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Unformatted text preview: of the series is (x − 2πm)2 for m ∈ Z such that (2m − 1)π ≤ x ≤ (2m + 1)π. SOLUTION (c) : If we let x = π , then cos(nπ ) = (−1)n for n = 1, 2, ... and so from part (a),
π2 3 +4 +4 4 ∞ (−1)2n n=1 n2 = π2 =⇒
π2 3 ∞ 1 n=1 n2 = π2
2π 2 3 =⇒
∞ 1 n=1 n2 = =⇒
∞ n=1 1 π2 =. n2 6 7 If we let x = 0, then cos(n0) = 0 for n = 1, 2, ... and so from part (a),
π2 3 +4 ∞ (−1)n n=1 n2 =0
2 =⇒ 4 −4 4
∞ ∞ (−1)n+1 n=1 n2 = − π3 = =
π2 3 π2 3 =⇒
∞ (−1)n n=1 n2 =⇒
∞ (−1)n+1 n=1 n2 =⇒ π2 (−1)n+1 =. n2 12 n=1 Problem 4 (a) Determine the Fourier cosine series of the function f (x) = cos(νx) where ν is an arbitrary real number. (b) From this series, deduce that for ν = n π sin πν = 1 + (−1)n ν n=1
∞ ∞ 0≤x≤π 1 1 + ν −n ν+n . , 1 π cot πν = + ν n=1 1 1 + ν −n ν+n (c) Integrate the last formula with respect to ν from ν = 0 to ν = θ with 0 < θ < 1, to show that ∞ sin πθ θ2 = 1− 2 . πθ n n=1 Note that the identities in part (b) represent meromorphic functions (functions that have only pole singularities) as sums over the poles. The identity of part (c) represents an entire function as a product of factors that indicate where the zeroes of the function are. These types of representations are known as MittagLefﬂer expansions. SOLUTION (a) : For a cosine...
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 Winter '09
 NilesA.Pierce

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