- #1

- 59

- 0

I was wondering. Since the Fourier Series coefficients can just be represented in the form of a Fourier Transform, what is the point of ever finding the Fourier coefficients and not doing the transform?

- Thread starter Rib5
- Start date

- #1

- 59

- 0

I was wondering. Since the Fourier Series coefficients can just be represented in the form of a Fourier Transform, what is the point of ever finding the Fourier coefficients and not doing the transform?

- #2

- 341

- 0

- #3

mathman

Science Advisor

- 7,889

- 460

Fourier series are defined for periodic functions (finite interval) while Fourier transforms are defined for functions integrable over the real line. They just have different contexts.

I was wondering. Since the Fourier Series coefficients can just be represented in the form of a Fourier Transform, what is the point of ever finding the Fourier coefficients and not doing the transform?

- #4

- 1,102

- 6

Fourier series are defined for periodic functions (finite interval) while Fourier transforms are defined for functions integrable over the real line. They just have different contexts.

Yes, but the OP's point is that all functions that can be expanded in fourier series can also be fourier transformed so why bother with fourier series at all. And I can't say I really have a reason to disagree.

- #5

- 59

- 0

I asked my teacher about it in class today, and he pretty much said the same thing. He said that the main reason we learn about fourier series is that it helps students understand the fourier transform more intuitively, and also for historical reasons about how the fourier transform was developed.Yes, but the OP's point is that all functions that can be expanded in fourier series can also be fourier transformed so why bother with fourier series at all. And I can't say I really have a reason to disagree.

- #6

- 488

- 2

How about, you do the Fourier series because it is easier.

- #7

mathman

Science Advisor

- 7,889

- 460

- #8

- 166

- 0

The OP's question concerns the other way around. Any function that has a Fourier series also has a Fourier transform. How are they related? The following argument has been taken from Brad Osgood's excellent course that's available online.^{-x2/2}. It has a Fourier transform, but you cannot expand it with a Fourier series.

Let [itex]\Phi[/itex] be a periodic function of period T. Then [itex]\Phi[/itex] can be written as

[tex]\Phi(t) = \phi(t) \ast \sum_{k = -\infty}^\infty \delta(t-kT) [/tex]

where [itex]\phi[/itex] is one period of [itex]\Phi[/itex] and [itex]\ast[/itex] denotes convolution. Now, by the convolution theorem (the Fourier tranform of a convolution is the product of the Fourier transforms),

[tex]

\begin{align*}

\mathcal{F}\Phi(s) &=

\Bigl(\mathcal{F}\phi(s) \Bigr) \frac{1}{T}\sum_{k = -\infty}^\infty \delta\left(s-\frac{k}{T}\right) \\

&= \frac{1}{T}\sum_{k = -\infty}^\infty \mathcal{F}\phi\left(\frac{k}{T}\right)\delta\left(s-\frac{k}{T}\right)

\end{align*}

[/tex]

Taking the inverse Fourier transform gives us back [itex]\Phi[/itex]. Since [itex]\mathcal{F}\phi\left(k/T\right)[/itex] are constants and the inverse Fourier transform of a shifted delta function is a complex exponential,

[tex]

\Phi(t) = \sum_{k = -\infty}^\infty \frac{1}{T}\mathcal{F}\phi\left(\frac{k}{T}\right) e^{2\pi i k t/T}

[/tex]

However,

[tex]

\frac{1}{T}\mathcal{F}\phi\left(\frac{k}{T}\right) =

\frac{1}{T} \int_{-\infty}^{\infty} e^{-2\pi i (k/T)t}\phi(t) \, dt \\

= \frac{1}{T} \int_{0}^{T} e^{-2\pi i (k/T)t}\Phi(t) \, dt

[/itex]

But this is the k-th Fourier coefficient [itex]c_k[/itex] of [itex]\Phi[/itex]. In other words, the Fourier series is a consequence of Fourier transform theory.

[tex]

\Phi(t) = \sum_{k = -\infty}^\infty c_k e^{2\pi ikt/T}, \qquad \text{where} \qquad

c_k = \frac{1}{T} \int_{0}^{T} e^{-2\pi i (k/T)t}\Phi(t) \, dt

[/tex]

Last edited:

- #9

- 675

- 4

Indeed, computational physicists use the terms "Fourier transform" and "Fourier series" interchangeably in casual conversation.

- #10

jasonRF

Science Advisor

Gold Member

- 1,369

- 435

Also, as noted by pellman, Fourier series show up in numerics all the time. And in the era of digitized data, the DFT (usually implemented as an FFT whenever possible) is used all the time to look at spectral content or to quickly do convolutions. What is the DFT? It effectively assumes your data are periodic and that you only have one period sampled, so it does the periodic extension of the data (like you learn about with simple Fourier series problems!) and finds the discrete Fourier series.

So of course you are right, and it is probably good for you to lump them together in your head (along with Laplace transforms, etc.), but I would claim that making yourself work through some of the Fourier series results in a class is still a useful exercise.

jason

- Replies
- 1

- Views
- 737

- Last Post

- Replies
- 1

- Views
- 2K

- Replies
- 1

- Views
- 1K

- Last Post

- Replies
- 1

- Views
- 1K

- Replies
- 3

- Views
- 1K

- Last Post

- Replies
- 1

- Views
- 1K

- Last Post

- Replies
- 1

- Views
- 2K

- Replies
- 1

- Views
- 3K

- Replies
- 4

- Views
- 2K

- Replies
- 1

- Views
- 1K