15. Power Series

Introduction to Power Series

15. Power Series

Introduction to Power Series

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Multiple Choice

Find the power series representation centered at $x equals 0$ of the following function. Give the interval of convergence for the resulting series.
$f (x) = \frac{1}{1 - x^{3}}$

$\sum_{n=0}^{\infty}x^{3n}$ ; converges on (-∞,∞)

$\sum_{n = 0}^{\infty} x^{3 n}$ ; converges on (-1,1)

$\sum_{n = 0}^{\infty} \frac{1}{x^{3 n}}$ ; converges on (-1,1)

$\sum_{n = 0}^{\infty} \frac{1}{x^{3 n}}$ ; diverges everywhere

Verified step by step guidance

Step 1: Recognize that the given function f(x) = 1 / (1 - x^3) resembles the geometric series formula, which is 1 / (1 - r) = ∑_{n=0}^{∞} r^n, valid for |r| < 1. Here, r = x^3.

Step 2: Rewrite the function as a geometric series using the formula. Substitute r = x^3 into the series expansion: f(x) = ∑_{n=0}^{∞} (x^3)^n = ∑_{n=0}^{∞} x^{3n}.

Step 3: Determine the interval of convergence for the series. The geometric series converges when |r| < 1. Since r = x^3, we require |x^3| < 1, which simplifies to |x| < 1.

Step 4: Conclude that the series representation of f(x) is ∑_{n=0}^{∞} x^{3n}, and the interval of convergence is (-1, 1).

Step 5: Verify the endpoints of the interval of convergence. At x = -1 and x = 1, the series ∑_{n=0}^{∞} x^{3n} does not converge because the terms do not approach zero. Therefore, the series converges only within (-1, 1).

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Struggling with Calculus?

Find the power series representation centered at x=0x=0 of the following function. Give the interval of convergence for the resulting series. f(x)=11−x3f\left(x\right)=\frac{1}{1-x^3}

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Introduction to Power Series practice set

Find the power series representation centered at $x equals 0$ of the following function. Give the interval of convergence for the resulting series.
$f (x) = \frac{1}{1 - x^{3}}$