# Thread: Parseval's identity: Real form

1. ## Parseval's identity: Real form

$\displaystyle \|f\|^2 = \sum_{n=-\infty}^{\infty}|c_n|^2$ where $\displaystyle c_n$ are the fourier coefficients.

What does Parseval's identity look like if you use the coefficients from the real form of the fourier series, i.e. $\displaystyle a_n = c_n + c_{-n}$, $\displaystyle b_n = i(c_n - c_{-n})$ and $\displaystyle f \sim \frac{a_0}{2}+ \sum_{n=1}^{\infty}(a_n\cos(n\Omega t) + b_n\sin(n\Omega t) )$

I've seen it used like this, but I can't figure out how you get here: $\displaystyle \|f\|^2 = \left|\frac{a_0}{2}\right|^2 +\frac{1}{2}\sum_{n=1}^{\infty}(|a_n|^2+|b_n|^2)$

2. Originally Posted by Mondreus
$\displaystyle \|f\|^2 = \sum_{n=-\infty}^{\infty}|c_n|^2$ where $\displaystyle c_n$ are the fourier coefficients.

What does Parseval's identity look like if you use the coefficients from the real form of the fourier series, i.e. $\displaystyle a_n = c_n + c_{-n}$, $\displaystyle b_n = i(c_n - c_{-n})$ and $\displaystyle f \sim \frac{a_0}{2}+ \sum_{n=1}^{\infty}(a_n\cos(n\Omega t) + b_n\sin(n\Omega t) )$

I've seen it used like this, but I can't figure out how you get here: $\displaystyle \|f\|^2 = \left|\frac{a_0}{2}\right|^2 +\frac{1}{2}\sum_{n=1}^{\infty}(|a_n|^2+|b_n|^2)$

As $\displaystyle c_n=\frac{1}{2}(a_n\pm ib_n)\Longrightarrow |c_n|^2=\frac{1}{4}(a_n^2+b_n^2)$ ...

Tonio

3. Both $\displaystyle a_n$ and $\displaystyle b_n$ are complex numbers though, so I don't think what you posted holds in general. $\displaystyle a_n^2 \neq |a_n|^2$ in general as well.

4. Originally Posted by Mondreus
Both $\displaystyle a_n$ and $\displaystyle b_n$ are complex numbers though, so I don't think what you posted holds in general. $\displaystyle a_n^2 \neq |a_n|^2$ in general as well.
If you write out the square integral of what you call the real form in extensive form you have terms of the form:

$\displaystyle \int_{t=-\pi/\Omega}^{\pi/\Omega} A_{n,m} \cos(n\Omega t) \sin( m\Omega t)\;dt=0$

$\displaystyle \int_{t=-\pi/\Omega}^{\pi/\Omega} B_{n,m} \sin(n\Omega t) \sin( m\Omega t)\;dt=\frac{1}{2}B_{n,m}\delta_{n,m}$

$\displaystyle \int_{t=-\pi/\Omega}^{\pi/\Omega} C_{n,m} \cos(n\Omega t) \cos( m\Omega t)\;dt=\frac{1}{2}C_{n,m}\delta_{n,m}$

CB

5. Originally Posted by Mondreus
Both $\displaystyle a_n$ and $\displaystyle b_n$ are complex numbers though, so I don't think what you posted holds in general. $\displaystyle a_n^2 \neq |a_n|^2$ in general as well.

This seems to be a problem of notation: I meant the real coefficients (real and imaginary parts) of the complex ones, and because of the same reason the coefficient 1/2 may change, according as what interval we're choosing to define our periodic functions on.

Tonio

6. Originally Posted by CaptainBlack
If you write out the square integral of what you call the real form in extensive form you have terms of the form:

$\displaystyle \int_{t=-\pi/\Omega}^{\pi/\Omega} A_{n,m} \cos(n\Omega t) \sin( m\Omega t)\;dt=0$

$\displaystyle \int_{t=-\pi/\Omega}^{\pi/\Omega} B_{n,m} \sin(n\Omega t) \sin( m\Omega t)\;dt=\frac{1}{2}B_{n,m}\delta_{n,m}$

$\displaystyle \int_{t=-\pi/\Omega}^{\pi/\Omega} C_{n,m} \cos(n\Omega t) \cos( m\Omega t)\;dt=\frac{1}{2}C_{n,m}\delta_{n,m}$

CB
Sorry, I don't understand that notation.

7. Originally Posted by Mondreus
Sorry, I don't understand that notation.
It is just a statement that the trig functions form an orthonormal basis for the space of square integrable functions on the interval $\displaystyle (-\pi,pi]$.

In particular I use Kronecker's delta:

$\displaystyle \delta_{a,b}=\left\{ \begin{array}{ll} 1&, {\text{ if }} a=b\\ 0&, {\text{ if }} a \ne b \end{array} \right.$

CB