Chem Differential Eq HW Solutions Fall 2011 4

# Chem Differential Eq HW Solutions Fall 2011 4 - 4 Chapter 2...

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4 Chapter 2 Fourier Series Solutions to Exercises 2.1 1. (a) cos x has period 2 π . (b) cos πx has period T = 2 π π = 2. (c) cos 2 3 x has period T = 2 π 2 / 3 = 3 π . (d) cos x has period 2 π , cos 2 x has period π , 2 π , 3 π , ˙ .. A common period of cos x and cos 2 x is 2 π . So cos x + cos 2 x has period 2 π . 5. This is the special case p = π of Exercise 6(b). 9. (a) Suppose that f and g are T -periodic. Then f ( x + T ) · g ( x + T ) = f ( x ) · g ( x ), and so f · g is T periodic. Similarly, f ( x + T ) g ( x + T ) = f ( x ) g ( x ) , and so f/g is T periodic. (b) Suppose that f is T -periodic and let h ( x ) = f ( x/a ). Then h ( x + aT ) = f x + aT a = f x a + T = f x a (because f is T -periodic) = h ( x ) . Thus h has period aT . Replacing a by 1 /a , we find that the function f ( ax ) has period T/a . (c) Suppose that f is T -periodic. Then g ( f ( x + T )) = g ( f ( x )), and so g ( f ( x )) is also T -periodic. 13. π/ 2 - π/ 2 f ( x ) dx = π/ 2 0 1 dx = π/ 2 . 17. By Exercise 16, F is 2 periodic, because 2 0 f ( t ) dt = 0 (this is clear from the graph of f ). So it is enough to describe F on any interval of length 2. For 0 <x< 2, we have F ( x ) = x 0 (1 - t ) dt = t - t 2 2 x 0 = x - x 2 2 . For all other x , F ( x +2) = F ( x ). (b) The graph of F over the interval [0
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