Question

Find the first four nonzero terms in each of two linearly independent power series solutions about the origin. What do you expect the radius of convergence to be for each solution? $$ y^{\prime \prime}+(\sin x) y=0 $$

Short answer

Question: Find two linearly independent power series solutions for the given second-order linear ordinary differential equation: $$ y''(x)+\sin(x)y(x)=0 $$ Find the first four nonzero terms of each solution and discuss the expected radius of convergence in each case. Answer: The two linearly independent power series solutions for the given differential equation are: $$ y_1(x) = 1 - \frac{1}{3!}x^2 + \mathcal{O}(x^3) $$ and $$ y_2(x) = x - \frac{1}{5!}x^3 + \mathcal{O}(x^4) $$ The expected radius of convergence for each solution is infinity.

Step-by-step solution

Convert the given equation into a power series equation

First, we convert the given differential equation into its power series form. We assume that the solution is of the form: $$ y(x) = \sum_{n=0}^{\infty}a_nx^n $$ Now, we need to find the first and second derivatives of y(x) to substitute into the given equation: $$ y'(x) = \sum_{n=1}^{\infty}na_nx^{n-1} \\ y''(x) = \sum_{n=2}^{\infty}n(n-1)a_nx^{n-2} $$ Similarly, we need to expand the sine function into a power series using its Maclaurin series expansion: $$ \sin(x) = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}x^{2n+1} $$ Substitute these power series expressions into the given equation: $$ \sum_{n=2}^{\infty}n(n-1)a_nx^{n-2} + \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}x^{2n+1} \sum_{n=0}^{\infty}a_nx^n = 0 $$

Calculate the coefficients of the first four nonzero terms

To find the two linearly independent power series solutions, let's take two cases for the initial terms. Case 1: $$ a_0 = 1 (Odd\; powers) \\ a_1 = 0 (Even\; powers) $$ Now, find the recurrence relation for the coefficients by setting the coefficient of the same power equal to zero: $$ (n-1)(n)a_n + \frac{(-1)^n}{(2n+1)!}a_{n+1} = 0 $$ Using the given initial terms, we can calculate the first 4 nonzero terms of the first solution: $$ a_0 = 1, \\ a_1 = 0, \\ a_2 = -\frac{1}{3!}(1), \\ a_3 = 0 $$ The first power series solution becomes: $$ y_1(x) = 1 - \frac{1}{3!}x^2 + \mathcal{O}(x^3) $$ Case 2: $$ a_0 = 0 (Odd\; powers) \\ a_1 = 1 (Even\; powers) $$ Using the same recurrence relation and the given initial terms, we can calculate the first 4 nonzero terms of the second solution: $$ a_0 = 0, \\ a_1 = 1, \\ a_2 = 0, \\ a_3 = -\frac{1}{5!}(1) $$ The second power series solution becomes: $$ y_2(x) = x - \frac{1}{5!}x^3 + \mathcal{O}(x^4) $$

Determine the radius of convergence for each solution

For each of the two solutions, the radius of convergence should be the same since they both come from the same differential equation. The radius of convergence is determined by the coefficients of the power series solutions. In our case, the coefficients for each solution are rational and converge to zero as n goes to infinity. Since the terms of the sine function converge to zero, we can expect the radius of convergence to be infinite. Therefore, the expected radius of convergence for each solution is infinity.

Key concepts

Differential Equations

Understanding differential equations is essential for solving a broad range of problems in mathematics and applied sciences. A differential equation is a mathematical equation that relates some function with its derivatives. In our exercise, the differential equation given is

\[\begin{equation}y^{\text{\prime \prime}} + (\text{sin } x)y = 0\text{\end{equation}\]}

Differential equations can be classified into several types. The given equation is a second-order linear homogeneous differential equation with variable coefficients. Homogeneous means that there is no term without the function y or its derivatives. The solution to such an equation is an expression for y in terms of the variable x, often represented as an infinite series. To find the solution, one must consider initial conditions or boundary conditions, which in the context of power series solutions, translate into choices for the coefficients of the series. In our exercise, we tackle the problem by proposing that the solution can be expressed as a power series centered at the origin.

Radius of Convergence

The radius of convergence is a vital concept when dealing with power series. It determines the interval within which the power series converges to a finite value. In simple terms, it tells us how far away from the center of the series we can go before the series stops being a good representation of our function.

In practice, finding the radius of convergence involves mathematical techniques such as the ratio test or the root test. For the differential equation:
\[\begin{equation}y^{\text{\prime \prime}} + (\text{sin } x)y = 0\text{\end{equation}\]}

we look at the coefficients of the power series solutions as n goes to infinity. If these coefficients approach zero, as is the case in our exercise, the radius of convergence can be infinite, meaning the series provides a valid solution for all x. This is a significant finding because it tells us that the behavior of the solution is well-behaved and predictable across the entire number line. Thus, the solution series gives insights into the function's behavior everywhere, not just near the origin.

Maclaurin Series

A Maclaurin series is a special kind of power series that represents a function as an infinite sum of terms calculated from the values of its derivatives at a single point, typically at zero.

\[\begin{equation}f(x) = \text{\sum}_{n=0}^{\text{\infty}} \frac{f^{(n)}(0)}{n!} x^n\text{\end{equation}\]}

Here, \[\begin{equation} f^{(n)}(0)\text{\end{equation}\]}
denotes the n-th derivative of the function evaluated at 0. The Maclaurin series is actually a special case of the Taylor series when the center is at zero. In our exercise, the Maclaurin series was used to expand the sine function, an essential step to finding the power series solution. This technique is particularly useful because it allows for the transformation of trigonometric, exponential, logarithmic, and other transcendental functions into an infinite polynomial, making them much easier to work with in the context of differential equations. Understanding the Maclaurin series not only provides a method to approximate complicated functions but also offers deeper insight into the function's properties, such as its derivatives at a specific point.