Ch_05 - Fourier Series - Exam_Soln

Ch_05 - Fourier Series - Exam_Soln - Exam Problems of...

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Exam Problems of Fourier Series, Integrals and Transforms
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Fourier Series, Integrals and Transforms Fourier Series Fourier Series 1. (’00) Derive the complex Fourier series from the trigonometric Fourier series. Solution: In the optional section, we show that the Fourier series 0 1 () ( c o s s i n ) nn n f xa an x bn x = =+ + can be written in complex form, which sometimes simplifies calculations. This is done by the Euler formula, cos sin it et i t and its companion sin it i t =− [obtained from cos( ) cos tt −= , sin( ) sin ] with tn x = , that is cos sin cos sin inx inx en x i n x x i n x By addition and division by 2 we get 1 cos ( ) 2 inx inx nx e e By subtraction and division by 2 i gives 1 sin ( ) 2 inx inx nx e e i From this using 1/ ii , we have 11 cos sin ( ) ( ) 22 ( ) ( ) inx inx inx inx n n inx inx x x a e e b e e i a i be −− += + + ++ With this, 0 1 ( ) inx inx n fx c ce ke = + where 00 ca = , and 1 ( ) ( )(cos sin ) ( ) 2 1 ( ) ( c o s s i n ) 2 inx n inx n c a ib f x nx i nx dx f x e dx ka i b f x n x i n x d x f x e d x ππ = = = + = ∫∫ ( 1, 2, n = " ) Finally, if we introduce the notation kc = − , we obtain 1 2 inx n n inx n cf x e d x π =−∞ = = ( 0, 1, 2, n = ±± " )
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Fourier Series, Integrals and Transforms Fourier Series 2. (’00) Let’s assume that () f x is a periodic function of period 2 π that can be represented by 0 1 () ( c o s s i n ) nn n f xa an x bn x = =+ + Show that how to determine 0 a , n a and n b . Solution:
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Fourier Series, Integrals and Transforms Fourier Series (continued)
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Fourier Series, Integrals and Transforms Fourier Series 3. (’00) If a periodic function () fx with period 2 π is piecewise continuous in the interval x −≤≤ and has a left-hand and right-hand derivatives at each point of that interval, then the Fourier series of () is convergent. Prove it. Solution: We prove convergence for a continuous function () having continuous first and second derivatives. Integrate Fourier coefficient n a term by parts, we get () s i n 1 ' ()c o s s i n n n x a f x nxdx nn ππ −− == ∫∫ The first term on the right is zero. Another integration by parts gives 22 o s 1 " o n n x af x n x d x =− The first term on the right is zero because of the periodicity and continuity of ' f x . Since " f is continuous in the interval of integration, we have " () f xM < for an appropriate constant M . Furthermore, cos 1 nx . It follows that 2 11 2 " o n M x n x d x M d x n =< = Similarly, 2 2/ n bM n < for all n . Hence the absolute value of each term of the Fourier series of () f x is at most equal to the corresponding term of the series 00 21 23 aab aM M ⎛⎞ + + = + +++ + ⎜⎟ ⎝⎠ "" which is convergent. Hence that Fourier series converges and the proof is complete. [ Ref: Weierstrass M -test or see Chapter 14.5 Theorem 3 ] Weierstrass M -test: Suppose {} n f is a sequence of real- or complex-valued functions defined on a set A , and that there exist positive constants n M such that f for all 1 n and all x in A .
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Ch_05 - Fourier Series - Exam_Soln - Exam Problems of...

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