0

I was working on a problem and I stumbled upon this equality:

$a_{n+2} = a_{n+1} + a_{n}$

And I'm trying to find the general expression of the sequence $a$.

For this, I wanted to show that $a$ can just be expressed as the sum of two geometric sequence so that I can rewrite this equality using the general expression of geometric sequences:

Edit: $q {\neq} 0$

$a_0*q^{n+2} = a_0*q^{n+1} + a_0*q^{n}$

but I didn't manage to prove that.

Could someone help me please

Beni
  • 1
  • 1
    What should $q$ be for that condition to be true? One condition suffices for all $n$. Can you match any initial conditions using sums of geometric sequences (not series)? – Jamie Radcliffe Mar 12 '23 at 15:47
  • 1
    These are the Fibonacci numbers (possibly with different initial conditions). They are well documented online. – lulu Mar 12 '23 at 15:48
  • yes my bad geometric sequence. And $q {\neq} 0$ – Beni Mar 12 '23 at 15:52
  • 1
    Depends on whether you wish to show that this is possible for any value of $q$, or instead wish to show that a specific value of $q$ exists, that makes this formula work. You would need $q$ to satisfy the equation $$q^2 = q + 1.$$ – user2661923 Mar 12 '23 at 15:57
  • yes I found this formula. But to prove this what should I write ? As you get this formula by using the general formula of a geometric sequence. Or can I just write that that $a_n$ can be written as a sum of two geometric series, with a q that exist such as $q^2 = q +1$ ? – Beni Mar 12 '23 at 16:00
  • @Beni See my answer here for a proof from scratch. – dxiv Mar 12 '23 at 17:40

3 Answers3

0

Second-order linear relations with constant coefficients are very well-documented problems.

In your case, the equation can be rewritten : $$a_{n+2} - a_{n+1} - a_n = 0$$

You need to study the characteristic polynomial equation $P(X) = X^2 - X - 1 = 0$. This one has two roots: $\varphi = \frac{1 + \sqrt 5}{2}$ and $\bar \varphi = \frac{1 - \sqrt 5}{2}$. (Note that $\varphi$ is the “golden ratio”.) Then, all the solutions to your recurrence relation are the sequences of form: $$a_n = A \varphi^n + B \bar\varphi^n$$ with $A, B \in \mathbf R$.

The values of $A$ and $B$ depend on your initial conditions.

Valentin Melot
  • 167
  • but shouldn't I prove first that $a_n$ can be written as the sum of two geometric sequences ? – Beni Mar 12 '23 at 16:06
  • This depends on what you assume and what you want to prove exactly. If you only need to prove that a sequence of form $A\varphi^n + B \bar\varphi^n$ is a solution to the problem, then it is enough to prove that it satisfies the recurrence relation. If you need to prove that these sequences are the only solutions, then look at my second answer. – Valentin Melot Mar 12 '23 at 16:30
  • Thanks a lot, how can I prove that it satisfies the recurrence relation though? – Beni Mar 12 '23 at 18:53
  • also I didn't really understand where do you get the characteristic polynomial equation from ? – Beni Mar 12 '23 at 19:13
  • If $a u_{n+2} + b u_{n+1} + c u_{n} = 0$, then the associated characteristic polynomial is $P(X) = a X^2 + bX + c$. Here, $a = 1$, $b = -1$ and $c = -1$. Concerning the recurrence relation, you can try to adapt the proof of my second answer. – Valentin Melot Mar 12 '23 at 20:00
0

Let $a,b$ be the roots of $$px^2+qx+r=0.....(1)$$ Then $$pa^2+qa+r=0$$ $$\implies Cpa^{n+2}+Cqa^{n+1}+Cra^n=0$$ Similarly, we have $$Dpb^{n+2}+Dqb^{n+1}+Drb^n=0$$ Adding these two equations, we get $$p(C a^{n+2}+D a^{n+2})+q(Ca^{n+1}+Db^{n+1})+r(Ca^n+Db^n)=0.$$ Introduce $$A_n=C a^n+D b^n.....(2)$$ Then we can write $$pA_{n+2}+qA_{n+1}+rA_n=0.....(3)$$

Finally (2) is the general solution (sum of two GPs) of the recurrence relation (3), where $a,b$ are the roots of (1) and $C,D$ are two arbitrary paramwteres to be determined by the given initial values.

Z Ahmed
  • 43,235
0

OK, now a solution if you do not assume the theorem about the closed form of recurrent linear sequences of order 2.

Consider any sequence $(u_n)_n$ such that $u_{n+2} = a u_{n +1} + b u_n$. Assume that polynomial $X^2 - a X - b$ has two complex roots $\alpha \neq \beta \in \mathbf C$.

We want to prove that there exists $\lambda, \mu \in \mathbf C$ such that for all $n \in \mathbf N$, $$ u_n = \lambda \alpha^n + \mu \beta^n$$

Necessary condition. For $n = 0$, one needs have $u_0 = \lambda + \mu$. For $n = 1$, one needs have $u_1 = \lambda \alpha + \mu \beta$. As $\alpha \neq \beta$, this system of equations has a single solution: $$\lambda = \frac{u_1 - \beta u_0}{\alpha - \beta}, \quad \mu = \frac{u_1 - \alpha u_0}{\beta - \alpha}$$

Sufficient condition. Now, we need to prove that the sequence: $$u_n = \lambda \alpha^n + \mu \beta^n$$ with $\lambda = \frac{u_1 - \beta u_0}{\alpha - \beta}$ and $\mu = \frac{u_1 - \alpha u_0}{\beta - \alpha}$ satisfies the recurrence condition. This can be proven by recurrence.

Indeed, the equality is matched for $n = 0$ and $n = 1$. Assume it is matched for some $n$ and for $n+1$. Then, at rank $n+2$: $$u_{n+2} = a u_{n+1} + b u_n$$ As the equality is assumed for ranks $n$ and $n+1$, $$u_{n+2} = a (\lambda \alpha^{n+1} + \mu \beta^{n+1}) + b(\lambda \alpha^n + \mu \beta^n)$$ $$u_{n+2} = \lambda \alpha^n (a \alpha + b) + \mu \beta^n (a \beta + b)$$ As $\alpha$ and $\beta$ are solutions to $X^2 = aX + b$, we have: $$u_{n+2} = \lambda \alpha^n \alpha^2 + \mu \beta^n \beta^2$$ $$u_{n+1} = \lambda \alpha^{n+2} + \mu \beta^{n+2}$$

This proves the equality at rank $n+2$. By recurence, for every $n \in \mathbf N$, one has: $$u_n = \lambda \alpha^n + \mu \beta^n.$$

QED.

Valentin Melot
  • 167