any way to rewrite
A(n)=A(n-1) +A(n-2)^2
so i dont have to figure out each number individually?
3,8,17,81,370... etc.
any way to rewrite
A(n)=A(n-1) +A(n-2)^2
so i dont have to figure out each number individually?
3,8,17,81,370... etc.
Last edited by bolshak25; 04-15-2011 at 01:18 AM.
F(1) = 3
F(2) = 8
F(x) = F(x - 2)^2 + F(x - 1)
dunno?
Is your account in an old-school random? Help SRL-OSR solve randoms!
F(0) = 3
F(1) = 8
F(n) = F(n-1) + F(n-2)^2
Something like that, perhaps?
edit: It seems we've thought of the same thing.
:-)
Yeah those work, but there has to be a way to write it so i dont have to find the other terms to get it. i want to be able to plug in 50 to get the 50th term without manually finding 48 and 49 terms
I'm pretty sure that's impossible.
Is your account in an old-school random? Help SRL-OSR solve randoms!
updated first post again. im not sure if this new part is possible, just wondering. thanks everyone for this equation as is.
what do you mean by having to figure out each number individually?
<TViYH> i had a dream about you again awkwardsaw
Malachi 2:3
well as of right now you cant figure out say the 10th term without calculating every term before it, so id need to figure out the 6th and 7th and 8th and 9th to get that 10th term
He just wants an exclusive equation, not a recursive one.
You may be able to for a specific sequence, however you must define a(1) and a(2) before you can try to find a closed-form expression like you want. (ie. give the first two terms of the sequence)
Also, there is no real set method for determining what the closed-form expression of a recursive sequence is, or even if such an expression exists as far as I know, so you'll just have to guess things I'm afraid.
Alternately you can try reading this - http://en.wikipedia.org/wiki/Recurre...lation#Solving.
Last edited by any_one; 04-15-2011 at 04:55 AM.
yeah our teacher told us today that you cant write it better than what we already figured out and we needed 3 and 8 as term 1 and 2 for it to work at that point. thanks for the help though
There are currently 1 users browsing this thread. (0 members and 1 guests)