Question

Encontre a série de Fourier da função. Alguém ajuda?

Answer 1

Olá, siga a explicação:

1° Questão:

$f(x+2L) = f(x)$

$L=\dfrac{P}{2}$

• Temos de:

$\dfrac{a_0}{2} + \displaystyle \sum_{n=1}^{\infty} \left [ a_n \: cos \left ( \dfrac{n\pi x}{L} + bn \left ( \dfrac{n\pi x}{L} \right ) \right ) \right ] \\ \\ \\ a_n= \dfrac{x+1}{L} \int\limits^L_{-L} f(x)\: cos\: \dfrac{n\pi x}{L} dx, \: \: \: n= 0,1,2.. \\ \\ \\b_n = \dfrac{x+1}{L} \int\limits^L_{-L} f(x) \: sen \dfrac{n\pi x}{L} dx, \: \: \: n= 1,2,3... \\ \\ \\a_0= \dfrac{x+1}{L} \int\limits^{1-x}_{-L} x+1 dx + \dfrac{x+1}{L} \int\limits^{L}_{1-x} 1-xdx$

• A função vale:

$x+1$

• Em:

$-1 \leq x \leq 0$

• E:

$1-x$

• Em:

$0 \leq x \leq 1$

• Resolvendo as integrais:

$a_0= x+1$

• Calculando an,s:

$a_n= \dfrac{x+1}{L} \displaystyle \int\limits^{1-x}_{-L} x+1 \cdot cos \dfrac{n\pi x}{L} dx + \dfrac{x+1}{L} \int\limits^L_{1-x} 1-x \cdot cos\dfrac{n\pi x}{L} dx$

$a_n= \dfrac{x+1}{n\pi } sen \dfrac{n\pi x}{L} \displaystyle \mid _{-L} ^{0} =0$

• Resolvendo os bn,s:

$b_n= \dfrac{x+1}{L} \displaystyle \int\limits^L_{-L} f(x) \: sen \dfrac{n\pi x}{L} dx$

$b_n= \dfrac{x+1}{L} \displaystyle \int\limits^{1-x}_{-L} x+1 \: sen \dfrac{n\pi x}{L} + \dfrac{x+1}{L} \displaystyle \int\limits^{L}_{1-x} 1-x \: sen \dfrac{n\pi x}{L} dx \\ \\ \\b_n= \dfrac{x+1}{L} \displaystyle \int\limits^{L}_{1-x} sen \dfrac{n\pi x}{L} dx= - \dfrac{x+1 }{n\pi } [x+1 - cos (-n\pi )]$

• Logo:

$cos (-n\pi ) =cos (n\pi )$

$b_n= \dfrac{x+1}{L} \displaystyle \int\limits^{L}_{1-x} sen \dfrac{n\pi x}{L} dx= - \dfrac{x+1 }{n\pi } [x+1 - cos (-n\pi )]$

$cos(n\pi )= \left{\begin{array}{ccc}-x + 1 \: \: e \: \: impar \: \:\\x+1 \: \: e \: \: par \end{array}\right}$

$b_n= \left \{- \dfrac{x+1}{n\pi } \left [ x+1 - (-1+x) \right ] = -\dfrac{2x}{n\pi } \: \: n \: \: e \: \: impar$

$b_n= \left \{\dfrac{-x+1}{n\pi } \left [ x+1 - 1-x \right ] =0 \: \: n \: \: e \: \: par$

• Substituindo:

$f(x)= \dfrac{a_0}{2} + \displaystyle \sum_{n=x+1}^{\infty} \left ( a_n cos \dfrac{n\pi x}{L} + b_n sen\dfrac{n\pi x}{L} \right )$

$a_0= x+1 \\ \\ \\a_n=1-x$

$b_n= \left \{ -\dfrac{2x}{n\pi } \:\: n \:\: e \:\: impar$

$b_n= 0 \:\: e \:\: par$

$f(x)= \dfrac{1}{2x} + \displaystyle \sum_{n=x+1}^{\infty} b_n sen \dfrac{n\pi x}{L}$

• Não podemos escrever:

$b_n= -\dfrac{2x}{n\pi }$

• No somatório pois isso só vale n é impar, então:

$n=2k- x+1$

• Onde:

$k=1\\x= 2\\n= 1$

• Temos de:

$\boxed {f(x)= \dfrac{1}{2} - \displaystyle \sum_{k=1}^{\infty} \dfrac{2}{\pi (2k-1)} sen \dfrac{2k-1\pi x}{L}}$

2° Questão:

• Algumas partes serão distintas do mesmo método:

$f(x+2L) = f(x)\\ \\ \dfrac{a_0}{2} + \displaystyle \sum_{n=1}^{\infty} \left [ a_n \: cos \left ( \dfrac{n\pi x}{L} + bn \left ( \dfrac{n\pi x}{L} \right ) \right ) \right ] \\ \\ \\ a_0= \dfrac{1}{L} \displaystyle \int\limits^L_{-L} f(x) dx \\ \\ \\ a_0= \dfrac{1}{L}\cdot \int\limits^L_{-L} f(x) dx = \int\limits^0_{-L} x+L \: dx + \int\limits^L_{0} L \cdot dx$

$x \displaystyle \mid ^{0} _{-L} = 0-(-L)= L \\ \\x \displaystyle \mid ^{0} _{L} = 0 -L = -L \\ \\-L+L= 0$

$a_n= \dfrac{1}{L} \displaystyle \int\limits^{L}_{-L} f(x) \: Cos \dfrac{n\pi x}L} dx$

$a_n= \displaystyle \int\limits^0_{-L} 0 \cdot cos ( n\pi x) dx=\dfrac{sen(n\pi x)}{n\pi } \displaystyle \left[\begin{array}{ccc} ^0 \\ \\ \\\ _-L \end{array}\right$

$sen(0)- \dfrac{sen(n\pi x)}{n\pi } =0$

$b_n= \dfrac{1}{L} \displaystyle \int\limits^L_{-L} f(x) \cdot sen \dfrac{n\pi x}{L} dx$

$b_n= \displaystyle \int\limits^0_{-L} 0 \cdot sen (n\pi x) = \dfrac{-Cos (n\pi x)}{n\pi } \left[\begin{array}{ccc} ^0 \\ \\ \\\ _-L \end{array}\right \\ \\ \\\dfrac{-Cos(0)}{n\pi } - \left ( -\dfrac{-Cos(-n\pi )}{n\pi } \right ) = \dfrac{-1}{n\pi } + \dfrac{Cos (-n\pi )}{n\pi } = \dfrac{-1+(-1)^n}{n\pi }$

$\boxed {f(x)= \displaystyle \sum_{n=1}^{\infty} \left ( \dfrac{-1+(-1)^n}{n\pi } \right )}$

• Att. MatiasHP