By Mejlbro L.

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27 The period is 2π. f (t) = |t|3 , t ∈ [−π, π]. com 48 Calculus 4b A list of problems in the Theory of Fourier series 30 25 20 15 10 5 –4 –2 0 2 4 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: � ∞ � � 2(1 − (−1)n ) (−1)n π3 cos nt. + + 6π f (t) = π 2 n4 n2 4 n=1 Pointwise: Nothing new. Parseval gives a mess. 28 The period is 2π. f (t) = {π|t| − t2 }2 , t ∈ [−π, π].

E. �∞ n=0 π2 1 . = 2 8 (2n + 1) ∞ � π4 1 . 11 The period is 8. f (t) = |t| for t ∈ [−4, 4]. Apart from the scaling the figure is the same as in a previous example. The function is continuous and piecewise C 1 and without any vertical half tangents. We can therefore write the Fourier series with “=” instead of “∼”: f (t) = 2 − ∞ π 1 16 � cos(2n + 1) t. 4 π 2 n=0 (2n + 1)2 By a substitution one is led back to a previous example. 12 The period is 2π.    2π − |t|, for |t| ≤ 2π , 3 3 f (t) = 2π   < |t| ≤ π.

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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