# Chapt_12 - Part IV Fourier Series and Partial Differential...

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Part IV Fourier Series and Partial Differential Equations 12 12 Orthogonal Functions and Fourier Series EXERCISES 12.1 Orthogonal Functions 1. 2 2 xx 2 dx = 1 4 x 4 2 2 = 0 2. 1 1 x 3 ( x 2 + 1) dx = 1 6 x 6 1 1 + 1 4 x 4 1 1 = 0 3. 2 0 e x ( xe x e x ) dx = 2 0 ( x 1) dx = 1 2 x 2 x 2 0 = 0 4. π 0 cos x sin 2 x dx = 1 3 sin 3 x π 0 = 0 5. π/ 2 π/ 2 x cos2 x dx = 1 2 1 2 cos2 x + x sin2 x π/ 2 π/ 2 = 0 6. 5 π/ 4 π/ 4 e x sin x dx = 1 2 e x sin x 1 2 e x cos x 5 π/ 4 π/ 4 = 0 7. For m = n π/ 2 0 sin(2 n + 1) x sin(2 m + 1) x dx = 1 2 π/ 2 0 cos2( n m ) x cos2( n + m + 1) x dx = 1 4( n m ) sin2( n m ) x π/ 2 0 1 4( n + m + 1) sin2( n + m + 1) x π/ 2 0 = 0 . 634

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12.1 Orthogonal Functions For m = n π/ 2 0 sin 2 (2 n + 1) x dx = π/ 2 0 1 2 1 2 cos2(2 n + 1) x dx = 1 2 x π/ 2 0 1 4(2 n + 1) sin2(2 n + 1) x π/ 2 0 = π 4 so that sin(2 n + 1) x = 1 2 π . 8. For m = n π/ 2 0 cos(2 n + 1) x cos(2 m + 1) x dx = 1 2 π/ 2 0 cos2( n m ) x + cos2( n + m + 1) x dx = 1 4( n m ) sin2( n m ) x π/ 2 0 + 1 4( n + m + 1) sin2( n + m + 1) x π/ 2 0 = 0 . For m = n π/ 2 0 cos 2 (2 n + 1) x dx = π/ 2 0 1 2 + 1 2 cos2(2 n + 1) x dx = 1 2 x π/ 2 0 + 1 4(2 n + 1) sin2(2 n + 1) x π/ 2 0 = π 4 so that cos(2 n + 1) x = 1 2 π . 9. For m = n π 0 sin nx sin mx dx = 1 2 π 0 cos( n m ) x cos( n + m ) x dx = 1 2( n m ) sin( n m ) x π 0 1 2( n + m ) sin( n + m ) x π 0 = 0 . For m = n π 0 sin 2 nx dx = π 0 1 2 1 2 cos2 nx dx = 1 2 x π 0 1 4 n sin2 nx π 0 = π 2 so that sin nx = π 2 . 10. For m = n p 0 sin p x sin p x dx = 1 2 p 0 cos ( n m ) π p x cos ( n + m ) π p x dx = p 2( n m ) π sin ( n m ) π p x p 0 p 2( n + m ) π sin ( n + m ) π p x p 0 = 0 . For m = n p 0 sin 2 p x dx = p 0 1 2 1 2 cos 2 p x dx = 1 2 x p 0 p 4 sin 2 p x p 0 = p 2 635
12.1 Orthogonal Functions so that sin p x = p 2 . 11. For m = n p 0 cos p x cos p x dx = 1 2 p 0 cos ( n m ) π p x + cos ( n + m ) π p x dx = p 2( n m ) π sin ( n m ) π p x p 0 + p 2( n + m ) π sin ( n + m ) π p x p 0 = 0 . For m = n p 0 cos 2 p x dx = p 0 1 2 + 1 2 cos 2 p x dx = 1 2 x p 0 + p 4 sin 2 p x p 0 = p 2 . Also p 0 1 · cos p x dx = p sin p x p 0 = 0 and p 0 1 2 dx = p so that 1 = p and cos p x = p 2 . 12. For m = n , we use Problems 11 and 10: p p cos p x cos p x dx = 2 p 0 cos p x cos p x dx = 0 p p sin p x sin p x dx = 2 p 0 sin p x sin p x dx = 0 . Also p p sin p x cos p x dx = 1 2 p p sin ( n m ) π p x + sin ( n + m ) π p x dx = 0 , p p 1 · cos p x dx = p sin p x p p = 0 , p p 1 · sin p x dx = p cos p x p p = 0 , and p p sin p x cos p x dx = p p 1 2 sin 2 p x dx = p 4 cos 2 p x p p = 0 . For m = n p p cos 2 p x dx = p p 1 2 + 1 2 cos 2 p x dx = p, p p sin 2 p x dx = p p 1 2 1 2 cos 2 p x dx = p, and p p 1 2 dx = 2 p so that 1 = 2 p , cos p x = p , and sin p x = p . 636

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12.1 Orthogonal Functions 13. Since −∞ e x 2 · 1 · 2 x dx = e x 2 0 −∞ e x 2 0 = 0 , −∞ e x 2 · 1 · (4 x 2 2) dx = 2 −∞ x 2 xe x 2 dx 2 −∞ e x 2 dx = 2 xe x 2 −∞ + −∞ e x 2 dx 2 −∞ e x 2 dx = 2 xe x 2 0 −∞ xe x 2 0 = 0 , and −∞ e x 2 · 2
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