I'm quite familiar with the procedure of, say, given a recurrence relation $f_n = a\times f_{n-1} + b\times f_{n-2}$, finding the general closed formula for the function $f_n$. The technique I'm familiar with is to algebraically rephrase the inital equation as $f_n - a\times f_{n-1} - b\times f_{n-2} = 0$, formulaically identify the characteristic equation to be $x^2 - ax - b = 0$, and solve for the characteristic roots, say, $\theta_1$ and $\theta_2$. After this, we finally acquire the general closed formula $a_n = a\times (\theta_1)^n + b \times (\theta_2)^n$; given the inital conditions we can then solve for $a$,$b$. However, I do not completely understand the reasoning of this method provided in my textbook. The author, after expressing the first-most equation as $f_n = a\times f_{n-1} + b\times f_{n-2}$, asks of us to assume that $a_n = r^n$. I understand the remaining procedure after this assumption; But I can hardly see how it is appropriate to assume such a claim. How is it justified that we may simply assume that $a_n = r^n$? Thank you in advance.
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See e.g. https://math.stackexchange.com/questions/955875/why-do-we-set-u-n-rn-to-solve-recurrence-relations – projectilemotion Feb 06 '23 at 18:38
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2It's a guess, not an assumption. We guess $f_n = cr^n$ (you already use $a$ and $b$ in your example). We find that this imposes relations on the $c$ and $r$. We solve them and check that $f_n$ with those solutions works. We need only find a solution that works, by whatever means. Once we have a solution, we make an additional argument that it is somehow the only possible solution. For linear recurrence relations, this is as easy as noting that solutions form a vector space of finite dimension ($2$ dimensional in this case for the two initial conditions). – Charles Hudgins Feb 06 '23 at 18:39
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See also linear algebra/operator proof, proof via generating functions, idea of proof via matrix exponentiation, proof by factoring polynomial right shift operator. – Mike Earnest Feb 06 '23 at 18:49