Chem Differential Eq HW Solutions Fall 2011 7

Chem Differential Eq HW Solutions Fall 2011 7 - g ( x ). By...

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Section 2.2 Fourier Series 7 s n_, x_ : Pi 2 4 Pi Sum 1 2k 1 ^2 Cos 2k 1 x , k, 0, n partialsums Table s n, x , n, 1, 7 ; f x_ x 2 Pi Floor x Pi 2 Pi g x_ Abs f x Plot g x , x, 3 Pi, 3 Pi Plot Evaluate g x , partialsums , x, 2 Pi, 2 Pi The function g(x) = | x | and its periodic extension Partial sums of the Fourier series. Since we are summing over the odd integers, when n = 7, we are actually summing the 15th partial sum. 9. Just some hints: (1) f is even, so all the b n ’s are zero. (2) a 0 = 1 π ± π 0 x 2 dx = π 2 3 . (3) Establish the identity ± x 2 cos( ax ) dx = 2 x cos( ax ) a 2 + ( - 2+ a 2 x 2 ) sin( ax ) a 3 + C ( a ± =0) , using integration by parts. 13. You can compute directly as we did in Example 1, or you can use the result of Example 1 as follows. Rename the function in Example 1
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Unformatted text preview: g ( x ). By comparing graphs, note that f ( x ) =-2 g ( x + ). Now using the Fourier series of g ( x ) from Example, we get f ( x ) =-2 n =1 sin n ( + x ) n = 2 n =1 (-1) n +1 n sin nx. 17. Setting x = in the Fourier series expansion in Exercise 9 and using the fact that the Fourier series converges for all x to f ( x ), we obtain 2 = f ( ) = 2 3 + 4 n =1 (-1) n n 2 cos n = 2 3 + 4 n =1 1 n 2 , where we have used cos n = (-1) n . Simplifying, we nd 2 6 = n =1 1 n 2 ....
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This note was uploaded on 12/22/2011 for the course MAP 3305 taught by Professor Stuartchalk during the Fall '11 term at UNF.

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