# Is Fourier transform still writing a function as a series of sines and cosines?

In the Fourier series we write a function as a series of sines and cosines. Fourier transform seems to me to be totally different, we are not finding a series but rather a function $\hat f(w)$. So what is the connection between them and when we write a Fourier transform are we still writing a function as a series of sines and cosines?

• @JessicaK Yes, but in the Fourier transform you are multiplying this by another function, unlike in the Fourier series where you are multiplying it by a constant. – Quantum spaghettification Apr 19 '15 at 11:16
• If I remember correctly: Fourier series is used to describe a signal in frequency domain with discrete frequencies, while Fourier transform is used with continuous frequencies. – Bhaskar Apr 19 '15 at 11:16
• Possible duplicates: Difference between Fourier series and Fourier transformation – Bhaskar Apr 19 '15 at 11:41
• Your Fourier integrals contain the function $\exp(\pm ikx) = \cos(kx) \pm i \sin(kx)$.. – mattos Apr 19 '15 at 11:51

I think I know what's bothering you! Yes, if I have understood your point of view, a Fourier series and a Fourier transform are two completely different things. Short answer: the Fourier transform is a continuous version of the coefficients in a Fourier series - and just as you have to sum the coefficients alongside with the sines and cosines to get back a function, you have to integrate the Fourier transform in a proper way to get back to the function. Let's recap. Leaving aside issues of convergence for a bit, and pathological examples, a Fourier series is a representation of some square integrable function $f\in L^2([-\pi,\pi])$ as a vector with the basis $\{\frac{1} {\sqrt2} \}\cup\{\cos{nx},\sin{nx}\}_{n\in\mathbb N}$. The most convenient scalar product on $L^2([-\pi,\pi])$ is given by $(f,g):=\frac{1}{\pi}\int_{-\pi}^{\,\pi}f(x)^*g(x)\mathrm dx$. It's with respect to this norm that the aforementioned basis vectors are orthonormal. Let's agree to call these basis vectors $e_n$, by assigning some $n\in\mathbb R$ to each one.

Now, a Fourier series amounts to claiming that $f(x)=\sum\limits_{n\in\mathbb N}(e_n,f)\,e_n$, or, defining Fourier coefficients $\hat{f_n}:=(e_n,f)$, $f(x)=\sum\limits_{n\in\mathbb N}\hat{f_n}\,e_n$.

By writing these Fourier coefficients suggestively, I'm trying to draw an analogy that might be helpful to you. Namely, it's the Fourier transform and the Fourier coefficients that are analogous!

To really see the analogy, we 'exponentialize' the series by writing the following: $a_n \cos{nx}+b_n\sin{nx}=\frac{a_n}{2}(e^{inx}+e^{-inx})+\frac{b_n}{2i}(e^{inx}-e^{-inx}):= \hat{f_n} e^{i n x}+ \hat{f}_{-n}e^{-inx}$, where now $\hat{f}_{\pm n}:=\frac{a_n}{2}\pm\frac{b_n}{2i}$. This enables us to write $f(x)=\sum\limits_{n=-\infty}^{\infty} \hat{f_n} e^{i n x}$. The point is that we can define a "coefficient function" $\hat{f}:\mathbb Z \to \mathbb R$ so that $\hat{f}(n)=\hat{f_n}$. On an arbitrary interval $[-L,L]$ this means that $\hat{f}(n)=\frac{1}{2 L}\int_{-L}^L f(k) e^{-in k\pi/L}\mathrm dk$.

Now the connection of these coefficients and the Fourier transform should be a bit more tangible. We can't just let $L\rightarrow \infty$ right away, and this is where the analogy breaks down, right near the end! The rest of the argument is really hand-wavy, but I will reproduce it. As you know, the problem is that if you let L go to infinity, you have functions that are "infinite" periodic, which doesn't really mean anything. (Does $f(x)=f(x+\infty)$ mean anything??)

Anyways, we could, say, define $\lambda_n:=n\pi/L$, and similarly $\Delta \lambda = \lambda_{n+1}-\lambda_{n}=\pi/L$, and do the following:

$f(x)=\sum\limits_{n=-\infty}^{\infty} \hat{f_n} e^{i n x \pi/L}=\sum\limits_{n=-\infty}^{\infty} \big(\lim\limits_{L\rightarrow\infty}\frac{1}{2 L}\int_{-L}^L f(k) e^{-in k\pi/L}\mathrm dk \big) e^{i n x \pi/L} \\ =\lim\limits_{L\rightarrow\infty}\sum\limits_{n=-\infty}^{\infty} \big(\frac{1}{2 \pi}\int_{-L}^L f(k) e^{-ik\lambda_n}\mathrm dk \big)e^{i x \lambda_n}\Delta\lambda \rightarrow \frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^{\infty}\hat{f}(\lambda)e^{i x \lambda} \mathrm d\lambda$.

Here $\hat{f}(\lambda)=\frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^{\infty} f(k) e^{-i k \lambda}\mathrm d k$ is the Fourier transform - not the function, but a generalized coefficient! Hopefully this helps!

A clear connection between Fourier series and the Fourier transform can be formulated in the framework of distribution theory. It turns out that the maximal domain of the Fourier transform is the space of so-called tempered distributions. These are linear functions from a space $S$, called the Schwartz class, into $\mathbb{R}$, with some continuity properties. The Schwartz class consists of smooth functions which have a certain strong decay property at infinity. The Schwartz class is somewhat larger than the space of smooth functions with compact support; for instance it includes $e^{-x^2}$.

One of the most basic kind of tempered distributions are integration against a function, where the function has enough integrability for this to make sense. For instance, given $f \in L^2$, we can define a tempered distribution $d$ by the equation $d(g)=\int_{-\infty}^\infty f(x) g(x) dx$. With this in mind we often write $\langle d,g \rangle$ instead of $d(g)$, by analogy to the $L^2$ inner product. We also usually identify the distribution given by integration against a function with the function itself.

The Fourier transform of a tempered distribution is defined so that the Fourier transform stays unitary. This means that $\langle \hat{d},\hat{g} \rangle$ should be $\langle d,g \rangle$. Replacing $g$ with $\check{h}$ (the inverse Fourier transform of $h$), we get a definition: $\langle \hat{d},h \rangle = \langle d,\check{h} \rangle$. Note that for various reasons, we often use Fourier transforms which are not normalized to be unitary. In this case the definition has an additional constant factor.

The connection is that the Fourier transform of a function $f$ with finite Fourier expansion $f(x)=\sum_{k=-n}^n a_k e^{i k x}$ is $\hat{f}(\xi) = \sum_{k=-n}^n a_k \delta(\xi-k)$, where $\delta$ denotes the Dirac delta. (Again there may be some omissions of normalization-dependent constants here.) In other words, the Fourier transform of a function with a finite Fourier expansion is a distribution which is entirely concentrated at the frequencies represented in the Fourier expansion, where the weight of each frequency is the coefficient $a_k$. After taking convergence issues into account, this can be extended to infinite Fourier series as well.