Fascinating Flat Function for Fun 

Yeah i love my bump functions. It makes sure I seldom needed to worry about things being smooth or well defined

Yeah, there exists a theorem like this. I agree that this is not obvious at all.

This is disappointing. I was going to recommend this video to my calculus class, but not with that gaping logical hole. You can prove his method is justified with the Mean Value Theorem (see for example Theorem 2.15.2 on p.207 of personal.math.ubc.ca/~CLP/CLP1/clp_1_dc_text.pdf) but this presentation is just misleading.

The "theorem" you are talking about is just the continuity of the derivative, right? That f'(0) = lim x->0 f'(x), I mean. It obviously depends on the function considered. Here the derivatives are continuious, but this is shown by computing the limit at 0 first, not the other way around, so you're right to point that problem.

@@Notthatkindofdr This is fine, the derivative is continuous. Just check the limits from the left and right. There is no "gaping hole"

@@theflaggeddragon9472 How do you know the derivative is continuous? Or that it exists at all? Without mentioning a theorem to justify what he did, it is a gaping logical hole to just assume the derivative exists.

I saw that too, a correction should be added to the video.

But are they analytic?

This is Theorem 2.15.2 on p.207 of personal.math.ubc.ca/~CLP/CLP1/clp_1_dc_text.pdf, for example. It is a short proof using the Mean Value Theorem.

@@Notthatkindofdr Well It certainly looked like it would be true but I had never seen a proof until now. Thanks!

It follows from the definition of the derivative as the limit of a slopes gradient

It isn't defined for non-real x. He's talking nonsense.

I agree that his description of the radius of convergence of the series is wrong. As you say, the Taylor series does converge for all x, but it just doesn't converge to f(x) for x>0.

It isn't defined on non-real x. He's talking nonsense.

You're exactly on it. Roughly speaking, a singularity is a divide by zero error. Translated to high school calc terms, a removable singularity is a hole on a function, a pole is something that makes a vertical asymptote, and an essential singularity is something where the idea of limits and indefinite integrals stop making sense entirely.

@@nathanbraswell2484 having done some rudimentary complex analysis a few years ago, I can also understand that another way to distinguish between poles and essential singularities is by looking at the associated Laurent series. However, it never was enough for me to hear "poles happen when there are finitely many nonzero negative power terms in the Laurent series, while poles have infinitely many such terms". Picard's great theorem helped shed some light on understanding the behavior near essential singularities, but it still remains a subject that keeps fascinating me. I also find interesting how poles can be dealt with by multiplication with some appropriate polynomial, but essential singularities seem to require more. Yes, multiplication with a polynomial does result in a new function, but what I want to stress is the removal of the pole. In a particular case, exp(1/z) has an essential singularity at z = 0. No matter what polynomial we would multiply by, z = 0 remains an essential singularity. However, multiplying by exp(-1/z) solves the issue. There are definitely many details I am not considering here or sweeping under the rug, but it's almost as if essential singularities don't require a polynomial to be fixed, they require an entire series (sort of). Then again, this is just an example, I'm confident there might be examples where this approach doesn't work. I personally think of essential singularities not as places where everything goes haywire, but as places where a function becomes "spicy", so to say. A kind of "I'll take on almost every possible value infinitely many times around this place" spicy function.

​​@@retired5548 not every possible value, there can be an exception. 0 in this case. 😅 ok i just found the word "almost". nevermind.

I'm curious now to know to what you refer in particular. -Stephanie MP Producer, Editor

@@MichaelPennMath the cookie monster effect on the hands at one point

OH YEAH!!! Haha. I edit videos pretty quickly and get through a lot at a time. I forget what I did in particular videos sometimes. Thanks :) I like to have fun. -Stephanie MP Producer, Editor

And the "upload!" Tshirt.

Why not? You can define it piecewisly (as an example) just as it is defined in the video. It is just that none of the definitions will be differentiable at zero.

​@@vladimir10 because "

​@@danielwarkentin6819 You can still define functions using < or > f(z)={ e^[-1/z] for Real(z) > 0 { 0 for Real(z)

​@@vladimir10 Because exp(-1/x) has an essential singularity at 0, because it oscillates wildly and infinitely often in the neighborhood of 0.

@@ianmathwiz7 none of that is enough to be a reason for not being defined as a function on complex plane including zero. Set f(z)=exp(-1/z) for all complex z except for 0 and any value you want for f(z=0). This function is defined for all complex numbers and infinitely differentiable for all z except 0.

Isn't that a definition of convex function?

You can get the same behaviour from exp(-1/x^2)

Why not? It may not be differentiable, but still definable :)

@@vladimir10 The relation ">" is not defined for complex numbers. You can define the function that we'd probably all think of as corresponding to this function, but you need to tweak the definition to make sense. Using Re(z) > 0 gives a function that's continuous but not differentiable/smooth/analytic like Michael said.