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https://www.youtube.com/watch?v=7mhvA5L7KqY&ab_channel=TrevTutor

So basically, I was watching video above which is on recurrence relations and I had a question about this statement (which is at timestamp 5:21):

$$a_n=a_{n-1}+6a_{n-2}\Longrightarrow a_n=\alpha(-2)^n+\beta(3)^n.$$

I understand how he got the $(-2)^n$ and $(3)^n$, but not about how he is adding them and then multiplying them by the variables $\alpha$ and $\beta$. He said that there is a proof online about why there is always going to be an $\alpha$ and a $\beta$ that will always make this statement true, but I wasn't able to find it and I was hoping that somebody could give me a step by step explanation about how this is. I was also wondering about the general case for getting a formula for recurrence relations in this form.

Note: I read somewhere that you can derive this from generating functions, but I don't have a strong background in them, so I was wondering if there is another way to derive the relation.

Banana Blitz Coding
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2 Answers2

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It comes from the fact that for a general, linear, homogenous recurrence (for simplicity, we'll say it's order 2), the solutions form a vector space.

Particularly, we are motivated that the general solution to something like $a_n=ra_{n-1}$ is $a_n=\alpha r^n$.

Let's say that the solution to the general recurrence $a_n=a_{n-1}+6a_{n-2}$ was of the form $ar^n$. We can substitute this to see that $$ar^n=ar^{n-1}+6ar^{n-2}$$ $$r^2=r+6$$ $$r=-2,3$$ So we can see that any scalar multiple of $(-2)^n$ or $3^n$ will satisfy the general recurrence. Moreover, it is pretty easy to see that any linear combination of these two will satisfy the recurrence. Hence, we can say that the solutions to the general recurrence will lie in the vector space with basis $\{(-2)^n,3^n\}$.

We then say that a particular solution is a linear combination of the elements of this basis and use the initial conditions to solve for the coefficients. This process is very similar when solving linear, homogenous differential equations.

Alan Abraham
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Let $px^2+qx+r=0$ has roots as $a,b$, then $$\alpha a^n(pa^2+qa+r)=0$$ and $$\beta b^n(pb^2+qb+r)=0$$ add them to get $$p(\alpha a^{n+2}+\beta b^{n+2})+q(\alpha a^{n+1}+ \beta b^{n+1})+(\alpha a^n+ \beta b^n)=0$$ Now define $A_n=\alpha a^n +\beta b^n$ So we get $$pA_{n+2}+qA_{n+1}+rA_n=0$$ This prvides a simple understandinf of what is going on here.

Z Ahmed
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