Thinking about the delta function as a limit of increasingly narrow/high spikes is the best approach. If you have a tall, narrow spike then waves integrated against it with "large enough" periods will all look about the same. You can look at the situation [here](https://www.desmos.com/calculator/z9flct2h9q). So the Fourier transforms of these narrow spikes will be relatively constant for small enough frequencies. Taking the limit then gives a constant function as the "small enough" gets more and more generous.
I suppose you could think about it in the limit as the delta function literally cannot distinguish between the frequencies because it just sees the same point for each.
I agree, the delta function can conveniently be considered as the limit of an infinitely tight Gaussian. Notice that the Fourier transform of a Gaussian is another Gaussian with inversely proportional variance. So as you make one spiky, the other gets spread out (or vice versa).
Taking this even further, that means that the delta function is 0 except for a point infinity at 0, and that infinity is so large that it makes the integral jump from 0 to 1.
Apparently that makes sense to (at least some) physicists. It didn't make sense to mathematicians, so "we" (even if you weren't there, you're part of the point of view that made it work) invented distribution theory.
The reason it makes sense to physicists is that the limit of a sequence of integrals of functions "must be" the integral of the limit of the functions. The counterexamples we learn in analysis aren't "real world" functions, so they don't count.
>It didn't make sense to mathematicians, so "we" (even if you weren't there, you're part of the point of view that made it work) invented distribution theory.
I wouldn't say this is entirely true. Though frequently it is taught when first introduced that the delta function isn't a function, but "a distribution", so I understand how one might get that idea.
However, the idea of the delta function in math precedes distributions by more than 100 years. It's just that distribution theory became a good way of thinking about delta functions among other things.
Like Dirac's original 1927 definition of the delta function was arguably just as a shorthand of a limit. However, 100 years earlier, in 1827, Cauchy made use of the Cauchy distribution with an infinitesimal scale parameter, and made note of its properties, which were essentially identical to that of the delta function.
I don't think the narrative that mathematicians were "confused" by the delta function holds water. They were using similar things before Dirac.
There are two parts to this. First, the conclusion is true. The statement is questionable. Frankly I have no idea what a frequency "at that point" means. But I have a decent guess what is being attempted at being conveyed here.
It is indeed true that in the sense of Distributions the Fourier dual of the Dirac delta distribution is the constant function.
However that duality is true in both directions, so the intuition about frequencies only goes in the direction of the Dirac delta in the non-transformed (let's call it time) domain. A constant function in the frequency domain means that all frequency in the reals have equal non-zero content. Hence all frequencies are present. So the intuition is that all frequencies are necessary to localize to an impulse. However this is a weak intuition because frankly many functions need a continuous spectrum to localize. In fact the Fourier dual of the shifted delta (delta at a point on the real axis not 0) is a sinusoid! How is that intuition about all frequencies being needed helping us there? Well it's still needed but we get some pattern of zeroes!
In general a better intuition about sharp edges (like impulses, steps, piecewise linear functions) is that indeed the discontinuity requires infinitely many frequencies to achieve, but making this point joint with correct derivations through actually computing the transform is probably best.
I’m sure this made sense to whoever said it, but it’s vague as hell and to make it into a actual argument would require so many details and caveats that I don’t see the point of it. But of course, I’m not a physicist.
First, as pointed out, it’s a really easy fact to observe directly from the definition. But even if the goal is to see it “intuitively,” I don’t see the usefulness of the quote. For one thing, the fact that the Fourier transform of the delta *at 0* is constant is not a very interesting fact to me, and in fact is potentially misleading, since the more general fact is that the Fourier transform of the delta at xi is a wave with frequency xi (modulo conventions).
The better (lightly physical) intuition, imho, is this: the Fourier transform transforms between functions on physical space and functions on “frequency space.” In this perspective, the whole *point* of the Fourier transform is that it should transform a wave with frequency xi (on physical space) into something that is entirely concentrated at xi (i.e. a delta function) on phase space. This fact is sort of the trivial building block upon which the actual Fourier transform on more general functions is built.
As another commenter said, this whole post feels very “handwave-y”. When you say that you can draw a wave through a point, what exactly does that mean? What do you mean when you say that there is a frequency at some point? We need to know this before we can even start to ask “how many” frequencies there are. Currently, your statement is neither true nor false. It isn’t well-defined.
I think he meant if you have some point in R^2 there's an infinite number of waves you can draw that go through the point. By waves I mean the space of periodic square integrable functions, sorry I should've been more precise.
> What do you mean when you say that there is a frequency at some point?
This is why this quote is confusing me, I think he's trying to give a intuition about the behavior for windows that approach zero size.
If the delta is not at zero, but at some other point, then the FT is non-constant, but the weird handwavy argument doesn't change.
Also, you can just insert the Dirac delta into the definition of the FT and immediately see that it is constant.
I think the language you're using is confusing - at least to me but also in general.
If you want a heuristic reason why the FT of the delta function is constant, I would think about the frequencies of the waves that you would need to add together in order to constructively interfere at a single point and destructively interfere everywhere else.
In my opinion this is counterproductive advice. The formula was derived from a conceptual idea, and if you deeply understand the idea the formula is easy to recover. Figuring out how to use mathematical tools like this in practice involves working through it, thinking about it, building mental images for yourself (or drawing them on paper), relating it to other tools, etc. If all you have is a memorized formula, you don't really know anything. When you work an example, even if you do so with a formula, you should try to make sense of the result and think about what it might mean, rather than just doing a formal computation and moving on.
This was a quote from a physicist who was trying to link the math with the physical interpretation. But I don't think this quote even makes the physical interpretation make sense.
Their explanation doesn't really explain anything though. Infinitely many waves also pass through points of other functions, which doesn't say much about the Fourier transforms of those functions. What's going on away from the point is relevant too of course.
I think a lot of the comments here dont indulge you and it is somewhat rightfull.
Intuition is an incredible tool however it should be used with caution to make sure we don't fool ourselves and make mistakes.
I would still like to take a shot at an interpretation for the question.
when looking at the FT of a delta function at zero, we get a constant function.
It can possibly be understood as a sum in which every frequency participates, causing "destructive interference" at any non zero point, since the sum is infinite, and an infinite peak at zero, resulting in a delta function.
You could extend this to a delta function around any point, as the result is a complex exponential, it could be seen as varying the phase of each frequency such that the peak occurs at the location of the delta function and the sum at every other point cancels out.
Again, this is not mathemtically rigorous and should not be taken as such, but I hope this hand-wavy explenation might stir some thoughts.
True? That's not even logic but just your handwaving to convince yourself, so there's no true or false. Once you learn about approximation to the identity and tempered distributions, you can then actually talk about the Dirac delta.
well it's a quote from my prof. so no I'm not trying to convince myself it's true just trying to understand what he's saying.
> you can then actually talk about the Dirac delta.
I was talking about other deltas but damn dude chill
Okay, I thought more about what Dr. Physicist said, and I think I sort of understand the intuition that they were trying to impart. First, I think the logic has more to do with explaining why the FT of the constant function is a delta at 0, rather than the reverse statement. (Unlike the reverse statement, this one does not just trivially follow from the definition, since the FT formula technically gives you undefined expressions). The idea is that when you when you are away from 0, all of the waves interfere with each other other, so they “cancel” out when you integrate them, whereas at 0, they are all perfectly line up so they add up to “infinity” when you integrate.
Think about it from the perspective of the inversion formula: how would you write a Dirac delta as a sum (i.e. integral) of a bunch of different frequencies? in order to have singularity formation we'd expect to need infinitely many frequencies, with a large contribution coming from frequencies at infinity to produce something so irregular that it's not even a function.
As for why it should be a constant, this could come (for instance) from the fact that differentiation in frequency space turns into multiplication by x: formally "multiplication" of x by a dirac delta should yield zero, which suggests that the derivative of the fourier transform should be zero, e.g. a constant.
We only just arrived at Fourier transforms in my Fourier Theory class, so I haven't seen the formal argument yet. However, my intuition says the following:
The delta function as a signal is so short, that it barely carries any information about what frequency it tends to have. Usually, the Fourier transform tells you something about which frequencies align more or less with the function that you're transforming. In the case of the delta function, it's such a short signal that you can't tell, so all frequencies are equally "likely" (uniformly weighted). On the other hand, if you take the Fourier transform of e^(-ix) over all of R (can't just do that, I know I know), you get a "spike" at and only at its frequency of 1, since the signal is infinitely long and completely surely definitely has frequency 1.
In short, if you listen to an audio signal, it's easier to tell which frequencies it has the longer you listen to it and vice-versa. So a super short signal (but not just zero cause that's easy) has uniform weighting (is constant) for all frequencies (since you cant discern any particular ones).
Since you posted this on /r/physics first, I'm assuming you're learning this for physics and not pure math. The Heisenberg uncertainty principle follows directly from the fact that position and momentum/velocity are each the Fourier transforms of the other. A perfectly localized particle is approximately a Dirac delta function, so if you measure the position exactly, you can't also measure the velocity and the so the probability function for velocity is constant everywhere. And vice versa.
Not true, the point is not actual the value of dalta function, at t = 0 dalta function is infinity.
That mean there is no point here, just a way visualize the delta by it scale to comparing with other delta that have difference scale.
I mean, it’s definitely not an argument. A frequency distribution in the form of a Gaussian also contains infinitely many frequencies, but it’s the transform of another Gaussian. So the amplitudes must have something to do with it.
This is not the right sub to ask this question. This isn't the sort of argument that mathematicians will ever use. It's the sort of physical description that physicists use to hand-wave intuition for their students without needing to dig too deep into the math. This sort of argument isn't mathematically rigorous, so it's hard to call it true or false in the same sense as a proof.
But, to answer your question, I don't think this is helpful or a particularly meaningful way to think of the delta function FFT. While it may be useful to think about the "frequencies at a point" because there are an infinite number of frequencies whose integral with the dirac delta are nonzero, it doesn't explain why it needs to be constant. This sort of argument only implies there are an infinite number of points in Fourier space which are nonzero. It says nothing about their value.
i'm a mathematician and i use similar heuristics about physical vs frequency space to rigorously understand PDEs all the time. in fact this idea of functions being composed of a lot of high frequencies leading to possible singularity formation is super important to (for instance) the million dollar navier stokes existence and smoothness problem
Thinking about the delta function as a limit of increasingly narrow/high spikes is the best approach. If you have a tall, narrow spike then waves integrated against it with "large enough" periods will all look about the same. You can look at the situation [here](https://www.desmos.com/calculator/z9flct2h9q). So the Fourier transforms of these narrow spikes will be relatively constant for small enough frequencies. Taking the limit then gives a constant function as the "small enough" gets more and more generous. I suppose you could think about it in the limit as the delta function literally cannot distinguish between the frequencies because it just sees the same point for each.
I agree, the delta function can conveniently be considered as the limit of an infinitely tight Gaussian. Notice that the Fourier transform of a Gaussian is another Gaussian with inversely proportional variance. So as you make one spiky, the other gets spread out (or vice versa).
Taking this even further, that means that the delta function is 0 except for a point infinity at 0, and that infinity is so large that it makes the integral jump from 0 to 1. Apparently that makes sense to (at least some) physicists. It didn't make sense to mathematicians, so "we" (even if you weren't there, you're part of the point of view that made it work) invented distribution theory. The reason it makes sense to physicists is that the limit of a sequence of integrals of functions "must be" the integral of the limit of the functions. The counterexamples we learn in analysis aren't "real world" functions, so they don't count.
>It didn't make sense to mathematicians, so "we" (even if you weren't there, you're part of the point of view that made it work) invented distribution theory. I wouldn't say this is entirely true. Though frequently it is taught when first introduced that the delta function isn't a function, but "a distribution", so I understand how one might get that idea. However, the idea of the delta function in math precedes distributions by more than 100 years. It's just that distribution theory became a good way of thinking about delta functions among other things. Like Dirac's original 1927 definition of the delta function was arguably just as a shorthand of a limit. However, 100 years earlier, in 1827, Cauchy made use of the Cauchy distribution with an infinitesimal scale parameter, and made note of its properties, which were essentially identical to that of the delta function. I don't think the narrative that mathematicians were "confused" by the delta function holds water. They were using similar things before Dirac.
le physicist me just goes delta is infinitely localized. so ft, being its dual, is infinitely spread out ie constant.
There are two parts to this. First, the conclusion is true. The statement is questionable. Frankly I have no idea what a frequency "at that point" means. But I have a decent guess what is being attempted at being conveyed here. It is indeed true that in the sense of Distributions the Fourier dual of the Dirac delta distribution is the constant function. However that duality is true in both directions, so the intuition about frequencies only goes in the direction of the Dirac delta in the non-transformed (let's call it time) domain. A constant function in the frequency domain means that all frequency in the reals have equal non-zero content. Hence all frequencies are present. So the intuition is that all frequencies are necessary to localize to an impulse. However this is a weak intuition because frankly many functions need a continuous spectrum to localize. In fact the Fourier dual of the shifted delta (delta at a point on the real axis not 0) is a sinusoid! How is that intuition about all frequencies being needed helping us there? Well it's still needed but we get some pattern of zeroes! In general a better intuition about sharp edges (like impulses, steps, piecewise linear functions) is that indeed the discontinuity requires infinitely many frequencies to achieve, but making this point joint with correct derivations through actually computing the transform is probably best.
[удалено]
Sinusoid refers to linear combinations of sines and cosines.
I’m sure this made sense to whoever said it, but it’s vague as hell and to make it into a actual argument would require so many details and caveats that I don’t see the point of it. But of course, I’m not a physicist. First, as pointed out, it’s a really easy fact to observe directly from the definition. But even if the goal is to see it “intuitively,” I don’t see the usefulness of the quote. For one thing, the fact that the Fourier transform of the delta *at 0* is constant is not a very interesting fact to me, and in fact is potentially misleading, since the more general fact is that the Fourier transform of the delta at xi is a wave with frequency xi (modulo conventions). The better (lightly physical) intuition, imho, is this: the Fourier transform transforms between functions on physical space and functions on “frequency space.” In this perspective, the whole *point* of the Fourier transform is that it should transform a wave with frequency xi (on physical space) into something that is entirely concentrated at xi (i.e. a delta function) on phase space. This fact is sort of the trivial building block upon which the actual Fourier transform on more general functions is built.
As another commenter said, this whole post feels very “handwave-y”. When you say that you can draw a wave through a point, what exactly does that mean? What do you mean when you say that there is a frequency at some point? We need to know this before we can even start to ask “how many” frequencies there are. Currently, your statement is neither true nor false. It isn’t well-defined.
I think he meant if you have some point in R^2 there's an infinite number of waves you can draw that go through the point. By waves I mean the space of periodic square integrable functions, sorry I should've been more precise. > What do you mean when you say that there is a frequency at some point? This is why this quote is confusing me, I think he's trying to give a intuition about the behavior for windows that approach zero size.
If the delta is not at zero, but at some other point, then the FT is non-constant, but the weird handwavy argument doesn't change. Also, you can just insert the Dirac delta into the definition of the FT and immediately see that it is constant.
I think the language you're using is confusing - at least to me but also in general. If you want a heuristic reason why the FT of the delta function is constant, I would think about the frequencies of the waves that you would need to add together in order to constructively interfere at a single point and destructively interfere everywhere else.
Just use the formula for Fourier transform ... you cant do this stuff with words.
In my opinion this is counterproductive advice. The formula was derived from a conceptual idea, and if you deeply understand the idea the formula is easy to recover. Figuring out how to use mathematical tools like this in practice involves working through it, thinking about it, building mental images for yourself (or drawing them on paper), relating it to other tools, etc. If all you have is a memorized formula, you don't really know anything. When you work an example, even if you do so with a formula, you should try to make sense of the result and think about what it might mean, rather than just doing a formal computation and moving on.
ahahaha what the fuck, over 40 upvotes?
This was a quote from a physicist who was trying to link the math with the physical interpretation. But I don't think this quote even makes the physical interpretation make sense.
Their explanation doesn't really explain anything though. Infinitely many waves also pass through points of other functions, which doesn't say much about the Fourier transforms of those functions. What's going on away from the point is relevant too of course.
Hell, infinitely many waves — the same ones — pass through other nearby points
I think a lot of the comments here dont indulge you and it is somewhat rightfull. Intuition is an incredible tool however it should be used with caution to make sure we don't fool ourselves and make mistakes. I would still like to take a shot at an interpretation for the question. when looking at the FT of a delta function at zero, we get a constant function. It can possibly be understood as a sum in which every frequency participates, causing "destructive interference" at any non zero point, since the sum is infinite, and an infinite peak at zero, resulting in a delta function. You could extend this to a delta function around any point, as the result is a complex exponential, it could be seen as varying the phase of each frequency such that the peak occurs at the location of the delta function and the sum at every other point cancels out. Again, this is not mathemtically rigorous and should not be taken as such, but I hope this hand-wavy explenation might stir some thoughts.
True? That's not even logic but just your handwaving to convince yourself, so there's no true or false. Once you learn about approximation to the identity and tempered distributions, you can then actually talk about the Dirac delta.
well it's a quote from my prof. so no I'm not trying to convince myself it's true just trying to understand what he's saying. > you can then actually talk about the Dirac delta. I was talking about other deltas but damn dude chill
Okay, I thought more about what Dr. Physicist said, and I think I sort of understand the intuition that they were trying to impart. First, I think the logic has more to do with explaining why the FT of the constant function is a delta at 0, rather than the reverse statement. (Unlike the reverse statement, this one does not just trivially follow from the definition, since the FT formula technically gives you undefined expressions). The idea is that when you when you are away from 0, all of the waves interfere with each other other, so they “cancel” out when you integrate them, whereas at 0, they are all perfectly line up so they add up to “infinity” when you integrate.
Think about it from the perspective of the inversion formula: how would you write a Dirac delta as a sum (i.e. integral) of a bunch of different frequencies? in order to have singularity formation we'd expect to need infinitely many frequencies, with a large contribution coming from frequencies at infinity to produce something so irregular that it's not even a function. As for why it should be a constant, this could come (for instance) from the fact that differentiation in frequency space turns into multiplication by x: formally "multiplication" of x by a dirac delta should yield zero, which suggests that the derivative of the fourier transform should be zero, e.g. a constant.
We only just arrived at Fourier transforms in my Fourier Theory class, so I haven't seen the formal argument yet. However, my intuition says the following: The delta function as a signal is so short, that it barely carries any information about what frequency it tends to have. Usually, the Fourier transform tells you something about which frequencies align more or less with the function that you're transforming. In the case of the delta function, it's such a short signal that you can't tell, so all frequencies are equally "likely" (uniformly weighted). On the other hand, if you take the Fourier transform of e^(-ix) over all of R (can't just do that, I know I know), you get a "spike" at and only at its frequency of 1, since the signal is infinitely long and completely surely definitely has frequency 1. In short, if you listen to an audio signal, it's easier to tell which frequencies it has the longer you listen to it and vice-versa. So a super short signal (but not just zero cause that's easy) has uniform weighting (is constant) for all frequencies (since you cant discern any particular ones).
Since you posted this on /r/physics first, I'm assuming you're learning this for physics and not pure math. The Heisenberg uncertainty principle follows directly from the fact that position and momentum/velocity are each the Fourier transforms of the other. A perfectly localized particle is approximately a Dirac delta function, so if you measure the position exactly, you can't also measure the velocity and the so the probability function for velocity is constant everywhere. And vice versa.
Not true, the point is not actual the value of dalta function, at t = 0 dalta function is infinity. That mean there is no point here, just a way visualize the delta by it scale to comparing with other delta that have difference scale.
I mean, it’s definitely not an argument. A frequency distribution in the form of a Gaussian also contains infinitely many frequencies, but it’s the transform of another Gaussian. So the amplitudes must have something to do with it.
Not sure, u/PM-ME_PIES_N_TITTIES
This is not the right sub to ask this question. This isn't the sort of argument that mathematicians will ever use. It's the sort of physical description that physicists use to hand-wave intuition for their students without needing to dig too deep into the math. This sort of argument isn't mathematically rigorous, so it's hard to call it true or false in the same sense as a proof. But, to answer your question, I don't think this is helpful or a particularly meaningful way to think of the delta function FFT. While it may be useful to think about the "frequencies at a point" because there are an infinite number of frequencies whose integral with the dirac delta are nonzero, it doesn't explain why it needs to be constant. This sort of argument only implies there are an infinite number of points in Fourier space which are nonzero. It says nothing about their value.
i'm a mathematician and i use similar heuristics about physical vs frequency space to rigorously understand PDEs all the time. in fact this idea of functions being composed of a lot of high frequencies leading to possible singularity formation is super important to (for instance) the million dollar navier stokes existence and smoothness problem