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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 11.4.79e
Chapter 11, Problem 11.4.79e

{Use of Tech} Fresnel integrals The theory of optics gives rise to the two Fresnel integrals
S(x) = ∫₀ˣ sin t² dt and C(x) = ∫₀ˣ cos t² dt
e. How many terms of the Maclaurin series are required to approximate C(−0.25) with an error no greater than 10⁻⁶?

Verified step by step guidance
1
Recall that the Fresnel integral \( C(x) = \int_0^x \cos(t^2) \, dt \) can be expanded into a Maclaurin series by expanding \( \cos(t^2) \) as a power series and integrating term-by-term.
Write the Maclaurin series for \( \cos(z) \), which is \( \cos(z) = \sum_{n=0}^\infty (-1)^n \frac{z^{2n}}{(2n)!} \). Substitute \( z = t^2 \) to get \( \cos(t^2) = \sum_{n=0}^\infty (-1)^n \frac{t^{4n}}{(2n)!} \).
Integrate the series term-by-term from 0 to \( x \) to find the series for \( C(x) \): \[ C(x) = \sum_{n=0}^\infty (-1)^n \frac{1}{(2n)!} \int_0^x t^{4n} \, dt = \sum_{n=0}^\infty (-1)^n \frac{x^{4n+1}}{(4n+1)(2n)!} \].
To approximate \( C(-0.25) \) with an error no greater than \( 10^{-6} \), use the alternating series remainder estimation. The error is less than or equal to the absolute value of the first omitted term in the series.
Find the smallest integer \( N \) such that the absolute value of the \( (N+1)^{th} \) term: \[ \left| (-1)^{N+1} \frac{(-0.25)^{4(N+1)+1}}{(4(N+1)+1)(2(N+1))!} \right| \leq 10^{-6} \]. This will tell you how many terms are needed to achieve the desired accuracy.

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Key Concepts

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

Fresnel Integrals

Fresnel integrals S(x) and C(x) are defined as integrals of sine and cosine functions with quadratic arguments, specifically S(x) = ∫₀ˣ sin(t²) dt and C(x) = ∫₀ˣ cos(t²) dt. They arise in optics and wave theory, representing diffraction patterns and are not expressible in elementary functions, often requiring series expansions or numerical methods for evaluation.
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Maclaurin Series Expansion

A Maclaurin series is a Taylor series expansion of a function about zero, expressing it as an infinite sum of derivatives at zero multiplied by powers of x. For functions like cos(t²), the Maclaurin series allows approximation by polynomials, facilitating numerical evaluation of integrals like C(x) by integrating term-by-term.
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Convergence of Taylor & Maclaurin Series

Error Estimation in Series Approximation

When approximating functions with series, the error is the difference between the true value and the partial sum. To ensure the error is below a threshold (e.g., 10⁻⁶), one uses remainder estimates or bounds on the next term in the series. This guides how many terms are needed for a desired accuracy.
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Related Practice
Textbook Question

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.

e. The Taylor series for an even function centered at 0 has only even powers of x.

Textbook Question

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.

d. Suppose f'' is continuous on an interval that contains a, where f has an inflection point at a. Then the second−order Taylor polynomial for f at a is linear.

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Textbook Question

{Use of Tech} Fresnel integrals The theory of optics gives rise to the two Fresnel integrals

S(x) = ∫₀ˣ sin t² dt and C(x) = ∫₀ˣ cos t² dt

d. How many terms of the Maclaurin series are required to approximate S(0.05) with an error no greater than 10⁻⁴?

Textbook Question

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.

d. If p(x) is the Taylor series for f centered at 0, then p(x−1) is the Taylor series for f centered at 1.

Textbook Question

Matching functions with polynomials Match functions a–f with Taylor polynomials A–F (all centered at 0). Give reasons for your choices.


f. e⁻²ˣ


A. p₂(x)= 1 + 2x + 2x²

B. p₂(x) = 1 − 6x + 24x²

C. p₂(x) = 1 + x − x²/2

D. p₂(x) = 1 − 2x + 4x²

E. p₂(x) = 1 − x + (3/2)x²

F. p₂(x) = 1 − 2x + 2x²

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