Smooth Interpolation Function in One Dimension | Smooth Interpolation Function E1
ฝัง
- เผยแพร่เมื่อ 6 มิ.ย. 2024
- #SoME2
This video gives a detailed construction of transition function for various levels of smoothness.
Sketch of proofs for 4 theorems regarding smoothness:
kaba.hilvi.org/homepage/blog/...
Faà di Bruno's formula:
en.wikipedia.org/wiki/Fa%C3%A...
Proof that e^(-1/x) is smooth:
en.wikipedia.org/wiki/Non-ana...
Chapters:
00:00 Intro
00:10 Definition of Smoothness
01:29 Hermite Interpolating Polynomials
04:32 Step Function
05:56 Auxiliary Theorems
08:43 Constructing Smooth Step Function
13:30 Main Problem
13:51 Outro
🎵Music provided by BGM President
🎵Track : Some Vintage Mood 7 - • [Royalty Free Music] S...
Corrections:
08:04 for the first theorem, f should be f'
08:15 for (2), it should be f(x) ≠ 0
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Is there a name for this kind of interpolation so that I can search more about?
I wish a name was given for the process but I do not know. In practice, matching boundaries up to certain order of derivatives is used all the time in numerical analysis (e.g. Hermite Polynomials) which are practical, but to require infinite differentiability in a closed formula is something you may or may not see in a course in smooth manifolds/functional analysis which are graduate level subjects, and thats why I thought it was appropriate to cover it since it is not very accessible at an elementary level in context.
But if I were to give the closest concept for this kind of "filling in the middle smoothly" process with a known name, I would say searching Smooth Urysohn's Lemma would get you most relevant results. My next video in the series is going to cover that topic
@@EpsilonDeltaMain I see! Thank you for your quick response. Another thing I would like to ask is: are there different solutions, that is, solutions other than taking e^(-1/x) ? If not, wouldn't it mean that the limit of Hermite polynomials when their degrees tend to infinity converges to the found solution? In the video, you said that using an infinite series would result in the Taylor series, but isn't this different from doing the previous limit? Because with limits you get what the polynomial approaches when it goes to infinity, not what it is at infinity.
@@pedrokrause7553 Very good questions. I like your questions and it adds so much value to some missing details in the video, so I pinned it if you don't mind
1. It doesn't have to be e^(-1/x). as long as the f in the f(-1/x) decays asymptotically a lot faster than -1/x shoots off to infinity as x -> 0+. For example f(x) = 5^(-x^2) will do it as well [*this is a gaussian], and if you are clever enough you can find an infinite family of these kinds of functions, including ones that does not directly use exponential function, such as erf(x) or 1/Γ(-x+2). But tail ends of functions like arctan(x) or 1/(x^10+1) will not decay fast enough to make derivatives of all order = 0 at x=0 for f(-1/x).
Plus, we dont even have to use f(-1/x), and use something like f(-1/x^2) as this example shows:
en.wikipedia.org/wiki/Flat_function
2. You are right, I assumed that if such Taylor series existed, it would fail to satisfy the left and the right simultaneously. e.g. if the left function was sin x, then taylor series uniquely defines the function extension to be the sin x.
3. But if we instead take a look at the limit of these hermite polynomials, the series wouldn't converge. Just take a look at first few hermite polynomials. for step interpolation.
en.wikipedia.org/wiki/Smoothstep
The coefficients blow up to tens of thousands fairly quickly, and the function only is bounded since the terms are alternating and pluses and minuses cancel each other. The limit of the polynomial would not exist since it would be like ∞x-∞x^2+∞x^3-∞x^4... if you look at the closed formula of coefficients of the hermite polynomial for each order
Look up partitions of unity.
Maybe look up "bump function" or "mollifier". If I understand it correctly, this sort of interpolation uses such bump-functions to "crossfade" between the two target functions.
The "Can we do any better?" with Lara Croft got me good
The subtle humor in this is incredible
Good stuff.
Kind of gives insight as to why finding an "analytic" continuation (but for this video not really as we are only dealing with reals, but more generally) can be difficult. Why "infinitely differentiable" is such a constraining condition.
(Hopefully) Constructive criticism:
I found myself losing track of which "Greek letter function" was modeling what parts of our goal.
Like it would be helpful to have a line like "φ will be the continuous step function used to interpolate" or something when you defined the function.
Same for the ψ function too.
If you did already describe it, there was enough time between when you and when you stated it and performing the proofs and derivations (8:30 ish) that it deserved having a reminder at that point.
I agree, proper distinction goes a long way into making whatever you're saying more understandable
lost me there as well
Usually the phrase "analytic continuation" applies to complex functions, and the Taylor series is _the unique_ analytic continuation. This construction introduces nasty essential singularities at x=0 & x=1 (not that real valued functions care).
The smooth continuation is in fact not analytic, at a and b. The function exp(-1/x), has all of its derivatives equal to 0 at x=0, so it would be equal to the null function in the neighborhood of 0 if it were analytic.
@@An-ht8so I mean, it doesn't need to be analytic, the fact that the transition is smooth already removes a lot of headaches when stitching functions together.
Safe to say, I’m confused
k on fused functions
@@PTAlisPT this got me 😂😂😂
It’s easy. He gradually and smoothly transitioned from one function to another function where the two functions are both smooth and chosen to meet perfectly with the ends of the two given functions.
@@terjeoseberg990ah, okay. personally, i’m just confused on the whole “C^k” and “psi” and “phi” stuff. what is “C”? all that kinda stumped me on my first time watching.
@@sans1331 C^k is a set, it is the set of functions that are k times differentiable with all those k derivatives being continuous. You can also use the notation C^k(Omega) which means that its functions are k times differentiable and the k-th derivative is continuous over the domain Omega.
Thank you so much! This has been in the back of my mind for a while now. Great explanation.
This is such quality content! I will be sure to send it to everyone I know who will be interested.
Woah. When I just saw this video in my feed I tried a few ideas on paper, and it's really cool to see what the actual solution is, and which ideas I had were in the right direction and which weren't.
How did you do it?
Just found this gem, brought back memories of an intro analysis class of a few years back. Thank you!
Very beautiful technique. I also love that that function e^(-1/x) is bonkers in the complex plane so this argument totally breaks down on the complexes.
Well, if a complex function has a first derivative on an open set, then it has derivatives of all orders on that set, and is even analytic there. (It's possible to construct a real function which is infinitely differentiable on all of R and yet is nowhere analytic. Real analysis is so good at crushing reasonable expectations.)
@@tomkerruish2982 e^(1/x) doesn't have a first derivative at 0. Looks flat in the real numbers but move the slightest bit in the imaginary direction and it's totally chaotic.
@@LukePalmer First, I'll admit that I glanced at your comment and read it as "exp(-1/x²)", erroneously inserting the exponent. Second, however, like you I was highlighting the (to me) main difference between the real and complex derivative. exp(-1/x²) has real derivatives of all orders at x=0, but is so badly behaved for complex values that it has an essential singularity.
I certainly confess to the twin sins of reading too quickly and writing too tersely.
Godddamn, you talked to me right on the cusp of my knowledge. I saw that interpolation a whole 5 minutes before you revealed it, but you built the conceptual framework so well that it basically taught itself. You made the knowledge jump out of the words and equations. Incredible!
What a wonderful video, one of the best in #SoME2! Keep on the good work!
I’ve actually always wondered about this. Thanks for sharing!
This video is just mindbowing, never thought that there would even be a way to construct truly smooth interpolation. Also, your visuals and presentation is really great, loved it. Keep going!
Interestingly enough, it was teached to us in the university when we were introduced into Dirac delta function. This was a part of some 3rd year math for physics students. These c-infinite functions are required to properly define and prove theorems which involve the Dirac delta, and by an extension, Green's functions which are "supercharged deltas" and therefore QED propagators, which are essentially Green's functions. This is why this was important for physics students - we actually need this to learn QED (which we learned afterwards).
I recently used the same thing in one of my projects and I ended up using the cubic interpolation approach.
I might implement something similar to what was shown at the end.
Thanks for the knowledge.
I 100% needed this, thank you so much
This is a very good video. I only ask that you look into stabilizing the volume of the voice over. I found that it was drifting up and down, occasionally to the point that I couldn't hear it over the music. You can probably do this with a single button press in your editing software.
Thank you for bringing this interesting math to the public eye. There's no way I would have seen something like this without you. I hope you keep making videos
turn it up and compress.
Amazing quality of mathematical argumentation, balancing rigor and pedagogy! I instantly subscribed
This channel truly has a future. Signed.
Nice vid. I cant wait to see what videos you post next
MAN, KEEP DOING WHAT UR DOING, YOU'LL GET A LOT OF SUBSCRIBERS IN NO TIME
I did not know I needed this, but glad I found it. Long live smooth transitions.
great video, and thanks for putting link to proofs in the description
This should have millions of views, this is incredibly useful in practice :D
So -- I tinker with math sometimes. And this might actually be exactly what I needed to take an idea to the next step. Great video!
Your use of color makes this much easier to follow. Subscribed.
Awesome topic with great presentation. I would not have guessed the solution is this elegant. Just great job on video.
OMG that cliffhanger at the end 😭 what a great video, congrats!
I had to play the video twice to finally understand that the solution you expose is what is known as "crossfading" in audio engineering, and that most of the video is devoted to how to easily build a decent S-shape transition signal. I appreciate when ideas and intentions come first in plain language, and the maths come after, it's more easy for me to follow. However, I subscribed to your channel. Please, keep the pace slow, and the music down ! :D
I had heard about analytic continuation, but hadn’t thought about how a function could be smooth and not analytic before now. Thanks
My intuition would say it's possible with real functions but not complex functions, but I'm not certain of that. Differentiable just has different consequences for each.
@@xinpingdonohoe3978that's true - complex differentiable (even just once) functions are *very* rigid. They're automatically infinitely complex differentiable (holomorphic) and all holomorphic functions are (complex) analytic
@@xinpingdonohoe3978complex differentiable functions are locally analytic, so you are correct that you cannot have a complex smooth function that isn't analytic
This is really surprising! When I first saw the title I figured smooth transitions would be impossible, but here we are lol
So cool! I had a feeling it should be possible to do this conceptually by taking the limit as k -> infinity of the k-differentiable approximations, but it's great to see a general construction of the infinitely differentiable version. Great video!
My intuition says the equivalent problem in 2D is impossible in general, but I can't wait for the video on it!
Well, an infinite series of 2K terms would be disappointing, but as he showed it can be done in closed form.
Very good videos, please continue!
Awesome video! Thank you!
Wow that was great!
amazing content !
I think it would be interesting didatically if you did a small recap at the end, but please keep doing this amazing work !
Great video man keep it up
Only 330 subscribers? This is a crime for such amazing content
Nice! This was surprisingly interesting (:
you are gonna go big keep it up
Awesome video! Learned a lot :)
Thanks. That smoothly connected several topics for me. You seemed to be approaching the halted problem when you halted.
That's really cool! The exercise of proving e^(-1/x) is smooth at x=0 must've come up in like five different math classes I took and now finally I see how that might be useful.
The real kicker is that even though it's smooth at 0, it doesn't have a Taylor series expansion around 0.
@@HilbertXVI That's what Laurent series are for.
@@HilbertXVI It does have a Taylor expansion at 0 (every smooth function does) - the kicker is that its Taylor series doesn't converge to it at (in any neighborhood of) 0.
@@schweinmachtbree1013 Not a very useful "Taylor expansion" if it doesn't converge to the function
You just revived my love for calculus🥺✨✨.
Great motion of mathematical thoughts!
your feelings are irrational
Would you like some pi with that?
I have been studying mathematics for over four years now and I had never seen Faà di Bruno's formula. Today, I've suddenly stumbled across it twice for completely unrelated reasons 😂
Just goes to show that there's always more to learn in mathematics
One small thought. The way I like to teach Taylor polynomials is by going "Okay, a tangent line is a good approximation but it doesn't approximate the derivative well, so what if we use a tangent line to approximate the derivative instead, and then take the integral ?"
Assuming both functions are smooth too, wouldn't that also be a possibility ? Take the derivatives, use a line to interpolate, and take the integral ?
It should, as long as you choose the constant terms in the integration process so that your function and one (should not need to be both) of the bounding functions match, up to their k-th derivative.
This will just construct a truncated Taylor series which will have the same problems as the Hermite and Taylor approaches.
@@tracyh5751 OP's post was about teaching how Taylor polynomials work. So ending up with a Taylor polynomial shows that it's a valid perspective on it.
A simplification of the phi function is 1/(1+e^((1-2x)/x(1-x)). Just for those who don’t want a ton of e^-1/x in the phi function
Nice. I guess, that will be useful for optimizing the code in a practical implementation because it reduces the number of calls to exp from 2 to 1.
Neat. Im not a math person, and im not good at math, but i love computer graphics and wish i could model certain processes, so i always end up having math questions out of my league.
This was one. And i wouldnt have known what to search.
Great video !!
This exactly what I wanted to find months ago when I created a function adder. A function that can add the graphs of two different functions. Unfortunately it was undefined at the cutting point because of a division.
wow! I remember thinking about this for ages when I was a student and I really thought no such method existed! should have thought of e^(-1/x) obviously...
GREAT !!! THANK YOU~~~
9:13 Psi needs to be *strictly* monotone increasing. Since otherwise the 0-function would satisfy your conditions, but phi could not be defined
monotonic increasing does mean that it's not the 0 function. The 0 function would be monotonic non-decreasing. So the video is correct.
@@SoumilSahu That depends on your specific definition I guess, if you use monotonic increasing for the usual definition of strictly monotonic increasing, then the 0-function is not monotonically increasing I suppose. Although monotonic non-decreasing is a weird term in my opinion.
@@SoumilSahu No the video isn't correct because it uses the condition f' ≥ 0 for a function f being 'monotone increasing', which is the condition for weak monotone increasingness; if strict monotone increasingness was meant then the condition f' > 0 would have been used (which would have ruled out the zero function)
7:56 “It is too big to fit in the margin” - Pierre de Fermat has entered the chat
Great video thank you
i think i remember this exact interpolation function coming up in thermodynamics somewhere, as a statistical distribution of energy states or something of that sort
it's been a while and it was never well explained at the time, but I've always remembered it as "that function that could probably interpolate two other functions _really_ nicely"
it could have just been one of so many variants of the logistic function, though; it's been ten years
this is highly useful for data compression!
Ive never been more thankful for python math modules that abstract this all into a function call i don't have to worry about lmao
My initial guess was to use interpolation between the two functions in the smoothing area with a shifted sine as the weight
Very cool 🔥💖
That's so interesting google showed this. I was actually working on something like this for a personal CS project and was just hammering out in mathematica. Ill be curious to plug this in abd see what it looks like
Thanks!
This is a great video
Nice and creative video ❤❤❤
That last part was important, it can be used to form a good number of questions even for basic calculus.
Great video. This topic reminds me of the notion of mollifier functions. It seems like you could use a mollifier function along with some arbitrary continuous interpolation to create a smooth interpolation, but I'm not actually sure if that's true.
This is indeed a “bump” or “mollifier” function. Not sure if the terms are equivalent
The tool I would immediately reach for is the error function. Define h(x)=(1+erf( (x-x0)/a ))/2 for some point x0 and width a, then your interpolated function is f(x)+(g(x)-f(x))*h(x).
Me: I need to go to bed earlier
Also me at 2 am: I wonder how to smoothly connect two functions
Hi, very interesting video/concept, did you ever end up making the video for higher dimensions ? I work on differential geometry for quantum physics for my PhD and am looking for similar stuff !
A little fast, and I had to take your word for a decent amount of it, but still very followable and intriguing, especially since I've thought about a very vaguely related thing before - how no 2 polynomials look the same over any interval, barring trivial exceptions like translations. I'm not even 100% sure it's true and I wouldn't have a clue how to prove it, but if it is true, it's fascinating to me that each polynomial shape is completely unique. This kinda links in to how it would be difficult to get 2 different functions to 'agree' with one another via a smooth transition function, though I admit it's a bit of a stretch.
Hmm I don’t think it would be possible to overlap two polynomials like that because every polynomial is analytic and has a unique Taylor series, which means that you can determine what it looks like over all the reals just by looking at all the derivatives at one point. It does feel crazy though, that with all the infinitely many polynomials, there aren’t two that line up for some interval.
I remember being very impressed when I found out about this
this is great
AWESOMEEE!!!!
Very cool! I have been looking for such a transition function but because of the Taylor series issue I thought it might not exist.
By the way: is it possible to write this function in a numerically stable way?
I mean without the infinities which occur temporarily in intermediate results during the computation due to the 1/x parts near 0 and 1.
Very cool!
Very Interesting topic!
The Lara craft pic while talking about jagged edges killed me
lol, you summed up 2 years of Calculus in the first 50 seconds of the video
Very interesting video. I hope I won’t need any of this as I try to build my own animation script, but it’s good to know. My only issue with the video is that your music is a bit loud.
The interpolation function you used here would actually be very useful for using bezier curves to construct smooth tracks, e.g. for rollercoasters. Cubic beizers are very intuitive to work with, but the acceleration (2nd deriv of the track position) is discontinuous. If you modify the beizer curve's interpolation function from stepped linear to this smoothstep, the acceleration will be continuous. You could use a gradient descent solver to minimize lateral G forces by manipulation of the curve control points!
the is actually a concept of geometric continuity (which is different to parameter continuitiy which is obviously to severe) of 1st and 2nd derivatives of bezier curves. You can google for it. Quite interesting.
Отличный контент!
My brain is smooth now
3:26 you just blew my mind
At 8:11 for the first statement if f' and not f. But still a great video with interesting topic
I love the line rider ref
This is used in motion control. I typically use 3rd and 5th order interpolation
bro really performed surgery on functions
Bro really left a comment on a TH-cam video
Awesome
And then that cliffhanger...
You Sir, know what you're doing.
Do you know about Exporational B-Splines?
Those are the limit of rational B-Splines for infinite degree.
Another fun variant is the Fabius function which can be defined through repeatedly integrating and rescaling an interval
I gained a follower!
Great video! I was just wondering if there isn't supposed to be f prime instead of f at 8:08. If not, it would mean that C^n+1 is always a subgroup of C^n, right?
At 2:24/14:04, the set of 4 equation (simultaneous), last equation f'(-1) = c1 -2c2 +3c3 = - 0.3678 (or -e^-1)
Nicely explained.
Thank you
I was like “wow this is really an analysis heavy question,” then I read the channel name.
there are smoothstep and smootherstep, behold, _the smootheststep_
Is this related to the concept of partitions of unity? And what if we want to do something else like minimize the length of the interpolation curve? Is there a unique solution? If not, is there some other reasonable constraint that guarantees a unique curve?
Great video, but please, slow down with your speaking and explanations.
You might know 3blue1brown, and his first videos were extremely rushed, making it very hard to follow. Nowadays, he talks slower, to let the information sink. That's something you could do, too. Maybe add some pauses.
It's not possible to procress so much information in such a short amount of time if you're not familiar with it. Even I, who took Calculus classes, had a hard time following. So great video, but please, slow down your talking and explanations
You can always watch it at 0.75, 0.5, or even 0.25 times speed.
@@olivergolde8435no, you can’t. It makes it harder to understand anything he’s saying in the first place
this was a lot of work to just simply interpolate the extensions over the timeframe of that extension. Meaning you have an assigned missing part of the graph in which you extend both functions into, you then go from one function to the other, similar to how spline works. If you want true smoothness, then it requires either doubling the missing length or using actual splines but could create 3 answers of the function at some x values. So we have for the simpler approach:
lerp(a, b, c) = (b-a)*c+a; line interpolation function
given A as x coordinate for first function f(x) cut off and B as second g(x), we have:
h(x) = lerp(f(x), g(x), (x-A)/(B-A)); for values between A and B
no derivative or difficult nonsense needed
linear interpolation. mollify.
if you are not happy that you have modified the original functions at the joining points, just let the given function extend a little further and mollify their linear interpolation with a function with support small enough.
I think I would also show what happens if you take the limit of the Hermite polynomials. I don't know if that was said in this video, but I didn't hear that or see that shown. I don't know for sure, but I guess that would be the power series of each extending to the middle with a sudden jump there
Check out the pinned comment, its answered there
Amazing Video!
How to interpolate between two arbitrary points at the end was a bit rushed tho... Perhaps you could go a little more in depth in the next video?