Question

In Exercises 49-54, show that the function represented by the power series is a solution of the differential equation. $$ y=\sum_{n=0}^{\infty} \frac{(-1)^{n} x^{2 n}}{(2 n) !}, y^{\prime \prime}+y=0 $$

Short answer

Indeed, the function represented by the power series \(y=\sum_{n=0}^{\infty} \frac{(-1)^{n} x^{2 n}}{(2 n) !}\) is a solution to the differential equation \(y'' + y = 0\)

Step-by-step solution

Identify The Summation Term

The first step will be to identify the summation term in the function, which is \(a_{n} = \frac{(-1)^{n} x^{2 n}}{(2 n) !}\)

Compute The First Derivative

To find the first derivative \(y'\), the power rule is applied, whereby the derivative of \(x^{k}\) is \(kx^{k-1}\). Therefore, the first derivative \(y' = \sum_{n=0}^{\infty} n\frac{(-1)^{n} x^{2 n - 1}}{(2 n - 1) !}\).

Compute The Second Derivative

In a similar manner, compute the second derivative \(y''\) by applying the power rule again. This gives \(y'' = \sum_{n=0}^{\infty} n(n-1)\frac{(-1)^{n} x^{2 n - 2}}{(2 n - 2)!}\).

Verify The Differential Equation

To verify the differential equation \(y'' + y = 0\), all terms in the power series of the second derivative and function \(y\) which have the same power should add up to zero. For the function \(y\) and its second derivative \(y''\), the terms with even powers are those with \(n=0,1,2,3,...\), where the terms of the function \(y\) are \(\frac{(-1)^{n} x^{2 n}}{(2 n) !}\) and those of \(y''\) are \(n(n-1)\frac{(-1)^{n} x^{2 n - 2}}{(2 n - 2)!}\). For all \(n=0,1,2,3,...\), \(- \frac{(-1)^{n} x^{2 n}}{(2 n) !} + n(n-1)\frac{(-1)^{n} x^{2 n - 2}}{(2 n - 2)!} = 0\), verifying the solution.

Key concepts

Power Series Solution

A power series is an infinite sum of the form \( \sum_{n=0}^{\infty} a_n x^n \), where \( a_n \) are coefficients and \( x \) is a variable, usually around a point often taken as zero (also known as a Maclaurin series). Power series solutions are particularly valuable when solving differential equations, as they can approximate solutions that are difficult to express with elementary functions.

For example, the equation \( y = \sum_{n=0}^{\infty} \frac{(-1)^{n} x^{2n}}{(2n)!} \) represents the power series solution to the differential equation \( y'' + y = 0 \) given in the exercise. The power series is centered at \( x = 0 \) and the \( a_n \) coefficients are determined to be \( \frac{(-1)^{n}}{(2n)!} \), which indicates an alternating series with a factorial in the denominator, often seen in expansions of exponential functions. This particular power series represents the cosine function, which is known to be the solution to the given differential equation. Identifying the pattern and convergence of a power series helps us to find exact or approximate solutions to complex differential equations.

First Derivative

In calculus, the first derivative of a function represents the rate of change of the function with respect to its variable. It is often denoted as \( y' \) or \( \frac{df}{dx} \). The process of finding this derivative is called differentiation. When we have a function represented as a power series, we find its derivative by differentiating term by term, following rules like the power rule.

In the given exercise, the first derivative \( y' \) of the power series is computed using the power rule: \( y' = \sum_{n=0}^{\infty} n\frac{(-1)^{n} x^{2n-1}}{(2n-1)!} \). This means each term \( x^{2n} \) in the series is differentiated to become \( (2n)x^{2n-1} \) and the coefficient is adjusted accordingly. The differentiation reduces the power by one and multiplies the term by the former exponent, leading to a new series that is still in an infinite sum form representing the first derivative of the original function.

Second Derivative

The second derivative of a function, denoted as \( y'' \) or \( \frac{d^2f}{dx^2} \) gives us the rate of change of the rate of change—essentially, it tells us how the slope of a function is changing. For a function represented as a power series, the second derivative is found by applying the differentiation process a second time.

In the context of the exercise, the second derivative \( y'' \) of the power series is determined by differentiating the first derivative term by term once again: \( y'' = \sum_{n=0}^{\infty} n(n-1)\frac{(-1)^{n} x^{2n-2}}{(2n-2)!} \). This derivative is crucial because the differential equation we're examining, \( y'' + y = 0 \) is a second-order differential equation—meaning that the highest derivative is the second derivative. Comparing the individual terms of \( y'' \) with those of the original function \( y \) allows us to verify whether the power series solution satisfies the differential equation, as performed in the step by step solution from the textbook. This verification is a testament to the symmetry and properties of power series when applied to differential equations, echoing underlying functions such as sine and cosine for their respective equations.