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**Home Assignment 5 / Problem 3**

« **on:**October 31, 2012, 11:14:52 PM »

**Part (b):**

The function is even, so $b_n = 0$.

\begin{equation*}

a_0 = \frac{2}{\pi} \int_{0}^{\pi} \sin x dx\\

= \frac{4}{\pi}.

\end{equation*}

\begin{equation*}

a_n = \frac{2}{\pi} \int_{0}^{\pi} \sin x \cos nx dx

\end{equation*}

Using $2\sin a \cos b = \sin(a+b) + \sin(a-b)$,

\begin{equation*}

=\frac{1}{\pi} \int_{0}^{\pi} \sin ((n+1)x) + \sin ((1-n)x) dx\\

\end{equation*}

Notice that for $n = 1$, the solution is $0$.

\begin{equation*}

=-\frac{1}{\pi}\left[\frac{\cos((n+1)x)}{n+1} - \frac{\cos((n-1)x)}{n-1} \right]_{0}^{\pi}\\

=-\frac{1}{\pi}\left[\frac{(-1)^{n+1}}{n+1} - \frac{1}{n+1} - \frac{(-1)^{n+1}}{n-1} + \frac{1}{n-1} \right]_{0}^{\pi}\\

=\frac{1}{\pi}\big((-1)^n + 1 \big) \big(\frac{1}{n+1} + \frac{1}{n-1} \big)\\

=\frac{-2}{\pi} \big((-1)^n + 1 \big) \big( \frac{1}{n^2 - 1} \big).\\

\end{equation*}

Let $n=2k$. Thus, for $n>1$,

\begin{equation*}

|\sin x| = \frac{2}{\pi} -\frac{4}{\pi} \sum_{k=1}^{\infty} \frac{1}{(2k)^2 - 1}\cos (2kx).

\end{equation*}

**Part (a):**

Again, the function $|\cos x|$ is even, so $b_n = 0$.

We proceed as above:

\begin{equation*}

a_0 = \frac{2}{\pi} \int_{0}^{\pi} |\cos x| dx \\

= \frac{2}{\pi} \int_{0}^{\frac{\pi}{2}} \cos x dx - \frac{2}{\pi} \int_{\frac{\pi}{2}}^{\pi} \cos x dx\\

= \frac{4}{\pi}\\

\end{equation*}

\begin{equation*}

a_n = \frac{2}{\pi} \int_{0}^{\pi} |\cos x|\cos{nx} dx \\

= \frac{2}{\pi} \int_{0}^{\frac{\pi}{2}} \cos x \cos{nx} dx - \frac{2}{\pi} \int_{\frac{\pi}{2}}^{\pi} \cos x \cos{nx} dx\\

= \frac{2}{\pi}\left(-2\frac{\cos{\frac{\pi n}{2}}}{n^2 - 1}\right)\\

\end{equation*}

Therefore,

\begin{equation*}

|\cos x| = \frac{2}{\pi} - \frac{4}{\pi}\sum_{n=1}^{\infty} \frac{\cos{\frac{\pi n}{2}}}{n^2 - 1} \cos{nx}.

\end{equation*}

Both functions are continuous, so the Fourier series will converge to each function at every point.