Problem 57 - Calculus: One and Several Variables

Question

Derive a series expansion in \(x\) for the function and specify the numbers \(x\) for which the expansion is valid. Take \(a>0\). $$f(x)=\ln (a+x)$$

Step-by-Step Solution

Answer

The series expansion for the given function \(f(x) = \ln(a+x)\) in terms of x is \[f(x) = \ln(a) + \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}\left(\frac{x}{a}\right)^n\] The expansion is valid for the range of \(x\) values: \(-a < x \le a\) where \(a > 0\).

Step 1: Rewrite the function

To find the Taylor series expansion, we need our function in the form of a power series around zero. One way to do this is by using the formula for the natural logarithm of 1 + x: \[\ln(1+x) = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}x^n\] We can rewrite our function \(f(x) = \ln(a+x)\) as: \[f(x) = \ln\left(a\left(1+\frac{x}{a}\right)\right)\] Now we can use the fact that \(\ln(ab) = \ln(a) + \ln(b)\) to rewrite our function as: \[f(x) = \ln(a) + \ln\left(1+\frac{x}{a}\right)\]

Step 2: Finding the Taylor series expansion for ln(1 + x/a)

Now, we need to write the second term, \(\ln\left(1 + \frac{x}{a}\right)\), as its power series representation. We can rewrite this term as: \[\ln\left(1+\frac{x}{a}\right) = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}\left(\frac{x}{a}\right)^n\]

Step 3: Combining the two terms

Now, we substitute the series expansion into our expression for \(f(x)\): \[f(x) = \ln(a) + \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}\left(\frac{x}{a}\right)^n\]

Step 4: Determining the range of convergence

The series expansion is convergent for the values of \(x\) that satisfies the convergence condition for the natural logarithm series representation. The original natural logarithm series \(\ln(1+x)\) converges for \(-1 < x \le 1\). Thus, for our case, we have: \[-1 < \frac{x}{a} \le 1\] This inequality gives us the range of \(x\): \[-a < x \le a\]

Step 5: Final Result

The series expansion for the given function \(f(x) = \ln(a+x)\) in terms of x is \[f(x) = \ln(a) + \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}\left(\frac{x}{a}\right)^n\] The expansion is valid for the range of \(x\) values: \(-a < x \le a\) where \(a > 0\).

Key Concepts

Natural Logarithm

The natural logarithm is a mathematical function that is central in calculus and mathematical analysis. It's often written as \( \ln(x) \) and is defined only for positive real numbers. The natural logarithm has a special base, known as Euler's number \( e \), approximately equal to 2.71828. This base makes it particularly useful for continuous growth processes, such as compound interest, population growth, and radioactive decay. Understanding \( \ln(x) \) also requires understanding its properties:

• Product Rule: \( \ln(ab) = \ln(a) + \ln(b) \)
• Quotient Rule: \( \ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b) \)
• Power Rule: \( \ln(a^b) = b \cdot \ln(a) \)

Recognizing these properties aids in simplifying and understanding logarithmic functions in various mathematical contexts. This series expansion exercise makes use of the property \( \ln(ab) = \ln(a) + \ln(b) \) to simplify \( \ln(a+x) \) into a form that allows us to apply the series expansion for \( \ln(1+x) \).

Power Series

A power series is a sum of terms in which each term is a power of a variable. It provides a way to express a function as an infinite sum of its derivatives at a single point. This makes power series a powerful tool in both theoretical and applied mathematics. For example, the power series representation of \( \ln(1+x) \) is:\[ \ln(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} x^n \]The derivation of this series involves considering the infinite sum of terms where each term \(\frac{(-1)^{n+1}}{n} x^n\) represents a fraction of the input variable \(x\) raised to increasing powers. A key characteristic of power series is their ability to converge to the function they represent within a certain radius of convergence.Using power series, we can effectively approximate complex functions and analyze their behavior within particular intervals. In the exercise, simplifying \( \ln(a+x) \) into \( \ln(a) + \ln(1+\frac{x}{a}) \) allows using the power series of \( \ln(1+x) \) to expand the expression.

Convergence

Convergence is a crucial concept when dealing with series, including power series. Simply put, a series is convergent if its partial sums approach a certain finite number as the number of terms increases. For the series expansion of \( \ln(1+x) \), convergence is determined by the interval \(-1 < x \le 1\). This means the series will accurately represent the \( \ln(1+x) \) function only for \( x \) values within this range.This idea extends to the function \( \ln(a+x) \) considered in our exercise. Here, the series derived is convergent when \( \frac{x}{a} \) lies within the interval \(-1 < \frac{x}{a} \le 1\), which simplifies to \(-a < x \le a\). Thus, checking convergence ensures the validity of the series expansion for the given \( x \) values. This step is crucial to avoid errors when using series expansions to approximate function values.