Chem Differential Eq HW Solutions Fall 2011 14

Chem Differential Eq HW Solutions Fall 2011 14 - 14 Chapter...

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14 Chapter 2 Fourier Series Solutions to Exercises 2.4 1. The even extension is the function that is identically 1. So the cosine Fourier series is just the constant 1. The odd extension yields the function in Exercise 1, Section 2.3, with p = 1. So the sine series is 4 π ± k =0 sin((2 k +1) πx ) 2 k +1 . This is also obtained by evaluating the integral in (4), which gives b n =2 ² 1 0 sin( nπx ) dx = - 2 cos nπx ³ ³ ³ 1 0 = 2 (1 - ( - 1) n ) . 9. We have b n ² 1 0 x (1 - x ) sin( nπx ) dx. To evaluate this integral, we will use integration by parts to derive the following two formulas: for a ± =0 , ² x sin( ax ) dx = - x cos( ax ) a + sin( ) a 2 + C, and ² x 2 sin( ax ) dx = 2 cos( ) a 3 - x 2 cos( ) a + 2 x sin( ) a 2 + C. So ² x (1 - x ) sin( ax ) dx = - 2 cos( ) a 3 - x cos( ) a + x 2 cos( ) a
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