Question 1167599: Find an explicit formula in terms of n for a sequence
b0, b1, b2,
that satisfies
bk = 2b_k − 1 − 10b_k − 2 for each integer k ≥ 2
with initial conditions
b0 = 5
and
b1 = 5.
(Your answer will involve complex numbers.)
bn = for every integer n ≥ 0
Discrete math homework question
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! The problem requires solving a second-order homogeneous linear recurrence relation with constant coefficients.
## Recurrence Relation and Initial Conditions
The recurrence relation is:
$$b_k = 2b_{k-1} - 10b_{k-2} \quad \text{for } k \ge 2$$
The initial conditions are:
$$b_0 = 5$$
$$b_1 = 5$$
***
## 1. The Characteristic Equation
The characteristic equation is derived by assuming a solution of the form $b_n = r^n$:
$$r^2 = 2r - 10$$
$$r^2 - 2r + 10 = 0$$
***
## 2. Finding the Roots
We use the quadratic formula to find the roots $r$:
$$r = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(10)}}{2(1)}$$
$$r = \frac{2 \pm \sqrt{4 - 40}}{2}$$
$$r = \frac{2 \pm \sqrt{-36}}{2}$$
$$r = \frac{2 \pm 6i}{2}$$
$$r_1 = 1 + 3i \quad \text{and} \quad r_2 = 1 - 3i$$
Since the roots are complex, we convert them to polar form, $r = \rho e^{i\phi}$:
$$\rho = |r_1| = \sqrt{1^2 + 3^2} = \sqrt{1 + 9} = \sqrt{10}$$
$$\phi = \arctan\left(\frac{3}{1}\right) = \arctan(3)$$
***
## 3. General Solution Form
The general solution for a recurrence with distinct complex roots $r_1$ and $r_2$ is:
$$b_n = c_1 r_1^n + c_2 r_2^n$$
Alternatively, using the polar form:
$$b_n = \rho^n (A \cos(n\phi) + B \sin(n\phi))$$
Using the roots $r_1 = 1 + 3i$ and $r_2 = 1 - 3i$, the general solution is:
$$b_n = c_1(1 + 3i)^n + c_2(1 - 3i)^n$$
***
## 4. Applying Initial Conditions
We use the initial conditions $b_0 = 5$ and $b_1 = 5$ to solve for $c_1$ and $c_2$.
**For $n=0$:**
$$b_0 = c_1(1 + 3i)^0 + c_2(1 - 3i)^0$$
$$5 = c_1(1) + c_2(1)$$
$$\mathbf{c_1 + c_2 = 5} \quad \text{(Equation 1)}$$
**For $n=1$:**
$$b_1 = c_1(1 + 3i)^1 + c_2(1 - 3i)^1$$
$$5 = c_1(1 + 3i) + c_2(1 - 3i)$$
$$5 = c_1 + 3ic_1 + c_2 - 3ic_2$$
$$5 = (c_1 + c_2) + 3i(c_1 - c_2)$$
Substitute $c_1 + c_2 = 5$ from Equation 1:
$$5 = 5 + 3i(c_1 - c_2)$$
$$0 = 3i(c_1 - c_2)$$
Since $3i \neq 0$, we must have:
$$c_1 - c_2 = 0$$
$$\mathbf{c_1 = c_2} \quad \text{(Equation 2)}$$
Substitute Equation 2 into Equation 1:
$$c_1 + c_1 = 5$$
$$2c_1 = 5$$
$$\mathbf{c_1 = \frac{5}{2}}$$
Therefore, $\mathbf{c_2 = \frac{5}{2}}$.
***
## 5. Explicit Formula
Substituting the values of $c_1$ and $c_2$ into the general solution:
$$b_n = \frac{5}{2}(1 + 3i)^n + \frac{5}{2}(1 - 3i)^n$$
$$b_n = \frac{5}{2} \left[ (1 + 3i)^n + (1 - 3i)^n \right]$$
This is the explicit formula.
$$b_n = \mathbf{\frac{5}{2} (1 + 3i)^n + \frac{5}{2} (1 - 3i)^n}$$
*(Note: While the formula involves complex numbers, the result for any integer $n \ge 0$ will always be a real number. This is because $b_n$ is the sum of a complex number and its conjugate, which always yields $2 \times \text{Re}(\text{complex number})$.)*
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