Note: The formulas for the Fourier transform on half intervals are often given in the form $\frac{2}{L}\displaystyle\int_{0}^Lf(x)cos\left(\frac{n\pi x}{L}\right) dx$, with the $\frac{2}{L}$ outside the integral. Computing these integrals will often involve u-substitutions, integration by parts, and other integration techniques that will produce all kinds of constants. With the way these problems are asked in WeBWorK, it would be hard to keep track of when constants should be factored out or not, so we will adopt the policy that constants are always part of the integrand. For example the formula for the cosine coefficient would be $\displaystyle\int_{0}^L\frac{2}{L}f(x)cos\left(\frac{n\pi x}{L}\right) dx$. When performing an integration by parts, all constants are included in the u term.

Compute the Fourier Cosine, Fourier Sine, and Fourier series of the function $f(x) = \begin{cases} 1 , & 0 < x < {\frac{3}{2}} \cr -6-x, & {\frac{3}{2}} \leq x < 3 \cr \end{cases}$

 $a_0$ $=$ $\displaystyle\int$ $dx +$ $\displaystyle\int$ $dx =$

 $a_n$ $=$ $\displaystyle\int$ $dx +$ $\displaystyle\int$ $dx$

$\hskip 250pt u =$$dv =$
$\hskip 250pt du =$$v =$

 $\hskip 10pt$ $=$ $\Bigg\vert$ $+$ $\Bigg\vert$ $+$ $\displaystyle\int$ $dx$
 $\hskip 10pt$ $=$ $+$ $\Bigg\vert$

 $\hskip 10pt$ $=$

 $b_n$ $=$ $\displaystyle\int$ $dx +$ $\displaystyle\int$ $dx$

$\hskip 250pt u =$$dv =$
$\hskip 250pt du =$$v =$

 $\hskip 10pt$ $=$ $\Bigg\vert$ $+$ $\Bigg\vert$ $-$ $\displaystyle\int$ $dx$
 $\hskip 10pt$ $=$ $-$ $\Bigg\vert$

 $\hskip 10pt$ $=$

Therefore on $[0,3]$

$\displaystyle f(x) =$ $+\displaystyle\sum\limits_{n=1}^\infty$ $\displaystyle\cos\left(\frac{n\pi x}{3}\right)$

$\displaystyle f(x) = \sum\limits_{n=1}^\infty$ $\displaystyle\sin\left(\frac{n\pi x}{3}\right)$

$\displaystyle f(x) =$ $+\sum\limits_{n=1}^\infty$ $\displaystyle\cos\left(\frac{2 n\pi x}{3}\right)$

$\displaystyle\hskip 70pt +$ $\displaystyle\sin\left(\frac{2 n\pi x}{3}\right)$