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Q0 for − 2 < t < 0,   � f (t) = q0 for 0 < t < ,  2   �   0 for < |t| < �. 2 Adjustments in the discontinuity points. Fourier series: � ∞ � π 1 − (−1)n π 1 2q0 � sin 2n t . sin(2n − 1) t + f∼ � 2n � π n=1 2n − 1 By investing some effort one can further reduce this result since 1 − (−1)n is either 0 or 2. Pointwise: Here we get some extremely ugly results which are not worth mentioning. Parseval: This formula is here reduced to the well-known �∞ n=0 π2 1 . 6 The period is 4�. Even function with  �   1 for 0 ≤ t ≤ ,   2  3� � 1 f (t) = for < t < ,  2 2 2     0 for 3� < t < 2�, 2 with adjustments in the discontinuity points.

2 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) = ∞ (−1)n+1 4� 1 sin(2n + 1)t. com 54 Calculus 4b A list of problems in the Theory of Fourier series 2 y –4 –2 1 0 2 4 x –1 –2 Pointwise: Nothing of interest. e. ∞ � 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?

N (2n+ 1)3 ∞ � π6 1 . = 6 960 (2n + 1) n=0 ♦ og lige. 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.