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week6 - Week 6 Lectures Math 716 Tanveer 1 Fourier Series...

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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 nπx l for x (0 , l ) (1) or f ( x ) = a 0 2 + summationdisplay n =1 a n cos nπx l for x (0 , l ) (2) and in other cases f ( x ) = a 0 2 + summationdisplay n =1 braceleftBig a n cos nπx l + b n sin nπx 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 Fourier-Series. 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 inner-product: ( 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 n -dimensional rectangle. 1

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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 = ℜ { ( X n , f ) } bardbl X n bardbl 2 Also, ∂E N ∂d n = i ( X n , f ) + i ( f, X n ) + 2 d n ( X n , X n ) = 0 implying d n = ℑ { ( X n , f ) } bardbl X n bardbl 2 Therefore A n = c n + id n is given by (7). Again with A n given by (7), the corresponding E N becomes E N = bardbl f bardbl 2 summationdisplay n N | A n | 2 bardbl X n bardbl 2 0 Taking the limit of N → ∞ , we obtain Bessel inequality.
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