By Mejlbro L.

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Com 33 Examples of Fourier series Sum function of Fourier series thus ∞ (−1)n−1 n=1 = 1 4 = 1 4 (2n)2 (2n − 1)2 ∞ ∞ ∞ ∞ N (−1)n−1 (−1)n−1 + + lim (−1)n−1 2 2 N →∞ (2n − 1) (2n + 1) n=1 n=1 n=1 (−1)n−1 (−1)n−1 − 2 (2n − 1) (2n − 1)2 n=1 n=2 + 1 lim 4 N →∞ N = 1 1 (−1)n−1 (−1)N + lim 2 −1+ 4 4 N →∞ 2n − 1 2N + 1 n=1 1 1 − 2n − 1 2n + 1 N (−1)n−1 + 2n − 1 n=1 N +1 n=2 (−1)n−1 2n − 1 ∞ = (−1)n−1 1 π 1 = Arctan 1 = . 17 The odd and periodic function f of period 2π is given in the interval [0, π] by ⎧ π ⎪ , ⎪ sin t, for t ∈ 0, ⎨ 2 f (t) = ⎪ ⎪ ⎩ − sin t, for t ∈ π , π 2 1) Find the Fourier series of the function.

NNE Pharmaplan is a company in the Novo Group. 16 Prove that for every n ∈ N, π 0 t2 cos nt dt = (−1)n · 2π . n2 Find the Fourier series for the function f ∈ K2π , given in the interval [−π, π] by f (t) = t2 sin t. Then write the derivative f (t) by means of a trigonometric series and ﬁnd the sum of the series ∞ (−1)n−1 · n=1 (2n)2 . (2n − 1)2 (2n + 1)2 4 2 –4 –2 2 x 4 –2 –4 We get by partial integration, π t2 cos nt dt = 0 1 2 t sin nt n π 0 − π 2 n t sin nt dt = 0+ 0 2t cos nt n2 π 0 − 2 n2 π 0 cos nt dt = (−1)n · 2π .

K k 2k 2 ∞ 1 2) Since n=1 2 is a convergent majoring series, the series of (3) can be split. Then by a π n rearrangement, ∞ ∞ 1 1 1 n n = 2+ cos2 cos 2 2 k 4k n n k n=1 n=1 If we especially choose k = 1 > ∞ for every k > 1 . π 1 , we get π ∞ 1 1 1 cos2 n = + cos n. 2 n 4 n=1 n2 n=1 3) Clearly, a0 (k) = 1 1 → π π for k → ∞. For n > 0 it follows by a Taylor expansion, an (k) = = n 2k 2 2k 2 n2 n 1 n2 1 − cos = + 2ε 1− 1− 2 2 2 k πn πn k k 2 k 1 n 1 +ε → for k → ∞. 20 Given the function f ∈ K2π , where t f (t) = cos , 2 −π < t ≤ π.

### Calculus 4c-l, Examples of Fourier series by Mejlbro L.

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