k and v = v 1 j + u 3 i + v 2


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1 ORTHOGONAL FUNCTIONS AND FOURIER SERIES Orthogonl functions A function cn e considered to e generliztion of vector. Thus the vector concets like the inner roduct nd orthogonlity of vectors cn e extended to functions. Inner roduct Consider the vectors u = u 1 i + u 2 j + u 3 k nd v = v 1 i + v 2 j + v 3 k in R 3, then the inner roduct or dot roduct of u nd v is rel numer, sclr, defined s (u,v) = u 1 v 1 + u 2 v 2 + u 2 v 2 = 3 u k v k k=1
2 The inner roduct ossesses the following roerties (u,v) = u 1 v 1 + u 2 v 2 + u 2 v 2 = 3 u k v k k=1 (i) (u,v) = (v, u) (ii) (ku, v) = k(u, v) (iii) (u, u) = 0 if u = 0 nd (u, u) > 0 if u 0 (iv) (u + v, w) = (u, w) + (v, w)
3 Suose tht f 1 nd f 2 re iecewise continuous functions defined on n intervl [, ]. The definite integrl on the intervl of the roduct f 1 (x) f 2 (x) ossesses roerties (i)  (iv) ove. Definition: Inner roduct of functions The inner roduct 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
4 Definition: Orthogonl functions Two functions f 1 nd f 2 re sid to e orthogonl on n intervl [, ] if ( f 1, f 2 ) = f 1 (x) f 2 (x) dx = 0 Exmle: f 1 (x) = x 2 nd f 2 (x) = x 3 re orthogonl on the intervl [ 1, 1] since ( f 1, f 2 ) = 1 1 x 2. x 3 dx = x6 1 = 0
5 Orthogonl sets We re rimrily interested in n infinite sets of orthogonl functions. Definition: Orthogonl set A set of relvlued functions {φ 0 (x), φ 1 (x), φ 2 (x),...} is sid to e orthogonl on n intervl [, ] if (φ m, φ n ) = φ m (x) φ n (x) dx = 0, m n
6 Orthonorml sets The norm u of vector u cn e exressed using the inner roduct: (u, u) = u 2 u = (u, u) Similrly the squre norm of function φ n is φ n 2 = (φ n, φ n ), nd so the norm is φ n = (φ n, φ n ). In other words, the squre norm nd the norm of function φ n in n orthogonl set {φ n (x)} re, resectively, φ n 2 = φ 2 n(x) dx nd φ n = φ 2 n(x) dx
7 Exmle 1: Orthogonl set of functions Show tht the set {1, cos x, cos 2x,...} is orthogonl on the intervl [, ]: φ 0 = 1, φ n = cos nx (φ 0, φ n ) = = 1 sin nx n φ 0 (x) φ n (x) dx = nd for m n, using the trigonometric identity (φ m, φ n ) = φ m (x) φ n (x) dx = cos nx dx = 1 [sin n sin( n)] = 0 n cos mxcos nx dx = 1 [cos(m + n)x + cos(m n)x] dx 2 = 1 sin(m + n)x sin(m n)x + = 0 2 m + n m n
8 Exmle 2: Norms Find the norms of the functions given in the Exmle 1 ove. φ 0 2 = dx = 2 φ 0 = 2 φ n 2 = cos 2 nx dx = 1 [1 + cos 2nx] dx = 2 φ n = Any orthogonl set of nonzero functions {φ n (x)}, n = 0, 1, 2... cn e normlized, i.e. mde into n orthonorml set. Exmle: An orthonorml set on the intervl [, ]: 1, cos x cos 2x,,... 2
9 Vector nlogy Suose v 1, v 2, nd v 3 re three mutully orthogonl nonzero vectors in R 3. Such n orthogonl set cn e used s sis for R 3, tht is, ny threedimensionl vector cn e written s liner comintion u = c 1 v 1 + c 2 v 2 + c 3 v 3 where c i, i = 1, 2, 3 re sclrs clled the comonents of the vector. Ech comonent cn e exressed in terms of u nd the corresonding vector v i : (u,v 1 ) = c 1 (v 1,v 1 ) + c 2 (v 2,v 1 ) + c 3 (v 3,v 1 ) = c 1 v c c 3.0 (u,v 2 ) = c 2 v 2 2 (u,v 3 ) = c 3 v 3 2
10 Hence nd c 1 = (u,v 1) v 1 2 c 2 = (u,v 2) v 2 2 c 3 = (u,v 3) v 3 2 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 n=1 (u,v n ) v n 2 v n
11 Orthogonl series exnsion Suose {φ n (x)} is n infinite orthogonl set of functions on n intervl [, ]. If y = f (x) is function defined on the intervl [, ], we cn determine set of coefficients c n, n = 0, 1, 2,... for which f (x) = c 0 φ 0 (x) + c 1 φ 1 (x) + c 2 φ 2 (x) c n φ n (x) +... (1) using the inner roduct. Multilying the exression ove y φ m (x) nd integrting over the intervl [, ] gives f (x) φ m (x) dx = = c 0 φ 0 (x) φ m (x) dx + c 1 = c 0 (φ 0, φ m ) + c 1 (φ 1, φ m ) c n (φ n, φ m ) +... φ 1 (x) φ m (x) dx c n φ n (x) φ m (x) dx +...
12 By orthogonlity, ech term on r.h.s. is zero excet when m = n, in which cse we hve f (x) φ n (x) dx = c n φ2 n(x) dx The required coefficients re then In other words f (x) = n=0 c n = c n φ n (x) = n=0 f (x) φ n (x) dx, n = 0, 1, 2... φ2 n(x) dx f (x) φ n (x) dx φ n (x) 2 φ n (x) = n=0 ( f, φ n ) φ n (x) 2 φ n(x)
13 Definition: Orthogonl set / weight function A set of relvlued functions {φ 0 (x), φ 1 (x), φ 2 (x),...} is sid to e orthogonl with resect to weight function w(x) on n intervl [, ] if w(x) φ m (x) φ n (x) dx = 0, m n The usul ssumtion is tht w(x) > 0 on the intervl of orthogonlity [, ]. For exmle, the set {1, cos x, cos 2x,...} is orthogonl w.r.t. the weight function w(x) = 1 on the intervl [, ].
14 If {φ n (x)} is orthogonl w.r.t. weight function w(x) on the intervl [, ], them multilying the exnsion (1), f (x) = c 0 φ 0 (x) + c 1 φ 1 (x)..., yw(x) nd integrting y rts yields c n = f (x) w(x) φ n (x) dx φ n (x) 2 where φ n (x) 2 = w(x) φ 2 n(x) dx
15 The series f (x) = n=0 c n φ n (x) (2) with the coefficients given either y c n = f (x) φ n (x) dx φ n (x) 2 or c n = f (x) w(x) φ n (x) dx φ n (x) 2 (3) is sid to e n orthogonl series exnsion of f or generlized Fourier series. Comlete sets We shll ssume tht n orthogonl set {φ n (x)} is comlete. Under this ssumtion f cn not e orthogonl to ech φ n of the orthogonl set.
16 Fourier series Trigonometric series The set of functions 1, cos x, cos 2 x, cos 3 x,..., sin x, sin 2 x, sin 3 x,... is orthogonl on the intervl [, ]. We cn exnd function f defined on [, ] into the trigonometric series f (x) = n cos n x + n sin n x n=1 (4)
17 Determining the ocefficients 0, 1, 2,..., 1, 2,... : We multily y 1 (the first function in our orthogonl set) nd integrte oth sides of the exnsion (4) from to f (x) dx = 0 dx + n cos n 2 xdx+ n sin n xdx n=1 Since cos(nx/) nd sin(nx/), n 1, re orthogonl to 1 on the intervl, the r.h.s. reduces s follows Solving for 0 yields f (x) dx = = 1 dx = 0 2 x = 0 f (x) dx (5)
18 Now, we multily (4) y cos(mx/) nd integrte f (x) cos m xdx = n=1 By orthogonlity, we hve cos m xdx n cos m x cos n xdx+ n cos m xdx= 0, m > 0 cos m x sin n xdx= 0 cos m x cos n xdx= δ mn where the Kronecker delt δ mn = 0 if m n, nd δ mn = 1 if m = n. cos m x sin n xdx
19 Thus the eqution (6) ove reduces to f (x) cos n xdx= n nd so n = 1 f (x) cos n xdx (6)
20 Finlly, multilying (4) y sin(mx/), integrting nd using the orthogonlity reltions we find tht n = 1 sin m xdx= 0, m > 0 sin m x cos n xdx= 0 sin m x sin n xdx= δ mn f (x) sin n xdx (7) The trigonometric series (4) with coefficients 0, n, nd n defined y (5), (6) nd (7), resectively re sid to e the Fourier series of the function f. The coefficients otined from (5), (6) nd (7) re referred s Fourier coefficients of f.
21 Definition: Fourier series The Fourier series of function f defined on the intervl (, ) is given y f (x) = n cos n x + n sin n x where n=1 0 = 1 n = 1 n = 1 (8) f (x) dx (9) f (x) cos n xdx (10) f (x) sin n xdx (11)
22 Exmle 1: Exnsion in Fourier series f (x) = With = we hve from (9) nd (10) tht 0 = 1 f (x) dx = dx + 0 n = 1 f (x) cos nx dx = 1 = 1 sin nx ( x) + 1 n 0 n 0 = 1 cos nx n n 0 cos n + 1 = n 2 = 1 ( 1)n n 2 0, < x < 0 x, 0 x < ( x) dx = 1 x x dx + ( x) cos nx dx 0 sin nx dx 0 = 2
23 Similrly, we find from (11) n = 1 n 0 ( x) sin nx dx = 1 n The function f (x) is thus exnded s f (x) = 1 ( 1) n 4 + n=1 n 2 cos nx + 1 n sin nx (12) We lso note tht 1 ( 1) n = 0, n even 2, n odd.
24 Convergence of Fourier series Theorem: Conditions for convergence Let f nd f e iecewise continuous on the intervl (, ); tht is, let f nd f e continuous excet t finite numer of oints in the intervl nd hve only finite discontinuities t these oints. Then the Fourier series of f on the intervl converges to f (x) t oint of continuity. At oint of discontinuity, the Fourier series converges to the verge f (x+) + f (x ), 2 where f (x+) nd f (x ) denote the limit of f t x from the right nd from the left, resectively.
25 Exmle 2: Convergence of oint of discontinuity The exnsion (12) of the function (Exmle 1) 0, < x < 0 f (x) = x, 0 x < will converge to f (x) for every x from the intervl (, ) excet t x = 0 where it will converge to f (0+) + f (0 ) 2 = = 2.
26 Periodic extension Oserve tht ech of the functions in the sis set 1, cos x, cos 2 x, cos 3 x,..., sin x, sin 2 x, sin 3 x,... hs different fundmentl eriod 2/n, n 1, ut since ositive integer multile of eriod is lso eriod, we see tht ll the functions hve in common the eriod 2. Thus the r.h.s. of f (x) = n cos n x + n sin n x n=1 is 2eriodic; indeed 2 is the fundmentl eriod of the sum. We conclude tht Fourier series not only reresents the function on the intervl (, ) ut lso gives the eriodic extension of f outside this intervl.
27 We cn now ly the Theorem on conditions for convergence to the eriodic extension or simly ssume the function is eriodic, f (x + T) = f (x), with eriod T = 2 from the outset. When f is iecewise continuous nd the right nd lefthnd derivtives exist t x = nd x =, resectively, then the Fourier series converges to the verge [ f ( )+ f (+)]/2 t these oints nd lso to this vlue extended eriodiclly to ±3, ±5, ±7, nd so on. Exmle: The Fourier series of the function f (x) in the Exmle 1 converges to the eriodic extension of the function on the entire xxis. At 0, ±2, ±4,..., nd t ±, ±3, ±5,..., the series converges to the vlues f (0+) + f (0 ) 2 = 2 nd f (+) + f ( ) 2 = 0
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