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finpractsol

# finpractsol - Solutions for practice problems for the Final...

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Solutions for practice problems for the Final, part 3 Note: Practice problems for the Final Exam, part 1 and part 2 are the same as Practice problems for Midterm 1 and Midterm 2. 1. Calculate Fourier Series for the function f ( x ), defined on [ - 2 , 2], where f ( x ) = ( - 1 , - 2 x 0 , 2 , 0 < x 2 . We have f ( x ) = a 0 2 + X n =1 a n cos πnx 2 + b n sin πnx 2 , where a 0 = 1 2 Z 0 - 2 ( - 1) dx + Z 2 0 2 dx = 1 , a n = 1 2 Z 0 - 2 ( - 1) cos πnx 2 dx + Z 2 0 2 cos πnx 2 dx = 1 2 ˆ ( - 1) 2 πn sin πnx 2 0 - 2 + 2 2 πn sin πnx 2 2 0 ! = 0 , n > 0 , and b n = 1 2 Z 0 - 2 ( - 1) sin πnx 2 dx + Z 2 0 2 sin πnx 2 dx = 1 2 ˆ - ( - 1) 2 πn cos πnx 2 0 - 2 - 2 2 πn cos πnx 2 2 0 ! = 1 πn (1 - cos πn ) - 2 1 πn (cos πn - 1) = 3 πn (1 - ( - 1) n ) . Therefore, we have f ( x ) = 1 2 + X n =1 3 πn (1 - ( - 1) n ) sin πnx 2 . An easy way to see that all of a n except a 0 are zero is to note that f ( x ) = 1 2 + g ( x ) , 1

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where g ( x ) is an odd function, g ( x ) = ( 3 / 2 , x > 0 , - 3 / 2 , x < 0 . 2. Calculate Fourier Series for the function f ( x ), defined on [ - 5 , 5], where f ( x ) = 3 H ( x - 2) . By a similar method, f ( x ) = 9 5 + X n =1 - 3 πn sin 2 πn 5 cos πnx 5 + 3 πn cos 2 πn 5 - ( - 1) n sin πnx 5 . 3. Calculate Fourier Series for the function, f ( x ), defined as follows: (a) x [ - 4 , 4], and f ( x ) = 5 . Comparing f ( x ) with the general Fourier Series expression with L = 4, g ( x ) = a 0 2 + X n =1 a n cos πnx 4 + b n sin πnx 4 , we can see that a 0 = 10, a n = b n = 0 for n > 0 will give f ( x ) = g ( x ). (b) x [ - π, π ], and f ( x ) = 21 + 2 sin 5 x + 8 cos 2 x. Again, for L = π , we have g ( x ) = a 0 2 + X n =1 ( a n cos nx + b n sin nx ) , and setting a 0 = 42, a 2 = 8, b 5 = 2 and the rest of the coefficients zero, we obtain f ( x ) = g ( x ). (c) x [ - π, π ], and f ( x ) = 8 X n =1 c n sin nx, with c n = 1 /n. 2
Similarly, we set b n = 1 /n for 1 n 8, and the rest of the coefficients zero. (d) x [ - 3 , 3], and f ( x ) = - 4 + 6 X n =1 c n (sin( πnx/ 3) + 7 cos( πnx/ 3)) , with c n = ( - 1) n . We have g ( x ) = a 0 2 + X n =1 a n cos πnx 3 + b n sin πnx 3 , so we set a 0 = - 8, a n = 7( - 1) n for 1 n 6 and b n = ( - 1) n for 1 n 6, and the rest of the coefficients zero. 4. (a) Let f ( x ) = x + x 3 for x [0 , π ]. What coefficients of the Fourier Series of f are zero? Which ones are non-zero? Why? f ( x ) is an odd function. Indeed, f ( - x ) = - x + ( - x ) 3 = - x - x 3 = - ( x + x 3 ) = - f ( x ) , therefore a n = 0, and b n can be nonzero. (b) Let g ( x ) = cos( x 5 )+sin( x 2 ). What coefficients of the Fourier Series of g are zero? Which ones are non-zero? Why? g ( x ) is an even function. Indeed, g ( - x ) = cos(( - x ) 5 )+sin(( - x ) 2 ) = cos( - x 5 )+sin( x 2 ) = cos( x 5 )+sin( x 2 ) = g ( x ) .

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finpractsol - Solutions for practice problems for the Final...

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