# Fourier Series of Absolute Value

I think I understand how to evaluate Fourier series, but I would love to have someone check this to make sure I am doing this correctly.

Find the Fourier series of the function $f(x) = 11 + \lvert 6x \rvert$.

First, we find $a_0$.

\begin{align} a_0 &= \bigg\langle f,\frac{1}{\sqrt{2}} \bigg\rangle \\ &= \frac{1}{\pi} \int_{-\pi}^{\pi}\frac{1}{\sqrt{2}} f(x) dx \\ &= \frac{1}{\pi} \int_{-\pi}^{\pi}\frac{1}{\sqrt{2}} (11 + \lvert 6x \rvert)dx \end{align}

\begin{align} a_k &= \bigg \langle f, \cos(kx) \bigg \rangle \\ &= \frac{1}{\pi} \int_{-\pi}^{\pi} \cos(kx) f(x) dx \\ &= \frac{1}{\pi} \int_{-\pi}^{\pi} \cos(kx)( 11 + \lvert 6x \rvert)dx \end{align}

\begin{align} b_k &= \bigg \langle f, \sin(kx) \bigg \rangle \\ &= \frac{1}{\pi} \int_{-\pi}^{\pi} \sin(kx) f(x) dx \\ &= \frac{1}{\pi} \int_{-\pi}^{\pi} \sin(kx)(11 + \lvert 6x \rvert)dx \end{align}

So, once you find these constants, you plug them into the following formula:

$$f \sim \frac{a_{0}}{\sqrt{2}} + \sum_{k=1}^{\infty} a_k \cos(kx) + b_{k} \sin(kx)$$

Is this right? If so, how do I solve the integral including the absolute value term?

Guide:

Note that in the $a_0$, don't forget the braces.

$$a_0 = \frac1{\pi}\int_{-\pi}^\pi \frac1{\sqrt2} \color{blue}(11+ |6x|\color{blue}) \, dx$$

Also, note that $11+|6x|$ is an even function, hence $b_k=0$.

For $a_k$, $$a_k = \frac{2}{\pi} \int_0^\pi \cos (kx) (11+6x)\, dx,$$

I think integration by parts should work.

Edit:

\begin{align}a_k &= \frac{2}{\pi} \int_0^\pi \cos (kx) (11+6x)\, dx,\\ &= \frac2\pi \left( \left.(11+6x)\frac{\sin(kx)}{k}\right|_{x=0}^{x=\pi}-6\int_0^\pi \frac{\sin(kx)}{k} \, dx\right)\\ &=\frac{-12}{\pi k}\int_0^\pi \sin(kx) \, dx\end{align}

• For the a_k integral, I keep getting a wildly complicated expression. Could you help me in going thru it? – Jess M Apr 18 '18 at 18:59
• edited to include more working, can you complete it? – Siong Thye Goh Apr 18 '18 at 19:28