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. 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 ³ cos2( n m ) x n + m x ´ dx = 1 4( n m ) sin2( n m ) x ¯ ¯ ¯ ¯ 2 0 1 4( n + m n + m x ¯ ¯ ¯ ¯ 2 0 . 634
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12.1 Orthogonal Functions For m = n Z π/ 2 0 sin 2 (2 n +1) xdx = 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. 8. m 6 = n Z 2 0 cos(2 n x cos(2 m = 1 2 Z 2 0 ³ 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 n x dx = 1 2 x ¯ ¯ ¯ ¯ 2 0 + 1 4(2 n n x ¯ ¯ ¯ ¯ 2 0 = π 4 so that k cos(2 n x k = 1 2 9. m 6 = n Z π 0 sin nx sin mx dx = 1 2 Z π 0 ³ 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 . m = n Z π 0 sin 2 nx dx = Z π 0 µ 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 µ 1 2 1 2 cos 2 p x dx = 1 2 x ¯ ¯ ¯ ¯ p 0 p 4 sin 2 p x ¯ ¯ ¯ ¯ p 0 = p 2 635
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12.1 Orthogonal Functions so that ° ° ° ° sin p x ° ° ° ° = r p 2 . 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 .
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This note was uploaded on 02/17/2010 for the course MATHEMATIC MAS201 taught by Professor Xingqin during the Spring '10 term at Korea Advanced Institute of Science and Technology.

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

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