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This Infinitely Differentiable Function Breaks Taylor Series

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Taylor Series are incredible. We can approximate a function with the Taylor Polynomial, and then hope that when we take the infinite series we get a series that is equal to the original function. However, this doesn't always work, for several reasons. The first is that we either can't compute the Taylor Series centered at some values (example ln(x) isn't defined at 0), or that the result series only converges on a limited interval of convergence as we see in our second example. The trickiest part however is that even when the Taylor Series converges, that doesn't immediately imply that it converges to the original function! We will see the very cool example of e^{-1/x^2} (with f(0) defined to be 0) that is infinitely differentiable (aka smooth) but whose taylor series is the zero series and thus clearly not the same as the original function. An analytic function is one whose taylor series convergences to the original function, so this example is an example of a smooth function that is non-analytic. Cool!
0:00 The big question about Taylor Series
0:30 Review of Taylor Series
2:20 Ln(x) and its Interval of Convergence
7:08 An infinitely differentiable but non-analytic example
9:55 Smooth vs Analytic
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15 авг 2024

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Комментарии : 205

@donaldhobson8873 2 года назад

exp(-1/x^2) may look like a nice function on the reals. But in the complex numbers, it explodes into total crazyness around 0.

@felipeopazo8375 Год назад

So in the complex numbers it isn't actually infinitely differentiable, right? I noticed that and thought that is probably why the Taylor series breaks, so it'd be nice to make sure

@donaldhobson8873 Год назад

@@felipeopazo8375 In the complex numbers it is totally infinitely differentiable everywhere except at 0. The image of any open set containing 0 is the whole complex plain except 0. The function is really wild around 0. And can't be defined continuously at 0.

@bradwilliams7198 Год назад

@@felipeopazo8375 The point he "added", f(0) = 0, is only valid if you approach zero on the real axis (from the positive or negative direction). Approaching zero on the imaginary axis (from either direction) gives a limit of infinity. And at a 45 degree angle on the complex plane (the real and imaginary parts are equal and both approach zero), there's no limit, because it oscillates with infinite frequency. In complex variable theory, this is referred to as an "essential singularity." These are trouble and best avoided!

@chimetimepaprika Год назад

Can anyone else drop some nice resources that talk about this behavior? Jejeje that sounds rad!

@user-pr6ed3ri2k Год назад

wait wat

@gamingbutnotreally6077 2 года назад

Wow, I learned about interval of convergence but didn't really connect the dots fully until I saw the actual graph and how it becomes inaccurate after 2. Awesome stuff!

@duckymomo7935 2 года назад

Really? That’s what interval of convergence means

@alejrandom6592 2 года назад

@@duckymomo7935 geez where are your friends

@manicmadpanickedman2249 2 года назад

Check out possible solution on my chan

@distrologic2925 Год назад

lol me too, I never understood radius of convergence properly

@colorx6030 Год назад

Same here. I never really gave much thought on what the interval of convergence means geometrically. I was just focused on getting the right answers. It's very nice to gain appreciation even if it's only after we discussed this topic that I learned it.

@zelda12346 2 года назад

When I was in analysis 1 and 2, a lot of my proofs would start with: "Let f be a *Nice* function on R to R" I never had to define what "Nice" meant or how it was different from "analytic" or "smooth". It didn't matter I had different professors from different universities; I could just say "Nice" and they instantly knew what I meant and thought it would be obvious what it meant for even undergrads to get. I still don't have a definition of "Nice function". I just know it's something that includes polynomials, sine functions, simple exponentials, and logs, but not square roots, complicated exponentials, infinitely discontinuous function, or those weird toy functions.

@mastershooter64 2 года назад

Nice = infinitely differentiable everywhere? I haven't properly studied real analysis yet so idk if that's a good definition lol

@stevend285 Год назад

@@mastershooter64 that's just smoothness

@SpartaSpartan117 Год назад

@@mastershooter64 I think by nice you mean a function that doesn't break my proof and obeys my implicitly assumptions 😂

@rupertmillard Год назад

Nice = analytic, isn’t it? But if you want nice, then ignore real analysis and only study complex analysis!

@scraps7624 2 года назад

I freaking love how you are able to break down your explanations into clear, short segments. I'm getting my PhD so that I may become a professor back in my country, your channel is also helping me learn about how these concepts should be taught

@RyanOManchester 2 года назад

This video is a great demonstration between the difference between mathematicians and engineers. As a grad student working in the data side of electrical engineering I would say that the linear Taylor Series approximation of exp(-x^(-2)) is actually pretty good around zero despite it being trivial and in a sense not related.

@DrTrefor 2 года назад

Hope you enjoyed this funky example of a Taylor Series can converge everywhere, but not to the original function! My thanks again to Maple for sponsoring today's video, don't forget to check out Maple Calculator for your phone: www.maplesoft.com/products/maplecalculator/download.aspx?p=TC-9857 or Maple Learn for your browser: www.maplesoft.com/products/learn/?p=TC-9857

@IanBLacy Год назад

I feel like this is kinda trivial once you think about it though- there’s no reason that a Taylor series for one part of a piecewise function should converge to the entire piecewise function… you just showed that the Taylor series of 0 is 0, which, yeah, it is. Convergence of the Taylor series for the ln(x) example was way more interesting

@chalkchalkson5639 2 года назад

This is a great reminder to be cautious. Especially as a physicist it is so easy to just pretend any function is as nice as you need it to be... One idea I think is interesting regarding this lack of caution relates to the way we teach analysis. Almost all of our analysis teaching is built on limits, so students get desensitised to the weirdness of them. Maybe that would be better if we taught non-standard calculus. If fewer things we did in calc classes were limits maybe it would be easier to treat limits with the appropriate respect. Besides I think it's easier to see that a->b and a=b are different when a and b are from completely different fields. (also great example of a topological but not metric space)

@RanEncounter 2 года назад

The problem is not everyone is a math major. They have to limit the concepts somewhere.

@decare696 2 года назад

exactly. Nonstandard analysis is actually easier to understand

@annaclarafenyo8185 2 года назад

This is not where people get confused, they get confused at the very idea of rigorous proofs. To fix that, there is a program of Homotopy Type Theory and proof assistants like Coq which turn writing proofs into a programming language in the natural way.

@mastershooter64 2 года назад

physicists not being cautious is exactly why they generate new mathematics, because they're scientists, the only thing they care about is describing and modelling nature and coming up with testable predictions. a great example of this is the feyman path integral, it doesn't even exist in mathematics, yet they can still use it to predict stuff which has been confirmed by experiment to incredible accuracy

@diarya5573 Год назад

I absolutely hated how hand wavey some of the arguments in my uni physics classes were, and I got screwed over so many times because you can ignore boundary terms in this, but not in that

@magic8ball237 Год назад

Wow I didn't know such broken functions existed. It is amazing that you can build a function using elementary functions that doesn't want to be analytic (ie. when we do the most natural completion, it isn't analytic)

@tomkerruish2982 Год назад

Something that blew my mind is that you can construct a function, defined and infinitely differentiable on all of R, yet nowhere analytic. (Wikipedia lists it under "Non-analytic smooth function.")

@Novak2611 2 года назад

Very good topic! that's why to understand it deeply one needs to see it from the big window of complex numbers. From the narrow window of real analysis, it is a smooth function, but from the wide window of complex analysis, we can notice that this function is really ugly. Real analysis does not give the full picture. The same can be said for example about the function f(x)=1/(1+x^2). It is smooth, but why its taylor series does not converge everywhere unlike the common functions e^x, polynomials, cos x ,etc...

@DrTrefor 2 года назад

Totally, I really want to do a mini series on complex numbers to help solidify so many of these concepts

@blinded6502 2 года назад

@@DrTrefor Don't forget to cover complex numbers as a part of geometric algebra. GA is an eye-opener to how complex numbers, quaternions, dual quaternions, elliptic and hyperbolic spaces, conformal transformations, as well many other geometric concepts can be fit smoothly together into an intuitive picture that is insanely useful and practical.

@Hermetics 2 года назад

@@DrTrefor ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-ZB-9f12zdEI.html complex numbers = cO|Or cartesian-system. Hov dumb can you be? for real... :)))

@thesecondderivative8967 2 года назад

This was explained really nicely. Thank you.

@SuperIdiotMan00 2 года назад

Would have been interesting to explore the Taylor series for e^1/x2 centered at values besides 0, and how those converge.

@piotrskalski1477 Год назад

They would probably have the radius of convergence equal to their distance from 0, since someone said here that 0 is the only singularity in the complex numbers

@randomdosing7535 2 года назад

hey Dr. Trefor amongst all the stuff available on YT, I love learning from you; and your personality. I studied biological sciences but my interest in maths and physics is more than that in biology. i have embedded your video in my website so other people from my country can also learn from your quality lectures. Respect from Pakistan

@martinepstein9826 2 года назад

Great video. This came up for me recently. The topic was whether it's a good idea to take e^x = 1 + x + x^2/2 +... as a definition (I say yes, absolutely). But people were describing this as "defining e^x by its Taylor series" which was leading to a lot of confusion. People thought the definition was circular - don't you need to take derivatives before you find the Taylor series? - and kept trying to analyze convergence using the Lagrange remainder and getting nowhere. Basically, people thought we were defining e^x as "the function with this Taylor series at 0" which is silly since, as you demonstrate, there are many such functions.

@dekippiesip 2 года назад

I prefer this definition: f(x) is the unique function where f'(x) = f(x) and f(0) = 1. Then proceed to prove that: 1. there is a number e such that f(x) = e^x, and define e as THAT special number. 2. prove the taylor series expansion.

@martinepstein9826 2 года назад

@@dekippiesip Also a good definition. It's easy to get one from the other.

@dekippiesip 2 года назад

@@martinepstein9826 yeah the 2 definitions are basically equivalent. But the first definition seems better pedagogically because the taylor series definition just seems to come out of nowhere. Asking if there is a function equal to it's own derivative is a natural question that could quickly arise after having learned the concept of derivatives.

@dekippiesip 2 года назад

Now I think of it, there is a subtle problem in my approach. The function e^x or a^x in general is only clearly defined on the rational numbers. The extension is based on taylor series in itself. And that is a bit circular, because taylor series are also generally proved by assuming the functions are already defined on the reals. So we should also need to come up with an alternative but equivalent way of extending a^x in general to all real values of x. But in order to do that you need to talk about the rationals as a 'dense subset on the reals', prove that every continious rational valued function has one and only one continious extension to the reals and use that to define a^x. But now it's even worse pedagogically, because this stuff is more advanced than taylor series in the first place...

@martinepstein9826 2 года назад

@@dekippiesip Hold up; the fact that irrational powers (and even worse, complex powers) are awkward to define in terms of repeated multiplication is _precisely_ the advantage of your approach (and mine). Let's call the solution to the initial value problem 'exp' and define e = exp(1). We know exp(a+x) = exp(a)*exp(x) because exp(a+x)/exp(a) solves the IVP and therefore equals exp(x). It follows that exp(p/q) = exp(1)^(p/q) = e^(p/q). For irrational or nonreal x writing exp(x) = e^x is just a formalism.

@geraldsnodd 2 года назад

I am currently learning Taylor expansions as part of highschool calculus. Thanks Prof .

@DrTrefor 2 года назад

Great timing!

@Hermetics 2 года назад

indoctrination you mean, not learning :))) Vatch this instead and spare your time vith stupid schO|Ol ;) ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-ZB-9f12zdEI.html

@ffggddss 2 года назад

A function can be of class C(∞) [infinitely differentiable] but not analytic. A mismatch that can happen in the real field, but not in the complex field. One of the gemstone-facts of analysis. [I was a little surprised you didn't mention this, even as a "teaser."] Fred

@martinepstein9826 2 года назад

If you define the derivative asymptotically it's much easier to show that e^(-1/x^2) has 0 n'th derivative for all n. A function f is n-times differentiable at 0 iff f(x) minus an n'th degree (or less) polynomial is in o(x^n). Clearly e^(-1/x^2) is in o(x^n) for all n.

@michaeljburt 2 года назад

Absolutely love your content. I am an electrical engineer who loves math, but work and life often get in the way. Almost every time I watch your videos my passion is reignited and I want to learn more. So thank you!

@DrTrefor 2 года назад

That’s awesome!

@MCLooyverse 2 года назад

Nitpick: In your natural log Taylor series, you have the `k = 0` still hanging around at the bottom. I think the `k = 0` term is a special case that gets dragged out from the sum, and noticed to be equal to 0 itself.

@MCLooyverse 2 года назад

@p s Yes, but it's just a typo, and changing it doesn't significantly change the video.

@Szynkaa 2 года назад

you are amazing teacher- yet again i learn something from your channel and i was pretty sure you can't really surprise me with something as simple as Taylor series. But now i feel like i need to rethink my understanding of whole idea of Taylor series, thank you

@justarandomdood 2 года назад

This is so cool, I wish I had seen this back when I was taking Calc 2, it would've cleared up a lot of confusion for me xD

@michaelharp2085 2 года назад

I would like to take a moment to share with you a special kind of smooth but not analytic function called a 'bump function': Define f(x) = exp( 1/( x - b ) - 1/( x - a ) ) (a real valued function) with the restriction that a

@victortitov1740 2 года назад

I have recently "invented" this function to use as a window function for a dsp FIR filter (or, to put it otherwise, a window function for spectral analysis). I was somewhat surprised such a function is not part of the standard library of window functions. It worked quite well, it turned out to be the best window function for my purposes compared to those defined in scipy.signal.window (tukey 0.3 was about as good, basically a draw).

@victortitov1740 2 года назад

oh, actually, my invented one was exp(-1/((x-a)*(b-x))), slightly different

@lydianlights 2 года назад

very cool, thanks for sharing!

@Sugarman96 2 года назад

That example is a good demonstration for why Taylor expansion isn't sufficient. When you expand past Taylor series and into Laurent series, that function that is only ever 0 as a Taylor series is very much not 0 as a Laurent series.

@MattMcIrvin 2 года назад

These infinitely differentiable but not analytic functions turn from annoyances to tools when we realize that, because they exist, it's possible to have an infinitely differentiable "bump function" that is only nonzero over a finite area. That can be really useful if you're using them to divide something into pieces in a smoothly feathered-off way that doesn't make any of the derivatives blow up. For instance, they're useful in differential topology when talking about describing a manifold in terms of an "atlas" of maps of pieces of it to Euclidean space. This is also a big difference between real analysis and complex analysis--in complex analysis, anything whose complex derivatives are all defined is analytic. (And analyticity also becomes an incredibly powerful constraint.)

@stackalloc7741 2 года назад

I've seen the terms "the series converges very slowly" or "very fast" used quite a lot but I don't usually see authors defining these properly. So I`ll leave this topic as a suggestion for future content (if you haven't done one already). Anyways, good video.

@DrTrefor 2 года назад

Ya this is a common issue. Ultimately it is answered in the next video linked at the end about remainders. We have an intuitive graphical sense, but when we are precise the remainder gives us the degree of the error

@purplenanite 2 года назад

With lnx, the radius of convergence was related to how far away you are from a singularity. with e^-1/x^2, you are at that singularity. it's surprising to me that it gives a radius at all there. also, with lnx, if you increase your "a" value, your radius increases, so I think I can do this! lnx = lim(b->infinity)[ b - sum(1 -> infinity) [ (-1)^n (x-e^b)^n / (n*e^nb)] ]

@OldSoulClimber 2 года назад

Maple on mobile really is the best thing for calculus. Step by step for derivative and such is a great way to learn. I will say that graphing and syntax are not the best so also have Desmos for your graphing needs

@lydianlights 2 года назад

Cool video -- I actually learned something! It's been a hot minute since I took calculus but seeing stuff like this explained so well is awesome.

@RSLT Год назад

Thanks for the great video. I wasn't aware of this before. I used to think that any infinitely differentiable function would have a Taylor series representation. However, after watching this and doing some research, I learned that this is not always the case. This video has expanded my understanding of the limitations and complexities of mathematical analysis and reminded me of the importance of continuous learning and exploration in the field of mathematics.

@johnphamlore8073 Год назад

It's great this function exists. This or analogs of it means the test function space of C^\infty with compact support is dense in any sense one wants, and can even consist of piecewise analytic functions?

@dinakosiliop8928 2 года назад

One of the craziest things that i learned on university is that on complex analysis smooth functions and analytic functions is the same , and not on real analysis. And the difference is just the multiplication we define on complex numbers. I understand the mathematics behind that but it is crazy ... Great video btw 😊

@dinakosiliop8928 2 года назад

Sorry for my terrible English 😅

@mtaur4113 2 года назад

The reason exp(-1/x^2) doesn't work for complex variable is that it blows up as x=it and t goes to zero. So the complex function has an essential singularity, not a removable singularity like it does as a real function. Then there's Picard's Theorem if you really want to get crazy... Their ongoing mission, to go everywhere, infinitely often, with at most one exception.

@eternal0paradox202 Год назад

0:07 I believe there's a mistake and in e^x series that last term before the 3 dots should be x^4/4! instead of x^5/4!

@davidgillies620 2 года назад

As a followup maybe you could introduce the corollary of this in complex analysis whereby functions that are complex-differentiable at every point in a set (holomorphic) _are_ necessarily analytic on that set. Then you can bring in the Cauchy-Riemann equations in a natural fashion.

@zunaidparker 2 года назад

If you did the Taylor series for exp(-1/x^2) around x=1 instead, what would be the interval of convergence and would it approximate the function itself?

@DrTrefor 2 года назад

Yes, but it would break down at x=0

@brilliantbutlazy3.14 2 года назад

We literally just learnt about analytic functions and series solutions of differential equations. Great timing 😅 Also i always wondered how do we prove whether the remainder will go to 0 as n->infinity. Since the remember term R(n+1) has a point other than the point we're evaluating the series about. If you can show an example of how to deal with these 2 variables in the limit.

@DrTrefor 2 года назад

There are a range of formulas for the remainder and sometimes the methods are ad hoc, I have done at least one example in the video linked add the end.

@brilliantbutlazy3.14 2 года назад

@@DrTrefor thanks!! I'll check it out 😄

@gibbogle Год назад

Very interesting. I didn't know about this. I only knew about analytic functions of complex variables.

@robertrobert5832 2 года назад

Thank you sir.... You have made me interested in algebra....

@giuthais 2 года назад

These are really high-quality videos!

@ShaunakDesaiPiano Год назад

0:43 slight issue but not really: we should separate out the first term from the series to avoid any notion of dividing by zero (as well as dealing with (-1)!).

@superhipposock 2 года назад

Not sure how you managed to talk about this without mentioning anything (even in passing) about Stoke's Wedges/problems - !! Probably the most important/interesting consequence of this. Next topic suggestion? :P

@andrewharrison8436 2 года назад

Better late than never - did a BA in maths over 50 years ago - never before seen a Taylor series misbehave like that. Thanks - maths is still fun.

@JojiThomas7431 2 года назад

Very good video. Nice explanation

@MathCuriousity Месяц назад

Please link the video where you show that Taylor of e^x converges for the function.

@platinummyrr Год назад

question though, doesn't the taylor series of e^(-1/x^2) always equal zero in some small range so the zero series taylor expansion does "approximate" it within that window where its always zero? Obviously this isn't really all that useful in practice, but...

@punditgi 2 года назад

Excellent!

@samuelmarger9031 2 года назад

Note: the series for ln(x) should start at k=1, not k=0. Other than that, nice video though!

@Alex_Deam 2 года назад

You mention looking at the remainder as a way to tell if the Taylor series actually converges to the function, but I'm curious if we can use complex analysis to also see this? For example, e^-1/z^2 has an essential singularity at z=0 and the real version e^-1/x^2 doesn't equal its Taylor series as you say. Log z has a branch cut, so non-isolated singularity, and we can't do a Taylor series at x=0 for log x. Is it true that all "problematic" real functions can be spotted this way by looking at the behaviour of the complex version of them, or is there a counterexample?

@DrTrefor 2 года назад

Absolutely! Complex analysis really solves many of these issues that come when we restrict to only reals

@chalkchalkson5639 2 года назад

You should look into "holomorphic" functions and the "identity theorem" for holomorphic functions. Then see if you can spot why holomorphic on a some domain might be related to the function being equal to it's own taylor series. Alternatively, if you don't feel like playing around with complex analysis, look at this wikipedia article: en.wikipedia.org/wiki/Analyticity_of_holomorphic_functions

@vascomanteigas9433 2 года назад

The exp(-x^2) have an infinite Laurent Series given by 1- 1/x^2 + 1/(2*x^4) - 1/(3*x^6) + ... Which means that exp(-x^2) have an essential singularity at x=0.

@AndrewCodeDev 2 года назад

Gotta say it, I love that shirt. I've gotta buy one of those and support the channel.

@DrTrefor 2 года назад

Thank you!! Would mean a lot:)

@AnitaSV 2 года назад

Due to existence of bump function, I always thought if analytic, then radius of convergence makes sense, otherwise no guarantee. I didn't realize exp(-1/x^2) is a simpler example. I always thought of analytic meaning differentiable in complex domain rather than it converges with taylor series.

@bobbob-nh6yd 2 года назад

Keep up the good work with videos! I’m only 14 and the fact I can keep up with your videos is amazing, and your lessons online are really helping me master calculus one step at a time so all I can say it thank you!

@DrTrefor 2 года назад

That’s awesome!

@Trizzer89 2 года назад

It would converge for all positive values and all negative values depending on which A you choose. You could choose a very very large A and essentially get the function for a very very large space

@protosstassadar20 2 года назад

In the taylor series in the first seconds of the video, why there is an "a" in the rhs but not in the lhs ?

@DrTrefor 2 года назад

The idea is we are approximating the function, but we can do that about any number of points. So there isn’t anything intrinsic to the function to have an a on the LHS

@mastershooter64 2 года назад

drr bazett will you make a series on differential geometry? like riemannian manifolds, differential forms, how vectors and vector fields are differential operators (or derivations?) affine connection, levi-cita connection, stuff like that! maybe even some complex differential geometry?

@ready1fire1aim1 2 года назад

3 sets of 3 dimensions. We're 4D, not 3D. 1D, 2D, 3D are spatial 4D, 5D, 6D are temporal 7D, 8D, 9D are spectral 1D, 4D, 7D line/length/continuous 2D, 5D, 8D width/breadth/emission 3D, 6D, 9D height/depth/absorption

@CppExpedition 2 года назад

I have one question, -> Why is there an Hipopotamus standing on the hypotenuse?

@carultch 2 года назад

Because the two words sound alike.

@CppExpedition 2 года назад

@@carultch you are a genuine born mathematician. No jokes only truths!

@monte314 2 года назад

Terrific treatment... as always!

@whatitmeans 2 года назад

If I defined a function f(x)=exp(x^2/(x^2-1))*(1-x^2+|1-x^2|)/2 is undefined at x=0 as your example exp(-1/x^2)?(so, it needs to be split in a piecewise definition) Or it is imply that the "hidden" step function is defining it as zero at x=0?

@alex_zetsu 2 года назад

Can we use this to make a function where before a point, the function is a constant 0, after a second value, the function is a constant 1, in between, the function is something, and the entire function is smooth at all points?

@DrTrefor 2 года назад

Absolutely this would work great for that

@JAzzWoods-ik4vv 2 года назад

for the function at 9:40 couldn't it be said that the radius of convergence is actually undefined since taking the ratio of a(k+1)/a(k) is actually undefined since it would imply dividing by 0?

@tapasmazumdar3831 2 года назад

If it had been any non-zero real divided by zero then yes, undefined. But 0/0 is unpredictable. Most times when you reach a 0/0 case, you have to find a different method to arrive at results.

@chobes1827 2 года назад

No, even though it may seem natural to think that, the ratio test actually doesn't work that way. If the limit of |a(k+1)/a(k)| exists and is less than 1, then we can conclude that our series converges, but the converse does not hold. There are infinite series that converge where the limit of |a(k+1)/a(k)| does not exist. Knowing that the limit used in the root test does not exist gives absolutely (or that it is equal to 1) gives us no information about the convergence of our series. In the specific case of the 0 series, we can use a stronger test called the root test to show that the radius of convergence is infinite.

@kilian8250 2 года назад

There is a more general way to compute the radius of convergence called the root test. The ratio test gives the same answer as the root test *when it converges*.

@orisphera 2 года назад

Is the function smooth in complex numbers?

@GhostyOcean 2 года назад

One crazy property of complex functions is that if a complex function is once differentiable, then it is automatically infinitely differentiable. Most functions that you deal with in a school or real world environment are complex differentiable.

@alexatg1820 2 года назад

except the point 0

@martinepstein9826 2 года назад

Let's plug in a small imaginary number like 0.1*i. If the function is smooth then the output should be close to f(0) = 0. e^(-1/(0.1*i)^2) = e^(-1/(-0.01)) = e^100 Nope.

@HadiLq Год назад

exp(-1/x^2) is not differentiable near zero! BTW, thank you for your other videos.

@sheksonson293 2 года назад

May you help and explain why the residue of Riemann Zeta may give complex number values at point near 0.9? Anything wrong or may be my mistake using software like mathematica or else or any suggestion? Here comes the code: For [ j = 0, j

@mariostelzner4530 Год назад

AH YES! IN MY ESTIMATION OF THIS PROBLEM, I HAVE CONCLUDED THAT YOU HAVE FORGOTTEN TO INCLUDE EINSTEIN'S THEORY OF RELATIVITY. AHAHAHA AHAHAHA LOL

@yqisq6966 Год назад

With the example, what if you take the limit of the sum rather than the sum of the limit?

@paulg444 Год назад

This is crazy, what is he going to tell us next: something insane like this function can be used for analysis to get fundamental features as a test function ?

@Siirxe Год назад

0:11 x^5 / 4!?

@odysseus9672 Год назад

Actually taking the derivatives to derive the infinite sum for the logarithm seems like too much effort. Just take the sum of the geometric series with r=-x and integrate both sides. Voila, an infinite sum for ln(1+x).

@gonzogil123 2 года назад

Position from which to impress: PhD....impressed.

@ready1fire1aim1 2 года назад

We're 4D. I keep hearing theories like "simulation", "holographic" or back to Leibniz' "contingent" universe. Those theories all make me think of the i, j, k in quaternions. Quaternion MATHEMATICS a complex number of the form w + xi + yj + zk, where w, x, y, z are real numbers and i, j, k are imaginary units that satisfy certain conditions. RARE/biblical a set of four parts, things or persons. (dimensions)

@ahsan4306 2 года назад

Sir please reply. Which books do u recommend for multivariable calculus and linear algebra, . for self-study

@jonasgajdosikas1125 2 года назад

the thing is, the taylor series perfectly approximates the function you defined at x=0; in a piecewise function the taylor series by definition cannot extract information from regions x is not

@GeoffryGifari 2 года назад

when i see the graph of exp(-1/x²) in the video, at some distance from zero (so far the graph is horizontal) the function starts to behave like an exponential, then like a logarithm. is the point where it starts to go up from being horizontal special?

@DrTrefor 2 года назад

Not particularly, we can compute the inflection point from second derivative fairly easily if we wish to

@egycg3569 2 года назад

That's what I am looking for, the caveats, and who shall explain them better than Dr. Trefor.

@DrTrefor 2 года назад

Glad you enjoyed!

@MrKnivan 2 года назад

Doesn’t this have more to do with how you defined the piecewise function though? It’s certainly a reasonable definition, since the limit is zero. But if you hadn’t defined the function that way, it wouldn’t be differentiable at zero (and hence not smooth at zero). Then we would know that no Taylor Series will approximate the function at all near zero. In fact, it’s no surprise that the Taylor Series of the piecewise function is zero everywhere (near zero), since you’re only approximating the constant function f(x) = 0, whose Taylor Series OUGHT to be just 0. Great video tho, I love the graphical demonstrations!

@peteschupp4545 2 года назад

So if I’m asked the radius of convergence of the function e^-1/x^2 is it infinity or 0 because 0 is the only point it is correct to the Taylor series

@DrTrefor 2 года назад

It only makes sense to ask the radius of convergence of the series, which in this case is infinite, but that’s a sort of useless observation

@peteschupp4545 2 года назад

@@DrTrefor well that’s exactly what I‘m ask to find in my calculus sheet. I am supposed to do it with the fact that the radius of convergence is at least the distance of z0 to the next singularity in complex plain so that Br(z0) does not include a singularity but I‘m a bit confused here because if Br(z0) should include no singularity and the distance between z0 and the next singularity is r then shouldn’t r be the maximum radius of convergence and not the lowest? And I‘m a bit confused that as you said the function is not defined for z=0 so the distance to the next singularity when you expand from z0=0 will be 0. would be nice if you can help

@ichigo_nyanko Год назад

What program were you using for those graphs at the beginning?

@kylecow1930 2 года назад

out of curiosity whats the limit of the radius of convergence as the centre of the taylor series goes to 0, it doesnt have to be constant right? cause ROC is dependant on the coefficients of a general power series, its not a taylor poly specific thing

@DrTrefor 2 года назад

Ya this is just the zero series regardless of its origins as a Taylor series, so infinite radius of convergence

@kylecow1930 2 года назад

@@DrTrefor really? the graph clearly has non 0 second and first derivatives in a lot of places so if you centre the series there then as far as i can tell you get a legitimate taylor no?

@nimennacnamme6328 2 года назад

00:15 Lol x⁵/4! seems like an off-by-one

@user-mr9lf4wb6i Год назад

Perhaps the point is that it is not the function and its derivatives that are equal to zero at the point x=0, but their limits.

@comdo777 6 месяцев назад

asnwer=(1a+ex)

@agrajyadav2951 2 года назад

Omg that was so cool

@SeasOfCheese929 Год назад

"Not all smooth functions are analytic" -Laughs in complex analysis

@justjacqueline2004 2 года назад

Succinct to the point.

@brooksbryant2478 2 года назад

You have a division by zero at 5:20

@DrTrefor 2 года назад

True! The k=0 term is zero so I should have just excluded it by starting at k=1

@lyndonellison1514 Год назад

Dude, your shirt's hilarious

@Phi1618033 2 года назад

This dude's Canadian accent is bordering on the Letterkenny-esque.

@maxp3141 2 года назад

Nice, could you plot this function near zero on the imaginary axis? :)

@DrTrefor 2 года назад

Absolutely, that would be good to do

@azzteke 2 года назад

We are talking about real x-values. No imaginary axis!

@martinepstein9826 2 года назад

@@azzteke But extending to the complex plane is revealing. We see that the function is actually not so well behaved near 0.

@Alex_Deam 2 года назад

@@martinepstein9826 Not so well behaved is underselling it - it takes on the value of any complex number (excepting zero itself) infinitely many times in any punctured neighbourhood of zero according to Great Picard's Theorem!

@Alex_Deam 2 года назад

Here's a visualisation of e^-1/z to give you an idea for e^-1/z^2: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-NaDhdCEvAxI.html

@JohnDlugosz 2 года назад

just after 9 minutes: You said the function is not flat.. you meant the function does not have a constant derivitive, which would make it a parabola.

@FunnyAndCleverHandle Год назад

WhAt program do you use?

@cristhianromero2504 Год назад

1^2

@IoEstasCedonta Год назад

0:09 - Uhhhhh...

@nikhilpathak2955 2 года назад

Hello i am form india and I like your video

@MaximusAurelius1987 Год назад

Let’s see some triad claws

@GeoffryGifari 2 года назад

so if exp(-1/x²) is not analytic, is it possible to approximate it with a series at all?

@vascomanteigas9433 2 года назад

Replace the exponential series with z=-1/x^2, and done. An infinite series with negative power terms, also known as Laurent Series of an essential singularity (The number of terms with negative power series are infinite.)

@GeoffryGifari 2 года назад

@@vascomanteigas9433 sounds like laurent series has more power

@DrTrefor 2 года назад

Yup with complex analysis we can give much more satisfying answers to these types of problems

@cristhianromero2504 Год назад

yea www

@alerikaisattera1465 Год назад

Misleading. This function has an essential singularity at x=0, and is not differentiable

@MrAlRats 2 года назад

What is so special about the ability to approximate a function using its Taylor series that an entire branch of mathematics is dedicated to it? Why not study functions that can be computed to arbitrary precision using any method involving arithmetic operations?

@bbsonjohn 2 года назад

I don’t know what this function. I bet it is some weirdo that can only appear in the instanton potential of a Yang-Mill theory or sth

@jamescoughlin8856 2 года назад

Is this correct? You say that your example is infinitely differentiable at x =0, but not analytic. Still it is discontinuous at zero. Granted it is a removeable discontinuity over the reals, which you removed with your piecewise definition, but still, one wonders if you are claiming the derivative to exist at a discontinuity. I am at least uncomfortable with the contrived nature of the example and I am not sure that it illuminates the issue for the student. On further reflection I think you hide the fact that the discontinuity is essential when viewed in the complex plane. It HAPPENS to have the same limit coming from the positive and negative real axis, but that is a coincidence, and a clear indication of non-differentiability in the complex plane. OTOH, 1/(1+x^2) is infinitely differentiable everywhere on the real line. Yet it's radius of convergence is 1. It is NOT infinitely differentiable or analytic at i, and this complex artifact induces a limited radius of convergence EVEN on the reals. This I think does help to illustrate the point of convergence inside a finite radius to the given function.

@DrTrefor 2 года назад

I do love 1/(1+x^2) too even if it shows in some sense the opposite problem (a weirdly small IOC from the perspective of the reals). I think the pedagogical challenge is that because when we do this on the complex numbers a complex differentiable function on a region is always analytic, so there sort of has to be some level of hiding the singularity to make the problem manifest in the reals where it is smooth but not analytic. I don’t know I particularly object to the piece wise definition to make this differentiable even if I agree it is contrived in some sense :/

@jamescoughlin8856 2 года назад

@@DrTrefor So maybe we should introduce the calculus in the complex plane sooner? This is not the only case where it makes calculus and analysis more elegant, and perhaps even more accessible!

@toastyPredicament 2 года назад

i wont turn on the subtitles to see better