312 therefore just as in the previous section we can

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Unformatted text preview: ch other. Indeed, under these conditions, f g is odd and hence f, g = 1 π π f (t)g (t)dt = −π 1 π 0 f (t)g (t)dt + −π 70 1 π π f (t)g (t)dt 0 = 1 π =− 0 f (−t)g (−t)(−dt) + π 1 π 1 π π −(f (t)g (t))(−dt) + 0 π f (t)g (t)dt 0 1 π π f (t)g (t)dt = 0. 0 The variable of integration has been changed from t to −t in the first integral of the second line. It follows that if f ∈ Wodd , a0 = f , 1 = 0, an = f , cos nx = 0, for n > 0. Similarly, if f ∈ Weven , bn = f , sin nx = 0. Thus for an even or odd function, half of the Fourier coefficients are automatically zero. This simple fact can often simplify the calculation of Fourier coefficients. Exercises: 3.2.1. Evaluate the inner product f, g = π 1 π f (t)g (t)dt, −π in the case where f (t) = cos t and g (t) = | sin t|. 3.2.2. Evaluate the inner product eπ f, g = 1 1 f (x)g (x)dx, x in the case where f (x) = cos(log(x)) and g (x) = 1, where log denotes the natural or base e logarithm. (Hint: Use the substitution x = eu .) 3.2.3. Det...
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This document was uploaded on 01/12/2014.

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