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Unformatted text preview: x − ct ) + u ( x + ct )] (20.5) Let c = 1. t = 1 2 : x r − 1 2 = 1 ⇒ x r = 3 2 x R + 1 2 = 1 x R = 1 2 x ` − 1 2 = − 1 ⇒ x ` = − 1 2 x L + 1 2 = − 1 x L = − 3 2 (20.6) t = 1: x r − 1 = 1 ⇒ x r = 2 x R + 1 = 1 ⇒ x R = 1 x ` − 1 = − 1 ⇒ x ` = 0 x ` + 1 = − 1 ⇒ x L = − 2 (20.7) t = 2: x r − 2 = 1 ⇒ x r = 3 x R + 2 = 1 ⇒ x R = − 1 x ` − 2 = − 1 ⇒ x ` = 1 x L + 2 = − 1 ⇒ x L = − 3 (20.8) 123...
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 Spring '11
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 Math, XR, The Domain, Sydney, D’Alembert’s Solution, SpaceTime Interpretation

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