[Pi])/6)]/3
or
Sum[((-1)^(n - k)) ((n - k)!/(k! ((n - k) - k)!)), {k, 0, n, 1}]
got that from oeis.org/search?q=1%2C-1%2C0%2C1%2C-1%2C0%2C1%2C-1%2C0&language=english&go=Search
i hope it helps...
i get the feeling this is unsolvable but i know nothing! thats why im here 
thx
#7 Help Me ! » Powers of 3 and integers » 2011-06-20 04:14:06
- wetsun
- Replies: 28
Im trying to find out the formula for the following..
i want to express every integer in powers of 3 with a formula and i only have college science math so i need help from you guys who went further than me
so its like 1 = 3^0
2 = 3^1 - 3^0
3 = 3^1
4 = 3^1 + 3^0
5 = 3^2 - 3^1 - 3^0
6 = 3^2 - 3^1
7 = 3^2 - 3^1 + 3^0
annnnnnnnnddd so forth
so i can see the pattern, i just think i dont know enouph math stuff to solve this... so if someone knows the function that will give you the powers of 3 needed for each integer.. tell me!!
THX for the help
math is fun!
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