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Ch. 11 - Power Series
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243당신이 사용하는 게 아니라요?교과서 변경
모든 교과서Briggs 3rd EditionCh. 11 - Power Series문제 11.2.54
11장, 문제 11.2.54

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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Start with the given function \( f(x) = \frac{1}{1 + x^2} \). Recognize that this is a geometric series of the form \( \frac{1}{1 - (-x^2)} \), which can be expanded as a power series centered at 0: \( f(x) = \sum_{n=0}^\infty (-1)^n x^{2n} \).
To find \( g(x) = \frac{x}{(1 + x^2)^2} \), notice that it can be expressed in terms of the derivative of \( f(x) \). Differentiate \( f(x) \) with respect to \( x \) to relate it to \( g(x) \). Use the chain rule: \( f'(x) = \frac{d}{dx} \left( \frac{1}{1 + x^2} \right) = -\frac{2x}{(1 + x^2)^2} \).
Rearrange the derivative expression to isolate \( g(x) \): \( g(x) = \frac{x}{(1 + x^2)^2} = -\frac{1}{2} f'(x) \). This means \( g(x) \) can be represented as \( -\frac{1}{2} \) times the derivative of the power series for \( f(x) \).
Differentiate the power series term-by-term: \( f(x) = \sum_{n=0}^\infty (-1)^n x^{2n} \) implies \( f'(x) = \sum_{n=1}^\infty (-1)^n 2n x^{2n-1} \). Then multiply by \( -\frac{1}{2} \) to get the power series for \( g(x) \): \( g(x) = -\frac{1}{2} f'(x) = \sum_{n=1}^\infty (-1)^{n+1} n x^{2n-1} \).
Determine the interval of convergence. Since the original series for \( f(x) \) converges for \( |x| < 1 \), and differentiation does not change the radius of convergence, the power series for \( g(x) \) also converges for \( |x| < 1 \).

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주요 개념

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Power Series Representation

A power series is an infinite sum of terms in the form a_n(x - c)^n, representing functions as polynomials of infinite degree centered at c. Understanding how to express functions as power series allows manipulation through differentiation and integration term-by-term, facilitating approximation and analysis.
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05:58
Intro to Power Series

Term-by-Term Differentiation and Integration of Power Series

Power series can be differentiated or integrated term-by-term within their interval of convergence. This property enables finding new series representations for related functions by applying calculus operations to known series, preserving convergence and simplifying complex function analysis.
추천 영상:
05:58
Intro to Power Series

Interval of Convergence

The interval of convergence is the set of x-values for which a power series converges to a finite value. Determining this interval is crucial after differentiation or integration, as these operations can alter convergence properties, ensuring the resulting series accurately represents the function within that domain.
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08:44
Interval of Convergence
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Limits Evaluate the following limits using Taylor series.

lim ₓ→∞ x sin(1/x)

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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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교과서 질문

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

∑ₖ₌₀∞ 2ᵏ x²ᵏ⁺¹

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교과서 질문

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

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교과서 질문

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

∑ₖ₌₀∞ (k²⁰ xᵏ)/(2k+1)!

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교과서 질문

In terms of the remainder, what does it mean for a Taylor series for a function f to converge to f?

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