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Ch. 11 - Power Series
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 11 - Power SeriesProblem 11.4.30
Chapter 11, Problem 11.4.30

Power series for derivatives


a. Differentiate the Taylor series centered at 0 for the following functions.
b. Identify the function represented by the differentiated series.
c. Give the interval of convergence of the power series for the derivative.


f(x) = (1 − x)⁻¹

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Start with the given function \( f(x) = (1 - x)^{-1} \). Recall that its Taylor series centered at 0 (Maclaurin series) is the geometric series \( \sum_{n=0}^{\infty} x^n \), valid for \( |x| < 1 \).
Differentiate the series term-by-term. The derivative of \( f(x) \) is \( f'(x) = \frac{d}{dx} (1 - x)^{-1} \). Using the power series, differentiate each term \( x^n \) to get \( n x^{n-1} \). So the differentiated series is \( \sum_{n=1}^{\infty} n x^{n-1} \).
Identify the function represented by the differentiated series. Differentiate \( f(x) = (1 - x)^{-1} \) directly using the chain rule: \( f'(x) = (1 - x)^{-2} \). This matches the sum of the differentiated series, confirming the function represented by the series.
Determine the interval of convergence for the differentiated series. Since the original series converges for \( |x| < 1 \), and differentiation does not change the radius of convergence, the interval remains \( (-1, 1) \).
Summarize: The differentiated power series is \( \sum_{n=1}^{\infty} n x^{n-1} \), representing the function \( (1 - x)^{-2} \), and it converges for \( |x| < 1 \).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Taylor Series and Power Series

A Taylor series represents a function as an infinite sum of terms calculated from the function's derivatives at a single point, often zero (Maclaurin series). It expresses functions as power series, allowing approximation and analysis of functions using polynomials.
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Term-by-Term Differentiation of Power Series

Power series can be differentiated term-by-term within their interval of convergence. Differentiating each term individually produces a new power series representing the derivative of the original function, preserving convergence properties inside the interval.
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Interval of Convergence

The interval of convergence is the set of x-values for which a power series converges. Differentiating a power series does not change its radius of convergence, so the interval for the derivative series is the same or possibly smaller, requiring verification at endpoints.
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Related Practice
Textbook Question

Manipulating Taylor series Use the Taylor series in Table 11.5 to find the first four nonzero terms of the Taylor series for the following functions centered at 0.


(1 + x⁴)⁻¹

Textbook Question

Functions to power series Find power series representations centered at 0 for the following functions using known power series. Give the interval of convergence for the resulting series.

f(x) = ln √(4 − x²)

Textbook Question

Shifting power series If the power series f(x)=∑ cₖ xᵏ has an interval of convergence of |x|<R, what is the interval of convergence of the power series for f(x−a), where a ≠ 0 is a real number?

Textbook Question

{Use of Tech} Best center point Suppose you wish to approximate cos (π/ 2) using Taylor polynomials. Is the approximation more accurate if you use Taylor polynomials centered at 0 or at π/6? Use a calculator for numerical experiments and check for consistency with Theorem 11.2. Does the answer depend on the order of the polynomial?

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Textbook Question

Radius and interval of convergence Determine the radius and interval of convergence of the following power series.


∑ₖ₌₁∞ (kx)ᵏ

Textbook Question

Suppose f(0)=1, f'(0)=0, f''(0)=2, and f⁽³⁾(0)=6. Find the third-order Taylor polynomial for f centered at 0 and use it to approximate f(0.2).