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We are given a recursive definition:

$a_1 = x,\\a_2=y, \\a_n= c_1a_{n-1}+c_2a_{n-2} \text{ for }n\ge3 $

where $x,y,c_1,c_2,n$ are natural numbers

we are to prove that $a_n \le c_3^n$ for all n

The base case is true

Assuming $a_k \le c_3^k$ is true

show $a_{k+1} \le c_3^{k+1}$ is true


We have:

  1. $c_3a_k \le c_3^{k+1}$

and

  1. $a_{k+1} = c_1a_k + c_2a_{k-1}$

where 1 can be written:

  1. $c_1a_k + ma_k$

where $m+c_1 = c_3$

which can be rewritten:

  1. $c_1a_k + mc_1a_k + mc_2a_{k-1}$

Now, 4 and 2 have the first term equal.

How can I determine whether $ mc_1a_k + mc_2a_{k-1}$ in 4 is bigger or not than $ c_2a_{k-1}$ in 2?

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    $\begingroup$ do you have any relation between $c_1$, $c_2$ and $c_3$ ? Or are you looking for an expression of $c_3$ using $c_1$ and $c_2$ that respect the inequality ? $\endgroup$
    – Optidad
    Sep 3 '20 at 11:31
  • $\begingroup$ $c_1,c_2,c_3$ are all constants. Yes I want them to respect the inequality. $\endgroup$ Sep 3 '20 at 21:40
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Let $c_3 = \max(1,x,y,c_1+c_2)$. Then $$ a_1 = x \stackrel{c_3 \geq x}\leq c_3^1 $$ and $$ a_2 = y \stackrel{c_3 \geq y}\leq c_3 \stackrel{c_3 \geq 1}\leq c_3^2. $$

Now suppose that $a_{n-2} \leq c_3^{n-2}$ and $a_{n-1} \leq c_3^{n-1}$. Then $$ a_n = c_1 a_{n-1} + c_2 a_{n-2} \stackrel{\text{assumption}}\leq c_1 c_3^{n-1} + c_2 c_3^{n-2} \stackrel{c_3 \geq 1}\leq c_1 c_3^{n-1} + c_2 c_3^{n-1} \stackrel{c_3 \geq c_1+c_2}\leq c_3^n. $$

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  • $\begingroup$ Note that $x$, $y$, $c_1$, $c_2$, $c_3$ are natural numbers, then the "1" in the maximum is not needed. Also, $c_1 c_3 + c_2 \le c_3^2$ is more permissive than $c_1+c_2 \le c_3$. $\endgroup$
    – Optidad
    Sep 3 '20 at 15:08
  • $\begingroup$ The bound isn't optimal anyhow. The correct rate of growth is $O(c^n)$, where $c$ is the larger root of $c^2 = c_1 c + c_2$. $\endgroup$ Sep 3 '20 at 16:39
  • $\begingroup$ @John L., all these constants may be 0. $\endgroup$
    – Optidad
    Sep 4 '20 at 7:35

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