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NNE Pharmaplan offers me freedom with responsibility as well as the opportunity to plan my own time. com NNE Pharmaplan is the world’s leading engineering and consultancy company focused exclusively on the pharma and biotech industries. NNE Pharmaplan is a company in the Novo Group. 1 We note that if we put t = 0 into (6), then f (0) = π 2 = ∞ (−1)n−1 2π 2 +4 , 3 n2 n=1 and hence by a rearrangement, ∞ (−1)n−1 π2 . 2 A function f ∈ K2π is given in the interval ]0, 2π] by f (t) = t2 . Notice the given interval!

24 1) Sketch the graph of the function f (t) = sin t , t ∈ R, in the interval [−2π, 2π]. 2 2) Prove that 2 4 f (t) = − π π ∞ cos nt , 2−1 4n n=1 t ∈ R. Hint: One may use without proof that sin t 4 cos nt dt = 2 4n2 −1 n sin t t 1 sin nt + cos cos nt , 2 2 2 for t ∈ R and n ∈ N0 . 3) Find, by using the result of (2), the sum of the series ∞ (a) 1 2−1 4n n=1 ∞ og (−1)n−1 . 4n2 − 1 n=1 1) The function f (t) is continuous and piecewise C ∞ without vertical half tangents. It is also even and periodic with the interval of period [−π, π[.

Since f is continuous, ∗ we have f = f . Since f is even, it follows that bn = 0, hence we have with equality sign by the main theorem that ∞ f (t) = 1 a0 + an cos nt, 2 n=1 where an = 2 π π f (t) cos nt dt = 0 2 π π/2 0 t− π 2 2 cos nt dt, n ∈ N0 . 5 –3 –2 0 –1 2 1 3 x 2) We have bn = 0 (an even function), and a0 = 2 π π/2 t− 0 π 2 2 dt = 2 1 π t− π 3 2 3 π/2 = 0 π2 . 12 For n ∈ N we get by partial integration an = π/2 2 π 0 = 0+ = − 4 πn t− π 2 t− 2 cos nt dt = π cos nt · 2 n 2 π t− π/2 0 − sin nt 4 π 1 4 − · − n πn 2 n πn2 4 πn2 π/2 = 0 π 2 2 · sin nt n π/2 0 − 4 πn π/2 0 t− π sin nt dt 2 π/2 cos nt dt 0 2 4 π − sin n .

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