- 18 Jun '06 11:40

No Sarahli meant that it reminded her of Pascal's triangle because of the fact both used numbers in them. No funny happy faces, curly-ques or basic wingdings are used in either. (You did represent some numbers with letters... but there are really numbers in behind so it doesn't count.)*Originally posted by TheMaster37***That would be**

1

11

121

1331

14641

15AA51

16FKF61

...using letters for higher numbers to keep it a bit compacter.

I don't see the resemblance yet

Good observation Sarah! - 18 Jun '06 20:03

Wow a soul-shun! I wonder how many solutions there are too? I didn't realize that souls even existed in a complex field but I guess that is why many feel they are imaginary.*Originally posted by aginis***... there should be 1 soultion in the complex field, but i can't prove it off hand.** - 19 Jun '06 16:18

what i meant is the recursive function a(n)=c1*a(n-1)+c2*a(n-2)...c6*a(n-6) with intial conditions a(0)=1 a(1)=12 a(2)=44 .... has a general solution with the form*Originally posted by TDR1***i dont think that matters this is a sequence not a polynomial function**

a(n)= k1*(r1^n) + k2*(r2^n) + ... + k6*(r6^n) when

r1 through r6 are solutions to the polynomial x^6-c1*x^5-c2*x^4-...-c6=0

k1 through k6 are constants that can be solved for using the initial conditions

all this is on the condition that there are no double roots.

of course every set of coefficients that satisfy the condition yield a general solution for a(n) then the 7th term in the sequence is a(6) although i can't guarentee that there wont be an overlap of solutions for different coefficients c1...c6

does that clear it up? - 19 Jun '06 16:34

It doesn't, because a rule for functions is that it is continuous not discrete, that is why you can differentiate a function but not a sequence or series (many math puzzles use this attribute). Therefore you cannot interpolate a function using these items in a sequence as points on the function, because the function requires that you have all the other bits in-between.*Originally posted by aginis***what i meant is the recursive function ...**

does that clear it up? - 20 Jun '06 00:15

not all functions are continuous*Originally posted by Gastel***It doesn't, because a rule for functions is that it is continuous not discrete, that is why you can differentiate a function but not a sequence or series (many math puzzles use this attribute). Therefore you cannot interpolate a function using these items in a sequence as points on the function, because the function requires that you have all the other bits in-between.**

any sequence is by definition a function from the set of natural numbers to some range be it real numbers or natural numbers or any other set (words, pairs, shapes, dates, anything) however the implication here is that there is some sort of logical connection between these points implying that they are not discrete. in a truly discrete example u can never predict the next element in a sequence because one elment is not realated to its neighbors.

Let us return to the problem at hand we are given six initial values of a sequence and we are asked to determine the 7th. this implies a recursive notation. I am assuming a linear function of six variables that means a sum of each value multiplied by a coefficient. that is

a(n)=sum (i,1,6, Ci*a(n-i)).

let us take a simple example a(n)=2*a(n-1) ; a(0)=5 if we assume that the explicit solution is of the form a(n)= c*x^n we see that

a(n)=2*a(n-1) => c*x^n=2*c*x^(n-1)=>

c*x^n - 2*c*x^(n-1)=0 => [c*x^(n-1)](x-2) {note the polynomial of degree 1: x-2 if we have one initial value and one step back in the recursion so it is of degree one; 6 initial values means 6 steps back and a polynomial of degree 6} =0 => x=2 (if c or x =0 then all the elements of the sequence are 0) thus

a(n)= c*x^n = c*2^n now we use the initial value a(0)=c*2^0=c =5 thus we have solved for c and found the explicit form a(n)=5*2^n that satisfies our conditions. a similar method can be used to solve a recursive notation of six initial values however we can solve for ANY recursion (it gets a bit complicated if we throw in multiple roots and a free term so we will avoid those)

i was never suggesting that the polynomial ITSELF would be the solution although this too is a viable option of course there is no need to limit oneself to polynomials any function(continuous or otherwise) on the domain of Reals of that goes through the points (0,1),(1,12) (2,44).... will also yield a sequence but this is already stretching reasonable boundries. - 20 Jun '06 00:24However, i believe the point of celticcounty's question is to find a recursive form of only three or four steps such that all six initial values correspond to those listed that narrows the field VERY much. if you throw in the constraint that a(n) is a monotnically rising set of natural numbers then the number of solutions is probably small enough to be counted on the fingers of one hand. and i personally am stumped.
- 20 Jun '06 10:00no a relation is a set of pairs for example the relation ">" over NxN (natural numbers squared) is the set of all pairs (a,b) s.t. a>b. A function is a very special relation f(x) from A to B then for every element of A there is exactly one related element of B

Continuous - is a property

A functions is continuous at point a IFF

lim (x -> a+) f(x)= lim (x -> a-) f(x)= f(a)

http://en.wikipedia.org/wiki/Continuous_function

http://en.wikipedia.org/wiki/Discontinuous

http://en.wikipedia.org/wiki/Relation_%28mathematics%29