b Compute the CFS for both f and f c What is the value of the CFS for f at x 1

B compute the cfs for both f and f c what is the

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(b) Compute the CFS for both f and f 0 . (c) What is the value of the CFS for f 0 at x = 1? (d) Verify that the CFS for f 0 can be obtained from the CFS for f by term-by- term differentiation (this is a general property, provided f 0 satisfies the Dirichlet conditions). (e) Note that the terms of the CFS for f have O (1 /n 2 ) while the terms of the CFS for f 0 have O (1 /n ). (It is important in this game to know roughly how fast terms decrease for | n | large. See RHS page 132 for a brief discussion of “Big Oh”.) Which series has faster convergence of its partial sums? Which exhibits a Gibbs phenomenon? (f) A general statement is that if f satisfies the Dirichlet conditions of RHB 12.1 and has no jumps, then its CFS has O (1 /n 2 ). On the other hand, if it satisfies the Dirichlet conditions and has at least one jump, then its CFS has O (1 /n ). 1
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3. The squared deviation between two functions f, g on [ - L, L ] is defined to be || f - g || 2 = 1 2 L Z L - L | f ( x ) - g ( x ) | 2 dx Think of this as the square of the “distance” between f and g . (a) Use the Parseval Theorem to show that if f and g (both real-valued functions on [ - L, L ]) have CFS with coefficients c n , d n respectively, then || f - g || 2 = | c 0 - d 0 | + 2 X n =1 | c n - d n | 2 (b) Suppose now that f is approximated by the truncated CFS g = M n = - M c n e πinx/L for some positive M . Show that the squared deviation error of this approxi- mation is given by || f - g || 2 = 2 X n = M +1 | c n | 2 4. Solve RHS problem 13.2. Comment about part (a): note any periodic function has R -∞ | f ( t ) | dt = (can you see why?). BONUS PROBLEM: show that if ka = 2 πn for some n Z then either | ˜ f ( k ) | = 0 or | ˜ f ( k ) | = . 2
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