I am confused about when to include a prefactor of $\frac{1}{2\pi}$ when dealing with integrals of functions that are expressed as fourier series. This is what I understand (please correct me if I'm wrong). Assume I have a square integrable function, $f\left(x\right)$. I can express it as a fourier series according to:

$$f\left(x\right)=\sum_{k=-\infty}^{\infty} c_k e^{ikx}$$

Where the coefficients, $c_k$, are obtained via the following inner-product:

$$c_k=\frac{1}{2\pi}\int_{0}^{2\pi} e^{-ikx} f\left(x\right)\text{d}x$$

My understanding of the standard inner-product between functions is, that it is defined by:

$$\left\langle h\left(x\right),g\left(x\right)\right\rangle \equiv \int_{\Omega} h^*\left(x\right) g\left(x\right) \text{d}\Omega$$

Where the $^*$ denotes complex conjugation, $\Omega$ is the domain, and $\text{d}\Omega$ is the invariant measure.

So, using this definition, I would intuitively think that the coefficients for a fourier series ought to be computed from:

$$c_k=\int_{0}^{2\pi} e^{-ikx} f\left(x\right)\text{d}x$$

i.e. without the prefactor of $\frac{1}{2\pi}$.

Question 1: So, why is the prefactor of $\frac{1}{2\pi}$ included in the special case when the basis functions are complex exponentials, i.e. for Fourier series'? I know that the complex exponentials are not orthonormal, only orthogonal, but why should the normalization convention of the basis functions change the definition of the inner-product? I don't think it comes from the invariant measure, since for a circle the invariant measure should be $\text{d}x$, not $\frac{1}{2\pi}\text{d}x$


Now, say that I have an equation that is defined using an integral whose arguments include functions expressed as Fourier series. Specifically:

$$\bar{S} = \int_{0}^{2\pi} f\left(x\right)S\left(x\right)\text{d}x$$

Where $\bar{S}$ represents the average value of $S\left(x\right)$ weighted by the probability density function $f\left(x\right)$. If I express $f\left(x\right)$ and $S\left(x\right)$ as Fourier series, and use the fact that $f\left(x\right)=f^*\left(x\right)$, then the equation should be:

$$\bar{S} = \sum_{k=-\infty}^{\infty} \sum_{m=-\infty}^{\infty} c_k^* s_m\int_{0}^{2\pi} e^{-ikx} e^{imx} \text{d}x=2\pi \sum_{k=-\infty}^{\infty} c_k^* s_k$$

Question 2: In this case the integral amounts to doing an inner-product but without the prefactor of $\frac{1}{2\pi}$. So, when doing integrals of functions that are expressed as Fourier series do you need to always throw in that prefactor or not? When should you?

up vote 3 down vote accepted

It depends on the inner product and the basis used.

Suppose you have an inner product $\langle \cdot, \cdot \rangle$ and an orthogonal (Schauder) basis $b_k$ for a Hilbert space $\mathbb{H}$.

If you represent some $f \in \mathbb{H}$ in terms of the basis, you will have $f = \sum_k \hat{f_k} b_k$, where $\hat{f_k}$ are the Fourier coefficients with respect to the basis $b_k$.

Then you have no choice over the $\hat{f_k}$, as they must satisfy $\langle b_k, f \rangle = \hat{f_k} \|b_k\|^2$, or $\hat{f_k} = \frac{\langle b_k, f \rangle}{\|b_k\|^2}$.

To answer your first question. Presumably you are dealing with $L^2[0,2 \pi]$, with $\langle f, g \rangle = \int_0^{2 \pi} \overline{f}(t)g(t) dt$, and the basis $b_k(t) = e^{i k t}$. Since $\|b_k\|^2 =\langle b_k, b_k \rangle = 2 \pi$, the coefficients are given by $\hat{f_k} = \frac{\langle b_k, f \rangle}{\|b_k\|^2} = \frac{1}{2\pi} \int_0^{2 \pi} e^{-ikt} f(t) dt$, whence the $2 \pi$.

For your second question, you have $\overline{S} = \int_{0}^{2\pi} f(x) S(x) dx$. In terms of the inner product from the previous question, this becomes $\overline{S} = \langle f, S \rangle$.

If $\hat{f_k}, \hat{S_k}$ are the Fourier coefficients (with respect to the basis and inner product above), then an easy computation (Parseval's theorem) shows that $\langle f, S \rangle = \sum_k \|b_k\|^2 \overline{\hat{f_k}} \hat{S_k}$, and so we obtain the result above (remember $\|b_k\|^2 = 2 \pi$).

  • Thanks, for that. How about other integrals though, that are not specifically inner-products, do you still need to divide by $||b_k||^2$? – okj Nov 8 '13 at 19:37
  • I added a paragraph to that effect. The norm squared of the basis functions appears, in this case it is $2 \pi$. – copper.hat Nov 8 '13 at 19:42
  • So, if I understand correctly, the definition of the inner-product is $\left\langle f,g \right\rangle = \int_{\Omega} \bar{f} g \text{d}\Omega$, period. However, we don't get the fourier coeffiecients from the inner-product, rather they are defined by $\left\langle b_k,f \right\rangle = \hat{f}_k {\left||b_k\right||}^2$, so that with an orthonormal basis the two definitions simply happen to coincide, because ${\left||b_k\right||}^2 = 1$. So then I take it that I don't have to artificially insert the ${\left||b_k\right||}^2$ with any other integrals? Is that correct? – okj Nov 8 '13 at 19:55
  • Well, not exactly. You can choose a different scaling for your inner product, but it washes out in the long run. The Fourier coefficients are given by that formula, but they are from the inner product, as such. If you normalize the basis, then essentially you are shifting constants from the coefficients to the basis. Does this make sense? – copper.hat Nov 8 '13 at 20:03
  • I think so, though it mostly seems like a "which came first, the chicken or the egg" type distinction. In any case though, am I correct in understanding that with other integrals that I may take of functions expressed as fourier series it is unnecessary to multiply the integral by $\frac{1}{{\left||b_k\right||}^2}$? – okj Nov 8 '13 at 20:19

The normalization doesn't change the inner product convention; what we are doing is projecting $f(x)$ onto the subspace spanned by $e^{ikx}$.

For instance, in a 'regular' vector space, if we wanted to expand a vector $v$ in terms of an orthogonal (not necessarily orthonormal) basis $\{e_k\}$, then we would write

$$ v = \sum_k c_k e_k; \quad c_k = \frac{\left<v,e_k\right>}{||e_k||^2} $$

In the same way, when we wish to expand a function in it's Fourier series, we have

$$ c_k = \frac{\left<f,e^{ikx}\right>}{||e^{ikx}||^2} = \frac{1}{2\pi} \int_0^{2\pi} f(x) e^{-ikx} dx. $$

If we wanted to, we could first normalize our basis functions, taking $e_k = \dfrac{e^{ikx}}{||e^{ikx}||} = \dfrac{1}{\sqrt{2\pi}} e^{ikx}$ so that $$ c_k = \frac{\left<f,e_k\right>}{||e_k||^2} = \left<f,e_k\right>= \frac{1}{\sqrt{2\pi}} \int_0^{2\pi} f(x) e^{-ikx} dx. $$ but then when we write our function in terms of the Fourier basis $$ f = \sum_k c_k e_k = \sum_k c_k \frac{1}{\sqrt{2\pi}}e^{ikx} = \sum_k \left(\frac{1}{\sqrt{2\pi}}c_k\right) e^{ikx} $$ the missing factor of $\dfrac{1}{\sqrt{2\pi}}$ comes back anyway.

Your second question is essentially about Parseval's theorem. Again, it is a matter of applying the weight $2\pi$ somewhere to properly normalize everything. If you took the basis to be $\dfrac{1}{\sqrt{2\pi}} e^{ikx}$ so that $c_k = \frac{1}{\sqrt{2\pi}} \int_0^{2\pi} f(x) e^{-ikx} dx$ (and similarly for $s_k$). Then you would have

$$ \bar S = \sum_k c_k^* s_k. $$

All in all, this is a reflection of the fact that the Fourier series isn't quite unitary. You either have to (somewhat awkwardly) normalize the basis functions, absorb the extra $2\pi$'s into the measure, or just live with 'basically unitary' and accept (and then subsequently forget about) the $2\pi$'s.

  • Thanks for identifying that the prefactor comes from the fact that the inner-product is a projection. How about other integrals though, like in my second example? – okj Nov 8 '13 at 19:19
  • @okj I've edited in an answer to your second question. – BaronVT Nov 8 '13 at 19:38

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