finpractsol - Solutions for practice problems for the...

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Unformatted text preview: 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 ≤ , 2 , < x ≤ 2 . We have f ( x ) = a 2 + ∞ X n =1 µ a n cos πnx 2 + b n sin πnx 2 ¶ , where a = 1 2 µZ- 2 (- 1) dx + Z 2 2 dx ¶ = 1 , a n = 1 2 µZ- 2 (- 1)cos πnx 2 dx + Z 2 2cos πnx 2 dx ¶ = 1 2 ˆ (- 1) • 2 πn sin πnx 2 ‚- 2 + 2 • 2 πn sin πnx 2 ‚ 2 ! = 0 , n > , and b n = 1 2 µZ- 2 (- 1)sin πnx 2 dx + Z 2 2sin πnx 2 dx ¶ = 1 2 ˆ- (- 1) • 2 πn cos πnx 2 ‚- 2- 2 • 2 πn cos πnx 2 ‚ 2 ! = 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 are zero is to note that f ( x ) = 1 2 + g ( x ) , 1 where g ( x ) is an odd function, g ( x ) = ( 3 / 2 , x > ,- 3 / 2 , x < . 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 µ 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 2 + ∞ X n =1 µ a n cos πnx 4 + b n sin πnx 4 ¶ , we can see that a = 10, a n = b n = 0 for n > 0 will give f ( x ) = g ( x ). (b) x ∈ [- π,π ], and f ( x ) = 21 + 2sin5 x + 8cos2 x. Again, for L = π , we have g ( x ) = a 2 + ∞ X n =1 ( a n cos nx + b n sin nx ) , and setting a = 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) + 7cos( πnx/ 3)) , with c n = (- 1) n . We have g ( x ) = a 2 + ∞ X n =1 µ a n cos πnx 3 + b n sin πnx 3 ¶ , so we set a =- 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....
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This note was uploaded on 01/12/2010 for the course MATH 421 taught by Professor Nataliakomarova during the Winter '07 term at San Jose State University .

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finpractsol - Solutions for practice problems for the...

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