AEM_3e_Chapter_12

# AEM_3e_Chapter_12 - Part IV Fourier Analysis and Partial...

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Part IV Fourier Analysis and Partial Differential Equations 12 12 Orthogonal Functions and Fourier Series EXERCISES 12.1 Orthogonal Functions 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 =0 6. Z 5 π/ 4 4 e x sin xdx = µ 1 2 e x sin x 1 2 e x cos x ¯ ¯ ¯ ¯ 5 4 4 9. For 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 . 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 , 205

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12.1 Orthogonal Functions and Z p p sin p x cos p xdx = Z p p 1 2 sin 2 p = p 4 cos 2 p x ¯ ¯ ¯ ¯ p p =0 . For 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 =2 p so that k 1 k = p 2 p, ° ° ° ° cos p x ° ° ° ° = and ° ° ° ° sin p x ° ° ° ° = p. 15. By orthogonality ± b a φ 0 ( x ) φ n ( x ) dx = 0 for n =1,2,3, ... ; that is, ± b a φ n ( x ) dx = 0 for n . 18. Setting 0= Z 2 2 f 3 ( x ) f 1 ( x ) dx = Z 2 2 ( x 2 + c 1 x 3 + c 2 x 4 ) dx = 16 3 + 64 5 c 2 and Z 2 2 f 3 ( x ) f 2 ( x ) dx = Z 2 2 ( x 3 + c 1 x 4 + c 2 x 5 ) dx = 64 5 c 1 we obtain c 1 = 0 and c 2 = 5 12 . 21. (a) The fundamental period is 2 π/ 2 π =1. (b) The fundamental period is 2 (4 /L )= 1 2 πL . (c) The fundamental period of sin x + sin2 x is 2 π . (d) The fundamental period of sin2 x + cos4 x is 2 2= π . (e) The fundamental period of sin3 x +cos4 x is 2 π since the smallest integer multiples of 2 3 and 2 4= 2 that are equal are 3 and 4, respectively. (f) The fundamental period of f ( x )is2 ( nπ/p )=2 p/n . EXERCISES 12.2 Fourier Series 3. a 0 = Z 1 1 f ( x ) dx = Z 0 1 1 dx + Z 1 0 = 3 2 a n = Z 1 1 f ( x )cos nπx dx = Z 0 1 cos nπx dx + Z 1 0 x cos nπx dx = 1 n 2 π 2 [( 1) n 1] b n = Z 1 1 f ( x )sin nπx dx = Z 0 1 sin nπx dx + Z 1 0 x sin nπx dx = 1 206
12.2 Fourier Series f ( x )= 3 4 + ± n =1 · ( 1) n 1 n 2 π 2 cos nπx 1 sin nπx ¸ 6. a 0 = 1 π Z π π f ( x ) dx = 1 π Z 0 π π 2 dx + 1 π Z π 0 ( π 2 x 2 ) dx = 5 3 π 2 a n = 1 π Z π π f ( x )cos nx dx = 1 π Z 0 π π 2 cos nx dx + 1 π Z π 0 ( π 2 x 2 ) cos nx dx = 1 π µ π 2 x 2 n sin nx ¯ ¯ ¯ ¯ π 0 + 2 n Z π 0 x sin nx dx = 2 n 2 ( 1) n +1 b n = 1 π Z π π f ( x )sin nx dx = 1 π Z 0 π π 2 sin nx dx + 1 π Z π 0 ( π 2 x 2 ) sin nx dx = π n [( 1) n 1] + 1 π µ x 2 π 2 n cos nx ¯ ¯ ¯ ¯ π 0 2 n Z π 0 x cos nx dx = π n ( 1) n + 2 n 3 π [1 ( 1) n ] f ( x 5 π 2 6 + ± n =1 · 2 n 2 ( 1) n +1 cos nx + µ π n ( 1) n + 2[1 ( 1) n ] n 3 π sin nx ¸ 9. a 0 = 1 π Z π π f ( x ) dx = 1 π Z π 0

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## This note was uploaded on 01/03/2011 for the course BIS 511 at Yale.

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AEM_3e_Chapter_12 - Part IV Fourier Analysis and Partial...

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