Fourier Series

**ynn6871** · March 29th 2009, 08:22 PM

I am having some trouble with the following problem:

Show that the Fourier series:

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

can be written in the form

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

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

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

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

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

**Opalg** · March 30th 2009, 12:13 AM

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

**ynn6871** · March 30th 2009, 10:38 AM

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

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

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

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

**Opalg** · March 30th 2009, 12:59 PM

Originally Posted by ynn6871

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

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

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 $a_n$ and $b_n$ 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 $a_n$ and $b_n$ are not relevant to this question.

Originally Posted by ynn6871

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

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

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

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