MHB Solve Recurrence: $s_n = 2_{s_n-1} - s_{n-2}$ | Discrete Math

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SUMMARY

The recurrence relation defined by $s_n = 2_{s_{n-1}} - s_{n-2}$ for $n \ge 2$, with initial conditions $s_0 = 4$ and $s_1 = 1$, can be solved using characteristic equations. The characteristic equation $r^2 - 2r + 1$ yields a double root $r = 1$. The closed-form solution is $s_n = c_1 + c_2n$. By applying the initial conditions, the parameters are determined as $c_1 = 4$ and $c_2 = -3$, leading to the final closed-form expression $s_n = 4 - 3n$.

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shamieh
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Derive an exact formula (solve the recurrence definition) for the following recursive sequence: $s_n = 2_{s_n-1} - s_{n-2}$ where $n \ge 2$, and $s_0 = 4$, $s_1 = 1$.

So I saw someone solving this by making it a differential equation or something?

How would you do that? should I do
let $\alpha = C_1$ let $\beta = C_2$ (because those symbols are ugly)

$r^2 - 2r + 1$ to get:

$r = 1$, $r = 1$

= $C_11^n + C_2n1^n$ ?
But how do I find my $C_1$ and $C_2$ ?

By the way this is a Discrete Mathematics course, not Calculus 4 course.
 
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For my solution I ended up with $S_n = 4*1^n - 3n1^n$
 
You have correctly identified that the characteristic root is $r=1$ of multiplicity 2, thus the closed-form is:

$$s_n=c_1+c_2n$$

Now, to determine the values of the parameters, you can use the initial conditions:

$$s_0=c_1+c_2\cdot0=4$$

$$s_2=c_1+c_2\cdot1=1$$

So, what do you get for the parameters when you solve the above system, and hence, what is the closed-form?
 
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