1. ## Fourier Series

I am having some trouble with the following problem:

Show that the Fourier series:

$\displaystyle \frac{1}{2} a_{0} +\sum^{\infty}_{n=1} (a_{n} \cos nx +b_{n} \sin nx)$

can be written in the form

$\displaystyle \frac{1}{2} \rho_{0} +\sum^{\infty}_{n=1} \rho_{n} \cos (nx + \theta_{n})$

where $\displaystyle \rho_{n} = \sqrt{a^{2}_{n} +b^{2}_{n} }$. Express $\displaystyle \theta_{n}$ in terms of $\displaystyle a_{n}$ and $\displaystyle b_{n}$.

Ok so I got down to the fourier series being equal to this:

$\displaystyle = \frac{1}{2\pi} \int^{\pi}_{-\pi} f(\theta) d\theta + \frac{1}{\pi} \int^{\pi}_{-\pi} f(\theta) \sum^{\infty}_{n=1} \cos n(x - \theta) d\theta$

which is not exactly what we want especially the $\displaystyle \cos n(x - \theta)$.
Any suggestion??
Also any idea how to do "Express $\displaystyle \theta_{n}$ in terms of $\displaystyle a_{n}$ and $\displaystyle b_{n}$."???

2. For each n, write $\displaystyle (a_n,-b_n)$ in terms of polar coordinates $\displaystyle \rho_n,\,\theta_n$. So $\displaystyle a_n = \rho_n\cos\theta_n$, $\displaystyle b_n = -\rho_n\sin\theta_n$. (For n=0, take $\displaystyle \rho_0=a_0,\ \theta_0=0$.) Then $\displaystyle a_n\cos nx + b_n\sin nx = \rho_n(\cos nx\cos\theta_n -\sin nx\sin\theta_n) = \rho_n\cos(nx+\theta_n)$.

3. I thought that in a Fourier Series the coefficients had to be:

$\displaystyle a_{n} = \frac{1}{\pi}\int^{\pi}_{- \pi}f(\theta)\cos n \theta d\theta$
$\displaystyle b_{n} = \frac{1}{\pi}\int^{\pi}_{- \pi}f(\theta)\sin n \theta d\theta$

So can I still just rewrite the way you show me??? Sorry I am just trying to make sure I have a good understanding of it all.

Also to express $\displaystyle \theta_{n}$ in terms of $\displaystyle a_{n}$ and $\displaystyle b_{n}$, I am guessing it would be something like:

$\displaystyle \theta_{n}= \tan^{-1} \frac{b_{n}}{a_{n}}$

4. Originally Posted by ynn6871
I thought that in a Fourier Series the coefficients had to be:

$\displaystyle a_{n} = \frac{1}{\pi}\int^{\pi}_{- \pi}f(\theta)\cos n \theta d\theta$
$\displaystyle b_{n} = \frac{1}{\pi}\int^{\pi}_{- \pi}f(\theta)\sin n \theta d\theta$

So can I still just rewrite the way you show me??? Sorry I am just trying to make sure I have a good understanding of it all.
Yes, but you are given $\displaystyle a_n$ and $\displaystyle b_n$ in the statement of the question (and you are not given the function f, except in terms of its Fourier coefficients). So those formulas for $\displaystyle a_n$ and $\displaystyle b_n$ are not relevant to this question.

Originally Posted by ynn6871
Also to express $\displaystyle \theta_{n}$ in terms of $\displaystyle a_{n}$ and $\displaystyle b_{n}$, I am guessing it would be something like:

$\displaystyle \theta_{n}= \tan^{-1} \frac{b_{n}}{a_{n}}$
$\displaystyle \theta_n$ will be either $\displaystyle \tan^{-1} \bigl(\tfrac{b_{n}}{a_{n}}\bigr)$ or $\displaystyle \tan^{-1} \bigl(\tfrac{b_{n}}{a_{n}}\bigr)+\pi$, depending on the signs of $\displaystyle a_{n}$ and $\displaystyle b_{n}$.