Chapter 7: Problem 2
The Pell sequence \(P_{n}\) is defined by \(P_{1}=1, P_{2}=2\), and \(P_{n+2}=2 P_{n+1}+\) \(P_{n}\). Show that $$ P_{n}=\frac{1}{2 \sqrt{2}}\left[(1+\sqrt{2})^{n}-(1-\sqrt{2})^{n}\right] . $$
Short Answer
Expert verified
The closed form for the Pell sequence is verified:
\( P_n = \frac{1}{2 \sqrt{2}}[(1+\sqrt{2})^n - (1-\sqrt{2})^n] \).
Step by step solution
01
Define the Characteristic Equation
The given recurrence relation is \( P_{n+2} = 2P_{n+1} + P_n \). Its characteristic equation can be found by substituting \( P_n = r^n \) into the recurrence relation: \[ r^{n+2} = 2r^{n+1} + r^n. \] Dividing throughout by \( r^n \), we get the characteristic equation: \[ r^2 = 2r + 1. \] This simplifies to \( r^2 - 2r - 1 = 0 \).
02
Solve the Characteristic Equation
The characteristic equation \( r^2 - 2r - 1 = 0 \) can be solved using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -2 \), and \( c = -1 \). Computing the discriminant, we find:\[ b^2 - 4ac = 4 + 4 = 8. \] Thus, the roots are: \[ r = \frac{2 \pm \sqrt{8}}{2} = 1 \pm \sqrt{2}. \] The solutions to the characteristic equation are \( r_1 = 1 + \sqrt{2} \) and \( r_2 = 1 - \sqrt{2} \).
03
Construct the General Solution
The general solution to the recurrence relation, based on the roots, is: \[ P_n = A(1+\sqrt{2})^n + B(1-\sqrt{2})^n, \] where \( A \) and \( B \) are constants determined by the initial conditions.
04
Apply Initial Conditions to Find Constants
Using the initial conditions \( P_1 = 1 \) and \( P_2 = 2 \), we substitute into the general solution:1. \( P_1 = A(1+\sqrt{2}) + B(1-\sqrt{2}) = 1 \).2. \( P_2 = A(1+\sqrt{2})^2 + B(1-\sqrt{2})^2 = 2 \). Solving these two equations gives \( A = \frac{1}{2\sqrt{2}} \) and \( B = -\frac{1}{2\sqrt{2}} \).
05
Verify the Closed Form Expression
With \( A = \frac{1}{2\sqrt{2}} \) and \( B = -\frac{1}{2\sqrt{2}} \), substitute back into the general solution:\[ P_n = \frac{1}{2\sqrt{2}}[(1+\sqrt{2})^n - (1-\sqrt{2})^n]. \] This matches the closed form given in the problem statement, confirming it is correct.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Characteristic Equation
The characteristic equation is a powerful tool in solving linear recurrence relations like those found in the Pell sequence. When dealing with a recurrence relation, we often assume solutions of the form \(P_n = r^n\). By substituting this assumed form into the recurrence equation \(P_{n+2} = 2P_{n+1} + P_n\), we can derive a polynomial equation known as the characteristic equation. For our specific case, this substitution leads to the equation \(r^{n+2} = 2r^{n+1} + r^n\). Dividing each term by \(r^n\), we simplify this to \(r^2 = 2r + 1\), resulting in the quadratic characteristic equation \(r^2 - 2r - 1 = 0\). This equation will allow us to determine the roots, which in turn will help in formulating a general solution for the sequence.
Recurrence Relation
A recurrence relation defines each term of a sequence using preceding terms. For the Pell sequence, this is given by the formula \(P_{n+2} = 2P_{n+1} + P_n\). It essentially prescribes how each term is generated from the previous two terms, setting into motion a predictable pattern.The relation starts with initial conditions, in this case, \(P_1 = 1\) and \(P_2 = 2\). As one computes further terms, these initial conditions serve as the foundation. Recurrence relations are important in mathematical sequences as they provide a structured way to understand and compute terms without having to explicitly outline every term in the sequence. Through the recurrence relation, a larger pattern or formula can often be derived, simplifying computation and analysis.
Quadratic Formula
The quadratic formula is crucial for solving the characteristic equation derived from our recurrence relation. Given a standard quadratic equation of the form \(ax^2 + bx + c = 0\), the quadratic formula enables us to find its roots as follows:\[\x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\\]For our characteristic equation \(r^2 - 2r - 1 = 0\), we plug in \(a = 1\), \(b = -2\), and \(c = -1\). First, we compute the discriminant, \(b^2 - 4ac\), which equals 8. Then we insert these values into the quadratic formula, providing us with the roots: \(r = 1 + \sqrt{2}\) and \(r = 1 - \sqrt{2}\). These roots play an essential role in constructing the general solution for the sequence.
General Solution
The general solution of a recurrence relation describes the overall pattern of the sequence based on the roots found from the characteristic equation. For the Pell sequence, the roots \(r_1 = 1 + \sqrt{2}\) and \(r_2 = 1 - \sqrt{2}\) yield a general solution of the form:\[\P_n = A(1+\sqrt{2})^n + B(1-\sqrt{2})^n\\]To determine the constants \(A\) and \(B\), we must use the initial conditions \(P_1 = 1\) and \(P_2 = 2\). Solving this system of equations gives us \(A = \frac{1}{2\sqrt{2}}\) and \(B = -\frac{1}{2\sqrt{2}}\). Hence, the full explicit formula for the Pell sequence becomes:\[\P_n = \frac{1}{2\sqrt{2}} [(1 + \sqrt{2})^n - (1 - \sqrt{2})^n]\\]This representation provides a closed formula, allowing us to compute terms of the sequence directly without recursive calculations.