Problem 15 - 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. \( f(x) = 2^x \)

Step-by-Step Solution

Answer

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

Step 1: Understand the Formula for the Maclaurin Series

The Maclaurin series is a special case of a Taylor series centered at 0. For a function \( f(x) \), the Maclaurin series 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 \( x=0 \).

Step 2: Calculate the Derivatives of the Function

Calculate the first few derivatives of \( f(x) = 2^x \) to identify the pattern:1. \( f(x) = 2^x \)2. \( f'(x) = 2^x \ln 2 \)3. \( f''(x) = 2^x (\ln 2)^2 \)4. \( f'''(x) = 2^x (\ln 2)^3 \)This pattern continues for higher derivatives: \( f^{(n)}(x) = 2^x (\ln 2)^n \).

Step 3: Evaluate the Derivatives at x=0

To find the coefficients of the Maclaurin series, evaluate the derivatives at \( x = 0 \):1. \( f(0) = 2^0 = 1 \)2. \( f'(0) = (2^0) \ln 2 = \ln 2 \)3. \( f''(0) = (2^0)(\ln 2)^2 = (\ln 2)^2 \)4. \( f'''(0) = (2^0)(\ln 2)^3 = (\ln 2)^3 \)This pattern also continues: \( f^{(n)}(0) = (\ln 2)^n \).

Step 4: Write the Maclaurin Series

Substitute the evaluated derivatives into the general formula for the Maclaurin series:\[ f(x) = \sum_{n=0}^{\infty} \frac{(\ln 2)^n}{n!} x^n \]This is the Maclaurin series for \( f(x) = 2^x \).

Step 5: Determine the Radius of Convergence

Use the ratio test to determine the radius of convergence for the series \( \sum_{n=0}^{\infty} \frac{(\ln 2)^n}{n!} x^n \).- The ratio of successive terms is: \[ \left| \frac{(\ln 2)^{n+1}}{(n+1)!} x^{n+1} \cdot \frac{n!}{(\ln 2)^n x^n} \right| = \left| \frac{(\ln 2)x}{n+1} \right| \]- As \( n \to \infty \), \( \left| \frac{(\ln 2)x}{n+1} \right| \to 0 \) for any \( x \), so the series converges for all \( x \).Therefore, the radius of convergence is infinite.

Key Concepts

Derivative

In calculus, a derivative represents the rate at which a function is changing at any given point. It is a core concept for understanding the behavior of functions. For the function given,

• The first derivative of the function \( f(x) = 2^x \) is calculated as \( f'(x) = 2^x \ln 2 \).
• This is done by applying the chain rule of differentiation, involving the natural logarithm derivative.

The above pattern identifies how differentiation can reveal function properties with respect to the input changes. Observing subsequent derivatives

• \( f''(x) = 2^x (\ln 2)^2 \)
• \( f'''(x) = 2^x (\ln 2)^3 \)

demonstrates a consistent pattern crucial for constructing the Maclaurin series.By understanding these derivatives,you can effectively determine the behavior of functions around a specific point such as \( x = 0 \), which is essential for power series expansions.

Radius of Convergence

The radius of convergence indicates the interval within which a power series converges to the function it represents. To find the radius of convergence for the Maclaurin series of \( f(x) = 2^x \),the ratio test is used.

• The general expression for a term in the series is \( \frac{(\ln 2)^n}{n!} x^n \).
• By considering the ratio of successive terms, \[ \left| \frac{(\ln 2)^{n+1}}{(n+1)!} \cdot \frac{x^{n+1}}{x^n} \cdot \frac{n!}{(\ln 2)^n} \right| \text{ simplifies to } \left| \frac{(\ln 2) x}{n+1} \right|. \]

As \( n \to \infty \), the term\( \left| \frac{(\ln 2)x}{n+1} \right| \to 0 \) for any finite \( x \).This indicates that the series converges for all real numbers \( x \),meaning the radius of convergence is infinite.This infinite radius shows the flexibility and usefulness of the exponential function's Maclaurin series across many applications.

Power Series Expansion

A power series expansion is an expression of a function as an infinite sum of terms calculated from its derivatives at a particular point. For a Maclaurin series,each coefficient in this series represents the value of the derivative of the function at \( x = 0 \).

• The Maclaurin series for \( f(x) = 2^x \) is obtained as: \[ \sum_{n=0}^{\infty} \frac{(\ln 2)^n}{n!} x^n. \]

This infinite sum allows us to approximate the function \( 2^x \) for values of \( x \) near zero. By efficiently converging to the function value in its radius of convergence, power series expansions provide a reliable model of functional behavior.

Taylor Series

A Taylor series is an infinite sum used to approximate functions, developed around any center \( x = a \). The Maclaurin series is a special case of the Taylor series where \( a = 0 \).

• For a function \( f(x) \) with derivatives of all orders, the Taylor series expansion around a point \( x = a \) is: \[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x - a)^n. \]
• The Maclaurin series simplifies this as it is calculated with all derivatives at \( x = 0 \), making it particularly advantageous for functions like \( 2^x \).

Using Taylor series, we can evaluate complex functions efficiently by substituting polynomial terms, making them easier to compute especially around known centers. This simplifies analyses and computations in calculus and differential equations.