330 HW 02

# 330 HW 02 - MAT 330 Fourier Series and Partial Differential...

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Unformatted text preview: MAT 330: Fourier Series and Partial Differential Equations Professor Sergiu Klainerman Problem Set 2 Rik Sengupta [email protected] February 13, 2010 1. Stein and Shakarchi, p. 58, problem 1 Suppose f is 2 π-periodic and integrable on any finite interval. Prove that if a,b ∈ R , then Z b a f ( x ) dx = Z b +2 π a +2 π f ( x ) dx = Z b- 2 π a- 2 π f ( x ) dx. Also prove that Z π- π f ( x + a ) dx = Z π- π f ( x ) dx = Z π + a- π + a f ( x ) dx. Solution. The first part of this problem is trivial, obtained by a simple change of variables. Set x = x +2 π , and then dx = dx , and when x = a , x = a +2 π , and when x = b , x = b +2 π . Furthermore, f is 2 π-periodic, and so by definition, f ( x ) = f ( x +2 π ) = f ( x ). So, by changing the variable, the integral becomes Z b a f ( x ) dx = Z b +2 π a +2 π f ( x ) dx = Z b +2 π a +2 π f ( x ) dx. Similarly, set x 00 = x- 2 π , and then dx 00 = dx , and when x = a , x 00 = a- 2 π , and when x = b , x 00 = b- 2 π . Furthermore, f is 2 π-periodic, and so by definition, f ( x 00 ) = f ( x- 2 π ) = f ( x ). So, by changing the variable, the integral becomes Z b a f ( x ) dx = Z b- 2 π a- 2 π f ( x 00 ) dx 00 = Z b- 2 π a- 2 π f ( x ) dx. So, adding the chains of equality, we get Z b a f ( x ) dx = Z b +2 π a +2 π f ( x ) dx = Z b- 2 π a- 2 π 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 =- π , x =- π + a , and when x = π , x = π + a . So, changing variables, the integral reduces to 1 Z π π f ( x + a ) dx = Z π + a- π + a f ( x ) dx = Z π + a- π + a f ( x ) dx. For the first equality in the second part, consider the number a . If we successively subtract 2 π from it (or add, if a is negative), at some point we must arrive in the interval [0 , 2 π ). Suppose the number we arrive at is a ∈ [0 , 2 π ), with a = 2 kπ + a , for some k ∈ Z . We can always do this uniquely. Now, our integral is Z π- π f ( x + a ) dx = Z π- π f ( x + a + 2 kπ ) dx = Z π- π f ( x + a ) dx [because f is periodic in 2 π .] = Z π- a- π f ( x + a ) dx + Z π π- a f ( x + a ) dx = Z π- a- π f ( x + a ) dx + Z π π- a f ( x + a- 2 π ) dx. Now, for the first half of this equation, set y = x + a , i.e. dx = dy , and notice that x =- π = ⇒ y =- π + a , and x = π- a = ⇒ y = π . For the second half, put z = x + a- 2 π , with dz = dx , and notice that x = π- a = ⇒ z =- π , and x = π = ⇒ z =- π + a . So now, we get Z π- π f ( x + a ) dx = Z π- a- π f ( x + a ) dx...
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## This note was uploaded on 05/07/2010 for the course MAT 330 taught by Professor Sergiuklainermann during the Spring '10 term at Princeton.

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330 HW 02 - MAT 330 Fourier Series and Partial Differential...

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