# 285hw9 - HW 91 1 Sec 9.1:21 We have f(t = t2 t 2 a0 = an =...

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a 0 = 1 & Z & & & t 2 dt = 2 & Z & 0 t 2 dt = 2 & ± t 3 3 j & 0 = 2 & 3 3 & = 2 & 2 3 : a n = 1 & Z & & & t 2 cos ntdt = 2 & Z & 0 t 2 cos ntdt = 2 &n Z & 0 t 2 d sin nt = 2 &n [ t 2 sin nt j & 0 & Z & 0 2 t sin ntdt ] = 4 &n 2 Z & 0 td cos nt = 4 &n 2 [ t cos nt j & 0 & Z & 0 cos ntdt ] = 4 &n 2 & & cos &n & 1 n sin nt j & 0 ± = 4 &n 2 & & ( & 1) n & 1 n (sin &n & sin 0) ± = 4 n 2 ( & 1) n : b n = 0 ; since t 2 is an even function. Thus f ( t ) = & 2 3 +4 1 X n =1 ( & 1) n n 2 cos nt = & 2 3 +4 ² & cos t + 1 4 cos 2 t & 1 9 cos 3 t + ::: ³ : 1 Note to the grader. Please grade the following problems: Sec. 9.1 :21, Sec. 9.1: 25, Sec. 9.2: 7, Sec. 9.2: 11, Sec. 9.2: 15 (20 pts for each problem). 1

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This is an even function. Hence b n = 0 . Now we calculate a n : a 0 = 1 1 Z 1 & 1 j t j dt = 2 Z 1 0 tdt = t 2 j 1 0 = 1 : a n = 1 1 Z 1 & 1 j t j cos n&tdt = 2 Z 1 0 t cos n&tdt = 2 n& Z 1 0 td sin n&t 2 1 Z 1 Thus f 0 0 .5 -8 -6 -4 -2 2 4 6 8 t 3

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b n = 0 since cos &t 2 is an even function. a 0 = 1 1 Z 1 & 1 cos &t 2 dt = 2 Z 1 0 cos &t 2 dt = 2 & 2 & sin &t 2 j 1 0 = 4 & & sin & 2 ± sin 0 ± = 4 & : a n = 1 1 Z 1 & 1 cos &t 2 cos n&tdt = 2 Z 1 0 1 2 ² cos & (2 n ± 1) t 2 + cos & (2 n + 1) t 2 ³ dt = Z 1 0 cos & (2 n ± 1) t 2 dt + Z 1 0 cos & (2 n + 1) t 2 dt = 2 & (2 n ± 1) sin & (2 n ± 1) t 2 j 1 0 + 2 & (2 n + 1) sin & (2 n + 1) t 2 j 1 0 =
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285hw9 - HW 91 1 Sec 9.1:21 We have f(t = t2 t 2 a0 = an =...

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