DEFINITION The inner product of two functions f 1 and f 2 on an interval [a, b] is the number. ( f 1, f 2 ) b DEFINITION 11.1.


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1 398 CHAPTER 11 ORTHOGONAL FUNCTIONS AND FOURIER SERIES 11.1 ORTHOGONAL FUNCTIONS REVIEW MATERIAL The notions of generlized vectors nd vector spces cn e found in ny liner lger text. INTRODUCTION The concepts of geometric vectors in two nd three dimensions, orthogonl or perpendiculr vectors, nd the inner product of two vectors hve een generlized. It is perfectly routine in mthemtics to think of function s vector. In this section we will exmine n inner product tht is different from the one you studied in clculus. Using this new inner product, we define orthogonl functions nd sets of orthogonl functions. Another topic in stndrd clculus course is the expnsion of function f in power series. In this section we will lso see how to expnd suitle function f in terms of n infinite set of orthogonl functions. INNER PRODUCT Recll tht if u nd v re two vectors in 3spce, then the inner product (u, v) (in clculus this is written s u v) possesses the following properties: (i) (u, v) (v, u), (ii) (ku, v) k(u, v), k sclr, (iii) (u, u) 0 if u 0 nd (u, u) 0 if u 0, (iv) (u v, w) (u, w) (v, w). We expect tht ny generliztion of the inner product concept should hve these sme properties. Suppose tht f 1 nd f 2 re functions defined on n intervl [, ]. * Since definite integrl on [, ] of the product f 1 (x) f 2 (x) possesses the foregoing properties (i) (iv) whenever the integrl exists, we re prompted to mke the following definition. DEFINITION Inner Product of Functions The inner product of two functions f 1 nd f 2 on n intervl [, ] is the numer ( f 1, f 2 ) f 1 (x) f 2 (x) dx. ORTHOGONAL FUNCTIONS Motivted y the fct tht two geometric vectors u nd v re orthogonl whenever their inner product is zero, we define orthogonl functions in similr mnner. DEFINITION Orthogonl Functions Two functions f 1 nd f 2 re orthogonl on n intervl [, ] if ( f 1, f 2 ) f 1 (x) f 2 (x) dx 0. (1) * The intervl could lso e (, ), [0, ), nd so on.
2 11.1 ORTHOGONAL FUNCTIONS 399 For exmple, the functions f 1 (x) x 2 nd f 2 (x) x 3 re orthogonl on the intervl [1, 1], since ( f 1, f 2 ) 1 x 2 x 3 dx x Unlike in vector nlysis, in which the word orthogonl is synonym for perpendiculr, in this present context the term orthogonl nd condition (1) hve no geometric significnce. ORTHOGONAL SETS functions. We re primrily interested in infinite sets of orthogonl DEFINITION Orthogonl Set A set of relvlued functions {f 0 (x), f 1 (x), f 2 (x),...} is sid to e orthogonl on n intervl [, ] if (m, n) m(x)n(x) dx 0, m Y n. (2) ORTHONORMAL SETS The norm, or length u, of vector u cn e expressed in terms of the inner product. The expression (u, u) u 2 is clled the squre norm, nd so the norm is u 1(u, u). Similrly, the squre norm of function f n is f n (x) 2 (f n, f n ), nd so the norm, or its generlized length, is f n (x) 1(n, n). In other words, the squre norm nd norm of function f n in n orthogonl set {f n (x)} re, respectively, B f n (x) 2 n 2 (x) dx nd f n (x) f 2 n(x) dx. (3) If {f n (x)} is n orthogonl set of functions on the intervl [, ] with the property tht f n (x) 1 for n 0, 1, 2,..., then {f n (x)} is sid to e n orthonorml set on the intervl. EXAMPLE 1 Orthogonl Set of Functions Show tht the set {1, cos x, cos 2x,...} is orthogonl on the intervl [p, p]. SOLUTION If we mke the identifiction f 0 (x) 1 nd f n (x) cos nx, we must then show tht We hve, in the first cse, ( 0, n) 0(x) n(x) dx 0, n 0, nd 1 n sin nx 0(x) n(x) dx cos nx dx m(x) n(x) dx 0, m n. 1 n [sin n sin(n)] 0, n 0,
3 400 CHAPTER 11 ORTHOGONAL FUNCTIONS AND FOURIER SERIES nd, in the second, (m, n) m(x)n(x) dx cos mx cos nx dx 1 [cos(m n)x cos(m n)x] dx ; trig identity 2 1 sin (m n)x sin (m n)x 2 m n m n 0, m n. EXAMPLE 2 Norms Find the norm of ech function in the orthogonl set given in Exmple 1. SOLUTION For f 0 (x) 1 we hve, from (3), f 0 (x) 2 so f 0 (x) 12. For f n (x) cos nx, n 0, it follows tht f n (x) 2 Thus for n 0, f n (x) 1. cos 2 nx dx 1 2 Any orthogonl set of nonzero functions {f n (x)}, n 0, 1, 2,... cn e normlized tht is, mde into n orthonorml set y dividing ech function y its norm. It follows from Exmples 1 nd 2 tht the set 1, cos x cos 2x,, is orthonorml on the intervl [p, p]. We shll mke one more nlogy etween vectors nd functions. Suppose v 1, v 2, nd v 3 re three mutully orthogonl nonzero vectors in 3spce. Such n orthogonl set cn e used s sis for 3spce; tht is, ny threedimensionl vector cn e written s liner comintion dx 2, 1 [1 cos 2nx] dx. u c 1 v 1 c 2 v 2 c 3 v 3, (4) where the c i, i 1, 2, 3, re sclrs clled the components of the vector. Ech component c i cn e expressed in terms of u nd the corresponding vector v i. To see this, we tke the inner product of (4) with v 1 : (u, v 1 ) c 1 (v 1, v 1 ) c 2 (v 2, v 1 ) c 3 (v 3, v 1 ) c 1 v 1 2 c 2 0 c 3 0. Hence c 1 (u, v 1) 'v 1 ' 2. In like mnner we find tht the components c 2 nd c 3 re given y c 2 (u, v 2) 'v 2 ' 2 nd c 3 (u, v 3) 'v 3 ' 2.
4 11.1 ORTHOGONAL FUNCTIONS 401 Hence (4) cn e expressed s u (u, v 1) 'v 1 ' 2 v 1 (u, v 2) 'v 2 ' 2 v 2 (u, v 3) 'v 3 ' 2 v 3 3 n1 (u, v n ) 'v n ' 2 v n. (5) ORTHOGONAL SERIES EXPANSION Suppose {f n (x)} is n infinite orthogonl set of functions on n intervl [, ]. We sk: If y f (x) is function defined on the intervl [, ], is it possile to determine set of coefficients c n, n 0, 1, 2,..., for which f (x) c 00(x) c 11(x) c nn(x)? (6) As in the foregoing discussion on finding components of vector we cn find the coefficients c n y utilizing the inner product. Multiplying (6) y f m (x) nd integrting over the intervl [, ] gives f (x)m(x) dx c 0 0(x)m(x) dx c 1 1(x)m(x) dx c n n(x)m(x) dx c 0 (0, m) c 1 (1, m) c n (n, m). By orthogonlity ech term on the righthnd side of the lst eqution is zero except when m n. In this cse we hve f (x)n(x) dx c n It follows tht the required coefficients re c n f (x)n(x) dx, n 0, 1, 2, n(x)dx In other words, f (x) c nn(x), (7) n0 where c n f (x)n(x) dx. 'n(x)' 2 (8) With inner product nottion, (7) ecomes f (x) ( f, n) n0 'n(x)' 2 n(x). (9) Thus (9) is seen to e the function nlogue of the vector result given in (5). 2 n(x) dx. DEFINITION Orthogonl Set/Weight Function A set of relvlued functions {f 0 (x), f 1 (x), f 2 (x),...} is sid to e orthogonl with respect to weight function w(x) on n intervl [, ] if w(x)m(x)n(x) dx 0, m n. The usul ssumption is tht w(x) 0 on the intervl of orthogonlity [, ]. The set {1, cos x, cos 2x,...} in Exmple 1 is orthogonl with respect to the weight function w(x) 1 on the intervl [p, p]. If {f n (x)} is orthogonl with respect to weight function w(x) on the intervl [, ], then multiplying (6) y w(x)f n (x) nd integrting yields c n f (x) w(x)n(x) dx 'n(x)' 2, (10)
5 402 CHAPTER 11 ORTHOGONAL FUNCTIONS AND FOURIER SERIES where f n (x) 2 2 w(x)n(x) dx. (11) The series (7) with coefficients given y either (8) or (10) is sid to e n orthogonl series expnsion of f or generlized Fourier series. COMPLETE SETS The procedure outlined for determining the coefficients c n ws forml; tht is, sic questions out whether or not n orthogonl series expnsion such s (7) is ctully possile were ignored. Also, to expnd f in series of orthogonl functions, it is certinly necessry tht f not e orthogonl to ech f n of the orthogonl set {f n (x)}. (If f were orthogonl to every f n, then c n 0, n 0, 1, 2,....) To void the ltter prolem, we shll ssume, for the reminder of the discussion, tht n orthogonl set is complete. This mens tht the only function tht is orthogonl to ech memer of the set is the zero function. EXERCISES 11.1 In Prolems 1 6 show tht the given functions re orthogonl on the indicted intervl. 1. f 1 (x) x, f 2 (x) x 2 ; [2, 2] 2. f 1 (x) x 3, f 2 (x) x 2 1; [1, 1] 3. f 1 (x) e x, f 2 (x) xe x e x ; [0, 2] 4. f 1 (x) cos x, f 2 (x) sin 2 x; [0, p] 5. f 1 (x) x, f 2 (x) cos 2x; [p2, p2] 6. f 1 (x) e x, f 2 (x) sin x; [p4, 5p4] In Prolems 7 12 show tht the given set of functions is orthogonl on the indicted intervl. Find the norm of ech function in the set. 7. {sin x, sin 3x, sin 5x,...}; [0, p2] 8. {cos x, cos 3x, cos 5x,...}; [0, p 2] 9. {sin nx}, n 1, 2, 3,...; [0,p] n 10. sin ; [0, p] p x, n 1, 2, 3, n 1, cos p x, n 1, 2, 3,... ; [0, p] n m 1, cos x, sin p p x, n 1, 2, 3,..., m 1, 2, 3,... ; [p, p] In Prolems 13 nd 14 verify y direct integrtion tht the functions re orthogonl with respect to the indicted weight function on the given intervl. 13. H 0 (x) 1, H 1 (x) 2x, H 2 (x) 4x 2 2; w(x) e x2, (, ) 14. L 0 (x) 1, L 1 (x) x 1, L 2 (x) 1 w(x) e x 2 x2 2x 1;, [0, ) Answers to selected oddnumered prolems egin on pge ANS Let {f n (x)} e n orthogonl set of functions on [, ] such tht f 0 (x) 1. Show tht n(x) dx 0 for n 1, 2, Let {f n (x)} e n orthogonl set of functions on [, ] such tht f 0 (x) 1 nd f 1 (x) x. Show tht (x )n(x) dx 0 for n 2, 3,... nd ny constnts nd. 17. Let {f n (x)} e n orthogonl set of functions on [, ]. Show tht f m (x) f n (x) 2 f m (x) 2 f n (x) 2, m n. 18. From Prolem 1 we know tht f 1 (x) x nd f 2 (x) x 2 re orthogonl on the intervl [2, 2]. Find constnts c 1 nd c 2 such tht f 3 (x) x c 1 x 2 c 2 x 3 is orthogonl to oth f 1 nd f 2 on the sme intervl. 19. The set of functions {sin nx}, n 1, 2, 3,..., is orthogonl on the intervl [p, p]. Show tht the set is not complete. 20. Suppose f 1, f 2, nd f 3 re functions continuous on the intervl [, ]. Show tht ( f 1 f 2, f 3 ) ( f 1, f 3 ) ( f 2, f 3 ). Discussion Prolems 21. A relvlued function f is sid to e periodic with period T if f (x T ) f (x). For exmple, 4p is period of sin x, since sin(x 4p) sin x. The smllest vlue of T for which f (x T ) f (x) holds is clled the fundmentl period of f. For exmple, the fundmentl period of f (x) sin x is T 2p. Wht is the fundmentl period of ech of the following functions? () f (x) cos 2px () f (x) sin 4 L x (c) f (x) sin x sin 2x (d) f (x) sin 2x cos 4x (e) f (x) sin 3x cos 2x (f) f (x) A 0 A n cos n n1 p x B n sin n p x, A n nd B n depend only on n
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