HW 8 solution - EGM 4313 S08 Homework Solution #8 EGM 4313...

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EGM 4313 S08 Homework Solution #8 1/17 EGM 4313 S08 - Homework#8 1. Problem Set 11.1, p.485, #21 Showing the details of your work, find the Fourier series of the given f(x), which is assumed to have the period of 2 π . Sketch or graph the partial sums up to that including cos5x and sin5x. 2 () , fx x x π = −<< Solution: From (6) on page 480, we calculate coefficients in Fourier series as 32 2 0 11 22 3 3 x ax d x ππ == = 2 2 2 2 2 1 cos 1 sin 2 sin 1 sin 2 cos 2cos 4 1 n n a x nxdx xn x x n x dx nn x nx x nx nx dx n n = ⎡⎤ ⎢⎥ =− ⎣⎦ =+ 2 1 sin 0 n b x nxdx (because x 2 is a even function, b n has to be zero) Therefore, 2 2 1 4 1 c o s 3 n n f x n = The following plot is generated based on f(x) with items up to n=5. f1=@(x) x^2; f2=@(x) pi^2/3 4*cos(x)+cos(2*x) 4/9*cos(3*x)+1/4*cos(4*x) 4/25*cos(5*x); figure; hold; fplot(f1,[ pi,pi]) fplot(f2,[ pi,pi])
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EGM 4313 S08 Homework Solution #8 2/17 -3 -2 -1 0 1 2 3 -2 0 2 4 6 8 10 x 2 Fourier series
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EGM 4313 S08 Homework Solution #8 3/17 2. Problem Set 11.2, p.490, #17 Using prob3, show that 2 11 1 1 1 ... 4 9 16 12 π −+− += Solution: In problem 3, 2 () , (1 1 ) , 2 , 1 fx x x p L =− < < == Using the formula (6) on page 487 and part of the results in previous problem in this homework, the coefficients for Fourier series are 32 2 0 11 1 22 3 3 3 L L L L xL ax d x LL = = 2 2 2 2 1 cos sin 2 sin 1 sin 2 cos 2 cos 1 44 L n L L L L L L L nn nx d x dx Ln n L L dx n n L ππ −− = ⎡⎤ ⎢⎥ ⎣⎦ =+ 2 1 sin 0 L n L bx d x Hence the Fourier series for 2 , 1 ) , 2 x p < < = , is 2 1 14 1 c o s 3 n n nL = + when x=0 is taken in above Fourier series, we obtain 1 01 3 n n n = hence
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EGM 4313 S08 Homework Solution #8 4/17 () 2 22 1 1 11 1 1 1 1 ... 12 4 9 16 n n n nn π = =− = − + − + −
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EGM 4313 S08 Homework Solution #8 5/17 3. Problem Set 11.3, p.496, #15 Is given function even or odd? Find its Fourier series. Sketch or graph the function and some partial sums. (Show the details of your work.) if 0 () if 0 x x x e fx x e π << = Solution: The given function is an even function. Hence b n are zeros in the resulting Fourier series. Use (2) of Theorem 1 on page 491, we obtain the coefficients 0 0 0 1 1 xx ae d x e e −− == = 00 2 cos 2 cos n a e nxdx e nxdx ππ ∫∫ Because 0 2 0 2 0 cos cos sin cos sin cos 11 c o s x x n x e nxdx e nx e n nxdx en x e n n x e n n x d x e n x d x ⎡⎤ =− + ⎢⎥ ⎣⎦ + =− − − + we obtain 2 0 cos 1 n x e x d x n = + , hence 2 0 2 2c o s 1 1 1 n x n a e nxdx e n + .
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HW 8 solution - EGM 4313 S08 Homework Solution #8 EGM 4313...

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