Answer to Question #128113 in Calculus for Vicky

Question

Find the Fourier series for the function
f(x) = (x-x)^2,-L < x < L

Answer 1

$$f(x)=x-x^2$$

$$a_0={1\over 2L}\displaystyle\int_{-L}^Lf(x)dx={1\over 2L}\displaystyle\int_{-L}^L(x-x^2)dx=$$

$$={1\over 2L}[{x^2 \over 2}-{x^3 \over 3}]\begin{matrix} L \\ -L \end{matrix}={1\over 2L}({L^2 \over 2}-{L^3 \over 3}-({L^2 \over 2}+{L^3 \over 3}))=$$

$$=-{L^2 \over 3}$$

$$a_n={1\over L}\displaystyle\int_{-L}^Lf(x)\cos ({n\pi x \over b})dx=$$

$$={1\over L}\displaystyle\int_{-L}^L(x-x^2)\cos ({n\pi x \over L})dx$$

$$\int x\cos({n\pi x \over L})dx={L \over n\pi}x\sin({n\pi x \over L})-{L \over n\pi}\int \sin({n\pi x \over L})dx=$$

$$={L \over n\pi}x\sin({n\pi x \over L})+{L^2 \over n^2\pi^2}\cos({n\pi x \over L})+C_1$$

$$\int x^2\cos({n\pi x \over L})dx={L \over n\pi}x^2\sin({n\pi x \over L})-{2L \over n\pi}\int x\sin({n\pi x \over L})dx=$$

$$={L x^2\over n\pi}\sin({n\pi x \over L})+{2L^2x \over n^2\pi^2}\cos({n\pi x \over L})-{2L^2 \over n^2\pi^2}\int \cos({n\pi x \over L})dx=$$

$$={Lx^2 \over n\pi}\sin({n\pi x \over L})+{2L^2x \over n^2\pi^2}\cos({n\pi x \over L})-{2L^3 \over n^3\pi^3}\sin({n\pi x \over L})+C_2$$

$$a_n={1\over L}\bigg[{L \over n\pi}x\sin({n\pi x \over L})+{L^2 \over n^2\pi^2}\cos({n\pi x \over L})\bigg]\begin{matrix} L\\ -L \end{matrix}-$$

$$-{1\over L}\bigg[{Lx^2 \over n\pi}\sin({n\pi x \over L})+{2L^2x \over n^2\pi^2}\cos({n\pi x \over L})-{2L^3 \over n^3\pi^3}\sin({n\pi x \over L})\bigg]\begin{matrix} L\\ -L \end{matrix}=$$

$$=0-(-1)^n {4L^2 \over n^2\pi^2}=-(-1)^n {4L^2 \over n^2\pi^2}$$

$$b_n={1\over L}\displaystyle\int_{-L}^Lf(x)\sin ({n\pi x \over b})dx=$$

$$={1\over L}\displaystyle\int_{-L}^L(x-x^2)\sin ({n\pi x \over L})dx$$

$$\int x\sin({n\pi x \over L})dx=-{L \over n\pi}x\cos({n\pi x \over L})+{L \over n\pi}\int \cos({n\pi x \over L})dx=$$

$$=-{L \over n\pi}x\cos({n\pi x \over L})+{L^2 \over n^2\pi^2}\sin({n\pi x \over L})+C_3$$

$$\int x^2\sin({n\pi x \over L})dx=-{L \over n\pi}x^2\cos({n\pi x \over L})+{2L \over n\pi}\int x\cos({n\pi x \over L})dx=$$

$$=-{L x^2\over n\pi}\cos({n\pi x \over L})+{2L^2x \over n^2\pi^2}\sin({n\pi x \over L})-{2L^2 \over n^2\pi^2}\int \sin({n\pi x \over L})dx=$$

$$=-{L x^2\over n\pi}\cos({n\pi x \over L})+{2L^2x \over n^2\pi^2}\sin({n\pi x \over L})+{2L^3\over n^3\pi^3}\cos({n\pi x \over L})+C_4$$

$$b_n={1\over L}\bigg[-{Lx \over n\pi}\cos({n\pi x \over L})+{L^2 \over n^2\pi^2}\sin({n\pi x \over L})\bigg]\begin{matrix} L\\ -L \end{matrix}-$$

$$-{1\over L}\bigg[-{Lx^2 \over n\pi}\cos({n\pi x \over L})+{2L^2x \over n^2\pi^2}\sin({n\pi x \over L})+{2L^3 \over n^3\pi^3}\cos({n\pi x \over L})\bigg]\begin{matrix} L\\ -L \end{matrix}=$$

$$=-{2L \over n\pi}(-1)^n-0=-(-1)^n{2L \over n\pi}$$

$$f(x)=-{L^2 \over 3}+\displaystyle\sum_{i=1}^n(-(-1)^n {4L^2\over n^2\pi^2})\cos({n\pi x \over L}) +$$

$$+\displaystyle\sum_{i=1}^n(-(-1)^n{2L \over n\pi})\sin({n\pi x \over L})$$