Problem 15 - Essential Calculus Early Transcendentals

Question

Find a power series representation for the function and determine the radius of convergence. $$ f(x)=\ln (5-x) $$

Step-by-Step Solution

Answer

The power series for \( f(x) = \ln(5-x) \) is \( \ln(5) - \sum_{n=1}^{\infty} \frac{x^n}{5^n n} \) with radius of convergence 5.

Step 1: Define the Function's Expansion

The function given is \( f(x) = \ln(5-x) \). To find a power series representation, we need to rewrite the argument as \( \ln(a) \) where \( a = 5 \), and express \( (5-x) \) in terms of \((1-u)\) such that \( u = \frac{x}{5} \). Therefore, \( 5-x = 5(1 - \frac{x}{5}) \).

Step 2: Use Taylor Series for \( \ln(1-u) \)

Utilize the Taylor series for \( \ln(1-u) \), which is \(- \sum_{n=1}^{\infty} \frac{u^n}{n} \). Here \( u = \frac{x}{5} \), thus:\[\ln(1 - \frac{x}{5}) = - \sum_{n=1}^{\infty} \frac{(\frac{x}{5})^n}{n}\]

Step 3: Transform to Match \( \ln(5-x) \)

The function \( \ln(5-x) \) can be expressed using the expansion for \( \ln(1-u) \):\[f(x) = \ln(5) + \ln(1 - \frac{x}{5}) = \ln(5) - \sum_{n=1}^{\infty} \frac{(\frac{x}{5})^n}{n}\]Expand to:\[\ln(5) - \sum_{n=1}^{\infty} \frac{x^n}{5^n n}\]

Step 4: Determine Radius of Convergence

The power series \(- \sum_{n=1}^{\infty} \frac{x^n}{5^n n}\) is based on the geometric series with \( u = \frac{x}{5} \).The radius of convergence \( R \) is determined from the condition \(|u| < 1\), which gives us \( |\frac{x}{5}| < 1 \) or \(|x| < 5\). Therefore, the radius of convergence \( R = 5 \).

Key Concepts

Taylor Series

The Taylor series is a powerful tool in calculus, used to approximate intricate functions with infinite sums of polynomial terms. It provides a way to express functions that might be difficult to handle directly, by representing them as an infinite sum based on the function's derivatives at a given point.
The general formula for the Taylor series of a function \( f(x) \) around a point \( a \) is:

• \( f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \ldots \)

This infinite series provides an exact representation of many types of functions if all orders of derivatives exist at the point \( a \). For the function \( \ln(5-x) \), Taylor series help in transforming the function into a power series representation, allowing us to analyze its behavior and characteristics.
Finding this series involves calculating derivatives, which means it can capture the essence of a function accurately using simpler, polynomial terms that are easier to manage.

Radius of Convergence

The radius of convergence is a crucial concept in the analysis of power series. It represents the interval within which a power series converges to the actual function it represents. Outside this interval, the series may not converge, leading to inaccurate results.
For a power series of the form \( \sum_{n=0}^{\infty} c_n(x-a)^n \), determining convergence involves using the ratio test, and the radius of convergence \( R \) can be derived from the condition:

• \( \lim_{n \to \infty} \left| \frac{c_{n+1}}{c_n} \right| = \frac{1}{R} \)

For the exercise, the series for \( \ln(5-x) \) was transformed to match a standard form by manipulation, leading to a power series expansion around a point \( a=5 \):

• \( - \sum_{n=1}^{\infty} \frac{x^n}{5^n n} \)

By substituting \( u = \frac{x}{5} \), and knowing the general rule that the series converges where \( |u| < 1 \), we derive that \( |x| < 5 \). Hence, the radius of convergence for the series is \( R = 5 \). This ensures convergence within the interval \(-5 < x < 5\).

Logarithmic Functions

Logarithmic functions, represented as \( \ln(x) \) for natural logarithms, are the inverses of exponential functions. They play a vital role in various mathematical contexts because they can transform multiplicative processes into additive ones, making complex calculations more manageable.
In dealing with differential calculus and series, logarithmic functions help simplify expressions, especially for functions like \( \ln(5-x) \). By knowing the series expansion of \( \ln(1-u) \), derived as:

• \( \ln(1-u) = - \sum_{n=1}^{\infty} \frac{u^n}{n} \)

the function \( \ln(5-x) \) could be expressed in a power series format, greatly simplifying the task of finding the power series representation.
This conversion not only provides easier methods for approximation but also assists in identifying convergence characteristics through the manipulation of the logarithmic terms into a series form. By visualizing these transformations, students can better grasp difficult concepts related to the relationship between exponential and logarithmic operations in calculus.