Ch_05 - Fourier Series - Exam_Soln

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

This preview shows pages 1–6. Sign up to view the full content.

Exam Problems of Fourier Series, Integrals and Transforms

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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 = ±± " )
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:

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Fourier Series, Integrals and Transforms Fourier Series (continued)
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 .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 04/27/2011 for the course CHEM 101 taught by Professor Suh during the Spring '11 term at University of Toronto.

Page1 / 40

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

This preview shows document pages 1 - 6. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online