Chem Differential Eq HW Solutions Fall 2011 144

Chem Differential Eq HW Solutions Fall 2011 144 - 144...

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144 Chapter 8 The Laplace and Hankel Transforms with Applications 49. We have sin t = ± k =0 ( - 1) k t 2 k +1 (2 k + 1)! for all t. So for t ± = 0, we divide by t both sides and get sin t t = ± k =0 ( - 1) k t 2 k (2 k + 1)! for all t ± =0 . As t 0, the left side approaches 1. The right side is continuous, and so as t 0, it approaches the value at 0, which is 1. Hence both sides of the equality approach 1as t 0, and so we may take the expansion to be valid for all t . Apply the result of the previous exercise, then L ( f ( t ))( s )= L ² ± k =0 ( - 1) k t 2 k (2 k + 1)! ³ ( s ) = ± k =0 ( - 1) k (2 k )! 1 s 2 k +1 (2 k + 1)! = ± k =0 ( - 1) k 1 s 2 k +1 (2 k +1) . Recall the expansion of the inverse tangent: tan - 1 u = ± k =0 ( - 1) k u 2 k +1 2 k +1 | u | < 1 . So tan - 1 1 s = ± k =0 ( - 1) k 1 s 2 k +1 (2 k | s | > 1 . Comparing series, we Fnd that, for s> 1, L ( sin t t ) = tan - 1 ( 1 s ) .
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