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other. Indeed, under these conditions, f g is odd and hence
f, g = 1
π π f (t)g (t)dt =
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 π −(f (t)g (t))(−dt) +
0 π f (t)g (t)dt
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
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 =
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 .)
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This document was uploaded on 01/12/2014.
- Winter '14