Talk:Sinc function

This is the talk page for discussing improvements to the Sinc function article.
This is not a forum for general discussion of the article's subject.

Put new text under old text. Click here to start a new topic.
New to Wikipedia? Welcome! Learn to edit; get help.

Article policies

Find sources: Google (books · news · scholar · free images · WP refs) · FENS · JSTOR · TWL

Archives: 1: 12 months

Computing (normalized) sinc(0)

If one didn't already know EmilJ's derivation for the limit of the (unnormalized) sinc at 0, how would one normally approach the problem of evaluating the simple expression:

\lim _{x\to 0}{\frac {\sin(\pi x)}{\pi x}}

?

If one knew the trick of turning it into the derivative of another function, after recognizing that the numerator and denominator both evaluate to 0 at the point where you're taking the limit (the same recognition that traditionally triggers application of l'Hopital's rule), one would write, as EmilJ did

\lim _{x\to 0}{\frac {\sin(\pi x)}{\pi x}}=\lim _{x\to 0}{\frac {\sin(\pi x)-\sin(\pi 0)}{\pi (x-0)}},

With a little rearrangement, this is easily recognized as the derivative of ${\frac {\sin(\pi x)}{\pi }}$ . For anyone with the most elementary skills at differentiation, this is clearly just $cos(\pi x)$ , which at x=0 is 1. So, same answer as before, with just slightly more complicated path; but I think it illustrates that the method is based on being able to intuit what to do with the subtractions, and on the ability to recognize a limit expression as a derivative even when it's not in exactly the right form. Is there any generality here, or is it just re-deriving l'Hopital's rule in a lucky case-specific way?

The way normal people (not you mathematicians) would do it is say: "whoa, ratio 0/0, how am I going to do that? Oh, I remember, they taught us about some French dude who had a rule that makes it work trivially...something about a hospital...oh, here it is...doh!"

The sinc function is very often used as an example of where l'Hopital's rule is useful; for example, in the article l'Hôpital's rule; or in this book; or in this book; or in this book; or in this book; in general, many more books than I found before, because they don't call it "sinc".

If there's a more straightforward way to evaluate such limits, I'd like to know it. Dicklyon (talk) 01:52, 17 October 2009 (UTC)[reply]

There is, in the sense that the limit is by a substitution that of sin(y)/y. Now why you know this is 1 is a matter of definitions, if it is not arguing in a circle. But we do know that limit. Charles Matthews (talk) 16:22, 17 October 2009 (UTC)[reply]

OK, that's just stupid. Replace one unsolved problem by another whose solution has not been well explained. I agree that there are ways to find a path to the limit via such substitutions. But that requires search, as opposed to the simple application of rule that is triggered by exactly this situation. Did you look at any the books I linked? Don't they all show the use of the l'Hopital's rule as the way to evaluate this limit? Don't th e other methods described here all require some messing around to find the path of transformations to get to an answer? That's the difference I'm talking about. Dicklyon (talk) 18:40, 17 October 2009 (UTC)[reply]

It seems that this discussion is no longer aimed at changing the article in any way, but at resolving this situation where two parties don't understand each other and don't know if they are talking past each other or what's going on. So far as I am concerned, that's fine. Here is how I feel about it, as a pure mathematician from a field where real numbers hardly occur at all: If you had asked me for this limit without giving me any hints or additional information, I would not have remembered l'Hôpital's rule. (Before EmilJ left a message at WT:WPM, this rule was hidden in a dusty corner of my brain.) Instead I would very likely have come up with the formula that I have given you above. (BTW: Sorry if the formula was nothing new for you, but at least it helped to clarify what we are talking about.) For a mathematician this is the natural way to do it; and part of our work in teaching university level maths is to make our students approach problems in the same way. If a significant portion of, say, second year mathematics students at a certain university would use l'Hôpital to compute this limit, this might be an indication that something went wrong in the course; that they don't think like mathematicians. For a mathematician doing a very small "backwards" step is no "messing around". We have to do it all the time and it becomes second nature. And many mathematicians find it harder to remember a theorem like l'Hôpital's rule than to remember vaguely that something like it exists and fill in the details by reinventing it. That's a result of our first year courses, in which the students must prove half a dozen theorems on their own every week. (Another result is that most people who drop out of a mathematics course do so in the first year, in contrast to most other subjects where people tend to drop out much later. At least that's the situation in Germany.) Hans Adler 01:05, 18 October 2009 (UTC)[reply]

I agree with all that entirely. I do a fair bit of math myself, though I'm primarily an engineer, so I know how to look around for the right transformations – or how to ask Mathematica to help me. But for "the rest of us" as Steve Jobs used to say of the non-nerds, it's handy to just apply the rule that's triggered by the situation; of course, for those who don't remember the rule, or if it doesn't get triggered, that's not going help a bit! Nonetheless, I was a bit miffed by all the people saying things like it's "completely pointless and brain-damaged"; or "no textbook would do it that way" when there's a ton of evidence that many do, and in fact use this particular limit as the canonical example; or saying "by definition" when they means it's possible to find a way... Sigh... Dicklyon (talk) 03:59, 18 October 2009 (UTC)[reply]

OK, I found the method, called "Recognizing a Given Limit as a Derivative", in this book. Example 52 explains exactly what you guys did above. Very nice. Then it concludes: "Also, l'Hopital's rule yields 1 immediately as the answer." Doh! Dicklyon (talk) 04:16, 18 October 2009 (UTC)[reply]

Forgive me, but what I suggested is a two-line proof, consisting of an observation and then a look-up. I realise that the ad hominem arguments above have hardly added to the tone of the discussion, but they weren't from me. I answered your question and hardly deserve to be called "stupid" for that. Charles Matthews (talk) 10:51, 18 October 2009 (UTC)[reply]

I'm not calling you stupid; the adjective applied to "that", which was a statement that the method involving a leap of recognition in the first step was "more straightforward" than simply applying the well-known rule. Dicklyon (talk) 15:18, 18 October 2009 (UTC)[reply]

Shrug. "How would one normally approach the problem of evaluating the simple expression ... ?". That was the question. First, recognise that π has no business in there complicating the issue. Second, "simple expression" is then "well-known expression" for the slope of the graph of sin at 0. Sorry, it does depend on who "one" is. It does depend on being comfortable with the limit concept. On realising that if you know the derivative of sin then a fortiori you know this limit. As has been said before, there is nothing to justify dressing up the procedure of evaluating the limit in an algorithmic garb just to produce the same answer by a longer way round. Anyway, if you ask an honest question it goes down badly if you react that way to a competent answer. Charles Matthews (talk) 15:57, 18 October 2009 (UTC)[reply]

What is the Sinc function and why should I care?

This article, like many others in the higher math domain, lack a root in reality. Wikipeida is a general encyclopedia not a PhD in Mathematics encyclopedia. Can an expert in the field please post why this is notable or relevant to those who are not math experts? Otherwise I propose it be deleted. 74.115.216.130 (talk) 20:20, 28 June 2010 (UTC)[reply]

Did you read the lead section? It is relevant in signal processing. Charles Matthews (talk) 20: 38, 28 June 2010 (UTC)

I am sorry the article lacks a root in YOUR reality, but you are wrong, Wikipedia IS an encyclopedia for math experts. And science experts, and art experts, and history experts and every kind of expert you can think of. Its also a general encyclopedia. You start out at the simplest level. If you want to learn more about a subject, hopefully there will be a link to help you instead of a stop sign that says "sorry, if you want more information on this, you will begin to become an expert, and we can't have that". Wikipedia is not intended to bolster your self-esteem about the level of your education. There is not one "expert" contributor to Wikipedia who isn't totally baffled by some other "expert's" contribution. Instead of proposing to delete what you don't understand, why not contribute what you do understand, and just hit the back button for the ones that you don't. PAR (talk) 01:28, 29 June 2010 (UTC)[reply]

sinc(x) Fourier Transform

It should be noted in the article that formally, sinc(x) has no fourier transform as it does not belong to G(R) (It is not absolutely integrable over the real axis). —Preceding unsigned comment added by 79.183.204.32 (talk) 20:07, 16 January 2011 (UTC)[reply]

I don't think that's a requirement for the existence of a Fourier transform. It's square integrable; isn't that enough? Dicklyon (talk) 20:09, 16 January 2011 (UTC)[reply]

By square integrable do you mean (sinc(x))^2 is integrable? that is definitely not a condition good enough for the Fourier transform to exist as (sinc(x))^2<abs(sinc(x)) for x>1. —Preceding unsigned comment added by 79.183.204.32 (talk) 20:18, 16 January 2011 (UTC)[reply]

79, you might want to check this out Fourier_transform#Square-integrable_functions. If you're correct, you'll have to fix it in more than one place. 70.109.187.2 (talk) 03:39, 17 January 2011 (UTC)[reply]

He does seem confused. I'm not absolutely certain that square integrable is a sufficient condition, but I am absolutely certain that this particular function's FT exists. Dicklyon (talk) 04:38, 11 February 2011 (UTC)[reply]

There are similar disputes about the dirac delta function and whether it can stand alone and be Fourier transformed. Personally, I'm willing to accept the duality property of the Fourier transform, which means if

\int _{-\infty }^{\infty }\mathrm {rect} (t)\,e^{-i2\pi ft}\,dt=\mathrm {sinc} (f),\,\!

then

\int _{-\infty }^{\infty }\mathrm {sinc} (t)\,e^{-i2\pi ft}\,dt=\mathrm {rect} (-f)=\mathrm {rect} (f),\,\!

71.169.191.4 (talk) 05:04, 11 February 2011 (UTC)[reply]

True, the Fourier-Integrals of sinc do not exist, as sinc is not absolutely integrable. However, L1(IR), the set of absolutely integrable functions, is a dense subset of L2(IR), the square integrable functions. On this dense subset, the Fourier integral defines a linear operator that is bounded in the L2 norm. Thus there exists a bounded extension of the Fourier integral to a bounded linear operator on the whole of L2 that is called the Fourier transform (operator). As such, the Fourier transform of sinc exists. -- For the delta distribution, look for Schwartz test functions and tempered distributions. The idea is similar, L1 and L2 may be interpreted as regular distributions and as such they are tempered distributions, and they are dense in the last set, so the Fourier transform operator may be extended as a bounded operator on the tempered distributions.--LutzL (talk) 10:26, 11 February 2011 (UTC)[reply]

Abbrevation for 'Cardinal Sine'?

Sinc stands for something like 'cardinal sine' (see french article). Moemin05 (talk) 00:25, 11 February 2011 (UTC)[reply]

Read the lead. Dicklyon (talk) 04:34, 11 February 2011 (UTC)[reply]

Limit of function

I've found intresting thing about generalized sinc function. If the function is given: n*(sin(nx)/nx) then function is limited by functions 1/x and -1/x. It is easy to see on function grapher — Preceding unsigned comment added by 83.23.6.202 (talk) 19:05, 17 June 2011 (UTC)[reply]

Relationship to the Dirac delta distribution

There is nothing special to the sinc function resulting in f(x/a)/a approaching the delta. This is true for a very wide class of functions. — Preceding unsigned comment added by 67.220.7.72 (talk) 03:04, 5 April 2012 (UTC)[reply]

sinc * sinc = sinc

I added "(k integer)" to where "orthonormal base" is mentioned. The orthonormality as well as values of all other scalar products (among functions sinc(t−k) where k is not necessarily integer, this time) follows immediately from the following nice property that you might want to include: sinc * sinc = sinc. (* denotes convolution.) This in turn follows via Fourier transform from rect²=rect. 90.180.192.165 (talk) 00:30, 19 August 2011 (UTC)[reply]

a graphic illustration would be nice (with sinc and sinc²)--46.115.99.134 (talk) 10:18, 8 September 2013 (UTC)[reply]

Incorrect definition of the Sinc function

The real definition of the Sinc function is $\mathrm {sinc} (x)={\frac {\sin(x)}{x}}\,\!$ if x ≠ 0 and Sinc 0 = 1. Blackbombchu (talk) 00:00, 2 December 2013 (UTC)[reply]

Extra property to add to the Properties section

$\mathrm {sinc} (x)=\int _{0}^{1}\cos(xt)\,dt$ Blackbombchu (talk) 01:25, 2 December 2013 (UTC)[reply]

Yes, but why is this interesting? And where can it be sourced? Dicklyon (talk) 05:51, 2 December 2013 (UTC)[reply]

I'm not sure if Wikipedia works this way but maybe it doesn't need to be sourced if it can instead be proven in the article. Blackbombchu (talk) 16:06, 2 December 2013 (UTC)[reply]

No, it doesn't work that way. Dicklyon (talk) 05:20, 3 December 2013 (UTC)[reply]

This is the same as saying that the sinc function is the inverse Fourier transform of a (area=1) box function:

\int _{\mathbb {R} }\left[{\tfrac {1}{2}}{1\!\!1}_{[-1,1]}(\omega )\right]e^{i\,\omega x}\,d\!\omega =\int _{0}^{1}{\tfrac {1}{2}}(e^{i\,\omega x}+e^{-i\,\omega x})\,d\!\omega =\ldots

so it should be covered in the Fourier transform section.--LutzL (talk) 10:51, 4 December 2013 (UTC)[reply]

History of the name

The name sinus cardinalis dates back by Edmund T. Whittaker in 1915, where he named the bandlimited or most simple function of a family of cotabular functions (sharing a function table, i.e., values at equally spaced sample points) the cardinal function of this family. One would have to check if he already named the basis functions or if that came later. See A History of interpolation.--LutzL (talk) 11:09, 4 December 2013 (UTC)[reply]

Put fork to Sinc filter on top of the article?

Since Sinc redirects here, it's very probable that some "audio freaks" will first search for sinc used in sound backends, but instead get a boatload of mathematical stuff they actually did not want to delve into so much. Of course, there is a link to the "Sinc filter", but it's on the very bottom, squeezed somewhere between antialiasing and lanczos stuff. However, there's already a "Redirect" template on top. In case you agree, could you give me a hand how to handle this best? (and remove the "Sinc filter" from the "See also" list in the process) -andy 2.242.236.31 (talk) 17:54, 11 October 2014 (UTC)[reply]

Infobox "Solution" year of 1952

The 'Infobox' that was introduced by the edit 2021 Nov 25, includes the heading "solution", and gives a year of 1952. That looks really dumb to me, as if that was the first year that the properties of sin(x)/x was first considered. Does anyone else think that putting "Solution" = "1952" is wrong-headed? — Preceding unsigned comment added by 70.20.25.26 (talk) 19:43, 29 March 2022 (UTC)[reply]

comment

I agree that "year of solution" should be "year of definition" or any close item provided by the infobox.

I feel that the series for the extrema should be an asymptotic one; if so this should be stated. 151.29.137.229 (talk) 19:11, 10 October 2023 (UTC)[reply]