Fourier-x

Fourier-x - Fourier Series of f (x) = x Given a real...

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Unformatted text preview: Fourier Series of f (x) = x Given a real periodic function f (x) , −π < x < π , one can find its Fourier series in two (equivalent) ways: using trigonometric functions: ∞ cos kx sin kx a0 f (x) = √ + ∑ ak √ + bk √ π π 2π k=1 or using the complex exponential ∞ f (x) = eikx ck √ . 2π k=−∞ ∑ Note that if f (x) is a real-valued function, we can take the real part of the complex exponential version to get the trigonometric version (caution: the coefficiants ck will probably be complex numbers). Here we will use complex exponentials. The Fourier coefficients are Zπ Z Zπ 1 1 −2i π −ikx xe dx = √ x sin kx dx ck = √ x[cos kx − i sin kx] dx = √ 2π −π 2π −π 2π 0 But Zπ Z −x cos kx π 1 π −π cos kπ π x sin kx dx = cos kx dx = = − (−1)k . + k k0 k k 0 0 Thus √ (−1)k 2i π ck = − √ . − (−1)k = i 2π k k 2π Consequently √ (−1)k eikx (−1)k ikx √ =i∑ x = i 2π ∑ e k k 2π k=0 k=0 ∞ (−1)k sin 2x sin 3x sin 4x sin kx = 2 sin x − + − +··· k 2 3 4 k=1 = −2 ∑ Finally we compute what the Phthagorean Theorem tells us: x 2 = ∑|ck |2 . Since Zπ 2 2 x= |x|2 dx = π3 , 3 −π and ∞ ∞ −1 1 1 1 |ck |2 = 2π ∑ 2 + ∑ 2 = 4π ∑ 2 ∑ −∞ k 1k 1k Therefore ∞ 1 23 π = 4π ∑ 2 , 3 1k Interesting! – and not obvious at all. ∞ that is, 1 1 1 ∑ k2 = π2 . 6 ...
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