Problem 6 - 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. $$f(x)=e^{-2 x}$$

Step-by-Step Solution

Answer

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

Step 1: Recall the Maclaurin Series Definition

The Maclaurin series for a function \(f(x)\) is given by \( f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots \). It is essentially the Taylor series expansion of a function at \(x=0\).

Step 2: Calculate Derivatives of the Function

Calculate the necessary derivatives of \(f(x) = e^{-2x}\):- \(f(x) = e^{-2x}\)- \(f'(x) = -2e^{-2x}\)- \(f''(x) = 4e^{-2x}\)- \(f'''(x) = -8e^{-2x}\)Notice that the \(n\)-th derivative of \(f\) is \((-2)^n e^{-2x}\).

Step 3: Evaluate Derivatives at x=0

Compute the values of these derivatives at \(x=0\):- \(f(0) = e^0 = 1\)- \(f'(0) = -2 \times 1 = -2\)- \(f''(0) = 4 \times 1 = 4\)- \(f'''(0) = -8 \times 1 = -8\) - Generally, \(f^{(n)}(0) = (-2)^n\).

Step 4: Write the Maclaurin Series

Substitute the values of the derivatives evaluated at zero into the Maclaurin series formula: \(f(x) = 1 - 2x + \frac{4}{2!}x^2 - \frac{8}{3!}x^3 + \cdots\)This simplifies to:\(f(x) = \sum_{n=0}^{\infty} \frac{(-2)^n}{n!}x^n\).

Step 5: Determine the Radius of Convergence

Use the general formula for the radius of convergence: \(\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \) where \(a_n\) is the general term of the series. Here, \(a_n = \frac{(-2)^n}{n!}\), so: \[ \left| \frac{(-2)^{n+1}/(n+1)!}{(-2)^n/n!} \right| = \left| \frac{-2}{n+1} \right| = \frac{2}{n+1} \].Taking the limit as \(n \to \infty\), we find the radius of convergence is \(\infty\).

Key Concepts

Power Series Expansion

A power series expansion is an infinite sum of terms in the form \( a_n x^n \), where \( a_n \) are coefficients and \( n \) is a non-negative integer. This series expands a function into a sum of polynomial functions around a point. For Maclaurin series, this point is 0. The power series representation gives us a way to express functions in a simpler form over certain intervals.

• Each term in the power series contains a power of \( x \) which increases with each term.
• The basic format of a power series is \( f(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \ldots \).
• Power series are used extensively in calculus and mathematical analysis for function approximation.

In our exercise, the function \( f(x) = e^{-2x} \) is expanded into its Maclaurin series, which is a specific type of power series centered at 0.

Radius of Convergence

The radius of convergence is crucial in determining where the power series represents the function effectively. It shows the interval around the center (in the case of Maclaurin series, \( x=0 \)) where the series converges to the actual function values.

• If the radius is infinite, the series converges everywhere on the real line.
• A finite radius indicates convergence within a certain interval, but not necessarily at its endpoints.
• The radius can be calculated using the ratio test: \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \).

For \( f(x) = e^{-2x} \), the radius of convergence was found using the ratio test, resulting in an infinite radius. This means the series converges for all real numbers, making it very powerful and flexible in approximating the function.

Taylor Series

The Taylor series is a general form of representing functions with infinite polynomials based on derivatives at a single point. Any function can be expressed as a Taylor series under certain conditions, and the Maclaurin series is a special case of this where the expansion is done at \( x=0 \).

• The Taylor series is expressed as \( f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \ldots \).
• It involves calculating derivatives of the function at the point \( a \).
• The series is powerful for analyzing functions locally near \( a \) and facilitates approximate calculations.

In the exercise, we utilized the Maclaurin series, a Taylor series centered at zero, to express \( f(x) = e^{-2x} \). By evaluating at zero, we derived a series that simplifies to a sum of the form \( \sum_{n=0}^{\infty} \frac{(-2)^n}{n!} x^n \), showcasing the beauty of this method in tackling complex functions.