Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have this difference equation:

$$c_0 x_n+c_1x_{n+1}+\cdots+c_m x_{n+m} = \sum\limits_{i=0}^m c_i x_{n+i} = 0 $$

And I have problem with understanding the dimension argument.

Dimension argument

Given $x_1,\ldots,x_m \Rightarrow x_{m+1} = -\dfrac{1}{c_m}\displaystyle\sum\limits_{i=0}^{m-1} c_i x_{n+i}$

Which means selection of $x_1,\ldots,x_m$ corresponds to picking a point in the m-dimensional space $\mathbb{R}^m$, so the solution space has dimension $m$.

And I don't understand the last part "Which means selection of ..". Someone who can explain it in details?

share|improve this question
    
Where did you see this? –  J. M. Oct 30 '11 at 13:43
    
From a lecture note. –  user12358 Oct 30 '11 at 13:47

1 Answer 1

up vote 1 down vote accepted

To get the recursion started, you pick arbitrary values for $x_1,\dots,x_m$. Once those values have been chosen, the difference equation uniquely determines the value of $x_{m+1}$, and then the value of $x_{m+2}$, and of $x_{m+3}$, and so on. So there are $m$ degrees of freedom in the solution; in other words, the solution space is $m$-dimensional.

share|improve this answer
    
Thank you. But how do you conclude that the solution space is m-dimensional, when there are m degrees of freedom? –  user12358 Oct 30 '11 at 17:09
    
@user12358: OK, to phrase it as in your lecture notes: assigning values $x_1,\dots,x_n$ arbitrarily is the same thing as choosing an arbitrary point $(x_1,\dots,x_m)$ in the $m$-dimensional space $\mathbb{R}^m$. –  Hans Lundmark Oct 30 '11 at 19:03

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.