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Orthogonal Functions 11.1 398 ● CHAPTER 11 ORTHOGONAL FUNCTIONS AND FOURIER SERIES 11.1 ORTHOGONAL FUNCTIONS REVIEW MATERIAL ● The notions of generalized vectors and vector spaces can be found in any linear algebra text. INTRODUCTION The concepts of geometric vectors in two and three dimensions, orthogonal or perpendicular vectors, and the inner product of two vectors have been generalized. It is perfectly routine in mathematics to think of a function as a vector. In this section we will examine an inner product that is different from the one you studied in calculus. Using this new inner product, we define orthogonal functions and sets of orthogonal functions. Another topic in a standard calculus course is the expansion of a function f in a power series. In this section we will also see how to expand a suitable function f in terms of an infinite set of orthogonal functions. INNER PRODUCT Recall that if u and v are two vectors in 3-space, then the inner product (u, v) (in calculus this is written as u ؒ v) possesses the following properties: (i)(u, v) ϭ (v, u), (ii)(ku, v) ϭ k(u, v), k a scalar, (iii)(u, u) ϭ 0 if u ϭ 0 and (u, u) Ͼ 0 if u 0, (iv)(u ϩ v, w) ϭ (u, w) ϩ (v, w). We expect that any generalization of the inner product concept should have these same properties. * Suppose that f1 and f2 are functions defined on an interval [a, b]. Since a definite integral on [a, b] of the product f1(x) f2(x) possesses the foregoing properties (i)–(iv) whenever the integral exists, we are prompted to make the following definition. DEFINITION 11.1.1 Inner Product of Functions The inner product of two functions f1 and f2 on an interval [a, b] is the number b ϭ ͵ ( f1, f 2) f1(x) f 2(x) dx. a ORTHOGONAL FUNCTIONS Motivated by the fact that two geometric vectors u and v are orthogonal whenever their inner product is zero, we define orthogonal functions in a similar manner. DEFINITION 11.1.2 Orthogonal Functions Two functions f1 and f2 are orthogonal on an interval [a, b] if b ϭ ͵ ϭ ( f1, f 2) f1(x) f 2(x) dx 0. (1) a *The interval could also be (Ϫϱ, ϱ), [0, ϱ), and so on. 11.1 ORTHOGONAL FUNCTIONS ● 399 2 3 For example, the functions f1(x) ϭ x and f2(x) ϭ x are orthogonal on the interval [Ϫ1, 1], since 1 1 1 ϭ ͵ 2 ؒ 3 ϭ 6 ϭ ( f1, f2) x x dx x ͉ 0. Ϫ1 6 Ϫ1 Unlike in vector analysis, in which the word orthogonal is a synonym for perpendic- ular, in this present context the term orthogonal and condition (1) have no geometric significance. ORTHOGONAL SETS We are primarily interested in infinite sets of orthogonal functions. DEFINITION 11.1.3 Orthogonal Set A set of real-valued functions {f 0(x), f1(x), f2(x),...} is said to be orthogonal on an interval [a, b] if b ␾ ␾ ϭ ͵ ␾ ␾ ϭ Y ( m, n) m(x) n(x) dx 0, m n. (2) a ORTHONORMAL SETS The norm, or length ʈuʈ, of a vector u can be expressed in terms of the inner product. The expression (u, u) ʈuʈ2 is called the square norm, and so the norm is ʈuʈ ϭ 1(u, u). Similarly, the square norm of a function fn 2 is ʈfn(x)ʈ ϭ (fn , fn), and so the norm, or its generalized length, is ␾ ␾ ʈfn(x)ʈ ϭ 1( n, n). In other words, the square norm and norm of a function fn in an orthogonal set {fn(x)} are, respectively, b b ʈ ʈ2 ϭ ͵ ␾2 ʈ ʈ ϭ ͵ 2 fn(x) n (x) dx and fn(x) fn(x) dx. (3) a B a If {fn(x)} is an orthogonal set of functions on the interval [a, b] with the property that ʈfn(x)ʈ ϭ 1 for n ϭ 0, 1, 2, . , then {fn(x)} is said to be an orthonormal set on the interval. EXAMPLE 1 Orthogonal Set of Functions Show that the set {1, cos x, cos 2x, . .} is orthogonal on the interval [Ϫp, p]. SOLUTION If we make the identification f0(x) ϭ 1 and fn(x) ϭ cos nx, we must ͵␲ ␾ ␾ ϭ ͵␲ ␾ ␾ ϭ then show that Ϫ␲ 0(x) n(x) dx 0, n 0, and Ϫ␲ m(x) n(x) dx 0, m n. We have, in the first case, ␲ ␲ ␾ ␾ ϭ ͵ ␾ ␾ ϭ ͵ ( 0 , n) 0(x) n(x) dx cos nx dx Ϫ␲ Ϫ␲ 1 ␲ 1 ϭ sin nx͉ ϭ [sin n␲ Ϫ sin(Ϫn␲)] ϭ 0, n 0, n Ϫ␲ n 400 ● CHAPTER 11 ORTHOGONAL FUNCTIONS AND FOURIER SERIES and, in the second, ␲ ␾ ␾ ϭ ͵ ␾ ␾ ( m , n) m(x) n(x) dx Ϫ␲ ␲ ϭ ͵ cos mx cos nx dx Ϫ␲ 1 ␲ ϭ ͵ [cos(m ϩ n)x ϩ cos(m Ϫ n)x] dx ; trig identity 2 Ϫ␲ 1 sin (m ϩ n)x sin (m Ϫ n)x ␲ ϭ ΄ ϩ ΅ ϭ 0, m n. 2 m ϩ n m Ϫ n Ϫ␲ EXAMPLE 2 Norms Find the norm of each function in the orthogonal set given in Example 1. SOLUTION For f0(x) ϭ 1 we have, from (3), ␲ ʈ ʈ2 ϭ ͵ ϭ ␲ f0 (x) dx 2 , Ϫ␲ so ʈf0(x)ʈ ϭ 12␲. For fn(x) ϭ cos nx, n Ͼ 0, it follows that ␲ 1 ␲ ʈ ʈ2 ϭ ͵ 2 ϭ ͵ ϩ ϭ ␲ fn (x) cos nx dx [1 cos 2nx] dx . Ϫ␲ 2 Ϫ␲ Thus for n Ͼ 0, ʈfn(x)ʈ ϭ 1␲. Any orthogonal set of nonzero functions {fn(x)}, n ϭ 0, 1, 2, . can be normalized—that is, made into an orthonormal set—by dividing each function by its norm. It follows from Examples 1 and 2 that the set 1 cos x cos 2x Ά , , , ...· 12␲ 1␲ 1␲ is orthonormal on the interval [Ϫp, p]. We shall make one more analogy between vectors and functions. Suppose v1, v2, and v3 are three mutually orthogonal nonzero vectors in 3-space. Such an orthogonal set can be used as a basis for 3-space; that is, any three-dimensional vec- tor can be written as a linear combination ϭ ϩ ϩ u c1v1 c2v2 c3v3, (4) where the ci , i ϭ 1, 2, 3, are scalars called the components of the vector. Each component ci can be expressed in terms of u and the corresponding vector vi . To see this, we take the inner product of (4) with v1: 2 .u, v1) ϭ c1(v1, v1) ϩ c2(v2, v1) ϩ c3(v3, v1) ϭ c1ʈv1ʈ ϩ c2 ؒ 0 ϩ c3 ؒ 0) (u, v ) Hence c ϭ 1 . 1 ' '2 v1 In like manner we find that the components c2 and c3 are given by (u, v ) (u, v ) c ϭ 2 and c ϭ 3 . 2 ' '2 3 ' '2 v2 v3 11.1 ORTHOGONAL FUNCTIONS ● 401 Hence (4) can be expressed as (u, v ) (u, v ) (u, v ) 3 (u, v ) u ϭ 1 v ϩ 2 v ϩ 3 v ϭ ͚ n v . (5) ' '2 1 ' '2 2 ' '2 3 ' '2 n v1 v2 v3 nϭ1 vn ORTHOGONAL SERIES EXPANSION Suppose {fn(x)} is an infinite orthogo- nal set of functions on an interval [a, b]. We ask: If y ϭ f (x) is a function defined on the interval [a, b], is it possible to determine a set of coefficients cn, n ϭ 0, 1, 2, . , for which ϭ ␾ ϩ ␾ ϩиииϩ ␾ ϩиии f (x) c0 0(x) c1 1(x) cn n(x) ? (6) As in the foregoing discussion on finding components of a vector we can find the coefficients cn by utilizing the inner product. Multiplying (6) by fm(x) and integrating over the interval [a, b] gives b b b b ͵ ␾ ϭ ͵ ␾ ␾ ϩ ͵ ␾ ␾ ϩиииϩ ͵ ␾ ␾ ϩиии f (x) m(x) dx c0 0(x) m(x) dx c1 1(x) m(x) dx cn n(x) m(x) dx a a a a ϭ ␾ ␾ ϩ ␾ ␾ ϩиииϩ ␾ ␾ ϩиии c0( 0, m) c1( 1, m) cn( n, m) . By orthogonality each term on the right-hand side of the last equation is zero except when m ϭ n. In this case we have b b ͵ ␾ ϭ ͵ ␾2 f (x) n(x) dx cn n(x) dx. a a It follows that the required coefficients are ͵b f (x)␾ (x) dx c ϭ a n , n ϭ 0, 1, 2, . n ͵b␾2 a n(x)dx ϱ ϭ ␾ In other words,f (x) ͚ cn n(x), (7) nϭ0 ͵b f (x)␾ (x) dx wherec ϭ a n . (8) n '␾ '2 n(x) With inner product notation, (7) becomes ϱ ( f, ␾ ) f (x) ϭ ͚ n ␾ (x). (9) '␾ '2 n nϭ0 n(x) Thus (9) is seen to be the function analogue of the vector result given in (5). DEFINITION 11.1.4 Orthogonal Set/Weight Function A set of real-valued functions {f0(x), f1(x), f2(x), . .} is said to be orthogonal with respect to a weight function w(x) on an interval [a, b] if b ͵ ␾ ␾ ϭ w(x) m(x) n(x) dx 0, m n. a The usual assumption is that w(x) Ͼ 0 on the interval of orthogonality [a, b]. The set {1, cos x, cos 2x, . .} in Example 1 is orthogonal with respect to the weight function w(x) ϭ 1 on the interval [Ϫp, p]. If {fn(x)} is orthogonal with respect to a weight function w(x) on the interval [a, b], then multiplying (6) by w(x)fn(x) and integrating yields ͵b f (x) w(x)␾ (x) dx c ϭ a n , (10) n '␾ '2 n(x) 402 ● CHAPTER 11 ORTHOGONAL FUNCTIONS AND FOURIER SERIES b ʈ ʈ2 ϭ ͵ ␾2 where fn(x) w(x) n(x) dx.
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