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So, we all know that the solution to a first grade, homogeneous, linear, with constant coefficients recurrence relation is $a_n = r^n * c$ where $c$ is a constant (given by the initial condition) and $r$ is the coefficient that accompanies the $a_{n-1}$ termn.

My question is: Why when we are working with a second grade difference equation we look for a solution of the same kind?

I mean, the characteristic equation you are left with is such as $r^n - C_1*r^{n-1} - C_2*r^{n-2} = 0$ meaning that you substituded $a_n$ for $r^n$

Can someone please help me? I don't get why we do this.

Thanks a lot!

EDIT: By "first grade, homogeneous, linear, with constant coefficients recurrence relation" I mean the following: $a_n = C_1 * a_{n-1} + C_2 * a_{n-2}$ where $C_1$ and $C_2$ are constant termns (they are not in fuction of $n$) and $C_1, C_2 ∈ R$

cbasitodx
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  • You should definitely write down what you mean by "first grade, homogeneous, linear, constant coefficients recurrence relation". – Chee Han Dec 26 '21 at 18:53
  • Already added it :) – cbasitodx Dec 26 '21 at 19:32
  • Check this out: https://people.math.umass.edu/~lr7q/ps_files/teaching/math456/Chapter3.pdf – Chee Han Dec 26 '21 at 19:44
  • Note that if $r^n-C_1r^{n-1}-C_2r^{n-2}=0$, then you have also $r^{n-2}(r^2-C_1r-C_2)=0$, hence, if $r\ne0$, $r^2-C_1r-C_2=0$. That is, if $r^n$ is a solution, then $r$ is a root of a trinomial. – Jean-Claude Arbaut Dec 26 '21 at 20:11
  • thats pretty useful! thanks. Definitely helped! – cbasitodx Dec 26 '21 at 21:28
  • @cbasitodx Second order recurrences can be reduced to a first order recurrence for a related sequence, then the formula follows naturally. There is a step by step derivation in my answer here for example. – dxiv Dec 27 '21 at 06:10
  • that was just amazing. Thank you so much for sharing. Such an elegant solution! I also liked the linear algebra approach, makes sense but I can't fully comprehend it (my first lenguage is spanish) – cbasitodx Dec 31 '21 at 00:42

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