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Unformatted text preview: Week 6 Lectures, Math 716, Tanveer 1 Fourier Series In the context of separation of variable to find solutions of PDEs, we encountered f ( x ) = summationdisplay n =1 b n sin nx l for x (0 ,l ) (1) or f ( x ) = a 2 + summationdisplay n =1 a n cos nx l for x (0 ,l ) (2) and in other cases f ( x ) = a 2 + summationdisplay n =1 braceleftBig a n cos nx l + b n sin nx l bracerightBig for x ( l,l ) (3) The general representation (3) is called the Fourier representation of f ( x ) in the ( l,l ) interval, while (1) and (2) are the Fourier sine and cosine representations of f ( x ) in the interval (0 ,l ). Our discussions revolve around some basic questions 1. When are represenations (1)(3) valid and in what sense? 2. How do we determine coefficients a n , b n in terms of f ( x ). 3. What conditions allow term by term differentiation of the FourierSeries. 2 General L 2 theory We need enough generality to be able to lay the framework for discussion of more general series representations of L 2 , i.e. square integrable functions than (1)(3), since they arise in other PDE problems. In the space L 2 ( a,b ) of generally complex valued functions in the interval ( a,b ), we introduce the L 2 innerproduct: ( f,g ) = integraldisplay b a f ( x ) g ( x ) dx (4) We note that the L 2 norm is related through bardbl f bardbl = ( f,f ) 1 / 2 = bracketleftBigg integraldisplay b a  f  2 ( x ) dx bracketrightBigg 1 / 2 (5) This may be generalized to any number of dimension, with x replaced by x and integration over the interval ( a,b ), replaced by integration over appropriate ndimensional rectangle. 1 Definition 1 A sequence { X n } n =1 L 2 ( a,b ) is orthogonal if ( X n ,X m ) = 0 iff m negationslash = n (6) This sequence is said to be orthonormal, if in addition ( X n ,X n ) = bardbl X n bardbl 2 = 1 . Theorem 2 Let { X n } n =1 L 2 ( a,b ) be a orthogonal set of functions. Let bardbl f bardbl < . Let N be a fixed positive integer. The choice of A n that minimizes mean square error E N = bardbl f N n =1 A n X n bardbl 2 is given by A n = ( f,X n ) bardbl X n bardbl 2 (7) Further, we have bardbl f bardbl 2 summationdisplay n =1  ( f,X n )  2 bardbl X n bardbl 2 Bessel inequality Proof. Define the square of the E N = bardbl f N summationdisplay n =1 A n X n bardbl 2 = parenleftBigg f N summationdisplay n =1 A n X n ,f N summationdisplay n =1 A n X n parenrightBigg Expanding the above using properties of inner product and the orthogonality of X n , we get E N = ( f,f ) N summationdisplay n =1 A n ( X n ,f ) n summationdisplay n =1 A n ( f,X n ) + N summationdisplay n =1 A n A n ( X n ,X n ) (8) We minimize E n as a function of 2 N real variables, { ( c n ,d n ) } N n =1 , where A n = c n + id n . On taking partial derivatives, we get E N c n = ( X n ,f ) ( f,X n ) + 2 c n ( X n ,X n ) = 0 implying c n = { ( X n ,f ) } bardbl X n bardbl 2 Also,...
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This note was uploaded on 07/26/2011 for the course MATH 716 taught by Professor Tanveer,s during the Spring '08 term at Ohio State.
 Spring '08
 Tanveer,S
 Fourier Series

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