Question

Use the power series representation $$f(x)=\ln (1-x)=-\sum_{k=1}^{\infty} \frac{x^{k}}{k}, \quad \text { for }-1 \leq x<1$$ to find the power series for the following functions (centered at 0 ). Give the interval of comvergence of the new series. $$f\left(x^{3}\right)=\ln \left(1-x^{3}\right)$$

Short answer

Question: Determine the power series representation and interval of convergence for the function \(f(x^3)=\ln(1-x^3)\). Answer: The power series representation for \(f(x^3)\) is \(-\sum_{k=1}^{\infty} \frac{x^{3k}}{k}\) and the interval of convergence is \(-1 \le x < 1\).

Step-by-step solution

Substitute \(x^3\) into the power series representation of \(f(x)\)

We have \(f(x) = -\sum_{k=1}^{\infty} \frac{x^k}{k}\), now substitute \(x\) with \(x^3\) to get \(f(x^3)\). So, $$f(x^3) = -\sum_{k=1}^{\infty} \frac{(x^3)^k}{k}$$

Simplify the power series for \(f(x^3)\)

By simplifying, we have: $$f(x^3) = -\sum_{k=1}^{\infty} \frac{x^{3k}}{k}$$

Determine the interval of convergence

For the original series, the interval of convergence is \(-1 \le x < 1\). Since we are trying to determine the interval of convergence for the power series of \(f(x^3)\), we need to substitute \(x^3\) instead of \(x\) in the inequality: $$-1 \le x^3 < 1$$ Since all the inequalities are monotonic, we can take the cubic root: $$-1 \le x < 1$$ Thus, the interval of convergence for the power series of \(f(x^3) = \ln(1-x^3)\) is: $$-1 \le x < 1$$ Hence the power series is $$f(x^3) = -\sum_{k=1}^{\infty} \frac{x^{3k}}{k}, \quad \text{for } -1 \le x < 1$$

Key concepts

Interval of Convergence

Understanding the interval of convergence is crucial when working with power series. It tells us where the series produces valid outputs. For a given power series, the interval of convergence defines the range of input values, usually expressed as \(-1 \leq x < 1\), for which the series converges to a finite sum. Without this knowledge, a power series might produce incorrect or undefined results.
In our exercise, the original power series for \(f(x) = \ln(1-x)\) converges when \(-1 \leq x < 1\). When substituting \(x^3\), the convergence remains unchanged at \(-1 \leq x < 1\), because substituting \(x^3\) makes the inequality \(-1 \leq x^3 < 1\). Taking the cube root, due to monotonic properties, does not alter the limits. Hence, it remains \(-1 \leq x < 1\).
This is essential for ensuring our newly formed series \(f(x^3)=\ln(1-x^3)\) holds true within this interval and reflects the original properties of the logarithmic function.

Series Representation

Power series are a fundamental tool in calculus, allowing complex functions to be expressed as a sum of simpler, polynomial terms. This representation is immensely powerful for both calculations and for approximating functions.
In the exercise, \(f(x)=\ln(1-x)\) is given in its series form as \(f(x)=-\sum_{k=1}^{\infty} \frac{x^k}{k}\), which represents the natural logarithmic function's expansion at its center, \(x=0\). Each term in the series \(\frac{x^k}{k}\) accounts for an increasingly higher degree of \(x\), and their sum gives the logarithmic function's value for all valid \(x\) within the interval of convergence.

• The series provides an approximation of \(\ln(1-x)\) by summing up its terms sequentially.
• It beautifully highlights how continuous functions can be broken down and analyzed using polynomial-like components.

By modifying this for \(f(x^3)=\ln(1-x^3)\), we adapt the representation to accommodate inputs that are cubed, transitioning our focus to how transformations affect power series.

Substitution Method

The substitution method in power series is a technique where one variable in a series is replaced with another expression. This strategy is useful for creating series representations for more complex functions based on existing simpler series.
To apply this to \(f(x^3)=\ln(1-x^3)\), start with the series for \(f(x)=\ln(1-x)\), and substitute every occurrence of \(x\) with \(x^3\). This effectively changes each term into \(\frac{(x^3)^k}{k} = \frac{x^{3k}}{k}\).

• This approach retains the convergent properties of the original series while modifying its structure to reflect the cubed input.
• Allows flexibility in interpreting and transforming functions with similar forms yet distinct variables.

Using substitution, we maintain the integrity of convergence and obtain \(\ln(1-x^3) = -\sum_{k=1}^{\infty}\frac{x^{3k}}{k}\). It’s a versatile method, ensuring you can adapt known series to new scenarios effortlessly.