Problem Set 4 Solution

N odd therefore the coefficients are cn 0

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Unformatted text preview: L ∫ 0 L [cos(nπ ) − cos(0)] nπ ȹ nπx ȹ L sinȹ ȹ dx = − [cos(nπ ) − 1] . ȹ L Ⱥ nπ Note that this integral equals 0 for n even, and is non-zero for n odd, € L ∫ 0 Ⱥ Ⱥ 0, Ⱥ ȹ nπx ȹ sinȹ dx = Ⱥ ȹ ȹ L Ⱥ Ⱥ 2 L , Ⱥ Ⱥ nπ Ⱥ n even Ⱥ Ⱥ Ⱥ . Ⱥ n odd Ⱥ Ⱥ Therefore the coefficients are € cn € Ⱥ Ⱥ 0, Ⱥ = Ⱥ Ⱥ 2 LC 2 ⋅ , Ⱥ Ⱥ nπ L Ⱥ n even Ⱥ Ⱥ Ⱥ . Ⱥ n odd Ⱥ Ⱥ 3 1 . continued The analytical forms of the first six coefficients in the expansion are given below, along with numerical results for the specific case L=2, C=1. c n ( analytic) n 1 2LC 2 ⋅ π L € € 2 3 0 0 2LC 2 ⋅ 5π L 0.2546 0 6 0.4244 0 € 0 2LC 2 ⋅ 3π L € 4 5 c n ( numerical) 1.2732 0 € Using the expansion coefficients listed above, the step function f ( x ) with L=2 and C=1 may be approximately expanded as ȹ πx ȹ ȹ 3πx ȹ ȹ 5πx ȹ f ( x ) ≈ 1.2732 ⋅ sinȹ ȹ + 0.4244 ⋅ sinȹ € ȹ + 0.2546 ⋅ sinȹ ȹ . ȹ 2 Ⱥ ȹ 2 Ⱥ ȹ 2 Ⱥ A graph of the six-term expansion of the step function given i...
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This document was uploaded on 12/05/2013.

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