To solve the recurrence relation $a_n = 2a_{n-1} - 2a_{n-2}$ for $n \geq 2$, with initial conditions $a_0 = 1$ and $a_1 = 2$, we will use generating functions.

Step 1: Define the generating function

Let the generating function $A(x)$ be the power series corresponding to the sequence $\{a_n\}$: $A(x) = \sum_{n=0}^{\infty} a_n x^n.$ Now, we multiply both sides of the recurrence relation $a_n = 2a_{n-1} - 2a_{n-2}$ by $x^n$ and sum over all $n \geq 2$.

Step 2: Formulating the recurrence with the generating function

The recurrence can be rewritten as: $\sum_{n=2}^{\infty} a_n x^n = 2 \sum_{n=2}^{\infty} a_{n-1} x^n - 2 \sum_{n=2}^{\infty} a_{n-2} x^n.$ Each of these sums can be expressed in terms of $A(x)$:

• For the left-hand side: $\sum_{n=2}^{\infty} a_n x^n = A(x) - a_0 - a_1 x$.
• For the first sum on the right-hand side: $\sum_{n=2}^{\infty} a_{n-1} x^n = x \sum_{n=1}^{\infty} a_n x^n = x(A(x) - a_0)$.
• For the second sum on the right-hand side: $\sum_{n=2}^{\infty} a_{n-2} x^n = x^2 \sum_{n=0}^{\infty} a_n x^n = x^2 A(x)$.

Thus, the recurrence relation in terms of the generating function becomes: $A(x) - a_0 - a_1 x = 2x(A(x) - a_0) - 2x^2 A(x).$

Step 3: Substituting the initial conditions

Substitute $a_0 = 1$ and $a_1 = 2$: $A(x) - 1 - 2x = 2x(A(x) - 1) - 2x^2 A(x).$ Simplifying this equation: $A(x) - 1 - 2x = 2xA(x) - 2x - 2x^2 A(x).$ $A(x) - 2xA(x) + 2x^2 A(x) = 1.$

Step 4: Solving for $A(x)$

Factor out $A(x)$ on the left-hand side: $A(x)(1 - 2x + 2x^2) = 1.$ Thus, the generating function is: $A(x) = \frac{1}{1 - 2x + 2x^2}.$

Step 5: Partial fraction decomposition (if needed)

To find the explicit form of $a_n$, we could decompose $A(x)$ using partial fractions and then expand it as a power series. However, the quadratic denominator suggests that the solution for $a_n$ might involve combinations of terms derived from solving the characteristic equation.

The characteristic equation corresponding to the recurrence relation is: $r^2 - 2r + 2 = 0.$ The roots of this quadratic are: $r = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(2)}}{2(1)} = \frac{2 \pm \sqrt{4 - 8}}{2} = \frac{2 \pm \sqrt{-4}}{2} = 1 \pm i.$ Thus, the general solution will involve terms of the form $a_n = c_1 (1 + i)^n + c_2 (1 - i)^n$.

Would you like further details on how to expand this generating function into a series for explicit terms of $a_n$, or help with specific steps?

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Here are 5 follow-up questions:

1. How do we use partial fractions to decompose a generating function?
2. What is the significance of solving the characteristic equation in recurrence relations?
3. How does the complex root structure affect the recurrence relation's general solution?
4. How can generating functions help solve other forms of recurrence relations?
5. How would we use initial conditions to solve for constants in the general solution?

Tip: Always check the discriminant of the characteristic equation to determine the nature of the roots, which influences the form of the solution.