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X
f.x/ D
c k sin.kx/
for x 2 .0; 1/:
(8.70)
k D 1
If this is the case, then we can use the methodology developed above to solve our
problem. Fourier realized this and it eventually made him find what today is referred
to as Fourier analysis.
Let us have a closer look at (8.70). It is important to note that, in this equation, f
is a given function and the constants c 1 ;c 2 ;c 3 ;::: are the unknowns that we want
to determine. Thus, this is one equation with infinitely many unknowns! How can
we solve this problem? Is it possible to solve? We will now see that a basic property
of the sine function leads to an amazingly simple solution of this problem.
Let k and l be two positive integers and consider the integral
Z 1
sin.kx/ sin.lx/ dx :
0
Recall the trigonometric identity that, for any real numbers a and b,
1
2 .cos.a b/ cos.a C b// :
sin.a/ sin.b/ D
(8.71)
By applying this identity, the reader can verify in a straightforward manner that
Z 1
sin.kx/ sin.lx/ dx D 0k ยค l;
1=2 k D l;
(8.72)
0
see Exercise 8.5 . How can this property be exploited in the present situation? Ideally,
we would like to have an explicit formula for the constants c 1 ;c 2 ;c 3 ;::: in (8.70).
It turns out that this can easily be accomplished by applying (8.72). If we multiply
the left- and right-hand sides of (8.70)bysin.lx/ and integrate, we find that
X
c k sin.kx/ ! sin.lx/ dx :
Z 1
f.x/sin.lx/ dx D Z 1
0
0
k D 1
Let us assume that we can interchange the order of integration and summation, 23
Z 1
Z 1
X
f.x/sin.lx/ dx D
c k
sin.kx/ sin.lx/ dx ;
0
0
k D 1
1
2 c l ;
D
23 This depends on the convergence properties of the involved series. The topic is beyond the scope
of this topic.
 
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