COD#SS-1**Integrals for fourier Series

Orthogonality:

I will show, by hand, that the following set is an orthogonal set:

\[
\{ 1, \cos (\pi x), \cos (2\pi x), \ldots, \sin (\pi x), \sin (2\pi x), \ldots  \}
\]

Recall that the scalar product that we are using is defined in the book :
\[
(f,g) = \int_{-1}^1 f(x)g(x) dx
\]

So, we just have to compute a lot of integrals.
Here is a list of the integrals that we need to compute:

\begin{eqnarray*}
(1,1) & = & \int_{-1}^1 1\cdot 1 \;dx \\
(1, \cos(n\pi x) & = & \int_{-1}^1 1\cdot \cos(n\pi x) \;dx \\
(1, \sin(n\pi x) & = & \int_{-1}^1 1\cdot \sin(n\pi x) \;dx \\
(\cos(m\pi x), \cos(n\pi x)) & = & \int_{-1}^1 \cos(m\pi x) \cdot \cos(n\pi x) \;dx \qquad m\neq n\\
(\cos(n\pi x), \cos(n\pi x)) & = & \int_{-1}^1 \cos(n\pi x) \cdot \cos(n\pi x) \;dx \\
(\cos(m\pi x), \sin(n\pi x)) & = & \int_{-1}^1 \cos(m\pi x) \cdot \sin(n\pi x) \;dx \\
(\sin(m\pi x), \sin(n\pi x)) & = & \int_{-1}^1 \sin(m\pi x) \cdot \sin(n\pi x) \;dx \qquad m\neq n \\
(\sin(n\pi x), \sin(n\pi x)) & = & \int_{-1}^1 \sin(n\pi x) \cdot \sin(n\pi x) \;dx
\end{eqnarray*}

The easy integrals:

You can do these, nothing too tricky here ($u$-substitution at the worst):

\begin{eqnarray*}
(1,1) & = & \int_{-1}^1 1\cdot 1 \;dx = 2 \\
(1, \cos(n\pi x) & = & \int_{-1}^1 1\cdot \cos(n\pi x) \;dx = 0 \\
(1, \sin(n\pi x) & = & \int_{-1}^1 1\cdot \sin(n\pi x) \;dx = 0
\end{eqnarray*}


The next level

These are not too bad.
Remember the trigonometric identities:
\[
\sin^2 u = \frac{1}{2}(1-\cos(2u)) \qquad \cos^2 u = \frac{1}{2}(1+\cos(2u))
\]

Applying these and using $u$-substitution gives:

\begin{eqnarray*}
(\cos(n\pi x), \cos(n\pi x)) & = & \int_{-1}^1 \cos(n\pi x) \cdot \cos(n\pi x) \;dx \\
  & = & \int_{-1}^1 \cos^2(n\pi x) \; dx
    = \int_{-1}^1 \frac{1}{2}(1+\cos(2n\pi x )) \; dx\\
  & = & \int_{-1}^1 \frac{1}{2} \; dx + \int_{-1}^1 \frac{1}{2}\cos(2n\pi x) \; dx \\
  & = & 1 + \int_{-1}^1 \frac{1}{2}\cos(2n\pi x) \; dx \\   
  & = & 1 + 0 = 1 \qquad (\mbox{$u$-substitution}) \\
\end{eqnarray*}

Showing that $(\sin(n\pi x), \sin(n\pi x)) = 1$ is similar and you should be able to do it.


The last ones:

Here are the ones we have left:

\begin{eqnarray*}
(\cos(m\pi x), \cos(n\pi x)) & = & \int_{-1}^1 \cos(m\pi x) \cdot \cos(n\pi x) \;dx \qquad m\neq n\\
(\cos(m\pi x), \sin(n\pi x)) & = & \int_{-1}^1 \cos(m\pi x) \cdot \sin(n\pi x) \;dx \\
(\sin(m\pi x), \sin(n\pi x)) & = & \int_{-1}^1 \sin(m\pi x) \cdot \sin(n\pi x) \;dx \qquad m\neq n
\end{eqnarray*}

These are handled by the following trigonometric identities:

\begin{eqnarray*}
\sin A \cos B & = & \frac{1}{2}[ \sin(A-B) + \sin(A+B) ] \\
\sin A \sin B & = & \frac{1}{2}[ \cos(A-B) - \cos(A+B) ] \\
\cos A \cos B & = & \frac{1}{2}[ \cos(A-B) + \cos(A+B) ]
\end{eqnarray*}

You can prove these identities to be true using the $\sin$ and $\cos$ addition formulas.

I'll use this and prove one of the harder integrals above and you can do the others:

\begin{eqnarray*}
(\cos(m\pi x), \cos(n\pi x)) & = & \int_{-1}^1 \cos(m\pi x) \cdot \cos(n\pi x) \;dx \\
   & = & \int_{-1}^1 \frac{1}{2} [ \cos((m\pi x) - (n\pi x)) + \cos((m\pi x)+(n\pi x)) ]  \;dx \\
   & = & \frac{1}{2} \int_{-1}^1 \cos((m-n)\pi x) + \cos((m+n)\pi x)  \;dx \\
   & = & \frac{1}{2} \int_{-1}^1 \cos((m-n)\pi x)\;dx + \frac{1}{2} \int_{-1}^1 \cos((m+n)\pi x)  \;dx \\
  & = & 0 + 0 \qquad \mbox{(by $u$-substitution)}
\end{eqnarray*}