Questões - Séries de Fourier e Desenho de Gráfico - 2023-2

1. Sejam W, V e U os três últimos algarismos do seu número de matrícula, respectivamente, o antepenúltimo, o penúltimo e o último. Seja a função periódica, cujo período é 2γ, f(x) = \frac{\alpha \cdot (x + \beta) \cdot (x - \beta \cdot \gamma)}{\gamma}, \text{com} \ x \in [-\gamma; \gamma], na qual \alpha = \begin{cases} -1, & \text{se} \ W \in \{0; 1\} \\ -1/2, & \text{se} \ W \in \{2;3;4\} \\ 1/2, & \text{se} \ W \in \{5;6;7\} \\ 1, & \text{se} \ W \in \{8;9\} \end{cases} \beta = \begin{cases} -1, & \text{se} \ V \in \{1;3;5;7;9\} \\ 1, & \text{se} \ V \in \{0;2;4;6;8\} \end{cases} \gamma = \begin{cases} 3, & \text{se} \ U \in \{0;1;2;3\} \\ 5, & \text{se} \ U \in \{4;5;6\} \\ 7, & \text{se} \ U \in \{7;8;9\} \end{cases} Lembrando de que essa função f(x) pode ser reescrita como uma série de senos e cossenos (Fourier) na forma f(x) = \frac{a_0}{2} + \sum_{m=1}^{\infty} \left(a_m \cos \left(\frac{m\pi x}{L}\right) + b_m \sin \left(\frac{m\pi x}{L}\right) \right) com coeficientes a_0, a_m e b_m a determinar, calcule esses coeficientes para a função f(x) correspondente ao seu número de matrícula. 2. Calcule o valor de cada coeficiente a_m e b_m para m \in \{0; 1; 2; 3; 4; 5\}, monte a série truncada grafos cada F(x) = \frac{a_0}{2} + \sum_{m=1}^{5} \left(a_m \cos \left(\frac{m\pi x}{L}\right) + b_m \sin \left(\frac{m\pi x}{L}\right) \right) e utilize o programa Geogebra para desenhar os gráficos de f e F, de acordo com o passo-a-passo mostrado no Campus Virtual. 1) W = 0 \rightarrow \alpha = -1 V = 8 \rightarrow \beta = 1 U = 6 \rightarrow \gamma = 5 f(x) = \frac{\alpha(x + \beta)(x - \beta \cdot \gamma)}{\gamma:5}, \ x \in [-L; L] f(x) = -((x + 1)\cdot(x - 5)) = \frac{-x^2 + 4x + 5}{5} Coeficientes: \frac{a_0}{2} = \frac{1}{2L} \int_{-L}^{L} f(x) dx = \frac{1}{2(5)} \int_{-5}^{5} \frac{-x^2 + 4x + 5}{5} dx \frac{a_0}{2} = \frac{1}{50} \left[-\frac{x^3}{3} + \frac{4x^2}{2} + 5x\right]_{-5}^{5} \frac{a_0}{2} = \frac{1}{50} \left[\frac{(5)^3}{3} + 2(5)^2 - 5(-5)\right] = \frac{-2}{3} \begin{align*} a_m &= \frac{1}{L} \int_{-L}^{L} f(x) \cos \left(\frac{m \pi x}{L}\right) dx = \frac{1}{5} \int_{-5}^{5} \frac{-x^2 + 4x + 5}{5} \cos \left(\frac{m \pi x}{5}\right) dx &= \frac{1}{25} \left[\int_{-5}^{5} x^2 \cos \left(\frac{m \pi x}{5}\right) dx + \int_{-5}^{5} 4x \cos \left(\frac{m \pi x}{5}\right) dx + \int_{-5}^{5} 5 \cos \left(\frac{m \pi x}{5}\right) dx \right] \end{align*} (\text{III}) \begin{align*}\int_{-5}^{5} 5 \cos \left(\frac{m \pi x}{5}\right) dx &= 5 \sin(m \pi) \left[\frac{m \pi x}{5}\right]_{-5}^{5}\\ &= \frac{25}{m \pi} \sin (m \pi)[\cancel{5}] \\ (II) & = 0 (II) = 4 \int_{-5}^{5} x \cos \left(\frac{m \pi x}{5}\right) dx = \frac{4}{m^2 \pi^2}\left[5 m \pi x \sin \left(\frac{m \pi x}{5}\right) + 25 \cos \left(\frac{m \pi}{5}\right) \right]_{-5}^{5} (II) = 0 (II) = \int_{-5}^{5} x^2 \cos \left(\frac{m \pi x}{5}\right) dx = \frac{1}{m^2 \pi^2}\left[50 m x^3 \cos \left(\frac{m \pi x}{5}\right) - 250 \sin \left(\frac{m \pi x}{5}\right)\right]_{-5}^{5} (II) = \frac{1}{m^2 \pi^2}(50 m)(5 \cos (m \pi) - (-5)\cos (-m \pi)) = -500 \frac{(-1)^m}{m^2 \pi^3} a_m = \frac{-20(-1)^m}{m^2 \pi^2} \end{align*} b_m = \frac{1}{L}\int_{-L}^{L}f(x)\sin\left(\frac{m\pi x}{L}\right)dx = \frac{d}{5}\int_{-5}^{5}\left(-\frac{x^2}{5} + \frac{4x}{5} + \frac{5}{5}\right)\sin\left(\frac{m\pi x}{5}\right)dx b_m = \frac{1}{25} \int_{-5}^{5}\left(x^2 + 4x + 5\right)\sin\left(\frac{m\pi x}{5}\right)dx b_m = \frac{1}{25}\left[-\int_{-5}^{5}x^2\sin\left(\frac{m\pi x}{5}\right)dx + 4\int_{-5}^{5}x\sin\left(\frac{m\pi x}{5}\right)dx + 5\int_{-5}^{5}\sin\left(\frac{m\pi x}{5}\right)dx\right] \downarrow = 0 = \frac{4}{m\pi}\left[-50\pi \sin(\cos(m\pi)) + 50 \sin(\sin(m\pi))\right] = -200(-1)^m = \frac{-200(-1)^m}{m\pi} b_m = \frac{1}{25}\left(\frac{-200(-1)^m}{m\pi}\right) b_m = \frac{-8(-1)^m}{m\pi} f(x) = \frac{-2}{3} + \sum_{m=1}^{\infty}\left[\left(\frac{-20(-1)^m}{\pi^2 m^2}\right)\cos\left(\frac{m\pi x}{5}\right) + \left(\frac{-8(-1)^m}{m\pi}\right)\sin\left(\frac{m\pi x}{5}\right)\right] F(x) = -\frac{2}{3} + \sum_{m=1}^{5}\left(a_m \cos\left(\frac{m\pi x}{5}\right) + b_m \sin\left(\frac{m\pi x}{5}\right)\right) \boxed{a_1 = \frac{-20(-1)^2}{(1^2)\pi^2} = \frac{20}{\pi^2}} \boxed{a_3 = \frac{-20(-1)^3}{(3^2)\pi^2} = \frac{20}{9\pi^2}} \boxed{a_5 = \frac{-20(-1)^5}{(5^2)\pi^2} = \frac{20}{25\pi^2}} a_2 = \frac{-20(-1)^2}{(2^2)\pi^2} = \frac{20}{4\pi^2} b_1 = \frac{-8(-1)^1}{\pi} = \frac{8}{\pi} \boxed{b_3 = \frac{-8(-1)^3}{3\pi} = \frac{8}{3\pi}} \boxed{b_5 = \frac{-8(-1)^5}{5\pi} = \frac{8}{5\pi}} b_2 = \frac{-8(-1)^2}{2\pi} = \frac{-8}{2\pi} F(x) = \frac{2}{3} + a_1 \cos\left(\frac{\pi x}{5}\right) + b_1 \sin\left(\frac{\pi x}{5}\right) + a_2 \cos\left(\frac{2\pi x}{5}\right) + b_2 \sin\left(\frac{2\pi x}{5}\right) + a_3 \cos\left(\frac{3\pi x}{5}\right) + b_3 \sin\left(\frac{3\pi x}{5}\right) + a_4 \cos\left(\frac{4\pi x}{5}\right) + b_4 \sin\left(\frac{4\pi x}{5}\right) + a_5 \cos\left(\frac{5\pi x}{5}\right) + b_5 \sin\left(\frac{5\pi x}{5}\right) F(x) = \frac{2}{3} + \frac{20}{\pi^2}\cos\left(\frac{\pi x}{5}\right) + \frac{8}{\pi}\sin\left(\frac{\pi x}{5}\right) + \frac{-20}{4\pi^2}\cos\left(\frac{2\pi x}{5}\right) + \frac{-8}{2\pi}\sin\left(\frac{2\pi x}{5}\right) + \frac{20}{9\pi^2}\cos\left(\frac{3\pi x}{5}\right) + \frac{8}{3\pi}\sin\left(\frac{3\pi x}{5}\right) + \frac{-20}{16\pi^2}\cos\left(\frac{4\pi x}{5}\right) + \frac{-8}{4\pi}\sin\left(\frac{4\pi x}{5}\right) + \frac{20}{25\pi^2}\cos\left(\frac{5\pi x}{5}\right) + \frac{8}{5\pi}\sin\left(\frac{5\pi x}{5}\right) 1) W=0 \hspace{20pt} \alpha = -1 \\ V = 8 \hspace{20pt} \beta = 1 \\ U = 6 \hspace{20pt} \gamma = \\ \hspace{20pt} \lambda = 5 \\ f(x) = \frac{\alpha}{\gamma} \cdot \frac{(x+\beta)(x-\beta)\ldots}{\lambda} \ldots \, x \in [-5, 5]\\ f(x) = \frac{-(x+1)(x-5)}{5} = \frac{-x^2 + 4x + 5}{5}\\ Coeficientes:\\ \frac{a_0}{2} = \frac{1}{2L} \int_{-L}^{L} f(x) dx = \frac{1}{2(5)} \int_{-5}^{5} \left( \frac{-x^2 +4x + 5}{5} \right) dx \\ \frac{1}{50} \int_{-5}^{5} \left( -x^2 + 4x + 5 \right) dx = \frac{1}{50} \left[ \frac{-x^3}{3} + 4 \frac{x^2}{2} + 5x \right]_{-5}^5\\ \frac{a_0}{2} = \frac{1}{50} \left[ -(5)^3 + 15 \right] 3 + 2(5^2)-(5^2) + 5(5-(-5)) = -\frac{2}{3} \\ a_m = \frac{1}{L} \int_{-L}^{L} f(x) \cos \left( \frac{m \pi x}{L}\right) dx = \frac{1}{5} \int_{-5}^{5} \left( \frac{-x^2 + 4x + 5}{5}\right) \cos \left(\frac{m \pi x}{5}\right) dx \\ a_m = \frac{1}{25} \left[ \int_{-5}^{5} -x^2 \cos \left( \frac{m \pi x}{5}\right)dx \right. + \int_{-5}^{5} 4x \cos \left( \frac{m \pi x}{5}\right)dx + \left. \int_{-5}^{5} 5 \cos \left( \frac{m \pi x}{5}\right) dx \right] \\ \hspace{20pt} \text{(I)} \hspace{90pt} \text{(II)} \hspace{95pt} \text{(III)} (III) \hspace{20pt} \text{es el caso más simple} \\ \text{(III)} = 5 \int_{-5}^{5} \cos \left( \frac{m \pi x}{5} \right) dx = 5 \sin \left(\frac{m\pi x}{5} \right) \cdot \frac{5}{m \pi} \\ = \frac{25}{m \pi} \sin \left( \frac{m \pi x}{5} \right)\bigg|_{-5}^5 = \frac{25}{m \pi} [\sin (m\pi) - \sin (-m\pi)] \\ \text{(III)} = 0 \\ \text{(II)}=4 \int_{-5}^{5} x \cos \left( \frac{m \pi x}{5}\right) dx = \frac{4}{m \pi} \left[ 5m \pi x \sin \left( \frac{m \pi x}{5}\right) \right.+25 \cos \left( \frac{m \pi x}{5}\right) \Bigg]_{-5}^{5} = \frac{4}{m \pi^2} [5m \pi (5) \sin (m \pi)] \\ + 25 \cos (m \pi) -5m \pi (-5) \sin(-m \pi) \\ = 0 - 25\cos(-m\pi) \UnderRightArrow \cos(m\pi) = \cos(-m\pi) \\ \text{(II)} = 0 \\ \text{(I)} = \int_{-5}^{5} x^2 \cos \left(\frac{m \pi x}{5}\right) dx = -\frac{1}{m^2 \pi^2} \Bigg[5\frac{m^2 x^2 x \sin \left(\frac{m \pi x}{5}\right)}{5} \right. \\ +50 m \pi x \cos \left(\frac{m \pi x}{5}\right) - 250 \sin \left(\frac{m \pi x}{5}\right)\Bigg]_{-5}^{5} \\ = -\frac{1}{m^2 \pi^2} \left[ 0\right. +5\cos(m\pi) -(-5)\cos(-m\pi) \\ = \frac{-500(-1)^m}{m^2 \pi^2} \\ a_m = \frac{-20(-1)^m}{m^2 \pi^2} b_m = \frac{1}{L} \int_{-L}^{L} f(x) \sin \left(\frac{m \pi x}{L}\right) dx = \frac{d}{5} \int_{-5}^{5} \left( \frac{-x^2}{5} + \frac{4x}{5} + \frac{5}{5} \right) \sin \left( \frac{m \pi x}{5} \right) dx \\ b_m = \frac{1}{25} \int_{-5}^{5}(x^2 + 9x + 5) \sin\left(\frac{m \pi x}{5}\right) dx \\ b_m = \frac{1}{25} \left( \int_{-5}^{5} x^2 \sin\left(\frac{m \pi x}{5}\right) dx + 4 \int_{-5}^{5} x \sin\left(\frac{m \pi x}{5}\right) dx + 5 \int_{-5}^{5} \sin\left(\frac{m \pi x}{5}\right) dx \right) \\ downarrow \Rightarrow = 0\\ \hspace{20pt} -\frac{4}{m^2 \pi^3} \left[50 \pi x \cos (m\pi) + (\pi x) \Rightarrow (-1)^m \right] = \frac{-200(-1)^m}{m \pi} \\ \Rightarrow \\ = \frac{1}{25} \left(\frac{-200(-1)^m}{\pi m}\right) \\ b_m = -\frac{8(-1)^m}{\pi m}\\ f(x) = \frac{-2}{3} + \sum_{m=1}^{\infty} \left[ \frac{-20(-1)^m}{\pi^2 m^2}\right] \cos\left(\frac{m\pi x}{5}\right)+ \left[ \frac{-8(-1)^m}{\pi m}\right] \sin \left(\frac{m\pi x}{5}\right) F(x) = -\frac{2}{3} + \sum_{m=1}^{5} \left( a_m \cos\left(\frac{m \pi x}{5}\right) + b_m \sin\left(\frac{m \pi x}{5}\right) \right) a_1 = \frac{-20(-1)}{(1)^2 \pi^2} = \frac{20}{\pi^2} \quad a_3 = \frac{-20(-1)^3}{(3)^2 \pi^2} = \frac{20}{9 \pi^2} \quad a_5 = \frac{-20(-1)^5}{(5)^2 \pi^2} = \frac{20}{25 \pi^2} a_2 = \frac{-20(-1)^2}{(2)^2 \pi^2} = \frac{-20}{4 \pi^2} \quad a_4 = \frac{-20(-1)^4}{(4)^2 \pi^2} = \frac{20}{16 \pi^2} b_1 = \frac{-8(-1)}{\pi} = \frac{8}{\pi} \quad b_3 = \frac{-8(-1)^3}{3 \pi} = \frac{-8}{3 \pi} \quad b_5 = \frac{-8(-1)^5}{5 \pi} = \frac{8}{5 \pi} b_2 = \frac{8(-1)^2}{2 \pi} = \frac{-8}{2 \pi} \quad b_4 = \frac{-8(-1)^4}{4 \pi} = \frac{-8}{4 \pi} F(x) = -\frac{2}{3} + a_1 \cos \left(\frac{\pi x}{5}\right) + b_1 \sin \left(\frac{\pi x}{5}\right) + a_2 \cos \left(\frac{2\pi x}{5}\right) + b_2 \sin \left(\frac{2\pi x}{5}\right) + a_3 \cos \left(\frac{3\pi x}{5}\right) + b_3 \sin \left(\frac{3\pi x}{5}\right) + a_4 \cos \left(\frac{4\pi x}{5}\right) + b_4 \sin \left(\frac{4\pi x}{5}\right) + a_5 \cos \left(\frac{5\pi x}{5}\right) + b_5 \sin \left(\frac{5\pi x}{5}\right) F(x) = \frac{-2}{3} + \frac{20}{\pi^2} \cos \left(\frac{\pi x}{5}\right) + \frac{8}{\pi} \sin \left(\frac{\pi x}{5}\right) + \frac{-20}{4 \pi^2} \cos \left(\frac{2\pi x}{5}\right) + \frac{-8}{2 \pi} \sin \left(\frac{2\pi x}{5}\right) + \frac{20}{9 \pi^2} \cos \left(\frac{3\pi x}{5}\right) + \frac{-8}{3 \pi} \sin \left(\frac{3\pi x}{5}\right) + \frac{-20}{16 \pi^2} \cos \left(\frac{4\pi x}{5}\right) + \frac{-8}{4 \pi} \sin \left(\frac{4\pi x}{5}\right) + \frac{20}{25 \pi^2} \cos \left(\frac{5\pi x}{5}\right) + \frac{8}{5 \pi} \sin \left(\frac{5\pi x}{5}\right) x -7 -7 -6 -6 -5 -5 -4 -4 -3 -3 -2 -2 -1 -1 1 2 3 4 5 6 7 8 9 10 10 11 11 12 12 y -9 -9 -8 -8 -7 -7 -6 -6 -5 -5 -4 -4 -3 -3 -2 -2 -1 -1 1 2 3 0 f(x) f(x) F(X) F(X)