Power series from the geometric series Use the geometric series a Σₖ ₌ ₀ ∞ (x)ᵏ = 1/(1 - x), for |x| < 1, to determine the Maclaurin series and the interval of convergence for the following functions.


ƒ(x) = 1/(1 - x²)

Verified step by step guidance

1. Recall the geometric series formula: \(\sum_{k=0}^{\infty} x^k = \frac{1}{1 - x}\) for \(|x| < 1\).
2. Notice that the given function is \(f(x) = \frac{1}{1 - x^2}\), which can be seen as \(\frac{1}{1 - (x^2)}\).
3. Apply the geometric series formula by substituting \(x\) with \(x^2\), so the series becomes \(\sum_{k=0}^{\infty} (x^2)^k = \sum_{k=0}^{\infty} x^{2k}\).
4. Write the Maclaurin series for \(f(x)\) as \(\sum_{k=0}^{\infty} x^{2k}\), which expands to \$1 + x^2 + x^4 + x^6 + \cdots$.
5. Determine the interval of convergence by applying the condition \(|x^2| < 1\), which simplifies to \(|x| < 1\).

Key Concepts

Geometric Series

A geometric series is an infinite sum of the form Σ x^k from k=0 to ∞, which converges to 1/(1 - x) when |x| < 1. This formula provides a foundation for expressing functions as power series by recognizing patterns similar to geometric series.

Maclaurin Series

The Maclaurin series is a special case of the Taylor series centered at zero, representing a function as an infinite sum of its derivatives at zero multiplied by powers of x. It allows us to express functions as power series expansions around x = 0.

Interval of Convergence

The interval of convergence is the set of x-values for which a power series converges to the function it represents. For geometric series, this interval is determined by the condition |x| < 1, and it must be adjusted when the series involves expressions like x².