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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

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