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Fourier Series and Partial Differential Equations Lecture Notes

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Chapter 2. <strong>Fourier</strong> series 23<br />

Note. For fo to really be odd we must have fo(0) = 0 <strong>and</strong> also fo(L) = −fo(−L) =<br />

−fo(L) (thelast equality follows fromperiodicity)givingfo(L) = 0<strong>and</strong>thereforefo(nL) =<br />

0 for all n ∈ Z. However, the value at of f at these isolated points does not affect the<br />

<strong>Fourier</strong> series.<br />

Example 2.4 Find the <strong>Fourier</strong> sine <strong>and</strong> cosine expansions of f(x) = x for x ∈ [0,L].<br />

Sine expansion The odd extension is defined by<br />

In this case<br />

fo(x) =<br />

0<br />

x x ∈ [0,L],<br />

−(−x) x ∈ (−L,0).<br />

−L<br />

y<br />

2L<br />

x<br />

(2.83)<br />

bn = 2<br />

L <br />

nπx<br />

<br />

xsin dx = (−1)<br />

L L<br />

n+12L<br />

, (2.84)<br />

nπ<br />

as in Example 2.3. For x ∈ [0,L) we therefore obtain<br />

x =<br />

∞<br />

(−1) n+12L<br />

nπ sin<br />

<br />

nπx<br />

<br />

. (2.85)<br />

L<br />

n=1<br />

Cosine expansion The even extension is given by<br />

Now,<br />

<strong>and</strong><br />

Thus for x ∈ [0,L] we get<br />

fe(x) =<br />

x x ∈ [0,L],<br />

−L<br />

a0 = 2<br />

L<br />

−x x ∈ [−L,0).<br />

y<br />

2L<br />

x<br />

(2.86)<br />

L<br />

xdx = L, (2.87)<br />

0<br />

an = 2<br />

⎧<br />

L <br />

nπx<br />

⎨0<br />

n = 2m even,<br />

xcos dx =<br />

L 0 L ⎩ −4L<br />

(2m+1) 2π2 n = 2m+1 odd.<br />

x = L<br />

2 +<br />

m=0<br />

(2.88)<br />

∞ 4L<br />

−<br />

(2m+1) 2 <br />

(2m+1)πx<br />

cos . (2.89)<br />

π2 L

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