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Chapter 8: Problem 38

a) Find the characteristic roots of the linear homogeneous recurrence relation \(a_{n}=2 a_{n-1}-2 a_{n-2}\) \([\text { Note: These are complex numbers. }]\) b) Find the solution of the recurrence relation in part (a) with \(a_{0}=1\) and \(a_{1}=2\)

Short Answer

Expert verified
The characteristic roots are \(1 + i\) and \(1 - i\). The solution is \( a_n = \frac{1-i}{2} (1+i)^n + \frac{1+i}{2} (1-i)^n \) with initial conditions.

Step by step solution

01

- Write the characteristic equation

For the recurrence relation given, write the characteristic equation as follows:\[ r^2 - 2r + 2 = 0 \]
02

- Solve the characteristic equation

We need to solve the quadratic equation to find the roots. Using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \(a = 1\), \(b = -2\), and \(c = 2\):\[ r = \frac{2 \pm \sqrt{4 - 8}}{2} = \frac{2 \pm \sqrt{-4}}{2} = \frac{2 \pm 2i}{2} = 1 \pm i \]So, the roots are \(1 + i\) and \(1 - i\).
03

- Form the general solution

Using the roots, the general solution of the recurrence relation is:\[ a_n = c_1 (1+i)^n + c_2 (1-i)^n \]
04

- Apply initial conditions

We use the given initial conditions \(a_0 = 1\) and \(a_1 = 2\) to solve for the constants \(c_1\) and \(c_2\).Starting with \(a_0 = 1\):\[ a_0 = c_1 (1+i)^0 + c_2 (1-i)^0 = c_1 + c_2 = 1 \]Next with \(a_1 = 2\):\[ a_1 = c_1 (1+i)^1 + c_2 (1-i)^1 = c_1 (1+i) + c_2 (1-i) = 2 \]
05

- Solve for constants

From \(c_1 + c_2 = 1\) and \(c_1 (1+i) + c_2 (1-i) = 2\), we can solve this system of equations:Subtracting the equations:\[ c_1 (1+i) + c_2 (1-i) - (c_1 + c_2 = 1) = 2 - 1 \]Simplify to get:\[ c_1i - c_2i = 1 \rightarrow c_1 - c_2 = 1/i = -i \]So, we have:\[ c_1 + c_2 = 1 \]\[ c_1 - c_2 = -i \]Adding these:\[ 2c_1 = 1 - i \rightarrow c_1 = \frac{1-i}{2} \]Subtracting them:\[ 2c_2 = 1 + i \rightarrow c_2 = \frac{1+i}{2} \]
06

- Write the final solution

Substitute the values of \(c_1\) and \(c_2\) back into the general solution:\[ a_n = \frac{1-i}{2} (1+i)^n + \frac{1+i}{2} (1-i)^n \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

characteristic roots
In linear homogeneous recurrence relations, characteristic roots are solutions to the characteristic equation. For the recurrence relation given in the problem, the characteristic equation was written as:
The characteristic equation helps us to find the constant multipliers in the general solution of the recurrence relation.
For our problem, the characteristic roots are complex numbers because of the discriminant in the quadratic formula.
As you solve the characteristic equation from the relation, the roots you find - like 1 + i and 1 - i - help specify the nature of the solution.
quadratic equation
A quadratic equation is a polynomial equation of degree 2. It generally has the form:





where a, b, and c are constants.

To solve it, we use the quadratic formula:
In our problem, the characteristic equation turned out to be a quadratic equation. By identifying a, b, and c, and plugging them into the quadratic formula, we found the characteristic roots.

We calculated:
This gave us the two roots: These roots were complex (involving i), leading to an oscillatory behavior in the solution.
complex numbers
Complex numbers are numbers that have a real and an imaginary part. They are usually written in the form a + bi, where a is the real part and b is the imaginary part. In our problem, the characteristic roots are complex: 1 + i and 1 - i.

Understanding complex numbers involves knowing that:
  • i is the imaginary unit where i虏 = -1
  • Complex conjugates: (a + bi) and (a - bi)
  • To combine, you use algebraic rules but include i

When solving recurrence relations, complex roots manifest oscillating (trigonometric) solutions. They combine elements to form a pattern over iterations.
initial conditions
Initial conditions are the values of the sequence at the beginning to ensure we get a specific solution. They anchor the general solution to the exact one we need. For our problem, the initial conditions were stated as:
  • a_0 = 1
  • a_1 = 2

We applied these initial values to solve for constants c_1 and c_2 that fit into the general solution:



Combining this with the characteristic roots, substituting the values, and solving yields specific constants. Through algebraic manipulations, equilibrates for constants.

Subsequently, these specific c_1 and c_2 concatenate to make our final solution nuanced and aligned with the problem鈥檚 conditions.

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