Combining power series Use the power series representation


f(x ) =ln (1 − x) = −∑ₖ₌₁∞ xᵏ/k, for −1 ≤ x < 1,


to find the power series for the following functions (centered at 0). Give the interval of convergence of the new series.


p(x) = 2x⁶ ln(1 − x)

Verified step by step guidance

1. Recall the given power series for \( f(x) = \ln(1 - x) \): \[ f(x) = \ln(1 - x) = -\sum_{k=1}^{\infty} \frac{x^k}{k}, \quad \text{for } -1 \leq x < 1. \]
2. To find the power series for \( p(x) = 2x^6 \ln(1 - x) \), multiply the entire series for \( \ln(1 - x) \) by \( 2x^6 \): \[ p(x) = 2x^6 \cdot \left(-\sum_{k=1}^{\infty} \frac{x^k}{k}\right) = -2x^6 \sum_{k=1}^{\infty} \frac{x^k}{k}. \]
3. Combine the powers of \( x \) inside the summation: \[ p(x) = -2 \sum_{k=1}^{\infty} \frac{x^{k+6}}{k}. \]
4. Rewrite the series with a new index if desired (for example, let \( n = k + 6 \)) to express the series in standard power series form centered at 0: \[ p(x) = -2 \sum_{n=7}^{\infty} \frac{x^n}{n - 6}. \]
5. Determine the interval of convergence. Since the original series for \( \ln(1 - x) \) converges for \( -1 \leq x < 1 \), and multiplication by \( 2x^6 \) does not affect the radius of convergence, the interval of convergence for \( p(x) \) remains \( -1 \leq x < 1 \).

Key Concepts

Power Series Representation

A power series expresses a function as an infinite sum of terms involving powers of x, typically centered at a point (here, 0). Understanding how to write and manipulate these series is essential for representing functions like ln(1 − x) as sums that converge within a specific interval.

Interval of Convergence

The interval of convergence is the set of x-values for which a power series converges to the function it represents. Determining this interval ensures the validity of the series representation and is crucial when modifying or combining series, such as multiplying by 2x⁶.

Operations on Power Series

Power series can be added, multiplied, or composed by manipulating their coefficients and powers of x. For example, multiplying ln(1 − x) by 2x⁶ involves shifting the powers and scaling coefficients, which requires careful term-by-term adjustment to find the new series.