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Ch. 11 - Power Series
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243당신이 사용하는 게 아니라요?교과서 변경
11장, 문제 11.R.56

Definite integrals by power series Use a Taylor series to approximate the following definite integrals. Retain as many terms as necessary to ensure the error is less than 10⁻³.
∫₀1/2 tan⁻¹ x dx

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Recall that the Taylor series expansion of \( \tan^{-1} x \) centered at 0 is given by the alternating series \( \tan^{-1} x = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{2n+1} \). Write out the first few terms explicitly: \( x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots \).
Set up the integral of the Taylor series term-by-term from 0 to \( \frac{1}{2} \): \( \int_0^{1/2} \tan^{-1} x \, dx = \int_0^{1/2} \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{2n+1} \, dx = \sum_{n=0}^{\infty} (-1)^n \frac{1}{2n+1} \int_0^{1/2} x^{2n+1} \, dx \).
Integrate each term inside the sum: \( \int_0^{1/2} x^{2n+1} \, dx = \left[ \frac{x^{2n+2}}{2n+2} \right]_0^{1/2} = \frac{(1/2)^{2n+2}}{2n+2} \). Substitute this back into the sum to get \( \sum_{n=0}^{\infty} (-1)^n \frac{(1/2)^{2n+2}}{(2n+1)(2n+2)} \).
Determine how many terms to keep by estimating the error. Since the series is alternating with decreasing terms in absolute value, the error after \( N \) terms is less than or equal to the absolute value of the \( (N+1)^{th} \) term. Calculate the magnitude of terms until the next term is less than \( 10^{-3} \).
Sum the terms up to the required \( N \) to approximate the integral \( \int_0^{1/2} \tan^{-1} x \, dx \) with an error less than \( 10^{-3} \).

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

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. For functions like arctan(x), the series allows approximation by polynomials, which simplifies integration and error estimation.
추천 영상:
08:42
Taylor Series

Definite Integration of Power Series

Integrating a power series term-by-term within its interval of convergence is valid and often easier than integrating the original function. This approach transforms the integral into a sum of integrals of polynomial terms, facilitating approximation.
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Intro to Power Series

Error Estimation in Series Approximations

To ensure the approximation error is below a threshold (e.g., 10⁻³), one uses remainder terms or bounds on the next omitted term in the Taylor series. This guarantees the truncated series provides a sufficiently accurate estimate of the integral.
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가이드 코스
04:57
Determining Error and Relative Error