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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
Chapter 11, Problem 11.4.25

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) = eˣ

Verified step by step guidance
1
Start with the Taylor series expansion of the function \(f(x) = e^x\) centered at 0, which is given by the infinite sum: \[f(x) = \sum_{n=0}^{\infty} \frac{x^n}{n!}\]
Differentiate the series term-by-term with respect to \(x\). Using the power rule for differentiation, the derivative of each term \(\frac{x^n}{n!}\) is: \[\frac{d}{dx} \left( \frac{x^n}{n!} \right) = \frac{n x^{n-1}}{n!} = \frac{x^{n-1}}{(n-1)!}\] Note that for \(n=0\), the term is a constant and its derivative is zero, so the summation index will start from \(n=1\) after differentiation.
Rewrite the differentiated series by adjusting the index to start from \(n=0\) for convenience. Let \(m = n - 1\), then: \[f'(x) = \sum_{m=0}^{\infty} \frac{x^m}{m!}\]
Recognize that the differentiated series is the same as the original Taylor series for \(e^x\). Therefore, the function represented by the differentiated series is also \(e^x\).
Determine the interval of convergence. Since the original series for \(e^x\) converges for all real \(x\), the differentiated series will have the same interval of convergence, which is \((-\infty, \infty)\).

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Key Concepts

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

Taylor Series Expansion

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). For eˣ, the series is ∑(xⁿ/n!), which converges for all real x. Understanding this expansion is essential to differentiate the series term-by-term.
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Taylor Series

Term-by-Term Differentiation of Power Series

Power series can be differentiated term-by-term within their interval of convergence. Differentiating each term of the Taylor series of eˣ involves applying the power rule to xⁿ/n!, resulting in a new series representing the derivative function. This process helps identify the derivative's power series.
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Intro to Power Series

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's series is the same as the original. For eˣ, the series converges for all real numbers, meaning the interval is (-∞, ∞).
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Related Practice
Textbook Question

Remainders Find the remainder Rₙ for the nth−order Taylor polynomial centered at a for the given functions. Express the result for a general value of n.


f(x) = sin x, a = π/2

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Textbook Question

Derivative trick Here is an alternative way to evaluate higher derivatives of a function f that may save time. Suppose you can find the Taylor series for f centered at the point a without evaluating derivatives (for example, from a known series). Then f⁽ᵏ⁾(a)=k! multiplied by the coefficient of (x−a)ᵏ. Use this idea to evaluate f⁽³⁾(0) and f⁽⁴⁾(0) for the following functions. Use known series and do not evaluate derivatives.


f(x) = ∫₀ˣ sin t² dt

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Textbook Question

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

∑ₖ₌₀∞ k(x−1)ᵏ

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Textbook Question

Representing functions by power series Identify the functions represented by the following power series.

∑ₖ₌₀∞ 2ᵏ x²ᵏ⁺¹

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Textbook Question

Differentiating and integrating power series Find the power series representation for g centered at 0 by differentiating or integrating the power series for f (perhaps more than once). Give the interval of convergence for the resulting series.


g(x) = x/(1 + x²)² using f(x) = 1/(1 + x²)

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Textbook Question

Does the accuracy of an approximation given by a Taylor polynomial generally increase or decrease with the order of the approximation? Explain.

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