Fourier Series and Partial Differential Equations Lecture Notes
Fourier Series and Partial Differential Equations Lecture Notes
Fourier Series and Partial Differential Equations Lecture Notes
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Chapter 2. <strong>Fourier</strong> series 18<br />
Theorem 2.2 (Convergence theorem) Let f be a periodic function with period 2π, with<br />
f <strong>and</strong> f ′ piecewise continuous on (−π,π). Then the <strong>Fourier</strong> series of f at x converges to<br />
the value 1<br />
2 [f(x+)+f(x−)], i.e.<br />
1<br />
2 [f(x+)+f(x−))] = 1<br />
2 a0 +<br />
∞<br />
[ancos(nx)+bnsin(nx)]. (2.53)<br />
Note that if f is continuous at x, then f(x+) = f(x−) = f(x) so the <strong>Fourier</strong> series<br />
converges to f(x).<br />
n=0<br />
Note that if a function is defined on an interval of length 2π, we can find the <strong>Fourier</strong><br />
series of its periodic extension <strong>and</strong> equation (2.53) will then hold on the original interval.<br />
But we have to be careful at the end points of the interval: e.g. if f is defined on (−π,π]<br />
then at ±π the <strong>Fourier</strong> <strong>Series</strong> of f converges to 1<br />
2 [f(π−)+f((−π)+)].<br />
−π<br />
y<br />
f(π−)<br />
f(π+)<br />
For Example 2.2 we have, by Theorem 2.2, that<br />
1<br />
2 [f(x+)+f(x−)] = 4<br />
π<br />
∞<br />
m=0<br />
π<br />
x<br />
1<br />
sin[(2m+1)x], (2.54)<br />
2m+1<br />
where both sides reduce to zero at x = 0, ±π. At x = π/2 we obtain<br />
<strong>and</strong> hence<br />
1 = 4<br />
π<br />
∞ 1<br />
2m+1 sin<br />
<br />
(2m+1)π<br />
2<br />
m=0<br />
−π<br />
π<br />
4 =<br />
∞<br />
m=0<br />
y<br />
1<br />
= 4<br />
π<br />
∞<br />
m=0<br />
(−1) m<br />
, (2.55)<br />
2m+1<br />
(−1) m<br />
. (2.56)<br />
2m+1<br />
−1<br />
π<br />
2<br />
π<br />
x