When Functions We Want to Interpolate Aren't Too Nice | Smooth Interpolation Function E1.1

This video is a direct follow up of the Part 1 of the series:
• Smooth Interpolation F...
Part 1 of the series was made for a submission for the Summer of Math Expositions 2:
• Summer of Math Exposit...
but since I had to meet a deadline, it was submitted with some missing details. In this video, we aim to address potential issues from pathological functions.
Chapters:
00:00 Intro
00:53 Problem Statement
01:20 Potential Issue 1
02:37 Potential Issue 2
🎵Music provided by Kevin MacLeod
🎵Track: Midnight Tale - • Midnight Tale
Corrections:
02:40 the discontinuity for f should be at x = -1, not x = 0
02:58 g(1) = 0
03:00 g'(1) = -1 and g^(k)(1) = 0 for all k greater than 1

[@tsawy6]

I feel like the very last claim is far and away the most interesting one of the video, and that it could do with more screen time

[@romajimamulo]

That was an incredibly abrupt ending, I feel like I needed some justification for these

[@denki2558]

You can just extrapolate the video using taylor series to see it's natural continuation.

[@ianweckhorst3200]

I want desmos graphs, seriously, it would be so fun to dig my roots into this

[@112BALAGE112]

I have a small suggestion to make regarding your style of presentation. It would greatly aid our understanding if at the end of the video, as a kid of recap, you would describe the solution in terms a fully fledged algorithm. All throughout the two videos so far you quickly state many assumptions, theorems and methods and it gets slightly hard to keep track of. It would be nice if we didn't have to remember everything in short term memory. You should write down more stuff, kinda like the way professors do on blackboards.

[@aslpuppy1026]

I would really like to see this as well!

[@abird9724]

Incredible, please continue!

[@redapplefour6223]

definitely the quickest followup from any of the random channels that sprouted up after SoME2 i subbed to!

[@bskim3860]

This is a very clear explanation. Please look forward to the next episode.

[@Epicfraxy]

Very eagerly awaiting the next installment. Very good content!

[@school6268]

So happy for part 2, wanted it so bad after the first one

[@atlasxatlas]

a follow up? hell yeah

[@frba9053]

Let’s go

[@TonyboyDK]

Subscribed. Videos are great, keep it up!

[@9WEAVER9]

Really 'smooth' follow up!

[@AJ-et3vf]

Awesome video! Thank you!

[@derblaue]

3:14 f(x) seems wrong. Shouldn't it be either e^(1/x) or 0 when x=1 instead of 0?

[@HaramGuys]

You are right

[@EpsilonDeltaMain]

Yep, I actually found more than one typo in that scene

[@Drevoed]

Brilliant!

[@quintium1]

This is basically what I suggested!

[@alex_zetsu]

I'm not sure why we want to interpolate two "not too nice" functions. If they aren't smooth in the transition area, it seems there is no "right" interpolation. I mean the method 2:03 works. Make any smooth function and then stich the 3 together. Change the middle one to whatever you want. Sure in this case f(x)= -X might look weird, but if you think about it since the functions we're trying to bridge aren't smooth during the whole transition, I don't really think any way is "right."

[@stewartzayat7526]

As he said, it depends on how you define your problem. If you only want smoothness in a mathematical sense, then you are right. But that little hump that was introduced looks very artificial and it feels like it is cheating the problem statement a little bit. The problem is more so that the original question was not concrete enough and allowed for something that technically works, but probably isn't what you wanted.

[@alex_zetsu]

@@stewartzayat7526 When the two functions you're trying to join aren't smooth, in my opinion anything is kind of artificial. Yes, the hump does look a bit odd. But at the f(0) of the final function is trying to smooth between negative infinity and 0, so I don't really think there is a natural answer.

[@misterlau5246]

It's a question of what works for the particular application. Audio, best function is sinc. 128 or more... But 128 is usually more than enough quality

[@paulkocyla1343]

My approach would be low-pass filtering the functions in the transition region (and some neighborhood) before interpolating. That´s just an intuitive idea, I guess they wouldn´t freak out if you set the cut-off frequencies right, as you´d limit their spectral bandwidth.

[@alexsere3061]

I am glad we have another video but I am confused. isn't the whole point that we developed the switch technique because interpolation using an infinite system didn't work? I am gonna have to sit down and think about this one Edit: okay i sat down for like and hour, and my solution was to build the basis of a vector space. You take the switch function f in the previous video. It has derivatives 0 at 1, it its value at 1 is 1. If you integrate it you end up with g such that its first derivative at 1 is 1, all higher derivatives are 0, and its image of 1 is a constant K. Leaving us with function h(x) = g(x)-K f(x). You can prove h(x) has all derivatives 0 at 0, and all derivativves (including 0th) at 1 equal zero, except for first derivative. Using this process you can build a basis of functions to interpolate smoothly.

[@giuseppepapari8870]

I like this video very much! Just a minor remark, at the end it is g(1) = 0 and g(k)(1) = 1, not the other way around. But said that, great work!!

[@willwu7353]

how about convolution with sin(x)/x at 0? integral of sin(x)/x from -inf to inf > lim 1/x from -inf to inf, and i believe it has bounded support.

[@frba9053]

Amazing

[@EmmanuelMessulam]

If I choose hermite interpolation to be the middle function, following from the concepts of part 1 (where one function is "activated" and the other "deactivated") shouldn't I choose a point a'

[@ryanlangman4266]

If I’m understanding your question correctly, I think the problem is that the last point where f and g are interpolatable frequently doesn’t exist in the codomain of the function. Like you said with 1/x there is no highest point that you can pick to make the interpolation the nicest. Infinity doesn’t exist in the codomain. It does seem nicer to extend it as far as possible, but you have to choose a point to stop because in the limit it’s no longer analytic. I do think this could be a better approach for something like sin(1/x) where you could assign g(0) to be the average value of sin(x) (so 0). But you still run into problems because you still have infinite oscillations around 0. So it’s still not analytic there. So, if you have to stop somewhere then why not just stop at the edge of the domain of the function. I do think, however, that when using the hermite polynomial method of interpolation, the domain of the hermite polynomial should be defined to be as small as possible in the limiting sense because that limit does exist. To extend this further, you could use 2 different domains for hermite polynomial on the left and right side interpolation. You could define the function on (a, (a+b)/2) as the smooth interpolation between f on its domain and the hermite polynomial defined from [(a+b)/2, b). Then, You could define the function on ((a+b)/2, b) as the smooth interpolation between the same hermite polynomial instead defined from [(a+b)/2, b) and g on its domain. This means that the only point that is not a natural continuation of the function is the point (a+b)/2.

[@ryanlangman4266]

Now that I’m thinking about it a little more, I think my plan might fall victim to the same problems as yours. If you define the function the way I stated, it won’t actually exist near (a+b)/2. It must be defined on a finite domain around that point and there is no smallest domain that is the most natural. I do think my method of limiting the hermite polynomials domain rather than extending the domain of the functions tends to give a smoother overall transition function, but they both have the same problem of not existing in the limit.

[@polyhistorphilomath]

The trick, I think, is to calculate a Schwarz function for a complex contour (bounding the region R, the middle [the non-disjoint union of regions in which your functions f and g are meromorphic or exhibit other pathological behavior]) that you can live with. Assume WLOG f is defined and well-behaved on the left, and g defined and well-behaved (without poles, etc.) on the right. Then simply continue f to F across the contour (moving to the right) and g to G, across the contour (toward the left) using the Schwarz reflection principle. You could have used h(f, g; a, b){x} as defined in the previous video, except that in framing this question we knew we had found or would find some fault with f and g. h(F, G; a, b)(x) is your friend, because it is analytic (except possibly at the points a and b, but I'm no Weierstrass and I don't actually care about the theoretical implications of a definition including the Heaviside step function).

[@polyhistorphilomath]

I suppose this approach (if it works for a single contour) would also work for two contours, so long as you don't have any branches interfering with the [a,b] interval on the real line. Then you can use a computationally or otherwise convenient Schwarz function for f and another for g.

[@EmmanuelMessulam]

@@polyhistorphilomath I don't think the proposed solution works well for sin(1/x) for example. Another problem (I think) it has is that it doesn't maintain monotonicity, which might actually break in some way the interpolation.

[@usernamenotfound80]

What do we do if the Taylor series' radius of convergence is 0? There are, in fact, smooth functions whose Taylor series has 0 radius of convergence everywhere.

[@omrizemer6323]

Isn't the last argument false? If f and g are only assumed to be smooth, then there are examples where the power series of f at some points has radius of convergence 0, so you can't extend the function at all...

[@EpsilonDeltaMain]

You are correct that there are such pathological function whose existence is proven by the Borel's Theorem. But I really wanted this video to be on applied calculus, not counterexamples in analysis, and showing that such function even exist via constructing one would be worth a big video on its own, and didn't think it was appropriate to consider at the moment (which I may be interested to make later down the line when I make a series in interesting counterexamples)

[@Mr0scotter]

@@EpsilonDeltaMain yes please

[@fellipetoffolo4226]

nice

[@bartekltg]

I may miss something, but it looks like it may not work for some nasty functions. Consider f(x) = \int_0^\inf exp(-t) cos(t^2 x) dt (and we choose the cutoff point at 0) it is smooth and f^(n) = +-(2n)! for odd n. So, the power series is c_n = +-(2n)!/n!, and lim sup |c_n|^(1/n) is infinite. The radius of convergence is 0. Our taylor series doesn't work for an extension of any length. Sure, this function seems smooth on the entire domain, so the first method work, but I do not see this would be the case always.

[@Bard_Gaming]

What do you think of the interpolation function "f(x) = 0.5 * (cos(pi*x +pi) + 1)"?

[@_mara_]

Sorry if this is silly, but is there a way to do this with out knowing the actual functions, but just a bunch of values, like with the waves of an audio file or the movement of a character in a game, or its animation, that kind of things, to make them more smooth?

[@EpsilonDeltaMain]

First of all, if you only have sets of data to begin with, your original "function" is discrete and isn't smooth to begin with. And smoothness is really a pure mathematics kind of thing, and in all practical application, we just have to match derivatives of first several degrees, and often matching up to second degree is good enough for most practical purpose. When we have many data points such as digital audio file as opposed to 2, we will frequently use splines instead of trying to interpolate using one function. In practice CUBIC SPLINE INTERPOLATION is used all the time since at each data point, up to second derivatives are matched.

[@_mara_]

@@EpsilonDeltaMain Ouh, that sounds interesting. Never thought about using splines in this kind of situation, I just recently learned about the Bézier Curves and how to use them, so it shouldn't be that hard to put them in practice. I'll research more about it, thank you!

[@ZomB1986]

What if you rotate function f about point a and rotate g about b? You get two new functions that are symmetric around a and b respectively and should still be smooth.

[@CatCat99998]

Rotating the function to continue it would not be smooth because the second derivative would not be continuous.

[@falcofurious]

So that’s why my premiere pro speed curve keeps crashing my computer

[@alexsere3061]

At 3:05 there is a mistake with the definition of g, it should be that g'(1)=-1, not g(1)=1

[@magicmulder]

x*sin(1/x) was one of my first favorite functions.

[@misterlau5246]

Sinc function is excellent for interpolation

[@EmmanuelMessulam]

Why don't you use the second method always? Doesn't it provide with better solutions than providing an arbitrary intermediate function?

[@michaelpristin9944]

The second method only works for functions that have a convergent Taylor expansion. If the right side function was something like log(x) there would be no such smooth extension past 0