Answer to Question #203679 in Differential Equations for rahimah

Question

1) Find the half range Fourier cosine and sine series for

𝑓(𝑤) = 1 − 𝑤; 0 < 𝑤 < 1

Answer 1

The Fourier series of the half range even function is given by:

$$f(w)=\dfrac{a_0}{2}+\displaystyle\sum_{n=1}^\infin a_n\cos (\dfrac{n\pi w}{1})$$

$$a_0=\dfrac{2}{1}\displaystyle\int_{0}^{1}(1-w)dw=2[w-\dfrac{w^2}{2}]\begin{matrix} 1 \\ 0 \end{matrix}=1$$

$$a_n=\dfrac{2}{1}\displaystyle\int_{0}^{1}(1-w)\cos (\dfrac{n\pi w}{1})dw$$

$$\int \cos(n\pi w)dw=\dfrac{1}{n\pi}\sin(n\pi w)+C_1$$

$$\int w\cos(n\pi w)dw$$

$$=\dfrac{1}{n\pi}w\sin(n\pi w)-\dfrac{1}{n\pi}\int\sin(n\pi w)dw$$

$$=\dfrac{1}{n\pi}w\sin(n\pi w)+\dfrac{1}{n^2\pi^2}\cos(n\pi w)+C_2$$

$$a_n=2\displaystyle\int_{0}^{1}(1-w)\cos (n\pi w)dw$$

$$=2[\dfrac{1}{n\pi}\sin(n\pi w)-\dfrac{1}{n\pi}w\sin(n\pi w)]\begin{matrix} 1 \\ 0 \end{matrix}$$

$$-2[\dfrac{1}{n^2\pi^2}\cos(n\pi w))]\begin{matrix} 1 \\ 0 \end{matrix}=\dfrac{4}{(2k+1)^2\pi^2}$$

$$b_n=0$$

$$f(w)=\dfrac{1}{2}+\displaystyle\sum_{k=0}^\infin \dfrac{4}{(2k+1)^2\pi^2}\cos ((2k+1)\pi w)$$

The Fourier series of the half range odd function is given by:

$$f(w)=\displaystyle\sum_{n=1}^\infin b_n\sin (\dfrac{n\pi w}{1})$$

$$b_n=\dfrac{2}{1}\displaystyle\int_{0}^{1}(1-w)\sin (\dfrac{n\pi w}{1})dw$$

$$\int \sin(n\pi w)dw=-\dfrac{1}{n\pi}\cos(n\pi w)+C_1$$

$$\int w\sin(n\pi w)dw$$ $$=-\dfrac{1}{n\pi}w\cos(n\pi w)+\dfrac{1}{n\pi}\int\cos(n\pi w)dw$$

$$=-\dfrac{1}{n\pi}w\cos(n\pi w)+\dfrac{1}{n^2\pi^2}\sin(n\pi w)+C_2$$

$$b_n=2\displaystyle\int_{0}^{1}(1-w)\sin (n\pi w)dw$$

$$=2[-\dfrac{1}{n\pi}\cos(n\pi w)+\dfrac{1}{n\pi}w\cos(n\pi w)]\begin{matrix} 1 \\ 0 \end{matrix}$$

$$-2[\dfrac{1}{n^2\pi^2}\sin(n\pi w))]\begin{matrix} 1 \\ 0 \end{matrix}=\dfrac{2}{n\pi}$$

$$f(w)=\displaystyle\sum_{n=1}^\infin\dfrac{2}{n\pi}\sin (n\pi w)$$