What's the definition of "periodic extension of a function"?

In summary, Fourier transformations aim to approximate a function by (periodic) trigonometric functions. Periodic extension is defined as repeating the function over a certain interval and is necessary for the Fourier series to approximate the original function. The Fourier series is considered a periodic extension of the original function and can be used to approximate the original function in its original range.
  • #1
CGandC
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TL;DR Summary
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I'm learning about Fourier theory from my lecture notes and I have a few questions that I wasn't able to concretely find answers to:

1. What's the definition of periodic extension? I think the definition is as follows ( Correct me if I'm wrong please ):
for ## f: [ a,b) \to \mathbb{R} ## its periodic extension is defined as ## \tilde{f}(x+n(b-a))=f(x) \quad ~~~ , \forall x \in[a, b), \quad n \in \mathbb{Z} ##.

2. Why is it necessary to periodically extend a function? In my lecture notes, before calculating ## \hat{f}(n) ## ( the Fourier coefficients of a periodic function ## f: [ a,b) \to \mathbb{R} ## with period ## T ## ) it is said that ## f ## should be periodically extended. But I still don't fully understand why a periodic extension is necessary.

3. Is Fourier series considered a periodic extension of a function?. I mean, is the following true?
Suppose ## H : \mathbb{R} \to \mathbb{R} ## is given. Also, suppose ## f: [ a,b) \to \mathbb{R} ## is given.
## H ## is Fourier series of ## f ## ## \iff ## ## H ## is a periodic extension of ## f ##

4. Does a function ## f ## must have a period in order to be periodically extendible?, according to my definition in question 1 above, the answer's no ( because the period can be defined as the length of the interval ), but still.

Thanks in advance for the help!
 
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  • #2
The goal of Fourier transformations is to approximate a function by (periodic) trigonometric functions. So either you have to make your function periodic or restrict the trig functions to a certain interval. The first case seems to be easier and reasonable since we do not have to "redefine" trig functions and can convert them into the exponential function at any time.
 
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  • #3
So I'm defining a sequence of functions ## \{ f_n \}_{ n \in \mathbb{Z} } ## where for every ## n \in \mathbb{Z} ##, ## f_n : [ a+n,b+n) \to \mathbb{R} ##, and this sequence is said to be the periodic extension of ## f ##? or is it that each such function is defined to be a perioidic extension of ## f ## ?

Or is it that I'm defining an entire function ## F : \mathbb{R} \to \mathbb{R} ## which is defined as the periodic extension of ## f ##?

I think it's the latter since I need a function defined on ##\mathbb{ R} ## so It can be approximated by a span of trigonometric polynomials ( which is then said to be the Fourier series of ## f ## ).
 
  • #4
I am not certain and would check it with the books. But I think we simply expand the domain of ##f## by repeating it: ##F\, : \,\mathbb{R}\longrightarrow \mathbb{R}## defined as ##F(x)=f(x-kb+ka)\text{ if } x\in [kb-(k-1)a,(k+1)b-ka).##

We won't get continuity at the interval limits, but that shouldn't be a problem.
 
  • #5
CGandC said:
I think it's the latter since I need a function defined on ##\mathbb{ R} ## so It can be approximated by a span of trigonometric polynomials ( which is then said to be the Fourier series of ## f ## ).
The Fourier series will approximate the extended function. You can use it to approximate the original function in the original range, but you should realize that the approximation is actually for the extended function. That explains why the approximation at each endpoint is heavily affected by the values at the other endpoint. If you want a better approximation for the original function in the original range, there are many that will go exactly through the endpoints. But they will not be as useful in analyzing the frequency content of the function.
 
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  • #6
CGandC said:
Summary:: x

I'm learning about Fourier theory from my lecture notes and I have a few questions that I wasn't able to concretely find answers to:

1. What's the definition of periodic extension? I think the definition is as follows ( Correct me if I'm wrong please ):
for ## f: [ a,b) \to \mathbb{R} ## its periodic extension is defined as ## \tilde{f}(x+n(b-a))=f(x) \quad ~~~ , \forall x \in[a, b), \quad n \in \mathbb{Z} ##.

2. Why is it necessary to periodically extend a function? In my lecture notes, before calculating ## \hat{f}(n) ## ( the Fourier coefficients of a periodic function ## f: [ a,b) \to \mathbb{R} ## with period ## T ## ) it is said that ## f ## should be periodically extended. But I still don't fully understand why a periodic extension is necessary.

3. Is Fourier series considered a periodic extension of a function?. I mean, is the following true?
Suppose ## H : \mathbb{R} \to \mathbb{R} ## is given. Also, suppose ## f: [ a,b) \to \mathbb{R} ## is given.
## H ## is Fourier series of ## f ## ## \iff ## ## H ## is a periodic extension of ## f ##

4. Does a function ## f ## must have a period in order to be periodically extendible?, according to my definition in question 1 above, the answer's no ( because the period can be defined as the length of the interval ), but still.

Thanks in advance for the help!
Ad 1: correct. A function that is continuous in a given interval ##[a,b)## is just copied and pasted infinite times so to speak towards ##\pm\infty##.

Ad 2: A periodic extension of a function is not necessary, but first of all just a concept. A better question would be: when is a periodic extension an interesting concept? An example where a periodic extension is applied is when the remainder term in the Euler--Maclaurin series is to be given in an explicit form: https://en.wikipedia.org/wiki/Euler–Maclaurin_formula

Ad 3: No. If a function H is given for all ##\mathbb{R}##, a periodic extension is simply speaking the restriction of that function on an interval ##[a,b)## and above-mentioned copy & paste patching.

Ad 4: No. The function only needs to be continuous in some half-open interval, which then defines the patch to be copied & pasted.
 
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  • #7
otennert said:
Ad 1: correct. A function that is continuous in a given interval ##[a,b)## is just copied and pasted infinite times so to speak towards ##\pm\infty##.

Ad 2: A periodic extension of a function is not necessary, but first of all just a concept. A better question would be: when is a periodic extension an interesting concept? An example where a periodic extension is applied is when the remainder term in the Euler--Maclaurin series is to be given in an explicit form: https://en.wikipedia.org/wiki/Euler–Maclaurin_formula

Ad 3: No. If a function H is given for all ##\mathbb{R}##, a periodic extension is simply speaking the restriction of that function on an interval ##[a,b)## and above-mentioned copy & paste patching.

Ad 4: No. The function only needs to be continuous in some half-open interval, which then defines the patch to be copied & pasted.

Helpful answer, thank you!
 
  • #8
otennert said:
Ad 4: No. The function only needs to be continuous in some half-open interval, which then defines the patch to be copied & pasted.
It does not need to be continuous inside the interval where it is defined.
 
  • #9
FactChecker said:
It does not need to be continuous inside the interval where it is defined.
I think it's usually expected to be in ##L^2[a,b)##, or whatever interval it's defined in, right? Maybe in ##L^p[a,b)## for some p? Edit: IIRC, the set ## \{ Cos(nx), Sin(nx) n=1,2,...\} ## is dense in ##L^2[a,b]##. Sorry for being lazy; it seems you have to filter a lot in web sources to get to this and similar.
 
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  • #10
WWGD said:
the set ## \{ Cos(nx), Sin(nx) n=1,2,...\} ## is dense in ##L^2[a,b]##.
I think you mean the set generated by that basis.
 
  • #11
Let ##f\in L^2(0,2\pi)##. Then the series ##\sum_{k\in\mathbb{Z}}f_ke^{ikx}## converges to a function ##g\in L^2_{loc}(\mathbb{R})## in ##L^2_{loc}(\mathbb{R})##. Here ##f_k## are the Fourier coefficients of ##f##.
The function ##g## has the following properties:
1) ##g(x)=f(x)## almost everywhere in ##(0,2\pi)## and
2) ##g(x+2\pi)=g(x)## for almost all ##x\in\mathbb{R}##.

(##L^p(0,2\pi)\subset L^2(0,2\pi)## for ##p\in[2,\infty]##)

Convergence in the space of continuous functions or in ##L^p(0,2\pi),\quad p\ne 2## is a much more tricky story
 
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  • #12
Yes, sorry, the span of ## aCos(nx)+ bSin(nx)## is dense in ##L^2[a,b]##. Yes, Hilbert spaces are just-about always nicer to work with. There is an algorithm to construct approximations in ##C[a,b]## using the dense subset ##\{1, x, x^2, ..., x^n,... \} ##. Using convolutions, I think.
 
  • #13
wrobel said:
Let ##f\in L^2(0,2\pi)##. Then the series ##\sum_{k\in\mathbb{Z}}f_ke^{ikx}## converges to a function ##g\in L^2_{loc}(\mathbb{R})## in ##L^2_{loc}(\mathbb{R})##. Here ##f_k## are the Fourier coefficients of ##f##.
The function ##g## has the following properties:
1) ##g(x)=f(x)## almost everywhere in ##(0,2\pi)## and
2) ##g(x+2\pi)=g(x)## for almost all ##x\in\mathbb{R}##.

(##L^p(0,2\pi)\subset L^2(0,2\pi)## for ##p\in[2,\infty]##)

Convergence in the space of continuous functions or in ##L^p(0,2\pi),\quad p\ne 2## is a much more tricky story
I thought inclusion applied only to bounded intervals. By Holder's theorem, I think.
 
  • #14
WWGD said:
I thought inclusion applied only to bounded intervals. By Holder's theorem, I think.
yes
WWGD said:
There is an algorithm to construct approximations in C[a,b] using the dense subset {1,x,x2,...,xn,...}. Using convolutions, I think.
the set of polynomials is dense , yes
 
  • #15
The set of trigonometric polynomials is dense in C[a,b] but not each function from C[a,b] can be expanded into the Fourier series

Gelbaum Olmsted Counterexamples in analysis
 

1. What is the definition of "periodic extension of a function"?

A periodic extension of a function is a mathematical concept where a given function is extended beyond its original domain by repeating the function values at regular intervals. This results in a function that repeats itself infinitely in both directions.

2. Why is periodic extension important in mathematics?

Periodic extension is important because it allows us to study and analyze functions over a larger domain, which can provide insights into the behavior of the function. It also helps in simplifying complex functions and making them more manageable to work with.

3. How is a periodic extension different from a periodic function?

A periodic extension is an extension of a function beyond its original domain, while a periodic function is a function that repeats itself at regular intervals within its original domain. In other words, a periodic extension is a way to extend a non-periodic function to make it periodic, while a periodic function is already periodic within its given domain.

4. Can any function be extended periodically?

No, not all functions can be extended periodically. For a function to have a periodic extension, it must satisfy certain conditions, such as being continuous and having a finite number of discontinuities within its original domain. Functions with infinite discontinuities or those that are not defined over a finite interval cannot be extended periodically.

5. How is periodic extension used in real-world applications?

Periodic extension has various applications in fields such as engineering, physics, and signal processing. It is used to model and analyze periodic phenomena, such as sound waves, electrical signals, and physical vibrations. It also plays a crucial role in Fourier analysis, which is used in image and signal processing, data compression, and data analysis.

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