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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 11.4.79c

{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
c. Use the polynomials in part (b) to approximate S(0.05) and C(−0.25).

Verified step by step guidance
1
Recall that the Fresnel integrals are defined as \(S(x) = \int_0^x \sin(t^2) \, dt\) and \(C(x) = \int_0^x \cos(t^2) \, dt\). To approximate these integrals, we use polynomial approximations (Taylor series) of the integrands \(\sin(t^2)\) and \(\cos(t^2)\) from part (b).
Write down the Taylor series expansions for \(\sin(t^2)\) and \(\cos(t^2)\) around \(t=0\). For example, \(\sin(t^2)\) can be expanded as \(t^2 - \frac{t^6}{3!} + \frac{t^{10}}{5!} - \cdots\), and \(\cos(t^2)\) as \(1 - \frac{t^4}{2!} + \frac{t^8}{4!} - \cdots\).
Substitute these polynomial approximations into the integrals for \(S(x)\) and \(C(x)\), so that \(S(x) \approx \int_0^x \left( t^2 - \frac{t^6}{3!} + \frac{t^{10}}{5!} - \cdots \right) dt\) and \(C(x) \approx \int_0^x \left( 1 - \frac{t^4}{2!} + \frac{t^8}{4!} - \cdots \right) dt\).
Integrate each term of the polynomial inside the integral with respect to \(t\) from 0 to \(x\). For example, \(\int_0^x t^2 dt = \frac{x^3}{3}\), \(\int_0^x t^6 dt = \frac{x^7}{7}\), and so on. This will give you polynomial expressions in terms of \(x\).
Finally, plug in the values \(x=0.05\) for \(S(0.05)\) and \(x=-0.25\) for \(C(-0.25)\) into the resulting polynomial expressions to get the approximate values of the Fresnel integrals.

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Key Concepts

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

Fresnel Integrals

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