Orthogonality Relations We now explain the basic reason why the remarkable Fourier coefﬁcent formulas work. We begin by repeating them from the last note: 1 L f (t) dt, L −L � 1 L π f (t) cos(n t) dt, an = L −L L � 1 L π f (t) sin(n t) dt. bn = L −L L a0 = � (1) The key fact is the following collection of integral formulas for sines and cosines, which go by the name of orthogonality relations: ⎧ ⎪ 1 n = m �= 0 ⎨ � 1 L π π L − L cos( n L t ) cos( m L t ) dt = ⎪ 0 n � = m ⎩ 2 n=m=0 � 1 L π π L − L cos( n L t ) sin( m L t ) dt = 0 � � 1 n = m �= 0 1 L π π L − L sin( n L t ) sin( m L t ) dt = 0 n �= m Proof of the orthogonality relations: This is just a straightforward calcu lation using the periodicity of sine and cosine and either (or both) of these two methods: iat −iat iat −iat Method 1: use cos at = e +2e , and sin at = e −2ie . Method 2: use the trig identity cos(α) cos( β) = 21 (cos(α + β) + cos(α − β)), and the similar trig identies for cos(α) sin( β) and sin(α) sin( β). Using the orthogonality relations to prove the Fourier coefﬁcient formula Suppose we know that a periodic function f (t) has a Fourier series expan sion � π � � π � ∞ a0 f (t) = + ∑ an cos n t + bn sin n t (2) 2 L L n =1 How can we ﬁnd the values of the coefﬁcients? Let’s choose one coefﬁcient, say a2 , and compute it; you will easily how to generalize this to any other coefﬁcient. The claim is that the right-hand side of the Fourier coefﬁcient formula (1), namely the integral � � π � 1 L f (t) cos 2 t dt. L −L L OCW 18.03SC is in fact the coefﬁcent a2 in the series (2). We can �replace � f (t) in this integral by the series in (2) and multiply through by cos 2 πL t , to get 1 L � π � � π � � π � � π � � π �� ∞ � cos 2 t + ∑ an cos n t cos 2 t + bn sin n t cos 2 t dt L 2 L L L L n =1 � L a0 −L Now the orthogonality relations tell us that almost every term in this sum will integrate to 0. In fact, the only non-zero term is the n = 2 cosine term 1 L � L � π � � π � a2 cos 2 t cos 2 t dt L L −L and the orthogonality relations for the case n = m = 2 show this integral is equal to a2 as claimed. a0 Why the denominator of 2 in ? 2 Answer: it is in fact just a convention, but the one which allows us to have the same Fourier coefﬁcent formula for an when n = 0 and n ≥ 1. (Notice that in the n = m case for cosine, there is a factor of 2 only for n = m = 0.) a0 Interpretation of the constant term . 2 We can also interpret the constant term a20 in the Fourier series of f (t) as the �L 1 average of the function f (t) over one full period: a20 = 2L − L f ( t ) dt.