Problem 18 - Calculus: Early Transcendentals

Question

Find the Maclaurin series for \( f(x) \) using the definition of a Maclaurin series. [ Assume that \( f \) has a power series expansion. Do not show that \( R_n (x) \to 0. \)] Also find the associated radius of convergence. \( f(x) = \cosh x \)

Step-by-Step Solution

Answer

The Maclaurin series for \( f(x) = \cosh x \) is \( \sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!} \), and the radius of convergence is \( \infty \).

Step 1: Identify the Function and Definitions

We are given the function \( f(x) = \cosh x \). The cosh function, or hyperbolic cosine, is defined as \( \cosh x = \frac{e^x + e^{-x}}{2} \). We're tasked with finding its Maclaurin series which is a Taylor series centered at \( x = 0 \).

Step 2: Write the Maclaurin Series Formula

The Maclaurin series for \( f(x) \) is given by \( f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n \), where \( f^{(n)}(0) \) are the derivatives evaluated at 0.

Step 3: Compute Derivatives of \( \cosh x \)

We compute the first few derivatives of \( \cosh x \) : - \( f(x) = \cosh x \) - \( f'(x) = \sinh x \) - \( f''(x) = \cosh x \) - \( f'''(x) = \sinh x \) Notice the pattern: Even derivatives of \( \cosh x \) are \( \cosh x \), and odd derivatives are \( \sinh x \).

Step 4: Evaluate Derivatives at \( x = 0 \)

Evaluate the series derivatives at zero:- \( f(0) = \cosh(0) = 1 \)- \( f'(0) = \sinh(0) = 0 \)- \( f''(0) = \cosh(0) = 1 \)- \( f'''(0) = \sinh(0) = 0 \)The even-indexed derivatives are 1, and the odd-indexed derivatives are zero.

Step 5: Substitute Derivatives in Series Formula

Substitute these values into the Maclaurin series formula:\[\cosh x = \sum_{n=0}^{\infty} \frac{f^{(2n)}(0)}{(2n)!} x^{2n} = \sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!}\]

Step 6: Find the Radius of Convergence

The radius of convergence \( R \) of a power series \( \sum a_n x^n \) is given by \( \frac{1}{\limsup_{n \to \infty} |a_n|^{1/n}} \). Here, \( a_n = \frac{1}{(2n)!} \), and using the ratio test or the formula, the series converges for all \( x \) because all entire functions (like exponentials) have infinite radii of convergence. Thus, \( R = \infty \).

Key Concepts

Radius of Convergence

To understand the radius of convergence, we need to focus on its role in determining where a power series converges. When you have a power series expression for a function, the radius of convergence is a boundary that tells us for which values of the variable the series converges. This is crucial in series representations because it helps establish the extent of usable values.

For a series \( \sum a_n x^n \), the radius of convergence \( R \) can be found using the formula:

\[ \frac{1}{\limsup_{n \to \infty} |a_n|^{1/n}}. \]

If \( R \) is infinite, like in the case of \( \cosh x \), the series converges for all real numbers \( x \). That’s because functions like \( e^x \) have no restrictions in real-number contexts, indicating an infinite radius of convergence.

Hyperbolic Functions

Hyperbolic functions are analogs of trigonometric functions but are based on hyperbolas rather than circles. These functions have applications in various fields like calculus and theory of complex numbers.

The hyperbolic cosine, \( \cosh x \), is defined as:

\[ \cosh x = \frac{e^x + e^{-x}}{2}. \]

This definition shows the symmetry in hyperbolic functions, which can be useful in solving certain types of differential equations and integrals. They are related to the exponential function and, by extension, to the Maclaurin and Taylor series.

• The hyperbolic sine, \( \sinh x \), is \( \frac{e^x - e^{-x}}{2} \).
• These functions often help model physical phenomena and mechanics.

Taylor Series

The Taylor series is a powerful tool in calculus, used to approximate functions by polynomials centered around a specific point. It gives us a way to express functions with infinite sums which can be computed with derivatives.

The Taylor series has the generic form:

\[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x-a)^n, \]

where \( f^{(n)}(a) \) represents the \( n \)-th derivative of \( f \) evaluated at \( a \).

A Maclaurin series is specifically a Taylor series centered at \( a = 0 \). This special case simplifies the computation because you only need derivatives evaluated at zero.

Derivatives

Derivatives are fundamental in calculus because they measure how a function changes as its input changes. They are essential for deriving Taylor and Maclaurin series, as they give the necessary coefficients.

When constructing the Maclaurin series for \( \cosh x \), you'll notice interesting patterns with derivatives.

• \( \cosh x \) and \( \sinh x \) alternate between even and odd derivatives.
• Evaluating these at \( x=0 \) helps specify which terms survive in the series expansion.

For \( \cosh x \), even derivatives remain, giving the series:

\[ \cosh x = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \cdots \]
Identifying this pattern is critical in systems where symmetry and oscillatory behaviors are considered.

Power Series Expansion

Using power series expansion allows us to express complicated functions as infinite sums of simpler polynomial terms. This simplifies mathematical analysis and computations in various fields.

A power series is expressed as:

\[ \sum_{n=0}^{\infty} a_n x^n. \]

For \( \cosh x \), the coefficients \( a_n \) are determined through the evaluation of derivatives at zero. The resulting expansion is:

• Based only on even powers of \( x \),
• Expressed as \( \sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!} \).

This series representation is extensively used in mechanics and wave equations due to its convergence for all real numbers, pinpointing its vast applicability.