# Arithmetic sequences in R^2.?

Hi all,

I was just wondering whether one could define arithmetic sequences in R^2 in a simmilar manner as in R.?

Here is what i see as a natural way of doing it, but neither have i read about it, nor heard.

$$\mbox{ Let } x_n \in R^2 \mbox{ be a sequence given as follows : } x_n=a+mb\\, \mbox{ where } a,b,m\in R^2.$$

$$\mbox{ That is, } a=(a_1,a_2),b=(b_1,b_2),m=(m_1,m_2). \mbox { So, } x_n=(a_1+m_1b_1,a_2+m_2b_2). \mbox{ We call such a sequence an arithmetic sequence in } R^2.$$

Would this definition be valid? If so, i believe one could define an arithmetic or geometric sequence in R^n as well. Right?

EDIT: Or maybe on a second thought i think that the following change would be better:

$$x_n=(a_1+nb_1,a_2+nb_2)=(a_{1n},a_{2n}). \mbox{ That is letting } m=(m_1,m_2)=(n,n).$$

Thnx

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## Answers and Replies

Hurkyl
Staff Emeritus
Gold Member
Your second idea is better. (can you explain why? I can name two reasons....)

Really, all you want is a recursion
xn+1 = xn + d​
right? This makes sense in any context where you have "addition". (you probably want "addition" to be commutative and associative)

In R2, you might wish to consider a sequence with two-dimensional indices satisfying
xm+1,n = xm,n + d1
xm,n+1 = xm,n + d2
which, of course, can be put in closed form
xm,n = x0,0 + m d1 + n d2

(Hrm. Maybe this is what you were trying to think about with your first idea?)

Your second idea is better. (can you explain why? I can name two reasons....)

Really, all you want is a recursion
xn+1 = xn + d​
right? This makes sense in any context where you have "addition". (you probably want "addition" to be commutative and associative)

Well, the reason why i decided to make that change was because previously my indicies were two-dimensional, and i wanted them to be one-dimensional(i.e. from Z+.) So, setting m=(n,n) took care of that, in some sense. Is this one of your two reasons?or?

In R2, you might wish to consider a sequence with two-dimensional indices satisfying
xm+1,n = xm,n + d1
xm,n+1 = xm,n + d2
which, of course, can be put in closed form
xm,n = x0,0 + m d1 + n d2

(Hrm. Maybe this is what you were trying to think about with your first idea?)

Any hints as how to put this in closed form? I can generally work with recurrence relations, but haven't worked before with systems of recurrence relations!

EDIT: Don't give me any hints for a couple more minutes! I will come back again. I think i have an idea that might work.

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Hurkyl
Staff Emeritus
Gold Member
Any hints as how to put this in closed form? I can generally work with recurrence relations, but haven't worked before with systems of recurrence relations!
Then work with one recurrence at a time! (e.g. what if the second index is a constant?)

The bigger problem I saw with your first idea is that it's constant -- n has no relation to anything else. The other one was a rather minor one -- while one can define multiplication of R-tuples "pointwise", it's not often something you mean to use.

Then work with one recurrence at a time! (e.g. what if the second index is a constant?).
THis was exactly my idea!
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However, i have another question, what if i want my outcome to be still in R^2, rather than simply R.

This is more what i was trying to do the first time.

In other words, let the indices be from Z+, but the outcome, that is the sequence $$\{x_n\}$$ be from $$R^2$$, where $$x_n=(a_{1n},a_{2n}).$$

EDIT: I got this part...then from the recurrence relation

$$x_{n+1}=x_n+d=>x_n=x_o+nd, \mbox{ with } x_o,d \in R^2.$$

Right?

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In R2, you might wish to consider a sequence with two-dimensional indices satisfying
xm+1,n = xm,n + d1
xm,n+1 = xm,n + d2
which, of course, can be put in closed form
xm,n = x0,0 + m d1 + n d2

I believe i got it. Here we go:

$$\mbox{ First let n=const, then } m=0=>x_{1,n}=x_{0,n}+d_1$$

$$m=1=> x_{2,n}=x_{0,n}+2d_1$$
...

$$x_{m,n}=x_{0,n}+md_1.$$

Now substituting this in the other relation we get:

$$x_{m,n+1}=x_{0,n}+md_1+d_2.$$

In a simmilar fashion as before, one eventually finds that :

$$x_{m,n}=x_{0,0}+md_1+nd_2.$$