Problem 17 - 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. \( f(x) = \sinh x \)

Step-by-Step Solution

Answer

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

Step 1: Define a Maclaurin Series

A Maclaurin series for a function \( f(x) \) is given by \( f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n \), where \( f^{(n)}(0) \) is the \( n \)-th derivative of \( f \) evaluated at 0.

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

The function \( \sinh x \) is defined as \( \sinh x = \frac{e^x - e^{-x}}{2} \). Calculate derivatives:- \( f^{(0)}(x) = \sinh x \)- \( f^{(1)}(x) = \cosh x \)- \( f^{(2)}(x) = \sinh x \)- \( f^{(3)}(x) = \cosh x \)The pattern shows that the derivatives alternate between \( \sinh x \) and \( \cosh x \).

Step 3: Evaluate Derivatives at x = 0

Calculate the values of the first few derivatives at \( x = 0 \):- \( f^{(0)}(0) = \sinh(0) = 0 \)- \( f^{(1)}(0) = \cosh(0) = 1 \)- \( f^{(2)}(0) = \sinh(0) = 0 \)- \( f^{(3)}(0) = \cosh(0) = 1 \) The values alternate between 0 and 1.

Step 4: Write the Maclaurin Series

For the Maclaurin series \( \sinh x = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n \), substitute the derivative evaluated values:\[\sinh x = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n = \frac{x}{1!} + \frac{x^3}{3!} + \frac{x^5}{5!} + \cdots = \sum_{k=0}^{\infty} \frac{x^{2k+1}}{(2k+1)!}\]

Step 5: Determine the Radius of Convergence

The Maclaurin series for \( \sinh x \) is similar to the power series of \( e^x \), which has a radius of convergence of \( \infty \). Therefore, the radius of convergence for \( \sinh x \) is also \( \infty \).

Key Concepts

Radius of Convergence

When dealing with series, the radius of convergence is a crucial concept. It tells us the range of inputs for which the series is valid or converges to a particular function.
For a power series, one typically writes it as \( \sum_{n=0}^{\infty} a_n (x - c)^n \). Here, \( c \) is the center of the series, and each \( a_n \) are the coefficients.

• The radius of convergence \( R \) is derived using the formula:

\[\frac{1}{R} = \limsup_{n \rightarrow \infty} \left| \frac{a_{n+1}}{a_n} \right|\]This formula helps in determining how far from the center \( c \), in each direction, the series will converge when \( x \) is substituted into it.
In our example with \( \sinh x \), this power series resembles that of \( e^x \), which has a radius of convergence of \( \infty \). This means no matter what real number is plugged into the series, it will always converge. That’s why we say the radius is infinite!

Power Series Expansion

Power series can be seen as polynomials with an infinite number of terms. They are written in the form \[ f(x) = \sum_{n=0}^{\infty} a_n (x - c)^n \].
Instead of having a finite number of terms like usual polynomials, this expansion never ends!
When dealing with functions like \( \sinh x \), we expand using a specific type of power series—a Maclaurin series. Here:

• The center \( c \) is 0.
• So its form becomes \( \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n \).

Calculating the derivatives of \( f(x) = \sinh x \) showed the values switch between \( \cosh \) and \( \sinh \). Evaluating these at 0 gives us the coefficients needed, namely 0 and 1 alternately.
Consequently, the series took shape as: \[\sinh x = \frac{x}{1!} + \frac{x^3}{3!} + \frac{x^5}{5!} + \ldots \] This pattern is then generalized into the sum: \[ \sinh x = \sum_{k=0}^{\infty} \frac{x^{2k+1}}{(2k+1)!} \].
Understanding power series and their expansions lets us approximate functions with great accuracy!

Hyperbolic Functions

Hyperbolic functions, such as \( \sinh x \) and \( \cosh x \), are analogues of trigonometric functions but for the hyperbola rather than the circle.
These functions are defined using exponentials:

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

They are used frequently in mathematics, physics, and engineering as they relate to solutions of some differential equations, and describe certain types of geometric shapes.
The derivatives of hyperbolic functions display intriguing patterns. For example:

• The derivative of \( \sinh x \) is \( \cosh x \).
• The derivative of \( \cosh x \) is \( \sinh x \).

These patterns are similar yet distinct to those found in regular trigonometric functions. Understanding hyperbolic functions and their properties aids in the analysis of growth-like processes and waves, giving insight into both natural and theoretical frameworks.