Question

Can I get help with finding the Fourier cosine series of F(x) = x - x^2

Answer 1

Assuming you want the cosine series expansion over an arbitrary symmetric interval
`[-L,L]`,
`L\\eq0`, the cosine series is given by


`f_C(x)=\frac{a_0}2+\displaystyle\sum_(n\ge1)a_n\cos nx`

You have


`a_0=\displaystyle\frac1L\int_(-L)^Lf(x)\,\mathrm dx`

`a_0=\frac1L\left(\frac{x^2}2-\frac{x^3}3\right)\bigg|_(x=-L)^(x=L)`

`a_0=\frac1L\left(\left(\frac{L^2}2-\frac{L^3}3\right)-\left(\frac{(-L)^2}2-\frac{(-L)^3}3\right)\right)`

`a_0=-\frac{2L^2}3`


`a_n=\displaystyle\frac1L\int_(-L)^Lf(x)\cos nx\,\mathrm dx`

Two successive rounds of integration by parts (I leave the details to you) gives an antiderivative of


`\displaystyle\int(x-x^2)\cos nx\,\mathrm dx=((1-2x)\cos nx)/(n^2)-((2+n^2x-n^2x^2)\sin nx)/(n^3)`

and so


`a_n=-(4L\cos nL)/(n^2)+((4-2n^2L^2)\sin nL)/(n^3)`

So the cosine series for
`f(x)` periodic over an interval
`[-L,L]` is


`f_C(x)=-\frac{L^2}3+\displaystyle\sum_(n\ge1)\left(-(4L\cos nL)/(n^2L)+((4-2n^2L^2)\sin nL)/(n^3L)\right)\cos nx`