# doc_ - .3#8 Asks if the function f(x = 1 x x3 x5 is even...

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Homework 01 11.3#8 Asks if the function f ( x ) = 1 - x + x 3 - x 5 is even odd or neither. The answer is neither. The 1 = x 0 ruins the oddnesss and - x + x 3 - x 5 ruins the evenness. Another way is to test a sample value say x = 1; f (1) = 1 - 1 + 1 - 1 = 0 and f ( - 1) = 1 + 1 - 1 + 1 = 2 and 0 = ± 2 11.1.#23 Asks for the Fourier series of the 2 π -periodic function f ( x ) = x 2 - π 2 < x < π 2 π 2 4 π 2 < x < 3 π 2 First lets see what function looks like on - π < x < π , the piece from - π to π / 2 is the “same” as the piece from π to 3 π / 2. We have f ( x ) = π 2 4 - π < x < - π 2 x 2 - π 2 < x < π 2 π 2 4 π 2 < x < π At this point it should be clear that the function is even. So all the b n = 0. Now for the a n . First n = 0. a 0 = 2 2 π π 0 f ( x ) dx = 1 π π 0 f ( x ) dx = 1 π π / 2 0 f ( x ) dx + π π / 2 f ( x ) dx = 1 π π / 2 0 x 2 dx + π π / 2 π 2 4 dx = 1 π x 3 3 π / 2 0 + π 2 4 π 2 = 1 π π 3 24 + π 3 8 = π 2 6 Next n > 0. We will use x 2 cos nx dx = x 2 sin nx n + 2 x cos nx n 2 - 2 sin nx n 3 + C which we will obtain from standard integration by parts trick: u

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Unformatted text preview: nx n-sin nx n 3 2 x cos nx n 2-2 sin nx n 3 a n = 2 π ± π f ( x ) cos nx dx = 2 π ² ± π/ 2 f ( x ) cos nx dx + ± π π/ 2 f ( x ) cos nx dx ³ = 2 π ² ± π/ 2 x 2 cos nx dx + ± π π/ 2 π 2 4 cos nx dx ³ = 2 π ² ´ x 2 sin nx n + 2 x cos nx n 2-2 sin nx n 3 µ ¶ ¶ ¶ ¶ π/ 2 + π 2 4 sin nx n ¶ ¶ ¶ ¶ π π/ 2 ³ = 2 π ´´ π 2 4 n sin nπ 2 + π n 2 cos nπ 2-2 n 3 sin nπ 2 µ-(0 + 0-0) + π 2 4 ´-1 n sin nπ 2 µ = 2 n 2 cos nπ 2-4 πn 3 sin nπ 2 = (-1) n/ 2 2 n 2 n even (-1) ( n +1) / 2 4 πn 3 n odd f ( x ) = π 6-4 π 1 3 cos x + 2 2 2 cos 2 x + 4 π 3 3 cos 3 x-2 4 2 cos 4 x-4 π 5 3 cos 5 x + 2 6 2 cos 6 x + . . ....
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