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F (t) = t(π − t), t ∈ [0, π] and odd. The function is continuous and piecewise C 1 without vertical half tangents. According to the main theorem the Fourier series can then be written with a pointwise equality sign instead of with “∼”: f (t) = ∞ 1 8� sin(2n + 1)t. π n=0 (2n + 1)3 Pointwise: For t = π we get 2 ∞ 8 � (−1)n π2 , = π n=0 (2n + 1)3 4 from which ∞ � π3 (−1)n . 19 The period is 2π.  � � π� π �2   , , t ∈ 0, t −  2 2 f (t) = �π �    ,π , 0, t∈ 2 �∞ 1 1 and n=1 3 . ♦ n (2n+ 1)3 ∞ � π6 1 .

N=0 1 (2n − 1)2 (2n + 3)2 = π2 . 64 ♦ Please click the advert what‘s missing in this equation? You could be one of our future talents MAERSK INTERNATIONAL TECHNOLOGY & SCIENCE PROGRAMME 53have you already graduated? Are you about to graduate as an engineer or geoscientist? P. Moller - Maersk. 35 The period is 2π.  π    | sin 2t|, |t| ≤ 2 , f (t) =  π   < |t| ≤ π. 5 x The function is continuous and piecewise C 1 and without vertical half tangents. We can according to the main theorem use pointwise equality instead of “∼” in the Fourier series: f (t) = � ∞ � 2 cos(4n + 1)t cos 4nt 2 cos(4n − 1)t 2� 4 1 .

5 x The function is continuous and piecewise C 1 and without vertical half tangents. According to the main theorem the Fourier series can then be written with a pointwise equality sign instead of with “∼”: � ∞ � � π 2 1 π2 cos nt. com 41 Calculus 4b A list of problems in the Theory of Fourier series Pointwise: For t = 0 we get ∞ ∞ � π 4� 1 1 π2 π2 sin n , − +2 = 3 2 2 π n=1 n n 24 4 n=1 hence by a rearrangement ∞ � π3 (−1)n . = 32 (2n + 1)3 n=0 It is possible to obtain the same result, though it is more difficult, for t = π.

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Calculus 4b, Fourier Series and Systems of Differential Equations and Eigenvalue Problems by Mejlbro L.


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