Talk:Fourier analysis

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This talk page has been archived at Talk:Fourier transform/Archive1. I moved the page, so the edit history is preserved with the archive page. I've copied back the most recent thread. Wile E. Heresiarch 23:58, 20 Sep 2004 (UTC)

[edit] Question about Notation

Wouldn't the correct notation be:

$f(t) = \mathcal{F}^{-1}\{F(\omega)\} = \frac{1}{\sqrt{2\pi}} \int\limits_{-\infty}^\infty F(\omega) e^{i\omega t}\,d\omega.$

$f(t) = (\mathcal{F}^{-1}\{F(\omega)\})(t) = \frac{1}{\sqrt{2\pi}} \int\limits_{-\infty}^\infty F(\omega) e^{i\omega t}\,d\omega.$

Since f(t) is in fact the inverse fourier transform of F, which is itself a function of $ω$ ?

Also, as a precedent, from the laplace transform article:

$f(t) = \mathcal{L}^{-1} \left\{F(s)\right\} = \frac{1}{2 \pi \imath} \int_{ \gamma - \imath \infty}^{ \gamma + \imath \infty} e^{st} F(s)\,ds,$

I'd say that the most correct notation, as Michael Hardy says below, is

$f(t) = (\mathcal{F}^{-1}\{F\})(t) = \frac{1}{\sqrt{2\pi}} \int\limits_{-\infty}^\infty F(\omega) e^{i\omega t}\,d\omega.$

The inverse Fourier transform operator works on a function, and the function is F. In contrast, F(ω) denotes the function F evaluated at a particular point, namely ω.

However, alternative notations are used quite frequently. -- Jitse Niesen (talk) 01:49, 13 March 2006 (UTC)

Yes, but couldn't one argue that ω is not a specific point, but rather a locus of all of the points ω such that ω extends from minus infinity to plus infinity? Also, it is helpful to show F(ω) rather than plain old F as a convenience so that we all can keep track of the fact that F is a function of ω and not, say, γ or β or whatever. Or t for that matter. -- Metacomet 02:00, 13 March 2006 (UTC)

It would also be helpful to use a letter other than F to denote the function of interest, since we are already using F (in calligraphy style) to represent the Fourier transform operator itself. Perhaps we could use, oh I don't know, maybe X or Y to represent the transformed function...

So, for example:

$X(\omega) = \mathcal{F} \{ x(t) \}$

seems less potentially ambiguous to me. -- Metacomet 02:10, 13 March 2006 (UTC)

I think using $(\mathcal{F}^{-1}\{F\})(t)$ is slightly ambiguous since it could be misinterpreted as the inverse fourier multiplied by t. It would seem extrenuous since the inverse fourier transform itself, as a result of the integral, generates a function of t... -- Tristan Jones (talk) 11:16, 13 March 2006 (MT)

Metacomet wrote: "Also, it is helpful to show F(ω) rather than plain old F as a convenience so that we all can keep track of the fact that F is a function of ω". This may well be a difference between mathematicians and engineers. For mathematicians, F is not a function of ω; F is a function which maps numbers to numbers. We can replace the dummy variable ω by another variable, say s:

$f(t) = (\mathcal{F}^{-1}\{F\})(t) = \frac{1}{\sqrt{2\pi}} \int\limits_{-\infty}^\infty F(s) e^{i\omega t}\,ds.$

I guess that for engineers or physicists, ω is not just a number but it also has a meaning (frequency, or something like that). In that context, I can see why you'd like to mention the argument.

Similarly, from a mathematical perspective, the inverse Fourier transform does not generate a function of t. It just generates a function.

Tristan is right that $(\mathcal{F}^{-1}\{F\})(t)$ is slightly ambiguous, but the same goes for

a (b + c)

: is this a times b+c or the function a applied to b+c? This ambiguity is just a feature of mathematical notation, a price we pay for its succinctness.

I agree that using f for the function to be Fourier transformed is indeed a bit confusing, giving that we use curly F for the Fourier transform operator. But it is rather customary to denote arbitrary functions by f. However, I won't complain if you were to change it. -- Jitse Niesen (talk) 07:07, 13 March 2006 (UTC)

In that case couldn't it be written as: $f = \mathcal{F}^{-1}F$ ? I think there is value in showing the omega (or any other variable) even if it is not "technically" correct, as the meaning is the same and it shows in a more explicit manner that the transform changes the function into a different 'space'... Also, since both methods could be considered correct (the omega merely shows that F is a function of some single variable, in this case valled omega, but it could be anything else), that to someone who has not previously seen the transform, writing $f(t) = \mathcal{F}^{-1}\{F(\omega)\}$ is more obvious and easier to understand... -- Tristan Jones (talk) 11:16, 13 March 2006 (MT)

In danger of ambiguity use a·(b+c) for multiplication and a(b+c) for evaluating the function a in the point (b+c). Jitse, you want to replace all three occurences of ω by s. And there is no need to use curly brackets. And the outer pair of parentheses is redundant.

$\mathcal{F}^{-1}(F)(t) = \frac{1}{\sqrt{2\pi}} \int\limits_{-\infty}^\infty F(s) e^{i s t}\,ds.$

Bo Jacoby 18:12, 13 March 2006 (UTC)

There are other forms of the continuous Fourier transform, which have different advantages and are favoured by different people because of their mathematical or practical simplicity, or because they make the inverse transform look more like the transform. I mean, it's not just a few recalcitrant nutbars with a chip on their shoulder; I'm talking widespread usage. Most discussions of the Fourier transform mention this, so maybe we ought to as well. For example,

$\mathcal{F}(f(t)) = \int\limits_{-\infty}^\infty f(t) e^{- i \pi s t}\,ds$

gets rid of the :: $\frac{1}{\sqrt{2\pi}}$ out the front, and gets s in terms of frequency rather than rads. This form is used in the course notes for Signal Processing here at Adelaide Uni. N-gon 10:08, 22 August 2006 (UTC)

If you had written $\mathcal{F}(f(t))(s)\,$ , you would have been more likely to notice that $ds\,$ is the wrong variable of integration. (Redundancy is a good thing.) I hope you also unintentionally omitted a factor of 2 with your $\pi\,$ . Thus:

$\mathcal{F}\{f(t)\}(s) = \int\limits_{-\infty}^\infty f(t)\cdot e^{- i 2\pi s t}\,dt$

But the suggestion to dispense with both the normalization factors and the radian/sec units (for cycles/sec) in one fell swoop is of course very sensible. If I were king, I would decree it and one more thing. I would switch $f\,$ and $s\,$ ; i.e. $f\,$ works well for frequency, leaving $s\,$ available for (say) signal (and $\mathcal{F}\{s\} = S(f)\,$ for Spectrum):

$\mathcal{F}\{s(t)\}(f) = \int\limits_{-\infty}^\infty s(t)\cdot e^{- i 2\pi f t}\,dt$

--Bob K 23:45, 22 August 2006 (UTC)

It's clear that widespread alternative conventions should be mentioned. However, it is best to keep this article a short summary and leave such details to continuous Fourier transform. Otherwise we end up with a lot of redundancy —Steven G. Johnson 16:03, 23 August 2006 (UTC)

[edit] Recommended Book

Delete this if you like but it will help lots of people. If you want to *truly* understand the Fourier Transform and where it really comes from then read the book "Who is Fourier? A mathematical adventure" ISBN 0964350408. It's excellent. Lecturers can't teach this subject for toffee. It's a shame I only found this book after my course involving Fourier but it more than makes up for it now. You can get it off Amazon.com Chris, Wales UK 16:14 25th November 2005

[edit] F notation

I don't know where else to ask. What's the difference between $\mathcal{F}$ and $F\,$ . Are they used correctly in this article? - Omegatron 01:56, Sep 19, 2004 (UTC)

Seems clear to me from the article.

F

is the function whose Fourier transform is to be found, and $\mathcal{F}(F)$ is the transformed function, so that $\mathcal{F}$ is the transform itself, i.e., the mapping from one space of functions to another. Michael Hardy 22:25, 19 Sep 2004 (UTC)

If you know it, explain it in the integral transform article, as well. - Omegatron 02:02, Sep 19, 2004 (UTC)

The function $F\,$ is the Fourier transform of the function $f\,$ . $\mathcal{F}$ is the Fourier transform operator.CSTAR 02:18, 19 Sep 2004 (UTC)

Ok. But when do you use the $\mathcal{F}$ ? $\mathcal{F}(f(x))$ ? hmm is there an article for operator? ... Yes, but it doesnt mention the notation. - Omegatron 02:28, Sep 19, 2004 (UTC)

Strictly speaking, $(\mathcal{F} F)(t)$ is the right way to parse the thing, and $\mathcal{F}(f(x))$ is a solecism, often used by engineers and sometimes used even by mathematicians. When you use $\mathcal{F}$ alone would include such things as when you say "the Fourier transform is a 90 ° rotation of the space of square-integrable functions". I seem to recall that the article titled operator was something of an Augean stable, but that was months ago; I don't know if it's improved. Michael Hardy 22:30, 19 Sep 2004 (UTC)

f(x) denotes the value of f at x. f denotes a function x denotes a real number. $\mathcal{F}(f(x))$ would make sense only if $\mathcal{F}$ were defined for numbers, but it;s defiend for functions.CSTAR 02:34, 19 Sep 2004 (UTC)

Oh oh I see. So

$F(\omega) = \mathcal{F}(f(t))$

Yeah. If you're picky you will note that t has no real purpose in the above formula. It really should be within the scope of a binding operator such as $t \mapsto f(t) \quad$ .; this however by rules of lambda-calclus reduces to the term f.CSTAR 02:47, 19 Sep 2004 (UTC)

$\mathcal{F}^{-1}(F)(t)$ is used in the article. is this the correct way to say the above, are they both incorrect, or is the t just extraneous but it doesn't really matter? - Omegatron 02:54, Sep 19, 2004 (UTC)

The variable t occurs on the right hand side as well as the left hand side. The RHS is a term (in this case an integral of an exponential); You can't get rid of the t. Try it. What would you get?CSTAR 03:12, 19 Sep 2004 (UTC)

In that case, I don't understand. I thought you meant that it was correct to use $F(\omega) = \mathcal{F}(f)$ . - Omegatron 03:30, Sep 19, 2004 (UTC)

If you think of $F(\omega) = \mathcal{F}(f)$ as meaning two things: (a) $F = \mathcal{F}(f)$ and (b) a suggestion that the symbol

ω

is reserved to name an independent variable to name the argument of the function F, since

ω

is often thought of (for instance by physicist or engineers) as frequency.CSTAR 03:40, 19 Sep 2004 (UTC)

I thought I understood, but I guess not. The term "scope of a binding operator" would probably help. You don't have to teach it to me if it's something I don't already know. Just point me where to look. Regardless, is the article notation right? - Omegatron 14:42, Sep 19, 2004 (UTC)

See variable-binding operator or some such thing. There is a link to this somewhere in wikipedia. CSTAR 15:36, 19 Sep 2004 (UTC)

It's at free variables and bound variables. Michael Hardy 22:14, 19 Sep 2004 (UTC)

I'm used to engineering notation, where $X(f) = \mathcal{F}(x(t))$ , so you can use f for frequency and avoid confusing Fs. Then of course there's $F(f) = \mathcal{F}(f(x))$ . :-) - Omegatron 02:40, Sep 19, 2004 (UTC)

In fact, I vote that we change f(t)->F(ω) into some other letter (I know you won't use X, but maybe g?), to avoid confusing newcomers to the Fourier transform. - Omegatron 02:42, Sep 19, 2004 (UTC)

I didn't make the choice of notation here. I'll leave your suggestion to somebody else.CSTAR 02:46, 19 Sep 2004 (UTC)

One thing the "engineering notation" does not allow for is the idea that functions have values. E.g., if f(x) = x³ for all values of x, then f(2) is the value of that function at 2, and is equal to 8. If you say f(ω) is the function to be transformed, and g(t) is the transformed function, then g(2) should be the value of the transformed function at the point t = 2. But if you use the "engineers' notation" and write $\mathcal{F}(f(\omega)) = g(t)$ , then you cannot plug 2 into the left side. But watch this: $(\mathcal{F} f)(2)$ is the result of plugging 2 into the transformed function.

One thing to be said for the difficulties introduced by the engineers' notation that are avoided by the cleaner, simpler, but more abstract "mathematicians' notation", is that perhaps sometimes one ought not to be evaluating these functions pointwise anyway! But that's a slightly bigger can of worms than what I want to open at this moment ... Michael Hardy 22:38, 19 Sep 2004 (UTC)

All I meant by "engineering notation" was not using the letter f as a function, since it would too easily get confused with frequency as a variable in lowercase and script F for the fourier transform in uppercase, leading to protracted discussions about why there are two capital Fs and why one is script and the other is not and what it all means. :-)

*Reads the recommended articles on bound variables* - Omegatron 03:40, Sep 20, 2004 (UTC)

Let me drop in w/ my $0.02 -- I agree w/ Mike Hardy that conventional engineering notation (which puts function arguments in inappropriate places) is imprecise & misleading, and the usual "pure math" notation is superior. That said, it is certainly confusing & unnecessary to have different kinds of "F" running around; making technical distinctions based on font types is problematic IMHO. So, Omegatron, I'm not opposed to replacing f by g, likewise F by G, throughout the article. I'll consider doing that myself. Regards & happy editing, Wile E. Heresiarch 15:11, 20 Sep 2004 (UTC)

Check Continuous Fourier transform as well. - Omegatron 15:31, Sep 20, 2004 (UTC)

[edit] involutory fourier transform

The formula for the discrete fourier transform

$x_k = \frac{1}{n} \sum_{j=0}^{n-1} f_j e^{2\pi ijk/n} \quad \quad k = 0,\dots,n-1$

can be simplified because

e 2π i = 1

into

$x_k = \frac{1}{n} \sum_{j=0}^{n-1} f_j 1^{\frac {j k}{n}} \quad \quad k = 0,\dots,n-1$

This convention that an n'th root of unity is written $1^{\frac{1}{n}}$ rather than $e^{\frac{2\pi i}{n}}$ is convenient. Introducing the analytical concepts $e$ and $i$ and $π$ into an algebraic context is confusing and pointless.

The transform can be modified a little by taking the square root of n and by taking the complex conjugate of f.

$x_k = \frac{1}{\sqrt{n}} \sum_{j=0}^{n-1} \bar{f_j} 1^{\frac {j k}{n}} \quad \quad k = 0,\dots,n-1$

Now this transformation is an involution. The procedure that transforms f into x is exactly the same as the procedure that transforms x into f. This is nice.

PS: I inserted the note "Moreover: reality in one domain implies symmetry in the conjugate domain."

Bo Jacoby 13:18, 8 September 2005 (UTC)

[edit] Why no involution ?

15:55, 21 September 2005 Stevenj (tricks to compute the inverse DFT in terms of the forward DFT are well known, but belong in the DFT article (they don't define a fundamentally different FT); "involutary" is just a general adjective)

Stevenj removed my entry on involutory fourier transform. I think it should be put back. It makes life easier if you don't have to worry about the difference between forwards and backwards. Note that it is not merely "a trick to compute the inverse DFT in terms of the forward DFT" but a true identification of the inverse and the forward transform. The fourier integral and fourier series should be considered limiting cases of the DFT, so the involutory version is perfectly general and not specific to the discrete fourier transform. Please explain your point of view more carefully. See involution. Bo Jacoby 12:23, 22 September 2005 (UTC)

It is closely related to well-known tricks to compute the forward in terms of the inverse transform, which are themselves trivial consequences of the close relationship between the FT and inverse FT due to unitarity. Moreover, I can't find any outside use of the term "Involutary Fourier Transform" (or the corresponding definition) so I don't think we should highlight this particular term as if it were widespread—do you have any references? (In any case, conceptually it doesn't really change the features of the transform.)

I added a mention of it, however, to the discrete Fourier transform article, which I also expanded to give several well-known tricks to compute the inverse DFT from the forward DFT. (Yes, other Fourier variants have similar properties, but the point is that this is essentially a property of the transforms, and belongs on their respective pages, rather than a fundamentally different transform.) What you called the "involutary Fourier transform" is a consequence of a standard conjugation trick. In fact, it is actually related by a factor of 1+i to the discrete Hartley transform, which is well known to be involutary. —Steven G. Johnson 18:41, 22 September 2005 (UTC)

[edit] Vote for new external link

Here's my site full of Fourier transform example problems. Someone please put it in the external links if you think it's helpful!

http://www.exampleproblems.com/wiki/index.php/PDE:Fourier_Transforms

[edit] Excuse me!

The introductory paragraph for any section in this encyclopedia should be accessible to all readers.

What was wrong with my input? Was it too easy to understand?:

" It is a mathematical technique for expressing a waveform as a weighted sum of sines and cosines."

Some introductory info is appropriate as:

It usually happens in mathematical analysis that certain given functions are to be operated on, and other functions or numbers are thus obtained. Often, the operations used involve integration, and here is found a very general class of operators, the so-called integral transforms. The simplest integral transform is the operation of integrating.

---Voyajer 04:34, 30 December 2005 (UTC)

What was wrong is that it was repetitive and redundant. The introduction already says that the transform re-expresses a function in terms of sinusoidal basis functions, i.e. as a sum or integral of sinusoidal functions multiplied by some coefficients ("amplitudes"). —Steven G. Johnson 19:41, 30 December 2005 (UTC)

Please try to make the intro more accessible like Voyajer's version. — Omegatron 19:51, 30 December 2005 (UTC)

---

I concur, the introduction hardly explains what FT is to the layman.

[edit] *rush to a square-wave

?can anyone tell me why the first term of Fourier series $a 0 = 0$ when expanding a square wave

I have done the exercise about it..

thanks a lot--HydrogenSu 14:53, 19 January 2006 (UTC)

a 0

is the DC (zero frequency) offset. If the average value of a square wave is zero,

a 0

will be zero. Madhu 19:37, 10 August 2006 (UTC)

[edit] DFT definition given is actually inverse DFT?

If you consider this article: [1], it appears that the definition of DFT you have here on this page is actually the inverse DFT. Whose page is wrong?

Answer: Neither. The DFT (or FT) can be defined with either sign in the exponent as long as the inverse is defined such that applying transform and inverse in succession gives the original sequence (or function). There is also some freedom as to wether to put the $1 / N$ , $1 / L$ , or $1 / 2π$ on the transform or inverse for discrete, finite, and infinte cases respectively. I suppose I should add this to the page.--Kevmitch 02:39, 1 April 2006 (UTC)
Yeah, its tough to say. This seems to be covered on most of the pages covering the specific types of Fourier transforms such as the beautiful table in Continuous Fourier transform How specific do we want to get on a the page that covers the Fourier Transform in general?--Kevmitch 03:04, 1 April 2006 (UTC)

[edit] definition and transformations

James Nilsson and Susan Riedel's book Electric Circuits seventh edition defines fourier series like this:

$f(w) = \int_{-\infin}^\infin f(t) e^{-jwt} dt$

and inverse:

$f(t) = \frac{1}{2 \pi} \int_{-\infin}^\infin f(w) e^{-jwt} dw$

I couldn't find a matching type of fourier transform on here, and they just call it "fourier transform" as if theres no other types. It goes on to give this table of transforms:

f(t)	F(w)
$δ(t)$	1
A (constant)	$2π A δ(w)$
sgn(t)	2/jw
u(t)	\pi delta(w) + 1/jw
etc	etc

Is there something I'm missing? Fresheneesz 02:16, 5 June 2006 (UTC)

You didn't see it because you didn't click on continuous Fourier transform. That paragraph does warn you that you have to follow that link to see it, but maybe it should be explicitly displayed here. I'll be back. Michael Hardy 18:49, 5 June 2006(UTC)

Actually, I did click on that link. Problem is, it defines it differently than I did above. Neither the tables, nor the definition quite match right. Everything except the inverse transform seems to be multiplied by $\sqrt{2 \pi}$ in the example I give above (compared to the one you gave a link to). I'm still a little confused as to why Nilsson and Riedel's definition is a constant multiple of the one on wikipedia. I'm even more confused as to why the inverse transform does not match that pattern. Fresheneesz 02:39, 7 June 2006 (UTC)

[edit] Strange introduction

Why does this article start off with with writing down the inverse transform?!?

It's not my doing, but I understand the thinking. What it actually starts off with is the assertion that $f(t)\,$ can be represented as a continuous sum of complex exponentials, and the amplitudes and phases of those exponentials are given by a function, $F(\omega)\,$ . It is of course appropriate to formalize that concept mathematically, which also happens to be the formula we call an inverse transform. Thus the inverse transform (for lack of a better name) is actually an equation to be solved (for $F(\omega)\,$ ), and the Fourier transform is the solution to that equation. It's actually more logical than the traditional approach of pulling a Fourier transform definition out of thin air. --Bob K 16:56, 13 July 2006 (UTC)

Also there is no mention of PDE's anywhere. This is where the subject found it's birth and much of it's most fruitful application. I will try to add something to this effect but input would be appreciated. Also as defined here (as with any other definition I am familiar with)the DFT takes a finite vector and produces a vector of the same length.

That is not how I read it. I know you are looking at the $k = 0, 1, \dots, N-1 \,$ part. But that is explained in the preceeding paragraph which states: "Due to periodicity [of the DTFT], the number of unique coefficients (N) to be evaluated is always finite, leading to ... ". --Bob K 04:39, 14 July 2006 (UTC)

And please sign your entries with "~~~~"

But in the section "Family of Fourier transforms" we include it on this list as periodic. There should be some explanation there as to what is meant. Are we thinking of the DFT living on $\mathbb{Z}_n$ ?

[edit] Notation once again

This article establishes that the Fourier transform of a function s(t) is a new function $S (ω)$ and this relation is written

$S(\omega) = \left(\mathcal{F}s\right)(\omega)= \mathcal{F}\{s\}(\omega)\,$

However, a frequently appearing notation (at several places in Wikipedia) is to write it simply as

$S(\omega) = \mathcal{F}\{s(t)\}\,$

There has already been a discussion above about the validity of the former notation. But it should also be recognized that the latter notation, while formally incorrect, has the advantage of allowing us to insert an arbitrary function expression into the transform operator. For example, in a compact way we can express that the Fourier transform of a sinc function is a rectangular as

$\mathcal{F}\{ \mathrm{sinc}(t) \} = \mathrm{rect}(\omega)$

The meaning of the $\mathcal{F}$ operator has here changed into a transform table lookup and this way of using $\mathcal{F}$ appears handy and is probably the motivation for its frequent use. There are limitations to this notations as well, for instance writing

$\mathcal{F}\{ \mathrm{sinc}(a t) \} = \mathrm{rect}(\omega/a)/|a|$

implicitly assumes that the function being transformed is a function of t, not of a.

Here are my points

Should THE NOTATION ISSUE be completely ignored in the article even though it appears to be a source of controversy on the talk pages of several articles in Wikipedia? Maybe a section which explains THE NOTATION ISSUE rather than to have it on this and other talk pages?

It is possible to use a notation according to
- $\mathcal{F}$ is the Fourier transform operator as defined in this article
- $\mathcal{F}\{f(t),t\}$ is a transformation lookup operator which explicitly gives the transform of f as a function of t? This implies that $\mathcal{F}\{f(t),t\} = \mathcal{F}\{f\}$ .

The advantage of the second form of the operator is that is allows a formalization of the lookup functionality which also makes explicit which variable is used in the transformation. --KYN 09:34, 4 August 2006 (UTC)

[edit] Suggestion

Using an arrow instead of a comma makes it more readable.

The function f, defined by y = f(x), can be written f = (y←x), pronounced: "y as a function of x".. This nice notation allows you to distinguish the power function, (x^y←x), from the exponential function, (x^y←y).

The transform

F = (S←s) = ((S(ω)←ω)←(s(t)←t))

means that

S = F(s)

(S(ω)←ω) = F(s(t)←t)

and

S(ω) = (F(s))(ω) = (F(s(t)←t))(ω)

This notation does not identify the function f with the function value f(x), using a special letter x to identify the independent variable. Bo Jacoby 12:07, 4 August 2006 (UTC)

Sure, this also solves the problem and if it is more established, I will go for it. Next issue: is it relevant to present this somewhere in the article, giving an example for a notation which both is formally correct and can be used for the occations when an explicit function of a specific variable needs to be transformed? --KYN 18:23, 5 August 2006 (UTC)

[edit] Where to put it

The natural place to put the notation is in the article function (mathematics), where presently the sum of two functions is defined by (f+g)(x) = f(x)+g(x) rather than by f+g = (f(x)+g(x)←x), thus defining the function value (f+g)(x) rather than the function f+g. For functions of several variables and for functions whose arguments or values are functions rather than simple variables, the notation y = f(x) instead of f = (y←x) is severely insufficient. The lambda calculus writes the independent variable to the left, f = λ x.y, rather than to the right as in f = (y←x). I prefer the latter convention because the formula for composition of functions,

(g o f)(x) = g(f(x)) for all x in X

is written without breaking the order like this:

g o f = (g(f(x))←x) for all x in X .

Anyway, when the arrow points from the independent to the dependent variable there is not much room for misunderstanding.

Note also the elegant notations for inverse functions such as the natural logaritm and for the square root:

f⁻¹ = (x←f(x))

log = (x←e^x)

sqrt = (x←x²)

I will make a note on Talk:Function (mathematics) to ask the advice of the mathematicians there.

Bo Jacoby 11:40, 7 August 2006 (UTC)

I don't mind that a more exensive discussion about the notation of functions and their values goes into function (mathematics), but I also believe that a short summary applied to the notation of Fourier transforms is useful in the Fourier transform article. The unorthodox notation is wide-spread and it would then be easier to justify changes by just refering to that summary. Otherwise the justification lies one hyper-link further away and is easier to ignore. --KYN 22:12, 7 August 2006 (UTC)

I agree. Is this understandable? :

The Fourier transform F = (S←s) transforms a function s = (s(t)←t) of time t into a function S = (S(ω)←ω) of frequency ω.

S = F(s)

The amplitude is

S(ω) = F(s)(ω)

Your examples look like this:

sinc = (sin(t)/t←t),

rect = ([−1<ω<+1]←ω), using Iverson bracket,

F(sinc) = rect .

There is no need to use special names for dummy variables, nor is there any need for special function names:

F(sin(x)/x←x) = ([−1<x<+1]←x).

Bo Jacoby 09:16, 8 August 2006 (UTC)

[edit] Wikipedia is not the place to promote new/obscure notations

The problem with the notation that Bo is suggesting for the Fourier transform, and for functions in general, is that it is much less common than the usual notations which are already employed in the articles. If the notation is established but obscure, it might be worth mentioning in function (mathematics), but a wholesale adoption in other articles is not a good idea.

Given that Bo has a history of trying to promote his own invented nonstandard notations in Wikipedia, however, I would also request that he provide a mainstream citation for his notation before inserting it into any article.

—Steven G. Johnson 17:03, 8 August 2006 (UTC)

Steven seems to have an excellent point here. I agree. Thenub314 23:50, 8 August 2006 (UTC)

To StevenJ the word standard means known to me, and the word nonstandard means unknown to me; he makes no reference to standardization documents. Stevenj has a history of deleting everything that is new to him. KYN's complaint that StevenJ's notation is unsatisfactory, makes no impression on StevenJ. Still StevenJ does me too much honor by believing that I am the original inventor of the trivial shorthand notation 1^x = (e^{2π i})^x for e^{2π i x}, or of the straightforward explicite notation f = (y←x) for the function defined by y = f(x), or of the involutary N'th order fourier transform: F_N = ( (N^−1/2Σ_j x_j* 1^jk/N←k) ← (x_j←j) ). The article is very bad, but, according to StevenJ, it is standard, and that is all that matters. Progress is suspended until StevenJ gets majorized by fellow wikipedians. Meanwhile, study the notation for functions used in the J programming language. (I am the inventor, however, of the ordinal fraction technology, and my WP article on Ordinal fraction was deleted on StevenJ's request). Bo Jacoby 07:30, 9 August 2006 (UTC)

Hi Bo, should I be surprised by the vehemence your invective? (Your protests seem rather strained, as in both the 1^x and "involutary" DFT cases, in addition to ordinal fraction, you admitted that you didn't know of any publication that used your proposed notations/definitions. Why is Wikipedia policy so hard for you to comprehend?) By the way, I'm aware that x^2←x is sometimes used as function notation (although I maintain that it is uncommon in this context), but I don't think one often sees it mixed with "=" in the way you suggest, as in f = (x^2 ← x) instead of f(x)=x^2. And notice that I simply asked you to provide mainstream citations, which would have supported your case far better than a diatribe. (If the best you can do is an uncommon programming language that appears to only loosely match the notation you suggest...) —Steven G. Johnson 04:55, 11 August 2006 (UTC)

Hi Steven. You are aware that x²←x is sometimes used as function notation although you maintain that it is uncommon in this context. Has the wikipedia policy requirement changed from known to common? If a is a mathematical expression, then of course you can name it: b=(a). So if x²←x is a function, of course you can name it if you want: f=(x²←x). The point is that you do not need to name is, as you do when you write f(x)=x². You can write a function value elegantly (x²+yx+1←x)(4) = 4y+17. (Oh I forgot, you are not interested in good or bad, you are interested in standard, although you never refer to standardization documents.) Bo Jacoby 11:07, 11 August 2006 (UTC)

[edit] I am unconvinced that we have a severe notational problem

I am unconvinced that we have a severe notational problem. It seems to me that in this case the proposed cure is worst than the disease (for the reasons StevenJ gave). Thank you for suspending progress. Also, I don't agree to the relevance of a programming language, because it is created for a completely different set of constraints. --Bob K 18:24, 9 August 2006 (UTC)

Hi Bob. What will you do to help KYN? Bo Jacoby 11:07, 11 August 2006 (UTC)

Hi. KYN is not proposing a change in notation (see below). No notation is perfect. I have suggested that we focus on specific formulas that seem unclear. I expect they can be remedied by supplementary annotation (unlike a computer language). KYN is free to ask my opinion. If I have one, I will give it. --Bob K 13:55, 11 August 2006 (UTC)

Yes, there is a notation problem. Severe or not depends on the reader/writer. What about writing something like

$\mathcal{F}\{rect(x t)\}$

This is,..., what? In the context of this article, it is a Fourier transform, but is it a transform of rect being a function of x, or of t, or it is a 2D transform of rect being a function of both x and t? One answer to the posed problem would be: Don't use this notation unless it is unambiguous which variables are involved in the transformation. Another answer would be, in the case that you want to transform a function of A SPECIFIC variable, use notation XYZ, where XYZ is whatever can be agreed upon. A third alternative, which appears to be the result of the above discussion, is that no established notation exists for this case, even though is appears to be useful (see example above). My point is that regardless of which answer this discussion provides it should be made explicit in the article simply because it appears to be a source of questions in various area of Wikipedia. --KYN 10:48, 11 August 2006 (UTC)

I agree with you. And no matter what we do there will always be confused readers. If we switch notations now, I will probably be one of them. The example you gave above is devoid of context, which never happens here. Or if it does, then that is the problem, not the notation. And let's not compare technical writing to computer languages, because compilers are ignorant of context, use a more limited symbol set, and ignore annotations. More helpful would be to cite specific articles that seem to be problematic, and discuss all the possible specific remedies. --Bob K 11:34, 11 August 2006 (UTC)

My point is not necessarily that we should change notation, but that this discussion should be included in the article. Since I havn't seen anyone in support of this idea, I guess we should't? --KYN 12:00, 11 August 2006 (UTC)

In general, yes, annotation and clarification is a good thing. In this case, my opinion is that the article is good enough. But that is based on my reading of the article, not on my reading of this page. So maybe I am missing an important point. What specifc formula are we trying to clarify? --Bob K 13:55, 11 August 2006 (UTC)

Here are some examples of the informal use of $\mathcal{F}$ (and other operators on function spaces) which can be found in Wikipedia:

In consequence of this confusion I suggest a section like:

Section About notation:

The Fourier transform is a mapping on a function space. This mapping is here denoted $\mathcal{F}$ and $\mathcal{F}\{s\}$ is used to denote the Fourier transform of the function s. This mapping is linear, which means that $\mathcal{F}$ can also be seen as a linear transformation on the function space and implies that the standard notation in linear algebra of applying a linear transformation to a vector (here the signal s) can be used to write $\mathcal{F} s$ instead of $\mathcal{F}\{s\}$ . Since the result of applying the Fourier transform is again a function, we can be interested in the value of this function evaluated at the value

ω

for its variable, and this is denoted either as $\mathcal{F}\{s\}(\omega)$ or as $(\mathcal{F} s)(\omega)$ . Notice that in the former case, it is implicitly understood that $\mathcal{F}$ is applied first to s and then the resulting function is evaluated at

ω

, not the other way around.

In mathematics and various applied sciences it is often necessary distinguish between a function s and the value of s when its variable equals t, denoted s(t). This means that a notation like $\mathcal{F}\{s(t)\}$ formally must be interpreted as the Fourier transform of the values of s at t, which must be considered as an ill-formed expression. Despite this flaw, the previous notation appears frequently, often when a particular function or a function of a particular variable is to be transformed. For example, $\mathcal{F}\{ \mathrm{rect}(t) \} = \mathrm{sinc}(\omega)$ is sometimes used to express that the Fourier transform of a rectangular function is a sinc function, or $\mathcal{F}\{s(t+t_{0})\} = \mathcal{F}\{s(t)\} e^{i \omega t_{0}}$ is used to express the shift property of the Fourier transform. Notice, that the last example is only correct under the assumption that the transformed function is a function of t, not of

t 0

. If possible, this informal usage of the $\mathcal{F}$ operator should be avoided, in particular when it is not perfectly clear which variable the function to be transform depends on.

Well, something along these lines, anyway. --KYN 21:47, 11 August 2006 (UTC)

Thank you. I like it. I think many readers will appreciate it. --Bob K 21:57, 11 August 2006 (UTC)

I agree that "informal usage of the $\mathcal{F}$ operator should be avoided, in particular when it is not perfectly clear which variable the function to be transform depends on."

The formula $\mathcal{F}\{s(t+t_{0})\larr t\} = \mathcal{F}\{s\} e^{(i \omega t_{0} \larr \omega)}$ tells which variable the function to be transformed depends on.

There is no need for calligraphy. The formula is: F(s(t+u)←t) = F(s) e^{(i ω u←ω)}.

Linearity is irrelevant in the context.

Bo Jacoby 14:31, 13 August 2006 (UTC)

Besides being unusual, that notation doesn't even make it clear for me what the F(s) part means. E.g., it could be a constant.

For the record, my preference is this:

$\mathcal{F}\{s(t)\} = S(\omega)\,$

$\mathcal{F}\{s(t+t_{0})\} = S(\omega)\cdot e^{i \omega t_{0}}\,$

And if the audience is really too green to understand that, we probably need more words rather than fewer definitions.

--Bob K 19:46, 13 August 2006 (UTC)

Hi Bob. The symbols t and t₀ enters symmetrically on the left hand side of $\mathcal{F}\{s(t+t_{0})\} = S(\omega)\cdot e^{i \omega t_{0}}\,$ , but asymmetrically on the right hand side. What is $\mathcal{F}\{s(u+v)\}$ ? The answer depends on whether u or v is the independent variable, and the readers are clueless. The explicit multiplication sign improves clarity: F(s(t+u)←t) = F(s)·(e^i·ω·u←ω). Rather than a formula for the function, you might prefer a formula for the function value: F(s(t+u)←t)(ω) = F(s)(ω)·e^i·ω·u. The expression F(s) = F(s(t)←t) is the transform of the function s. People are aquainted to expressions giving function values, while the fourier transform is about functions. You need some notation for a function f apart from the function value f(x) or f(t) or f(ω). Bo Jacoby 09:04, 14 August 2006 (UTC)

Hi Bo. Yes, I understand that $\mathcal{F}\{s(u+v)\}$ is ambiguous. All I am saying is that the ambiguity can usually be resolved by context and/or by some commentary. And I think the latter is a fairly rare necessity. Also, most of the articles here use $t\,$ instead of u or v, and that is sufficient for most readers. But if stand-alone non-ambiguity is really necessary, then I would prefer something like this:

$\mathcal{F}_v\{s(u+v)\} \,$	$\equiv \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty s(u+v)\cdot e^{- i\omega v}\,dv$
	$= \mathcal{F}_v\{s(v)\}\cdot e^{i \omega u}\,$

Also, if I had to choose between F(s)(ω)·e^i·ω·u and F(s)·(e^i·ω·u←ω), I would choose the former. I just don't like that darn arrow. :-)

And by the way, these html expressions are nasty to type!

Have a great day.

--Bob K 11:52, 14 August 2006 (UTC)

Thanks, Bob. When it comes to liking and disliking there is no point arguing. I trust that time will settle the question, so that either you will learn to love the arrow (which you seriously doubt because it looks strange), or I will learn to love the ambiguity (which I seriously doubt because unambiguous notation makes commentary unnecessary, which is lovely). The html is nasty mainly because of the italics. I write first without the italics, F(s)(ω)·e^i·ω·u , and then I insert two apostrophes around each variable name: F(s)(ω)·e^i·ω·u . Bo Jacoby 12:14, 14 August 2006 (UTC)

Actually, for an encyclopedia article, I think most concepts should be explained in commentary if possible, even if it is redundant with the mathematics. And especially Wikipedia, where anybody can write an article or edit one. You've heard the old saying... "If you think you understand something, try explaining it to someone who doesn't." --Bob K 17:14, 14 August 2006 (UTC)

[edit] The dot notation

Hi, the topic of this thread is twofold: (1) is there an established notation which can be used for saying, eg, the FT of sinc is rect, in a formally correct way, and (2) should the notation issue be mentioned in the article. Bo has proposed a notation which addresses (1) which it OK with me, but since I havn't seen any evidence that it is established I don't want to write about it since Wikipedia should not be the place to introduce new notations which does not exist in the literature. An alternative to Bo's arrow can be found in Daubechies "Ten Lectures on Wavelets" where the anonymous variable is simply denoted with a dot: $\mathcal{F}\{sinc(\cdot)\}=rect(\cdot)$ . To my surprise the informal notation can be found in hard core math text book, for example see Debnath & Mikusinsky "Introduction to Hilbert Spaces with Applications" (2ed, page 194). About (2), I will try to figure out something based on the text I wrote above and put it into the article shortly. --KYN 15:48, 14 August 2006 (UTC)

Hi KYN. A dot as a placeholder for the independent variable is worse than x or t or ω or k. The formula F(s(t+u)←t) = F(s)·(e^i·ω·u←ω) becomes F(s(·+u)) = F(s)·e^i···u which is spooky. Bo Jacoby 09:15, 15 August 2006 (UTC)

The dot notation has a place. But it needs to know that place and stay there. When the point is simply that the Fourier tranform of sinc is rect, it avoids the irrelevant question of whether we're talking about time or space or some other dimension. --Bob K 11:31, 15 August 2006 (UTC)

When stating that the Fourier tranform of sinc is rect, no dot is needed: F(sinc) = rect. I know of no place where the dot notation is sufficient. It is limited to functions of one variable only, and it looks like a multiplication sign. It has no future. Bo Jacoby 12:52, 15 August 2006 (UTC)

F(x(·)) confirms for me that x is a function. I am not so sure about F(x). --Bob K 14:42, 15 August 2006 (UTC)

Standard notation for F(x(·)) is F(x(t)). You cannot tell, however, whether you are refering to the value F(x(t)) or the function (F(x(t))←t) or the operator value F(x) = F(x(t)←t) or the operator itself F = (F(x)←x) = (F(x(t)←t)←(x(t)←t)) . Bo Jacoby 10:08, 16 August 2006 (UTC)

"the value F(x(t))" ... that ambiguity is avoided by F(x(·)). In certain situations, F(x(·)) is better than either F(x(t)) or F(x). --19:49, 16 August 2006 (UTC)

[edit] Conclusion

My conclusion of this discussion is:

there is no consensus about notation, in particular about how, e.g., "Fourier transform of rect is sinc" should be written in a concise and formally correct way. (Yes, we have seen proposals, but no consensus).
the informal/ill-formed $\mathcal{F}\{\mathrm{rect}(t)\}=\mathrm{sinc}(\omega)$ appears to be frequently used, even in math text books. Becomes meaningful only with proper context.
there are several proposals for a formal way to rewrite this expression, e.g., Bo's arrow and Daubechies' dot. The arrow notation, although perfectly valid, appears not be in use in the literature, and the dot notation is also not in frequent use (Daubechies' book is the only example that I have encountered). Given that Wikipedia explicitly should have a verifiable content, which I interpret in this particular case as "describe notations which are established in literature or in common use", I suggest to leave the arrow and the dot until more evidence of their usages can be demonstrated.

--KYN 19:50, 15 August 2006 (UTC)

The following notation is supposed to be standard, $\mathcal{F}\{t \mapsto \mathrm{rect}(t)\}=\omega\mapsto\mathrm{sinc}(\omega)$ , according to the recommandation at Talk:Function (mathematics)#Standard notation. The maps-to-arrow must have the special shape $\mapsto$ , and it must point to the right, otherwise it is not standard. Bo Jacoby 12:30, 17 August 2006 (UTC)

Here is my favorite excerpt from that reference: "In most written mathematics, it is enough to just say (in prose) what the bound variable is when confusion could arise. Since the entire community of mathematicians seems to have agreed that this is precise enough, new notation is apparently not necessary." --Bob K 14:04, 17 August 2006 (UTC)

Bob, you cannot be serious. Which reference are you talking about? Bo Jacoby 12:25, 18 August 2006 (UTC)

Yours! (Talk:Function (mathematics)#Standard notation. ) --Bob K 13:50, 18 August 2006 (UTC)

Don't you consider me and KYN as part of the community of mathematicians? Don't you think that wikipedia should be readable to others than the few mathematicians who can reliably guess the meaning of say $\mathcal{F}\{\mathrm{rect}(t)\}$ ? Bo Jacoby 12:25, 18 August 2006 (UTC)

I did not claim to know the community of mathematicians. Your reference did that. --Bob K 13:50, 18 August 2006 (UTC)

Sorry. I agree that there is no point in solving a nonexisting problem. It is no problem to write f(x)=x² rather than f=λx.x². However, the notation f(x(t)) is ambiguous, as has been pointed out. We cannot pretend that there is no problem of notation. Bo Jacoby 14:31, 20 August 2006 (UTC)

Is someone pretending that? Nothing is without disadvantages. So finding one is not sufficient justification for change. That just leads us into circles. --Bob K 15:44, 20 August 2006 (UTC)

Yes, BobK, you wrote: "I am unconvinced that we have a severe notational problem". Now we agree that there is a severe notational problem. KYN suggests "to leave the arrow and the dot until more evidence of their usages can be demonstrated". Now we have evidence for the maps-to arrow and the lambda-dot. Is there any reason why we should not start clarifying the article using these notations? Bo Jacoby 16:50, 20 August 2006 (UTC)

Clearly, I did not agree to that. I said nothing is without disadvantages, and that includes your proposals. What I also said to you, 13:55, 11 August 2006, and subsequently echoed by your own reference (CMummert), is to use supplementary annotation (aka prose) where clarification is needed. --Bob K 17:48, 20 August 2006 (UTC)

Somehow, people in mathematics and engineering have learned to live with this "severe notational problem," and the reason is obvious. The meaning of $\mathcal{F}\{x(t)\}$ is perfectly clear, taken in context, since it only has one reasonable interpretation if you know that $\mathcal{F}$ stands for the Fourier transform. In the rare cases where clarification is needed, it can be given in prose. This is no reason to depart from the most established and widely understoon notation for this subject. "A foolish consistency is the hobgoblin of little minds." —Steven G. Johnson 17:53, 20 August 2006 (UTC)

No reason at all, except to make the article comprehensible to new readers, the so-called 'little minds'. Bo Jacoby 05:59, 21 August 2006 (UTC)

[edit] graphs and pictures

It would be nifty if there were a small graph of sinc function, etc. in the tables of equations. The continuous graphs plotted with nice smooth curves to emphasize that it is continuous, while discrete graphs plotted as more of a bar-chart or set of spikes, to emphasize its discreteness. Do we really need 4 sets of equations and graphs:

continuous-time/continuous-frequency in Continuous_Fourier_transform#Table_of_important_Fourier_transforms
discrete-time/discrete-frequency in Discrete Fourier transform
discrete-time/continuous-frequency in Discrete-time Fourier transform
discrete-frequency/continuous-time in Fourier series

? (If there is some significant difference in the equations and graphs, it would be more visible if they were plotted all on one page, rather than scattered across all 4 articles).

But there are also advantages to scattering, and they outweigh the disadvantages. One advantage is divide-and-conquer. I.e., it allows better focus on small parts of a large picture. Scattering, linking, and search-engines are the advantages of online technology. "web" is a good thing. --Bob K 15:25, 10 August 2006 (UTC)

[edit] Normal distribution?

I understand that the normal distribution is an eigenfunction of the Fourier transform. Is this the case? Should the article mention this? —Ben FrantzDale 19:32, 9 August 2006 (UTC)

To me, that seems perfectly appropriate for inclusion in this article. Michael Hardy 23:19, 9 August 2006 (UTC)

More precisely: the density function of the normal distribution is an eigenfunction of the Fourier transform. Michael Hardy 23:19, 9 August 2006 (UTC)

Added. —Ben FrantzDale 00:53, 10 August 2006 (UTC)

just keep in mind that there are an infinite number of eigenfunctions in the Fourier transform. like sinc()+rect(). any even-symmetry function added to the F.T. of that (which is also even symmetry) is an eigenfunction. but the Gaussian function is a nice simple and intrinsic eigenfunction and i can't think of another that is. r b-j 05:27, 10 August 2006 (UTC)

We have also polynomials times Gaussian which as a class provide eigenfunctions to the FT. ~~Or maybe an irreducible representation?~~ There is also the shah or comb function. Anyway, if this subject is to be developed to some length, I suggest to do it in a separate article "Eigenfunctions of the Fourier transform" rather than putting it into this article since it is rather long as it is. --KYN 10:30, 11 August 2006 (UTC)

Rbj is correct in that there are uncountably many choices of eigenfunctions, because there are only four eigenvalues (the 4th roots of unity), each with infinite multiplicity. However, arguably the most important eigenfunctions of the continuous Fourier transform are the Hermite functions (of which the gaussian is a special case). These functions are orthogonal and have some optimal localization properties I believe. The most appropriate place to mention them, however, would be under continuous Fourier transform. Unfortunately, they don't generalize in an easy way to the other Fourier variants. For example, it is not obvious what is the appropriate discrete analogue of the Hermite functions for the DFT, and this is still being debated in the literature. —Steven G. Johnson 17:59, 15 August 2006 (UTC)

[edit] Asymptotic expansion ?

Hello my question is why nobody has pointed the "asymptotic" behavior for a Fourier transform?.. in the sense:

$\int_{-\infty}^{\infty}dx f(x)e^{i\omega x}$ as $\omega \rightarrow \infty$

[edit] Linear Combination?

I wish to raise an issue about the use of the expression "linear combination" in relation to the contunuous Fourier transform. Much as I appreciate that the transformed function is simmilar to the coefficients of a linear combination, it's not what the transformed function IS. For one thing, we could change many of the values of the transformed function and as long as we did it over finitely many values, or over a simmilarly null set, we would change really nothing about the essential properties of the transformed function, ie would generate the same function when the inverse transform was appled. So we're not really talking about the individual contributions made by different frequencies, because each frequency taken in isolation contributes nothing. Also, and perhaps this is a pointless thing to argue about, I have repeatedly had it drilled into me by more than one person that linear combinations have to be finite, not infinite and definitely not continuous.

My bad. Thanks. I removed that phrase. If you can improve on the new intro, please feel free. --Bob K 01:55, 5 September 2006 (UTC)

[edit] Merge?

See Talk:Continuous Fourier transform --catslash 12:19, 4 September 2006 (UTC)

[edit] Fourier integral

Fourier integral redirects here, and I was reading Partial differential equation which suggested it is somewhat different from a Fourier transform. Can anyone explain this? - Rainwarrior 16:53, 27 September 2006 (UTC)

I seems strange to me Fourier Integral redirects hear. Often in Partial Differential Equations the term Fourier Integral Operator comes up. Which are indeed different the Fourier Transforms, could this be what the book was discussing? Thenub314 00:27, 28 September 2006 (UTC)

It wasn't in a book, it was in the Partial differential equation article. Specifically, it was in the Heat equation in one space dimension section. Anyhow, from the way it was described it sounds that the Fourier integral is related to, but distinct from, the Fourier transform. I would like to know more about it, but there is no information about the Fourier integral here.

I went and read it, here they mean fourier transform. I will edit the article to make that more clear. Thenub314 23:08, 28 September 2006 (UTC)

I'm suggesting that someone who knows about it should amend this article to explain the difference, or at least acknowledge the existance of a "Fourier integral", otherwise the redirect isn't very useful. - Rainwarrior 03:36, 28 September 2006 (UTC)

An article on Fourier Integral Operators would be nice, but would be an endevor to write. Prehaps sometime when I am feeling brave. :) Thenub314 23:08, 28 September 2006 (UTC)

[edit] Variants of Fourier analysis

This section used to be called "Variants of Fourier transforms", when the whole article was called "Fourier transform". Is it really the analysis that differs, or is it only the transforms?

On another note, I feel that much of this section is redundant, which is discussed in Talk:Fourier transform. I'll see if I can change the section so that it expresses what I meant in that discussion. — Sebastian (talk) 00:25, 8 October 2006 (UTC)

[edit] Consistency between formulas

I was trying to standardize the formulas so the comparison is easy. Here's what I ended up with:

Name	transform	inverse transform
(Continuous) Fourier transform	$f(t) = c \int\limits_{-\infty}^\infty F(\nu) \cdot e^{i 2 \pi \nu t}\,d\nu$	$F(\nu) = C \int\limits_{-\infty}^\infty f(t) \cdot e^{-i 2 \pi \nu t}\,dt$
Fourier series	$f(t) = c \sum_{\nu=-\infty}^{\infty} F_\nu \cdot e^{i 2 \pi \frac{\nu}{T} t}$
Discrete-time Fourier transform	$F(\nu) = C \sum_{t=-\infty}^{\infty} f_t \cdot e^{-i 2 \pi \frac{\nu}{N} t}$
Discrete Fourier transform	$F_\nu = C \sum_{t=0}^{T-1} f_t \cdot e^{-i 2 \pi \frac{\nu}{T} t}$	$f_t = c \sum_{\nu=0}^{N-1} F_\nu \cdot e^{i 2 \pi \frac{\nu}{N} t}$

Now I feel gypped: The formula for DTFT is basically be same as that for the Fourier series. Both transforms go from a continuous, periodic function to a discrete and aperiodic. Calling one variable "time" and the other "frequency" is an arbitrary convention that has nothing to do with mathematical reality. Is that really the way it is defined, or are we only looking at the inverse transformation here? I rather expected something like this:

Name	transform	inverse transform
(Continuous) Fourier transform	$f(t) = c \int\limits_{-\infty}^\infty F(\nu) \cdot e^{i 2 \pi \nu t}\,d\nu$	$F(\nu) = C \int\limits_{-\infty}^\infty f(t) \cdot e^{-i 2 \pi \nu t}\,dt$
Fourier series	$f(t) = c \sum_{\nu=-\infty}^{\infty} F_\nu \cdot e^{i 2 \pi \frac{\nu}{T} t}$	$F_\nu = C \int\limits_{0}^T f(t) \cdot e^{-i 2 \pi \nu t}\,dt$
Discrete-time Fourier transform	$f_t = c \int\limits_{0}^N F(\nu) \cdot e^{i 2 \pi \nu t}\,d\nu$	$F(\nu) = C \sum_{t=-\infty}^{\infty} f_t \cdot e^{-i 2 \pi \frac{\nu}{N} t}$
Discrete Fourier transform	$f_t = c \sum_{\nu=0}^{N-1} F_\nu \cdot e^{i 2 \pi \frac{\nu}{N} t}$	$F_\nu = C \sum_{t=0}^{T-1} f_t \cdot e^{-i 2 \pi \frac{\nu}{T} t}$

— Sebastian (talk) 03:28, 8 October 2006 (UTC)

Yes, the DTFT is just an inverse transform. I always wondered why wikipedia made this distinction but decided not to fight about it too much. Also from a mathematical point of view I would reverse the titles on your table. Thenub314 12:42, 8 October 2006 (UTC)

The reason for the distinction is that the DTFT is a forward transform (time-to-frequency), as its name correctly states. It converts a discrete-time function into a periodic frequency function. Fourier series converts a periodic [time] function into a set of discrete [frequency] components. --Bob K 22:03, 9 October 2006 (UTC)

I understand this, my point was given a discrete function simply forget you call it's domain time, decided instead to call it frequency, then the transform we are talking about is the the same as the inverse transform for fourier series. I am not suggesting any change to the article, just letting Sebastian I agree that "Calling one variable 'time' and the other 'frequency' is an arbitrary convention that has nothing to do with mathematical reality." But I understand it may be useful for intuition to make such a convention. Thenub314 23:47, 9 October 2006 (UTC)

Sebastian, if you are just learning about this stuff for the first time, and you don't even know which one is the forward transform and which is the inverse, I'm not sure if you should be the one to make wholesale changes in notation. And it's not Wikipedia that makes the distinction between the DTFT and the Fourier series, it's the real world — because the practical applications are quite different. I'm not sure we should try so hard for consistency in this article that we adopt notations that conflict with what everyone uses in practice and with what is used in the main Wikipedia articles on each subject. —Steven G. Johnson 14:22, 8 October 2006 (UTC)

Note that the comparison table already in the article is now messed up, too, because the third column mixes up the forward and the inverse transforms. —Steven G. Johnson 16:06, 8 October 2006 (UTC)

This is why I didn't make a fuss. Stevenj is more of an expert then myself on what the real world does. I didn't mean to imply this article doesn't make the distinction for a good reason, I simply never understood what it was.Thenub314 23:47, 9 October 2006 (UTC)

[edit] Informal Language

Considering this is such a technical article - Why do I come across the word "jaggies" 82.25.174.136 19:22, 24 November 2006 (UTC)

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Category: Wikipedia articles that are too technical