ABSTRACT
At the end of the previous chapter, we were given an arbitrary function f(x) defined on 0 ≤ x ≤ L, and we wanted to know if there is a linear combination (possibly with infinitely many terms) of the functions sin nπx
L that is identical to f(x) on 0 ≤ x ≤ L. That is, can we find coefficients cn, n = 1, 2, 3, . . ., such that
f(x) =
cn sin npix
L (3.1)
for 0 ≤ x ≤ L? Fourier’s affirmative answer to this question in 1807, while not quite correct,
marks a pivotal moment in the history of mathematics. Indeed, this surprising and profound result not only disturbed Fourier’s mathematical contemporaries but, ultimately, rocked the very foundations of mathematical analysis. What we would like to do is to assume that (3.1) is true and actually find
the values of the coefficients cn for which it is, in fact, true. Let’s do this and worry about serious mathematical issues afterwards. (We say that we proceed formally.) To this end, we will fix positive integer N , multiply both sides of (3.1) by sin Nπx
L and then integrate both sides from x = 0 to x = L, resulting in ∫ L
f(x) sin Npix
L dx =
cn
sin npix
L sin
Npix
L dx. (3.2)
Then, it is easy enough to calculate the integrals in (3.2). It turns out that
sin npix
L sin
Npix
L dx =
{ 0, if n = N , L 2 , if n = N ,
(3.3)
so (3.2) becomes
f(x) sin Npix
L dx = c1 · 0 = c2 · 0 + · · ·+ cN−1 · 0 + cN · L
2 + cN+1 · 0 + · · ·
or
cN = 2
∫ L
f(x) sin Npix
dx. (3.4)
n true. have we? Many readers will notice that we assumed the truth of (3.1) and then cal-
culated the values that the coefficients must take on. This, of course, says nothing about the truth of (3.1). Further, the series in (3.1) is an infinite series of functions. As we know, an infinite series of constants may not even converge; an infinite series of functions may, then, converge for some values of x and diverge for others. Finally, there is a more subtle problem with the above argument. We
actually skipped a step in going from (3.1) to (3.2)—we assumed that we could integrate the right side term-by-term, i.e., that
( ∞∑ n=1
cn sin npix
L sin
Npix
L
) dx =
cn
sin npix
L sin
Npix
L dx.
