Chem Differential Eq HW Solutions Fall 2011 13

# Chem Differential Eq HW Solutions Fall 2011 13 - F you can...

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Section 2.3 Fourier Series of Functions with Arbitrary Periods 13 33. The function F ( x ) is continuous and piecewise smooth with F ( x ) = f ( x ) at all the points where f is continuous (see Exercise 25, Section 2.1). So, by Exercise 26, if we differentiate the Fourier series of F , we get the Fourier series of f . Write F ( x ) = A 0 + n =1 A n cos p x + B n sin p x and f ( x ) = n =1 a n cos p x + b n sin p x . Note that the a 0 term of the Fourier series of f is 0 because by assumption 2 p 0 f ( x ) dx = 0. Differentiate the series for F and equate it to the series for f and get n =1 - A n p sin p x + p B n cos p x = n =1 a n cos p x + b n sin p x . Equate the n th Fourier coefficients and get - A n p = b n A n = - p b n ; B n p = a n B n = p a n . This derives the n th Fourier coefficients of F for n 1. To get A 0 , note that F (0) = 0 because of the definition of F ( x ) = x 0 f ( t ) dt . So 0 = F (0) = A 0 + n =1 A n = A 0 + n =1 - p b n ; and so A 0 = n =1 p b n . We thus obtained the Fourier series of F in terms of the Fourier coefficients of f ; more precisely, F ( x ) = p π n =1 b n n + n =1 - p b n cos p x + p a n
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Unformatted text preview: F , you can integrate the Fourier series of f term by term. Furthermore, the only assumption on f is that it is piecewise smooth and integrates to 0 over one period (to guarantee the periodicity of F .) Indeed, if you start with the Fourier series of f , f ( t ) = ∞ ± n =1 ² a n cos nπ p t + b n sin nπ p t ³ , and integrate term by term, you get F ( x ) = µ x f ( t ) dt = ∞ ± n =1 ² a n µ x cos nπ p tdt + b n µ x sin nπ p tdt ³ = ∞ ± n =1 ² a n ¶ p nπ · sin nπ p t ¸ ¸ ¸ x dt + b n ¶-p nπ · cos nπ p t ¸ ¸ ¸ x ³ = p π ∞ ± n =1 b n n + ∞ ± n =1 ²-p nπ b n cos nπ p x + p nπ a n sin nπ p x ³ , as derived earlier. See the following exercise for an illustration....
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