Problem 9 - Essential 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 \(R_{n}(x) \rightarrow 0 . ]\) Also find the associated radius of convergence. $$f(x)=\sinh x$$

Step-by-Step Solution

Answer

The Maclaurin series for \(\sinh x\) is \(\sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!}\) with an infinite radius of convergence.

Step 1: Understanding Maclaurin Series

A Maclaurin series is a Taylor series expansion of a function about 0. It is given by the formula: \\[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n \] \where \(f^{(n)}(0)\) is the \(n\)-th derivative of \(f(x)\) evaluated at \(x=0\).

Step 2: Calculate Derivatives of \( \sinh x \)

The hyperbolic sine function is defined as \( \sinh x = \frac{e^x - e^{-x}}{2} \). We need to calculate the derivatives at \( x = 0 \).\1. \( f(x) = \sinh x \). So, \( f(0) = \sinh(0) = 0 \).2. \( f'(x) = \cosh x \), calculate \( f'(0) = \cosh(0) = 1 \).3. \( f''(x) = \sinh x \), calculate \( f''(0) = \sinh(0) = 0 \).4. \( f'''(x) = \cosh x \), calculate \( f'''(0) = \cosh(0) = 1 \).5. Continue alternating derivatives for pattern.

Step 3: Construct Maclaurin Series

Substituting the derivatives into the Maclaurin series formula:\[f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n = 0 + \frac{1}{1!} x + 0 \frac{x^2}{2!} + \frac{1}{3!}x^3 + 0 \frac{x^4}{4!} + \frac{1}{5!}x^5 + \cdots\]This simplifies to \\[\sinh x = \sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!}\]

Step 4: Radius of Convergence

The radius of convergence \( R \) of a power series can be found using the ratio test. In this series, every term is \(\frac{x^{2n+1}}{(2n+1)!}\). Applying the ratio test, we find:\[\lim_{n \to \infty} \left| \frac{x^{2n+3}}{(2n+3)!} \cdot \frac{(2n+1)!}{x^{2n+1}} \right| = \lim_{n \to \infty} \frac{x^2}{(2n+2)(2n+3)} = 0\]For any finite \(x\), this approaches 0 as \(n \to \infty\), thus the series converges for all \(x\). Consequently, the radius of convergence is \( R = \infty \).

Key Concepts

Taylor series

The Taylor series is a powerful way to express a wide variety of functions as an infinite sum of terms, calculated from the values of their derivatives at a particular point. This allows us to approximate complex functions with polynomials, making them easier to work with under various conditions.The general formula for the Taylor series of a function \( f(x) \) about a point \( a \) is:

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

For the special case where \( a = 0 \), this becomes the Maclaurin series:

• \( f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n \)

In our example with the hyperbolic sine function, \( \sinh x \), we calculated the derivatives at \( x = 0 \) to construct its Maclaurin series. The beauty of the Taylor series, and especially the Maclaurin series, is in its ability to simplify complicated expressions into understandable terms by systematically breaking them down into polynomial terms.

Hyperbolic functions

Hyperbolic functions are analogs of the trigonometric functions, but they are based on hyperbolas rather than circles. The most common hyperbolic functions are the hyperbolic sine (\( \sinh \)) and cosine (\(\cosh\)).The hyperbolic sine function, \( \sinh x \), is defined as:

• \( \sinh x = \frac{e^x - e^{-x}}{2} \)

This function exhibits properties similar to those of the trigonometric sine function, yet differs significantly in its growth and periodicity.For instance, as \( x \) tends to infinity, \( \sinh x \) behaves similarly to \( e^x/2 \), leading to its exponential-type growth, unlike the oscillatory nature of the sine function. Understanding these properties allows for their application in various fields including calculus and complex analysis, where they often serve to simplify problems involving exponential growth.

Radius of convergence

The radius of convergence of a power series is the distance from the center of the series where the series converges, or sums to a finite value. It is vital in determining the domain of valid inputs for which the series provides an accurate representation of the function.The radius of convergence is commonly determined through the ratio test or the root test, both of which are methods to determine the convergence behavior of series terms as the number of terms approach infinity.For the Maclaurin series of \( \sinh x \), the power series we derived has the form:

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

Applying the ratio test here, we find that the limit approaches zero as \( n \) approaches infinity for any finite \( x \). This reveals that the series converges for all real values of \( x \), yielding an infinite radius of convergence.Knowing the radius of convergence helps greatly in practical applications, as it guides us in understanding the extent to which we can rely on the series expansion in approximating the original function within specified bounds.