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18.03 Topic 22: Sine and cosine series; calculation tricks. Read: EP §8.3 (up to subsection ’Fourier Series Solutions of DE’s’). Calculation shortcuts 1. Use even-odd as discussed before. 2. Make new series from old ones. 3. Use diﬀerentiation and integration. Even and odd functions Example 1: Used this last time to compute the Fourier series for the period 2π square wave and period 2π continuous sawtooth 4 1 1 π = f (t) = sin(πt) + sin(3πt) + sin(5πt) + · · · (square π 3 5 wave) cc c c c cc cc c cc cc c π 4 cos 3t cos 5t = g(t) = − cos t + + + · · · . (sawtooth or −2π 2π 2 π 32 52 triangular wave) New series from old ones –shifting and scaling 1 4 sin nt f (t) = square wave = π t = . 2π π n n odd 2 ˜ ˜ 4 sin nt f (t) = π t ⇒ f (t) = 1 + f (t) = 1 + . 2π π n n odd 2 ˜ ˜ 8 sin nt f (t) = π t ⇒ f (t) = 2f (t) = . 2π π n n odd −2 1 ˜ ˜ 1 1 2 sin nt f (t) = π t ⇒ f (t) = (1 + f (t)) = + . 2π 2 2 π n n odd shift and scale time: 1 4 sin nπt ˜ f (t) = t ˜ ⇒ f (t) = f (πt) = . 1 2 π n −1 n odd (continued) 1 18.03 topic 22 2 1 ˜ ˜ 4 sin n(t + π/2) f (t) = t ⇒ f (t) = f (t + π/2) = . −π/2 π/2 π n odd n −1 ˜ 4 sin(3t + 3π/2) 4 cos 3t ⇒ f (t) = sin(t + π/2) + + ... = cos t − + ... . π 3 π 3 Diﬀerentiation and integration Can diﬀerentiate term-by-term. Example: Let f (t) be the continuous sawtooth from example 1 it’s derivative is the square wave. π 4 cos 3t cos 5t f (t) = − cos t + + + ... 2 π 32 52 4 sin 3t sin 5t ⇒ f (t) = sin t + + + ... . π 3 5 Note f (t) has a corner and its coeﬃcients decay like 1/n2 . f (t) has a jump and and its coeﬃcients decay like 1/n. Can also integrate term-by-term. We will do examples of integration and diﬀerentiation of discontinuous functions next time. Even and odd extensions, sine and cosine series Suppose f (t) is deﬁned on the interval [0, L], we give a graph as an example. We will need functions like this when we study the wave and heat equations. R oo o o RRR RR f (t) = t L Emphatically: this is not periodic since it is only deﬁned on an interval. So, it doesn’t have a Fourier series. But, it does have periodic ’extensions. Periodic extension: period = L, agrees with f (t) on [0, L]. mU mU mU mm UU mm UU mm UU mm UU mm UU mm UU UU UU UU U U U t −L L 2L Even periodic extension: period = 2L, even, agrees with f (t) on [0, L]. mUU mUU mUU §§ mmmm UU §§ mmmm UU §§ mmmm UU §§§ UU §§ UU §§ UU UU §§ UU §§ UU ˜ §§§ §§ §§ fe (t) = t −L L 2L (continued) 18.03 topic 22 3 Odd periodic extension: period = 2L, odd, agrees with f (t) on [0, L]. mU mU mU mm UU mm UU mm UU mm UU mm UU mm UU UU UU UU U U U ˜ fo (t) = UU UU UU UU t UU UU UU UU UU −L U L U 2L UU UU mmm UU m UU m UU mmm U mmm U mmm mm m m mm ˜ ˜ A0 π fe is periodic and even ⇒ fe (t) = + An cos nt . 2 L ˜ ˜ π fo is periodic and odd ⇒ fo (t) = Bn sin nt . L ˜ ˜ On [0, L] they all agree: f (t) = fe (t) = fo (t). A0 π ˜ π ⇒ on [0, L], f (t) = + An cos nt . = fo (t) = Bn sin nt . 2 L L These are the Fourier cosine and Fourier sine series for f (t). (Emphatically, the series only agree with f (t) on [0, L].) Computing sine and cosine series Directly from the deﬁnition of Fourier coeﬃcients and the integration of even and odd functions we get: L L 2 π 2 π An = f (t) cos( nt) dt, Bn = f (t) sin( nt) dt L 0 L L 0 L Important: 1. Sine and cosine series are about functions deﬁned on an interval. 2. The sine and cosine series agree with f on (0, L) (assuming f is continuous). Since f is only deﬁned on [0, L] this is usually what we want. ˜ ˜ 3. Repeating (2): fe (t) = fo (t) for 0 < t < L. 4. Computing An and Bn only depends on f . 5. We will make use of sine and cosine series when we do the wave equation. Example: (Computing sine and cosine series.) Let f (t) = sin t on [0, π]. Find its Fourier sine and cosine series. Sine series: f (t) = sin t on [0, π]. 4 π for n = 0 2 π Cosine series: L = π, An = sin t cos nt dt = 0 for n = odd π 0 −4 π(n2 −1) for even n > 0 2 4 cos 2t cos 4t cos 6t 2 4 cos nt ⇒ sin(t) = − + + + ... = − . π π 3 15 35 π π n>0, even n2 − 1 (Only on [0, π].)

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posted: | 5/30/2010 |

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