Question

Find the Fourier series of f on the given interval. f(x) = 1, ?7 < x < 0 1 + x, 0 ? x < 7

Answer 1

`f(x)=\begin{cases}1&\text{for }-7<x<0\\1+x&\text{for }0\le x<7\end{cases}`

The Fourier series expansion of
`f(x)` is given by


`\frac{a_0}2+\displaystyle\sum_(n\ge1)a_n\cos\frac{n\pi x}7+\sum_(n\ge1)b_n\sin\frac{n\pi x}7`

where we have


`a_0=\displaystyle\frac17\int_(-7)^7f(x)\,\mathrm dx`

`a_0=\displaystyle\frac17\left(\int_(-7)^0\mathrm dx+\int_0^7(1+x)\,\mathrm dx\right)`

`a_0=\frac{7+\frac{63}2}7=\frac{11}2`

The coefficients of the cosine series are


`a_n=\displaystyle\frac17\int_(-7)^7f(x)\cos\frac{n\pi x}7\,\mathrm dx`

`a_n=\displaystyle\frac17\left(\int_(-7)^0\cos\frac{n\pi x}7\,\mathrm dx+\int_0^7(1+x)\cos\frac{n\pi x}7\,\mathrm dx\right)`

`a_n=(9\sin n\pi)/(n\pi)+(7\cos n\pi-7)/(n^2\pi^2)`

`a_n=(7(-1)^n-7)/(n^2\pi^2)`

When
`n` is even, the numerator vanishes, so we consider odd
`n`, i.e.
`n=2k-1` for
`k\in\mathbb N`, leaving us with


`a_n=a_(2k-1)=(7(-1)-7)/((2k-1)^2\pi^2)=-(14)/((2k-1)^2\pi^2)`

Meanwhile, the coefficients of the sine series are given by


`b_n=\displaystyle\frac17\int_(-7)^7f(x)\sin\frac{n\pi x}7\,\mathrm dx`

`b_n=\displaystyle\frac17\left(\int_(-7)^0\sin\frac{n\pi x}7\,\mathrm dx+\int_0^7(1+x)\sin\frac{n\pi x}7\,\mathrm dx\right)`

`b_n=-(7\cos n\pi)/(n\pi)+(7\sin n\pi)/(n^2\pi^2)`

`b_n=(7(-1)^(n+1))/(n\pi)`

So the Fourier series expansion for
`f(x)` is


`f(x)\sim\frac{11}4-(14)/(\pi^2)\displaystyle\sum_(n\ge1)\frac1{(2n-1)^2}\cos\frac{(2n-1)\pi x}7+\frac7\pi\sum_(n\ge1)\frac{(-1)^(n+1)}n\sin\frac{n\pi x}7`