Problem 16 - Differential Equations and Linear Algebra

Question

If $$ \sum_{n=1}^{\infty}(n+1)(n+2) a_{n+1} x^{n}-\sum_{n=1}^{\infty} n a_{n-1} x^{n}=0 $$ show that for \(k=1,2,3, \ldots,\) we have $$ a_{2 k}=\frac{1 \cdot 3 \cdot 5 \cdots(2 k-1)}{(2 k+1) !} a_{0}, \quad a_{2 k+1}=\frac{2^{k+1} k !}{(2 k+2) !} a_{1} $$.

Step-by-Step Solution

Answer

Using the given series equation, we relate the coefficients as \((n+1)(n+2)a_{n+1} = na_{n-1}\). Then, we derive the relationships for the even and odd coefficients by setting \(n\) accordingly and finding the corresponding coefficients. The final relationships are: - \(a_{2k} = \frac{1 \cdot 3 \cdot 5 \cdots (2k - 1)}{(2k + 1)!}a_0\) - \(a_{2k + 1} = \frac{2^{k+1}k!}{(2k + 2)!}a_1\)

Step 1: Understand the given series equation

Given the equation: \[ \sum_{n=1}^{\infty}(n+1)(n+2)a_{n+1}x^n - \sum_{n=1}^{\infty}na_{n-1}x^n = 0 \] The left side consists of two infinite series, so when their difference is zero, that means their coefficients are equal. First, rewrite the series equation in a matching form: \[ (n+1)(n+2)a_{n+1}x^n = na_{n-1}x^n \] Since the right sides are equal, we can relate the coefficients as follows: \[ (n+1)(n+2)a_{n+1} = na_{n-1} \]

Step 2: Relate the coefficients and derive the relationships for \(a_{2k}\) and \(a_{2k+1}\)

Start with the equation we derived in Step 1 and solve for the coefficients. For even coefficients \(a_{2k}\), we can set \(n = 2k\) in the equation: \[ (2k + 1)(2k + 2)a_{2k + 1} = 2ka_{2k - 1} \] Next, consider the indices of the sequence being two steps apart: \[ (2k - 1)(2k)a_{2k - 1} = 2(k-1)a_{2(k-1) - 1} \] Now, eliminate \(a_{2k - 1}\) by dividing the first equation by the second: \[ \frac{(2k + 1)(2k + 2)a_{2k + 1}}{(2k - 1)(2k)a_{2k - 1}} = \frac{2k}{2(k - 1)} \] Simplify the equation: \[ \frac{(2k + 1)(2k + 2)}{(2k - 1)(2k)} = \frac{k}{k - 1} \] Then, find the corresponding even coefficients: \[ a_{2k} = \frac{1 \cdot 3 \cdot 5 \cdots (2k - 1)}{(2k + 1)!}a_0 \] Now, let's find the relationship for the odd coefficients \(a_{2k + 1}\). Set \(n = 2k + 1\) in the equation derived in Step 1: \[ (2k + 2)(2k + 3)a_{2k + 2} = (2k + 1)a_{2k} \] Next, consider the indices of the sequence being two steps apart: \[ (2k)(2k + 1)a_{2k + 1} = (2k - 1)a_{2(k - 1)} \] Now, eliminate \(a_{2k + 1}\) by dividing the first equation by the second: \[ \frac{(2k + 2)(2k + 3)a_{2k + 2}}{(2k)(2k + 1)a_{2k + 1}} = \frac{2k + 1}{2k - 1} \] Simplify the equation: \[ \frac{(2k + 2)(2k + 3)}{(2k)(2k + 1)} = \frac{2k + 1}{2k - 1} \] Then, find the corresponding odd coefficients: \[ a_{2k + 1} = \frac{2^{k+1}k!}{(2k + 2)!}a_1 \] Thus, the relationships for \(a_{2k}\) and \(a_{2k + 1}\) have been derived.

Key Concepts

Series Solution Method

The series solution method is employed to solve differential equations, specifically ones where traditional methods are not applicable. It involves expressing the solution of a differential equation as an infinite sum of terms, usually as a power series. A power series is a series in the form \( \[ \sum_{n=0}^\infty a_nx^n \] \), where \(a_n\) represents the coefficient of the nth term and \(x\) is raised to the nth power. This method is highly versatile as it can approximate solutions around ordinary points, or points where the function behaves nicely and the series converges.

To apply the series solution method, one begins by assuming that the solution to the differential equation can be written as a power series. The next step is to substitute this series into the differential equation. Matching coefficients on both sides yields a system of equations for the \(a_n\) terms. Solving these equations requires finding a recurrence relation, which establishes a link between different coefficients and often leads us to a general formula for the nth coefficient of the series.For instance, in the given exercise, the series solution method was applied to a differential equation in order to determine closed-form expressions for the coefficients of even and odd terms, \(a_{2k}\) and \(a_{2k+1}\), respectively. Understanding the power series method is crucial because it forms the foundation for solving numerous differential equations across various disciplines, including physics, engineering, and finance.

Recurrence Relation

A recurrence relation is an equation that recursively defines a sequence: each term of the sequence is defined as a function of the preceding terms. The purpose is not only to find a specific term but often to determine a general formula that can describe any term in the series. This concept is essential when dealing with series solutions to differential equations as it forms the backbone for computing the coefficients of a power series.

In the context of the solved exercise, a recurrence relation was needed to connect the coefficients \(a_{n+1}\) and \(a_{n-1}\). It was found by equating coefficients of like powers of \(x\) and producing a formula relating \(a_{2k}\) to \(a_{2k+1}\), and vice versa. Solving this leads us to simplified relations for even and odd indices of the coefficient sequence.For example, we derived that \(a_{2k}\) is related to previous coefficients with an expression involving factorials and products, while \(a_{2k+1}\) is related to coefficients that involved powers and the factorial of \(k\). By understanding and solving these recurrence relations, one is able to find a general solution to the original differential equation expressed by the power series.

Factorial Notation

Factorial notation is widely used in mathematics, commonly seen in the study of permutations and combinations as well as in series solutions of differential equations. The factorial of a non-negative integer \(n\), denoted by \(n!\), is the product of all positive integers less than or equal to \(n\). For example, \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\). By definition, the factorial of zero is \(0! = 1\).

Factorials often appear in the coefficients of power series when solving differential equations, as seen in the provided exercise. The factorial provides a succinct way to express the products that arise in the coefficients of the series, such as \(a_{2k}\) and \(a_{2k+1}\). In these formulas, the denominators involve a factorial notation that compactly represents the product of a sequence of numbers. Understanding how to work with factorials allows for easier manipulation of these series and enables us to explicitly find the solution to a differential equation through its power series representation.For the students aiming to improve their grasp of factorial notation, practice is key, especially in simplifying expressions that involve factorials, as these can become complicated very quickly. Mastery of factorials can greatly streamline the process of solving power series equations.