Problem 61 - Essential Calculus

Question

Let \(f(x)=\sum_{n=0}^{\infty} \frac{(-1)^{n} x^{2 n+1}}{(2 n+1) !} \quad\) and \(\quad g(x)=\sum_{n=0}^{\infty} \frac{(-1)^{n} x^{2 n}}{(2 n) !}\). (a) Find the intervals of convergence of \(f\) and \(g\). (b) Show that \(f^{\prime \prime}(x)=g(x)\). (c) Show that \(g^{\prime}(x)=-f(x)\). (d) Identify the functions \(f\) and \(g\).

Step-by-Step Solution

Answer

The intervals of convergence for both functions \(f\) and \(g\) are the entire real line. \(f''(x) = g(x)\) and \(g'(x) = -f(x)\). The functions \(f\) and \(g\) are the sine and cosine functions, respectively.

Step 1: Finding Convergence Intervals

We first find the intervals of convergence for \(f(x)\) and \(g(x)\). These are power series representations and their interval of convergence is the entire real line, since the coefficients are \( (-1)^{n} / (2 n!) \) and \( (-1)^{n} / (2 n+1)! \) which get smaller and smaller, allowing the series to converge for all \(x\). So the intervals of convergence for both functions is \( (-\infty, +\infty) \).

Step 2: Prove \( f''(x) = g(x) \)

Next, we show that the second derivative of \( f \) equals \( g \). The derivative of \( f \) is \( f'(x) = \sum_{n=0}^{\infty} (-1)^n (2n+1)x^{2n}/(2n)(2n+1)! = \sum_{n=0}^{\infty} (-1)^n x^{2n}/(2n)! = g(x) \). Thus \( f''(x) = g(x) \).

Step 3: Prove \( g'(x) = -f(x) \)

To show that the derivative of \( g \) equals negative \( f \), we compute \( g'(x) = \sum_{n=0}^{\infty} (-1)^n 2nx^{2n-1}/(2n)! = -\sum_{n=0}^{\infty} (-1)^n x^{2n+1}/(2n)! = -f(x) \). Thus, \( g'(x) = -f(x) \).

Step 4: Identify the Functions

Finally, we identify the functions. These power series representations represent the sine and cosine functions. \( f(x) = \sin(x) \) and \( g(x) = \cos(x) \) as these are their power series representations. Thus the original functions are sine and cosine functions.

Key Concepts

Interval of Convergence

The interval of convergence for a power series is the set of all values of \( x \) for which the series converges. For both functions \( f(x) \) and \( g(x) \), which are written as infinite power series, it’s important to understand how these intervals work. The coefficients for \( f(x) \) and \( g(x) \) follow a pattern: \( (-1)^n / (2n+1)! \) and \( (-1)^n / (2n)! \), respectively, which decrease as \( n \) increases.

This behavior tells us that both series converge for all real numbers \( x \). Thus, the interval of convergence for both series is from \( -\infty \) to \( +\infty \). In simple terms, no matter what real number you plug into these series, they will produce a finite sum. This results from the fact that the factorial in the denominator grows very rapidly, outpacing the growth of \( x^{2n+1} \) and \( x^{2n} \), ensuring convergence.

Derivative

Derivatives are mathematical operations that tell us the rate of change of a function. Here, they play a crucial role in connecting the functions \( f(x) \) and \( g(x) \).

To demonstrate that \( f''(x) = g(x) \), we perform differentiation twice on \( f(x) \). The first derivative of a power series involves taking the term's exponent, multiplying it, and then reducing the exponent. Doing this twice, the terms in \( f(x) \) rearrange to become those of \( g(x) \). Similarly, finding \( g'(x) \) involves performing the derivative operation once on \( g(x) \) and seeing how its series terms then align with \(-f(x)\).

These steps reveal the fascinating interplay between derivatives and power series, connecting our functions more closely and elegantly.

Sine Function

The sine function, denoted as \( \sin(x) \), is a well-known trigonometric function. It can be represented as an infinite series, known as its power series representation, particularly useful in calculus.

In the given exercise, \( f(x) \) was identified to be the sine function. This is because the series expansion of \( f(x) \) perfectly matches that of \( \sin(x) \), which is given by:

• \( \sin(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} \).

This representation is one of the easiest methods to calculate sine, especially for small values of \( x \), and is widely used in computing because it approximates \( \sin(x) \) very well.

Cosine Function

The cosine function, \( \cos(x) \), is another fundamental trigonometric function closely linked to the sine function. It too can be expressed as a power series, facilitating calculations in calculus and engineering.

From the exercise, \( g(x) \) is revealed to be the cosine function.

• This is based on the power series expansion: \( \cos(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} \).

This series is very useful for evaluating \( \cos(x) \) efficiently without a calculator. It is especially accurate for small values of \( x \) and ensures that this fundamental function can be estimated well using just pen and paper.