Chapter 12

Chapter 12 - 12 2 Orthogonal Functions and Fourier Series...

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12 Orthogonal Functions and Fourier Series Exercises 12.1 1. 592 Z 2 2 xx 2 dx = 1 4 x 4 ¯ ¯ ¯ ¯ 2 2 =0 2. Z 1 1 x 3 ( x 2 +1) dx = 1 6 x 6 ¯ ¯ ¯ ¯ 1 1 + 1 4 x 4 ¯ ¯ ¯ ¯ 1 1 3. Z 2 0 e x ( xe x e x ) dx = Z 2 0 ( x 1) dx = µ 1 2 x 2 x ¯ ¯ ¯ ¯ 2 0 4. Z π 0 cos x sin 2 xdx = 1 3 sin 3 x ¯ ¯ ¯ ¯ π 0 5. Z π/ 2 2 x cos2 = 1 2 µ 1 2 x + x sin2 x ¯ ¯ ¯ ¯ 2 2 6. Z 5 4 4 e x sin = µ 1 2 e x sin x 1 2 e x cos x ¯ ¯ ¯ ¯ 5 4 4 7. For m 6 = n Z 2 0 sin(2 n x sin(2 m = 1 2 Z 2 0 [cos2( n m ) x cos2( n + m x ] dx = 1 4( n m ) sin2( n m ) x ¯ ¯ ¯ ¯ 2 0 1 4( n + m n + m x ¯ ¯ ¯ ¯ 2 0 . m = n Z 2 0 sin 2 (2 n + 1 ) = Z 2 0 µ 1 2 1 2 cos2(2 n x dx = 1 2 x ¯ ¯ ¯ ¯ 2 0 1 4(2 n sin2(2 n x ¯ ¯ ¯ ¯ 2 0 = π 4 so that k sin(2 n x k = 1 2 π. 562
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Exercises 12.1 8. For m 6 = n Z π/ 2 0 cos(2 n +1) x cos(2 m xdx = 1 2 Z 2 0 [cos2( n m ) x + cos2( n + m x ] dx = 1 4( n m ) sin2( n m ) x ¯ ¯ ¯ ¯ 2 0 + 1 4( n + m n + m x ¯ ¯ ¯ ¯ 2 0 =0 . m = n Z 2 0 cos 2 (2 n = Z 2 0 µ 1 2 + 1 2 cos2(2 n x dx = 1 2 x ¯ ¯ ¯ ¯ 2 0 + 1 4(2 n sin2(2 n x ¯ ¯ ¯ ¯ 2 0 = π 4 so that k cos(2 n x k = 1 2 π. 9. m 6 = n Z π 0 sin nx sin mxdx = 1 2 Z 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 ) n + m ) x ¯ ¯ ¯ ¯ π 0 . m = n Z π 0 sin 2 nxdx = Z 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 k sin nx k = r π 2 . 10. m 6 = n Z p 0 sin p x sin p = 1 2 Z p 0 µ 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 . m = n Z p 0 sin 2 p = Z 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 so that ° ° ° ° sin p x ° ° ° ° = r p 2 . 563
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Exercises 12.1 11. For m 6 = n Z p 0 cos p x cos p xdx = 1 2 Z p 0 µ 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 . m = n Z p 0 cos 2 p = Z p 0 µ 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 Z p 0 1 · cos p = p sin p x ¯ ¯ ¯ ¯ p 0 = 0 and Z p 0 1 2 dx = p so that k 1 k = p and ° ° ° ° cos p x ° ° ° ° = r p 2 . 12. m 6 = n , we use Problems 11 and 10: Z p p cos p x cos p =2 Z p 0 cos p x cos p Z p p sin p x sin p Z p 0 sin p x sin p . Also Z p p sin p x cos p = 1 2 Z p p µ sin ( n m ) π p x + sin ( n + m ) π p x dx , Z p p 1 · cos p = p sin p x ¯ ¯ ¯ ¯ p p , Z p p 1 · sin p = p cos p x ¯ ¯ ¯ ¯ p p , and Z p p sin p x cos p = Z p p 1 2 sin 2 p = p 4 cos 2 p x ¯ ¯ ¯ ¯ p p . m = n Z p p cos 2 p = Z p p µ 1 2 + 1 2 cos 2 p x dx = p, Z p p sin 2 p = Z p p µ 1 2 1 2 cos 2 p x dx = p, and Z p p 1 2 dx p so that k 1 k = ± 2 p, ° ° ° ° cos p x ° ° ° ° = and ° ° ° ° sin p x ° ° ° ° = p.
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Chapter 12 - 12 2 Orthogonal Functions and Fourier Series...

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