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Unformatted text preview: MAT 330: Fourier Series and Partial Differential Equations Professor Sergiu Klainerman Problem Set 2 Rik Sengupta rsengupt@princeton.edu February 13, 2010 1. Stein and Shakarchi, p. 58, problem 1 Suppose f is 2 periodic and integrable on any finite interval. Prove that if a,b R , then Z b a f ( x ) dx = Z b +2 a +2 f ( x ) dx = Z b 2 a 2 f ( x ) dx. Also prove that Z  f ( x + a ) dx = Z  f ( x ) dx = Z + a + a f ( x ) dx. Solution. The first part of this problem is trivial, obtained by a simple change of variables. Set x = x +2 , and then dx = dx , and when x = a , x = a +2 , and when x = b , x = b +2 . Furthermore, f is 2 periodic, and so by definition, f ( x ) = f ( x +2 ) = f ( x ). So, by changing the variable, the integral becomes Z b a f ( x ) dx = Z b +2 a +2 f ( x ) dx = Z b +2 a +2 f ( x ) dx. Similarly, set x 00 = x 2 , and then dx 00 = dx , and when x = a , x 00 = a 2 , and when x = b , x 00 = b 2 . Furthermore, f is 2 periodic, and so by definition, f ( x 00 ) = f ( x 2 ) = f ( x ). So, by changing the variable, the integral becomes Z b a f ( x ) dx = Z b 2 a 2 f ( x 00 ) dx 00 = Z b 2 a 2 f ( x ) dx. So, adding the chains of equality, we get Z b a f ( x ) dx = Z b +2 a +2 f ( x ) dx = Z b 2 a 2 f ( x ) dx, as required. For the second part of the problem, it is easy to show the equality between the first and third terms by a simple change of variables. Set x = x + a , then precisely as before, dx = dx , but when x = , x = + a , and when x = , x = + a . So, changing variables, the integral reduces to 1 Z f ( x + a ) dx = Z + a + a f ( x ) dx = Z + a + a f ( x ) dx. For the first equality in the second part, consider the number a . If we successively subtract 2 from it (or add, if a is negative), at some point we must arrive in the interval [0 , 2 ). Suppose the number we arrive at is a [0 , 2 ), with a = 2 k + a , for some k Z . We can always do this uniquely. Now, our integral is Z  f ( x + a ) dx = Z  f ( x + a + 2 k ) dx = Z  f ( x + a ) dx [because f is periodic in 2 .] = Z  a f ( x + a ) dx + Z  a f ( x + a ) dx = Z  a f ( x + a ) dx + Z  a f ( x + a 2 ) dx. Now, for the first half of this equation, set y = x + a , i.e. dx = dy , and notice that x = = y = + a , and x =  a = y = . For the second half, put z = x + a 2 , with dz = dx , and notice that x =  a = z = , and x = = z = + a . So now, we get Z  f ( x + a ) dx = Z  a f ( x + a ) dx...
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 Spring '10
 SERGIUKLAINERMANN
 Differential Equations, Equations, Partial Differential Equations, Fourier Series

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