# Thread: Parseval's identity: Real form

1. ## Parseval's identity: Real form

$\|f\|^2 = \sum_{n=-\infty}^{\infty}|c_n|^2$ where $c_n$ 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. $a_n = c_n + c_{-n}$, $b_n = i(c_n - c_{-n})$ and $f \sim \frac{a_0}{2}+ \sum_{n=1}^{\infty}(a_n\cos(n\Omega t) + b_n\sin(n\Omega t) )$

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

2. Originally Posted by Mondreus
$\|f\|^2 = \sum_{n=-\infty}^{\infty}|c_n|^2$ where $c_n$ 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. $a_n = c_n + c_{-n}$, $b_n = i(c_n - c_{-n})$ and $f \sim \frac{a_0}{2}+ \sum_{n=1}^{\infty}(a_n\cos(n\Omega t) + b_n\sin(n\Omega t) )$

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

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

Tonio

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

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

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

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

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

CB

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

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

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

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

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 $(-\pi,pi]$.

In particular I use Kronecker's delta:

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

CB