MATH 129 UC Davis

MAT 129 Midterm Practice Exam The actual midterm exam will be conducted in class on Wednesday Nov. 8, 2006 Name: Student ID #: Read each problem carefully. Write every step of your reasoning clearly. Usually, a better strategy is to solve the easi

Score of this page:_ Problem 1 (20 pts) Consider a function f () = e on [0, ). (a) (10 pts) Expand f into its Fourier cosine series. Answer: Using the formula for the Fourier cosine coefcients over [0, ], we have: an = = = = = = 2 2 1 0 0 1 e

Score of this page:_ Problem 1 (20 pts) Consider the following periodic function with period 2 dened as f (x) = 0 x 1 for 1 x 0; for 0 x 1 (a) (10 pts) Expand f into its Fourier series. Answer: Because the interval is of length 2, the Fourier

MAT 129: Fourier Analysis: Supplementary Notes I by Naoki Saito A Brief History of the Convergence of the Fourier Series Theorem 1 (Dirichlet, 1829) Suppose f is 1-periodic, piecewise smooth on R. Then, nth partial sum, Sn [f ](x) = n ck e2ikx , sat

MAT 129: Fourier Analysis Supplementary Notes II by Naoki Saito The Fourier Inversion Theorem The Fourier transform F was dened initially on L1 (R), a space of integrable functions, and F : L1 (R) BC(R) = C(R) L (R). However, f , the Fourier tran

= 0 (otherwise, the solution becomes the trivial solution), we must have sin = 0, i.e., = n where n Z. Thus

MAT 129: Fourier Analysis Supplementary Notes II by Naoki Saito The Fourier Inversion Theorem The Fourier transform F was dened initially on L1 (R), a space of integrable functions, and F : L1 (R) BC(R) = C(R) L (R). However, f , the Fourier tran

Score of this page:_ Problem 1 (20 pts) Consider a function f () = e on [0, ). (a) (10 pts) Expand f into its Fourier cosine series. Answer: Using the formula for the Fourier cosine coefcients over [0, ], we have: an = = = = = = 2 2 1 0 0 1 e

MAT 129: Fourier Analysis: Supplementary Notes I by Naoki Saito A Brief History of the Convergence of the Fourier Series Theorem 1 (Dirichlet, 1829) Suppose f is 1-periodic, piecewise smooth on R. Then, nth partial sum, Sn [f ](x) = n ck e2ikx , sat