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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?

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

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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

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