322 the integral in this formula is said to be

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Unformatted text preview: nt]|π 2 π/ 2n n −π sin(nπ/2) cos(nπ/2) 1 + = − 2 cos(nπ ). 2 2n n n = Thus when n ≥ 1, an = 2 [2 cos(nπ/2) − 1 − 1(−1)n ], πn2 and the Fourier sine series of f (t) is f (t) = π 2 2 2 − cos 2t − cos 6t − cos 10t − . . . . 4 π 9π 25π (3.17) Note that we have expanded exactly the same function f (t) on the interval [0, π ] as either a superposition of sines in (3.16) or as a superposition of cosines in (3.17). Exercises: 3.3.1.a. Find the Fourier sine series of the following function defined on the interval [0, π ]: 4t, for 0 ≤ t < π/2, f (t) = 4π − 4t, for π/2 ≤ t ≤ π . b. Find the Fourier cosine series of the same function. 3.3.2.a. Find the Fourier sine series of the following function defined on the interval [0, π ]: f (t) = t(π − t). b. Find the Fourier cosine series of the same function. 3.3.3. Find the Fourier sine series of the following function defined on the interval [0, 10]: t, for 0 ≤ t < 5, f (t) = 10 − t, for 5 ≤ t ≤...
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