Problem 36 - Calculus Early Transcendentals

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 convergence of the new series. $$g(x)=x^{3} \ln (1-x)$$

Step-by-Step Solution

Answer

Answer: The power series representation for the function \(g(x) = x^3 \ln(1-x)\) is: $$g(x) = -\sum_{n=4}^{\infty} \frac{x^n}{n-3}$$ The interval of convergence for the new series is \(-1 \leq x < 1\).

Step 1: Write given power series representation of f(x)

We are given the power series representation of the function \(f(x) = \ln(1-x)\) as: $$f(x) = -\sum_{k=1}^{\infty} \frac{x^k}{k}, \quad \text{for } -1 \leq x <1 $$

Step 2: Find the power series representation of g(x)

To find the power series representation of \(g(x)=x^3 \ln(1-x)\), we will multiply the series representation of \(f(x)\) by \(x^3\): $$g(x) = x^3f(x) = -x^3 \sum_{k=1}^{\infty} \frac{x^k}{k} = -\sum_{k=1}^{\infty} \frac{x^{k+3}}{k}$$

Step 3: Rewrite the power series in terms of n

To simplify the power series further, let's use a new index: \(n = k + 3\). Then, \(k = n - 3\). Our power series becomes: $$g(x) = - \sum_{n=4}^{\infty} \frac{x^n}{n-3}$$

Step 4: Find the interval of convergence of the new series#g(x)

Since in both cases the series is a ratio test series, the interval of convergence for the new series will be the same as for the original series \(f(x)\). Therefore, the interval of convergence for the new series is: $$-1 \leq x < 1$$

Step 5: Final Answer

The power series representation for the function \(g(x) = x^3 \ln(1-x)\) is: $$g(x) = -\sum_{n=4}^{\infty} \frac{x^n}{n-3}$$ And the interval of convergence for the new series is \(-1 \leq x < 1\).

Key Concepts

Logarithmic Functions

Logarithmic functions are a fundamental concept in mathematics, widely used in various fields like engineering, physics, and economics. The logarithmic function, denoted as \( \ln(x) \), is the inverse of the exponential function. In simpler terms, if \( b^y = x \), then \( y = \log_b(x) \), where \( b \) is the base of the logarithm. For the natural logarithm \( \ln(x) \), the base is \( e \), Euler's number, which is approximately 2.71828.

In solving calculus problems, particularly those involving series, the property \( \ln(1-x) \) is quite useful. This expression can be expanded into a power series:

• \( \ln(1-x) = - \sum_{k=1}^{\infty} \frac{x^k}{k} \), valid for \( -1 \leq x < 1 \)

This identity helps in creating a workable form that can be manipulated in various calculations.

Logarithmic functions are essential for handling problems that involve changes in multiplication to addition and are especially useful in the field of calculus.

Interval of Convergence

The interval of convergence is a crucial aspect to understand when dealing with power series. It specifies the set of \( x \) values for which the series converges to a finite value. For any given power series, determining this interval is necessary to ensure that the series representation accurately reflects the function it's intended to model.

In our problem, the power series \( \ln(1-x) = -\sum_{k=1}^{\infty} \frac{x^k}{k} \) converges within \( -1 \leq x < 1 \). This means the series will converge for all \( x \) in that range. When we multiply the series by \( x^3 \) to find the power series of \( g(x) = x^3 \ln(1-x) \), this interval of convergence remains the same:

• \(-1 \leq x < 1\)

The reasoning is straightforward: multiplying by a power of \( x \) does not change where the series behaves nicely and converges. Identifying this interval is an essential step in ensuring any power series manipulation results in valid and meaningful expressions.

Index Substitution

Index substitution is a mathematical technique used to simplify series expressions and make them easier to work with. Essentially, it involves shifting the indices of summation, which can reshape the series and make further calculations more straightforward.

In our exercise, we started with the power series for \( g(x) = x^3 \ln(1-x) \) given by:

• \( -x^3 \sum_{k=1}^{\infty} \frac{x^k}{k} \)

To simplify, we performed index substitution by setting \( n = k + 3 \), which transforms the original summation into:

• \( - \sum_{n=4}^{\infty} \frac{x^n}{n-3} \)

This shift helps in aligning terms for further analysis or comparison. Index substitution acts like a flexible adjustment tool, ensuring the series becomes more manageable while preserving its core properties. It also helps when integrating series, adding terms, or finding new configurations.