Problem 4 - Essential Calculus Early Transcendentals

Question

Find the Taylor series for \(\int\) centered at 4 if $$f^{(n)}(4)=\frac{(-1)^{n} n !}{3^{n}(n+1)}$$ What is the radius of convergence of the Taylor series?

Step-by-Step Solution

Answer

The Taylor series is \( \sum_{n=0}^{\infty} \frac{(-1)^n}{3^n(n+1)} (x-4)^n \); the radius of convergence is 3.

Step 1: Understanding the Function

The problem involves finding the Taylor series of a function based on its derivatives at a certain point. Here, we want to determine the Taylor series centered at 4 for a function whose nth derivative at 4 is given by: \( f^{(n)}(4) = \frac{(-1)^n n!}{3^n (n+1)} \).

Step 2: Write the Taylor Series Formula

The general Taylor series of a function \( f(x) \) centered at a point \( a \) is given by: \[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x-a)^n \]. In this problem, the center \( a \) is 4, so the series is: \[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(4)}{n!} (x-4)^n \].

Step 3: Substitute the Derivatives

Substitute the expression for the derivatives into the Taylor series formula: \[ f(x) = \sum_{n=0}^{\infty} \frac{(-1)^n n!}{3^n (n+1)} \cdot \frac{1}{n!} (x-4)^n = \sum_{n=0}^{\infty} \frac{(-1)^n}{3^n(n+1)} (x-4)^n \].

Step 4: Find the Radius of Convergence

To find the radius of convergence, use the formula for a power series \( \sum a_n (x-a)^n \), where the radius \( R \) is \( \frac{1}{\limsup_{n \to \infty} |a_n|^{1/n}} \). Here, \( a_n = \frac{(-1)^n}{3^n(n+1)} \). So, \[ \limsup_{n \to \infty} \left| \frac{1}{3^n(n+1)} \right|^{1/n} = \lim_{n \to \infty} \left( \frac{1}{3^n} \right)^{1/n} = \frac{1}{3} \]. Thus, \( R = 3 \).

Step 5: Conclusion

The Taylor series for the function centered at 4 is: \[ f(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{3^n(n+1)} (x-4)^n \]. The radius of convergence for this series is 3.

Key Concepts

Radius of Convergence

The radius of convergence is a key concept when working with Taylor series. It determines the interval around the center point where the series converges and thus accurately represents the function. To find it, we employ a special formula related to power series. The formula is given by:
\[ R = \frac{1}{\limsup_{n \to \infty} \sqrt[n]{|a_n|}} \]where \( a_n \) represents the coefficients of the series.
This formula helps us by calculating the "limiting behavior" of the coefficients as the series progresses. In simpler terms, it examines the behavior of the coefficients for large \( n \). If they tend to zero quickly, the radius of convergence will be large.

• In the exercise, \( a_n = \frac{(-1)^n}{3^n(n+1)} \).
• We found the limit as \( \frac{1}{3^n} \) tends to zero fast enough that the series converges in the radius \( R = 3 \).

This means the series accurately represents the function when \( |x - 4| < 3 \). Understanding the radius of convergence is crucial for determining the intervals where series-based approximations are valid.

Power Series

Power series are a fundamental concept in calculus, providing a way to express functions as infinite sums of terms of increasing powers. A general power series centered at \( a \) is written as:
\[ \sum_{n=0}^{\infty} a_n (x-a)^n \]These series are like infinite polynomials, and much of their usefulness comes from the ability to choose any center \( a \) and adjust the coefficients \( a_n \) based on the function we wish to approximate.
In the exercise, the problem asked for a Taylor series centered at 4. The form of the series is:

• \( f(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{3^n(n+1)} (x-4)^n \)
• The terms \( (x-4)^n \) shift the "center" of the series to \( x = 4 \).

Power series are powerful tools for approximating complex functions with simpler polynomial-like expressions. By exploring these expansions, we gain insight into functions' behavior near specific points.

Derivatives

Derivatives are fundamental in calculus and are key to understanding how functions change. In Taylor series, derivatives help determine the coefficients \( a_n \). They essentially describe the "rate of change" of a function at various levels. Here's how it connects to Taylor series:
Each coefficient \( f^{(n)}(a) \) in a Taylor series is derived from the nth derivative of the function centered at point \( a \). Each coefficient measures the rate of change at that point.

• In the given exercise, \( f^{(n)}(4) = \frac{(-1)^n n!}{3^n(n+1)} \) represents the nth derivative at 4.
• This function value helps define each series term (along with factorial adjustments and powers of \((x-4)\)).

Derivatives help us capture dynamic aspects of functions, describing their behavior around central points to build accurate and meaningful approximations.

Calculus Concepts

Calculus is a branch of mathematics that focuses on limits, functions, derivatives, integrals, and infinite series. Understanding calculus is essential for grasping how Taylor series work. Let's break down some relevant calculus concepts involved in the exercise:
Limits

• Essentially, limits help describe how functions behave as they approach a particular point.

Infinite Series

• These are sums of infinitely many terms and crucial for expressing functions like polynomials, as seen in the Taylor series.

Factorials (!)

• Factorials appear in calculus to handle permutations, combinations, and derivatives—the ! symbol means to multiply all smaller positive integers.

Together, these concepts capture the essence of how functions can be expressed and approximated through series like the Taylor series, helping enrich students' understanding and application of calculus.