Talk:Fourier analysis

This talk page has been archived at Talk:Fourier transform/Archive1.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)

---

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

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

Seems clear to me from the article. ${\displaystyle F}$ is the function whose Fourier transform is to be found, and ${\displaystyle {\mathcal {F}}(F)}$ is the transformed function, so that ${\displaystyle {\mathcal {F}}}$ 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 ${\displaystyle F\,}$ is the Fourier transform of the function ${\displaystyle f\,}$. ${\displaystyle {\mathcal {F}}}$ is the Fourier transform operator.CSTAR 02:18, 19 Sep 2004 (UTC)

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

Strictly speaking, ${\displaystyle ({\mathcal {F}}F)(t)}$ is the right way to parse the thing, and ${\displaystyle {\mathcal {F}}(f(x))}$ is a solecism, often used by engineers and sometimes used even by mathematicians. When you use ${\displaystyle {\mathcal {F}}}$ 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. ${\displaystyle {\mathcal {F}}(f(x))}$ would make sense only if ${\displaystyle {\mathcal {F}}}$ were defined for numbers, but it;s defiend for functions.CSTAR 02:34, 19 Sep 2004 (UTC)

Oh oh I see. So

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

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 ${\displaystyle t\mapsto f(t)\quad }$.; this however by rules of lambda-calclus reduces to the term f.CSTAR 02:47, 19 Sep 2004 (UTC)

${\displaystyle {\mathcal {F}}^{-1}(F)(t)}$ 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 ${\displaystyle F(\omega )={\mathcal {F}}(f)}$. - Omegatron 03:30, Sep 19, 2004 (UTC)

If you think of ${\displaystyle F(\omega )={\mathcal {F}}(f)}$ as meaning two things: (a) ${\displaystyle F={\mathcal {F}}(f)}$ and (b) a suggestion that the symbol ${\displaystyle \omega }$ is reserved to name an independent variable to name the argument of the function F, since ${\displaystyle \omega }$ 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 ${\displaystyle X(f)={\mathcal {F}}(x(t))}$, so you can use f for frequency and avoid confusing Fs. Then of course there's ${\displaystyle F(f)={\mathcal {F}}(f(x))}$.  :-) - 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 ${\displaystyle {\mathcal {F}}(f(\omega ))=g(t)}$, then you cannot plug 2 into the left side. But watch this: ${\displaystyle ({\mathcal {F}}f)(2)}$ 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)

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

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

${\displaystyle \int _{-\infty }^{\infty }dxf(x)e^{i\omega x}}$ as ${\displaystyle \omega \rightarrow \infty }$