Exercícios Resolvidos Series de Fourier - Calculos e Desenvolvimento

1 sen α x T 2πα sen 5x T 2π5 2 cos2πx T 2π2π 1 4 sin πL x T 2ππ LL 2L 5 tan α x T πα tanπ x T ππ 1 14 fx a₀2 n1 to an cosnπxL bn sennπxL L P2 2L L2 a₀ 1L L to L fx dx 12 1 to 0 1 dx 12 0 to 1 0 dx 12 λ₀¹ 12 an 1L L to L fx cosnπxL dx 1L 1 to 0 1cosnπxL dx an 1L 2mπ senmπxL₀¹ 0 bn 1L L to L fx sennπxL dx 1L Lnπ cosnπxL₀¹ bn 1nπ cos 0 cos nπ 1nπ 1 cos nπ bn 2nπ n é impar 0 n é par as a₀ 1 an 0 bn 2nπ n é impar 0 n é par fx 12 n2 to bn sen nπx2 n 2n1 por ser impar fx 12 n2 to 2 2n1π sen 2n1πxL fx 12 2π n2 to sen 2n1πxL 2n1 15 fx x π x 0 0 0 x π fx 2π fx L P2 2π2 π a₀ 12π π to π fx dx 12π π to 0 x dx 12π 0 to π 0 dx π4 an 1π π to 0 x cosnx dx u x du dx dθ cos nx dx θ 1n sennx u dθ uθ θ du an 1π 1n sennxx₀π 1n π to 0 sennx dx an 1π 1n sennxx ₀π 1n² cosnx ₀π an 1π 1n² cosnπ 2 cos nπ π n² 1 1ⁿ π n² an 2n²π n é impar 0 n é par bn 1π π to 0 x sennx dx u x du dx dθ sennx dx θ 1n cosnx u dθ uθ θ du bn x 1n cosnx ₀π 1n π to 0 cosnx dx 1π bn 1π 1n x cosnx 1n² sennx ₀π bn 1π πn cos nπ 1ⁿ n Fx π4 n1 n impar to 2n² cosnx n1 to 1ⁿ n sennx 16 fx x1 2 x 0 1x 0 x 1 T 2 L 22 1 Fx a₀ n1 to an cosnπxL bn sennπx2 a₀ 12 2 to 0 x1 dx 12 0 to 1 1x dx 12 12 12 12 an 1 1 to 1 fx cosnπx dx 1 to 0 x1 cosnπx dx 0 to 1 1x cosnπx dx an 1 to 0 x cosnπx dx u dv ux dudx dvcosnπx dx ux dudx dθcosnπx dx θ1nπ sennπx udu x sennπxnπ from 1 to 0 1nπ sennπx dx from 1 to 0 x sennπxnπ cosnπxnπ2 from 1 to 0 cos0nπ2 cosnπnπ2 1nπ2 1nnπ2 1n 1nπ2 ux dudx dθcosnπx dx θ1nπ sennπx udu x sennπxnπ from 0 to 1 1nπ sennπx dx from 0 to 1 from 0 to 1 1x cosnπx dx 1nπ2 1nnπ2 an 1n 1nπ2 1nπ2 1nnπ2