Exactly what makes this famous function non-elementary??

[@Nickesponja]

Official petition for a video on the proof of that theorem :)

[@danielgarrison7463]

Agree!!

[@mekbebtamrat817]

Liouville's theorem. I have not seen its proof but seems a bit complicated to do so in a video without some knowledge of fields and advanced algebra.

[@ethandole2218]

I'd love that too !

[@redwanekhyaoui7232]

I jumped to comments just to asl for the same, 😊

[@l.a.s8274]

Up

[@goodplacetostop2973]

2:27 For anyone interested for that theorem, you can search for "Liouville’s Theorem on Integration in Terms of Elementary Functions". Thanks @yoav elbaz 10:52 Happy birthday Vojtěch Jarník… and Srinivasa Ramanujan!

[@lukasmoudry9973]

Nice, I wouldn't expect mentioning of a Czech mathematician here :D

[@lowbudgetmaths]

The non-elementary examples you mentioned were all from composition. I like to bring up sin(x)/x because both sin(x) and 1/x are easily anti-differentiated, but the product of the two is not. So there is obviously no product rule for integrals.

[@CraigNull]

The composition examples could also be written as products anti-differentiable functions, such as x^2*sqrt(1 - x^3) and 1/x^2. Granted they're not as simple. Hypothesis: elementary functions with nonelementary integrals can be expressed as a product of (two?) functions with elementary integrals. Or maybe expressed as a sum of functions with that property

[@pacolibre5411]

Funnily enough (and probably related to the reason for the theorem), solving that differential equation directly involves the integral of e^(x^2) [r(x)=e^(-x^2)*(int(e^(x^2))+C)]

[@redcrafterlppa303]

I'm a bit confused isn't (e^x)^2 the same as e^2x meaning that the integral of it is 0.5e^2x + c?

[@niallloughran5149]

@@redcrafterlppa303 correct but the function in question is exp(x^2), not exp(x)^2.

[@samwalko]

Might be worth noting on the last board m=0 implies alpha is not a root of the denominator of r(x), which is implied to be contradiction by our setup.

[@faraday4160]

pin

[@pierreabbat6157]

I'm a surveyor, and surveyors occasionally run into Euler spirals, which are used for parts of centerlines of freeways and parkways. The Euler spiral is non-elementary for the exact same reason the normal distribution function is. It is, in fact, the normal distribution function turned 45° in the complex plane. I've been working on a surveying CAD program called Bezitopo, which calculates the Euler spiral with the Taylor series.

[@smartube4828]

In addition to the brilliant content, writing on the board with the chalk is what I really loved! Thanks Mr. Penn.

[@edmundwoolliams1240]

Maybe then he should be called Mr. Chalk

[@General12th]

Dr. Penn*

[@bradwilliams7198]

I always found integral calculus deeply disconcerting, because even though there's always an infinite number of antiderivatives for a given function, sometimes you can't find any of them!

[@Bennici]

On a related note, if you watch the Numberphile video on "all the numbers", they encounter the realization that most numbers are not computable, but normal. And we currently know exactly zero of those numbers. Proper existential crisis for me when I watched that.

[@Nicholas_Buck]

They can be found (as power series, for example, in some cases), but cannot be expressed in finite terms using algebraic and transcendental functions.

[@landsgevaer]

@@Nicholas_Buck Nope, those are transcendental numbers. The non-computable numbers cannot be computed by any deterministic algorithm. Otherwise they were "computable". They all look like your typical but specific random number, with infinite decimals.

[@Nicholas_Buck]

@@landsgevaer I was replying to the original comment that some antiderivatives cannot be found, not to the comment on normal numbers.

[@landsgevaer]

@@Nicholas_Buck Ah, oops. 👍

[@davidgillies620]

Given the recent posts on hypergeometric functions it might be fun to show how integral of sqrt(1 - x^3) can be expressed in terms of 2F1.

[@jamiepianist]

I’ve wanted to see this for years! TY!

[@kamerplant3957]

I'm a little confused on the use of the fundemental theorem of algebra here. It would imply that x is a complex variabele, so do all the rules for differentiation still hold? Also, the root is probably not a real number, yet it is crucial for the proof. If we're just interested in e^(x^2) with a real variable x, how does proof tell us anything? The proof of r not existing relies on something that happens when pluggin in a complex number, which wouldn't happen for real x.

[@kevinmartin7760]

I think a bit more explanation of why alpha cannot be a zero or pole of p is required. This requirement boils down to (x-alpha) not being a factor of either the numerator or the denominator. In other words we have extracted all the (x-alpha) factors from p.

[@stephenbeck7222]

I think he explained it well enough though it was quick. We know r(x) is rational and since it’s not polynomial then its denominator must have at least one root. Call that root alpha and then rewrite r(x) as another rational function p(x) divided by as many factors of (x-alpha) that were in r(x).

[@MoodyG]

@@stephenbeck7222 But shouldn't we prove it for the general case of many roots in the denominator of r(x) (i.e. many different alphas) ?

[@hayimshamir8426]

@@MoodyG thats why P(x) is a rational function and not a polinomial

[@RexxSchneider]

I think there's a little bit of ambiguity in what we're calling a rational function. If a rational function is the quotient of any two polynomials P and Q, then the method used can fail. I believe that most of the time we're considering the rational function P/Q to be the quotient of two polynomials where their polynomial greatest common divisor is 1. That's just equivalent to saying we can divide P and Q by their polynomial greatest common divisor to obtain an equivalent reduced rational function. For example P/Q = (x³ + 5x² - 12x - 4) / (2x³ - 3x² - x - 2) has a polynomial gcd of the numerator and denominator equal to (x-2), so we can reduce P/Q to (x² + 7x + 2) / (2x² + x 1). In that case, the resulting polynomials in the numerator and denominator do not share a common root. Once you have that, then any root of the denominator cannot be a zero of the numerator and hence α cannot be a zero of the rational function p in the equation r(x) = p(x) / (x - α)^m. Of course, we defined m to be the maximum power of (x - α) in the denominator of r(x), and so α cannot be a zero of the denominator of p(x), i.e. it cannot be a pole of p(x).

[@MoodyG]

@@hayimshamir8426 Oh yeah you're right... the form he works with is indeed the most general form (according to the definition of course).

[@danielgarrison7463]

Make this a series!! I was actually reading about this the other day😁

[@manucitomx]

Wow, that was very cool. Thank you, professor.

[@diddly4929]

5:00 you didn't group like terms, so this claim that am=0 is incomplete. You instead need to show that m*am + 2a[m-2] = 0 for m>1 Then we check the highest degree terms--only in the 2xr(x) component--to see that an,a[n-1] = 0, and work from there to show the others to be 0

[@nikolasscholz7983]

Thought the same. For the slightly less mathematically inclined reader: there is two terms with x² in the expression, one coming from r (2a1, on the table), one from r' (3a3, not on the table). Added together they need to be zero, not individually.

[@diddly4929]

@Nikolas Scholz thanks for the rewording! To clarify how the proof proceeds from here: just like there are two x² terms, every other exponent has two terms that have to add to 0. The exception is the two highest degrees, which only come from 2x*r(x). Those higher coefficients must be 0; then, because they're also paired with another term--they show up again in r'(x) --those paired terms are also 0. And that continues down, making the entire r(x) polynomial equal to 0.

[@oida10000]

Even if it would be an hour long video, I would be very interested in a video proofing this theorem from scratch.

[@stephenbeck7222]

I think it would be more than an hour, or would itself have more shortcuts not proved to the viewer.

[@oida10000]

@@stephenbeck7222 well I wouldn't mind a series of such proofs down to parts the average viewer can show themselves.

[@amethystcluster7271]

Multiplying by (x-a)^(m+1) seems okay but then setting x=a It causes that we just multiplied both sides of eq. by 0 I think a is out of domain in this situation. if im wrong i'm sorry correct me :)

[@martinepstein9826]

That doesn't matter much. If two rational functions are equal for all x =/= α [Edit: and defined at x = α] then they're also equal at x = α by continuity.

[@reeeeeplease1178]

@@martinepstein9826 they dont have to be continous, r(x) is not continous at a in this case

[@reeeeeplease1178]

Had the same thought as op

[@martinepstein9826]

@@reeeeeplease1178 "p(x) is not continous at a in this case" Did you mean r(x)? It's assumed that α is not a pole of p(x). At the end we have two rational functions that are defined at α and equal for all x near α. Hence they are equal at α by continuity.

[@reeeeeplease1178]

@@martinepstein9826 yeye, will fix it

[@kkanden]

thanks for bringing me down a 2-hour long rabbit hole of reading wikipedia pages on Liouville, differential fields, galois theory and Galois himself lol

[@matkach200]

Is there a name for that theorem?

[@yoav613]

Liouville's theorem on integration

[@emiltonklinga3035]

1:29 I will be famous and rich now since I actually just found an antiderivative of 1 - x³.

[@lowbudgetmaths]

I missed that at first. 😀

[@carultch]

Seriously? That's a trivial application of the power rule. Everyone who knows what an antiderivative is, knows how to find that one.

[@thatdude_93]

5:50 i think that could have just been solved by looking at the degrees of both sides

[@anshumanagrawal346]

Yup

[@anshumanagrawal346]

That's what I thought

[@Happy_Abe]

Why is the denominator for r(x) limited to (x-a)^m? Why can’t there be more terms

[@japanada11]

There can be more terms - they just get included in b(x)

[@Happy_Abe]

@@japanada11 thanks!

[@neomooooo]

Here is a minimal proof. If we write f(x)e^g(x) =[r(x)e^g(x) ]’ then expand the RHS by the product rule, and cancel the common exponential term. Thus f(x)= r’(x)+r(x)g’(x) Looking at the top expression, integrating both sides gives Int( f(x) e^g(x) ) dx= r(x) e^g(x), Omitting the constant of integration. This is what we desired.

[@ddognine]

Wow! That was great! I was wondering how one proves that elementary solutions of certain integrals don't exist. There is a small discussion in my undergraduate calculus text about non-elementary integrals along with several listed, but no proof is offered.

[@giack6235]

Thank you, but there's a point not clear at the end: we cannot multiply by (x - alpha)^m+1 (supposing x != alpha) both sides of the equation and then pose x = alpha...

[@danielpetersen3184]

Hold on, at 8:15 you multiply the equation by (x-alpha)^(m+1) which means that x-alpha cannot be zero so x cannot be equal to alpha. In the next step you suppose that x=alpha, which we can’t do. I suppose then that maybe the proof can be saved by the fact that it is supposed to be true for all x?

[@japanada11]

If f(x) and g(x) are continuous, and f(x)/(x-alpha)^(m+1) = g(x)/(x-alpha)^(m+1) for all x not equal to alpha, then you can conclude that f(x)=g(x) for all x not equal to alpha. You're right that this by itself says nothing about the values at x=alpha. However, by _continuity_ you can conclude that they must also be equal at x=alpha.

[@_Blazing_Inferno_]

@@japanada11 Thank you so much for the explanation!

[@Dooley13]

Man, i had a Diff Eq assignment a couple weeks back where I ended up trying to evaluate the integral of e^[(-2/3)x] and I drove myself bonkers trying to figure out a solution for it. I get it now, but at the time i was like "why is finding this FINITE AREA so hard??" lmao

[@CraigNull]

Pedantic quibble: That α is not a zero of p(x) was a stipulation that went unutilized until we derived p(α) = 0, so that doesn't really feel like a contradiction. Either define α as a pole of r(x) with order exactly m >= 1, or highlight the impossibility of the discovery that r(x) has a zero wherever it has a pole.

[@jamesfortune243]

I wasn't aware of that theorem.

[@JamesLewis2]

It would be good to point out why you can't just use the usual solution method (an integrating factor) to prove this (that is, the naïve proof attempt would say that the solution to the broader differential equation is not a rational function), which is that that method might not find *every* solution.

[@nirmankhan2134]

(√π/2)erfi(x)+C 🙂🙃🙂🙃

[@mekbebtamrat817]

Nice shortcut, bruv. But naming it does not help a lot.

[@nirmankhan2134]

@@mekbebtamrat817 😅😅😅

[@MichaelRothwell1]

The reason we have to invent a new function (erf) for this integral is because it is not an elementary function.

[@DiegoTuzzolo]

awesome video!

[@gavintillman1884]

Never seen anything like the theorem used, would be good to see a proof :-)

[@StatelessLiberty]

Is this theorem true? g(x) = 0 and f(x) = 1/x seems to be a counterexample. The antiderivative of f(x) is log, which is an elementary function, yet there is no rational function r(x) such that r'(x) = f(x).

[@kamerplant3957]

g(x) needs to be non-constant for the theorem to hold

[@_P_a_o_l_o_]

Is there an intuition for why that theorem is true?

[@yoav613]

I don't think so,i found the proof and it has 26 pages with fields and alot of details..

[@random19911004]

It looks very similar to the integrating factor solution method for first order ODEs

[@goodplacetostop2973]

@@yoav613 Mind to share it? Or at least a way people could find this proof? 😀

[@yoav613]

Search for liouville's theorem on integration

[@goodplacetostop2973]

@@yoav613 Thanks!

[@edwardlulofs444]

OK, that all made sense. But I was hoping to increase my intuitive understanding of why it wasn't an elementary function.

[@schweinmachtbree1013]

I think the problem with intuition for what is and isn't elementary is that the definition of an elementary function is very arbitrary

[@nelsonnavarrete410]

Love this video! I’ve always wondered what ‘non-elementary’ meant, and this is really informative!

[@matthieumoussiegt]

very fun problem I don't know if it is me that indirectly suggest this problem via the comment section but that is very fun. It would be interesting if it is possible to look at the theorem some day in a dedicated longer video

[@CTJ2619]

I have never hear the term ‘pole’ in a math class before! (Would that be the same as asymptotic?)

[@stephenbeck7222]

It’s the complex version of a vertical asymptote, basically. The rational function r(x) = 1/(x^2+1) does not have any vertical asymptotes in the real plane but it does have a complex pole at i and -i. If my r(x) was the proposed solution to Michael’s differential equation, then he would have me rewrite r(x) with p(x) = 1/(x+i), then divide p(x) by (x-i) and say alpha = i.

[@ethandole2218]

It's the term we use in a complex context- it's essentially just an asymptote.

[@michaelleue7594]

What is the reasoning behind the statement at 6:02?

[@canrex7540]

I get a little smarter each day when I learn something that should have been obvious. Rational numbers can be represented as a ratio(nal). Sometimes I need it said aloud before I get it, thank you.

[@tarmotaipale5704]

I would like to know more about the proof of the theorem used to prove integral of e^x^2 is not writable by elementary functions: especially I would like to know if the proof tells anything about how to use the r(x) to calculate the integral. In other words, if we know the r(x) satisfying the condition (thus due to equivalence we know it's possible to integrate the function analytically), can we deduce from the r(x) what the actual integral function is?

[@pawemarsza9515]

r' + g' * r = f => int (f * exp(g)) = r * exp(g) + C

[@tarmotaipale5704]

@@pawemarsza9515 oh of course, that comes just by multiplying by exp(g) and using derivative of product. Thanks!

[@artashesasoyan6272]

Proving something using another thing that we dont understand doesn't make us feel satisfied 🧐

[@anarchostalinprimitividiag1030]

Why can you divide by (x-alpha)^m and then set x=alpha afterwards? Isnt the problem here that you are dividing by zero?

[@japanada11]

He cleared denominators first, so by the time he plugged in x=alpha there wasn't any x-alpha in the denominator

[@reeeeeplease1178]

@@japanada11 ye but putting x=a is not allowed in the last step because it was allowed in the step before If you define a rationl fuction f as f(x) = 1/x then f(0) is undefined Now you can say x*f(x) = 1 But you mustnt plug in x=0 although you cleared the denominator If you did: 0 * f(0) = 1 Essentially, you have multiplied both sides by 0 and weird stuff happens

[@elvisaguero9976]

I believe he actually cant do that. The expression at 8:15 is not defined at x = alpha, you cant just multiply by a polynomial and obtain an expression which is defined at x = alpha, then conclude something about your previous expression.

[@craig4320]

@@reeeeeplease1178 Back in complex analysis course, "There's a pole at the value..." (I fell asleep... waking up...) "...therefore the division is acceptable." Sorry, that's all I got.

[@japanada11]

@@elvisaguero9976 it's technically true that the functions aren't defined at x=a even after clearing denominators, but this is an example of a "removable discontinuity:" if f(x)/(x-a) and g(x)/(x-a) are equal at all x≠a, *and* f(x) and g(x) are continuous at x=a, then you can conclude that f(a)=g(a). This is used all the time in calculus when calculating derivatives, computing partial fraction decompositions, etc.

[@neilgerace355]

6:18 How do we know that the denominator has m identical roots?

[@dansheppard2965]

This caught me out as well and found it in "J L"''s comment below. Notice that p(x) is a rational function, *not* a plain polynomial. All he has done is pulled out one root (doesn't matter which one) which has at least degree one (it could be repeated), the number of times it's repeated being m (which is introduced here for that purpose), it could well be 1. Everything else in the denominator of the of the rational function goes into b(x) which is part of p(x), shown in the numerator of the fraction.

[@neilgerace355]

@@dansheppard2965 thanks!

[@Mosux2007]

Is the set of all non-elementary functions countable or uncountable?

[@carultch]

What do you consider to be a single function? Even with just one family of functions, there are an infinite number of variations on it, because the constants could be any real number. Do you consider every possible function, of a given family of functions to be a different function? Or are you asking about separate families of functions? Because it is trivial to make more variations on erf(x), than the total number of natural numbers. Even if you were asking about separate families of functions, I'd guess that it is an uncountable number.

[@edcoad4930]

Why can't one simply say that the integral of the function which is then differentiated doesn't result back at the original integrand? Example: exp(x^2): using the standard rules = 1/(2x)exp(x^2) which of course doesn't differentiate back.

[@carultch]

Because that wouldn't be the integral of a function. You are attempting to use u-substitution to solve this one, by letting u=x^2. But since you cannot generate the derivative of u, outside the exponential function, you won't be successful taking the integral. Because you cannot rearrange the original integral to equal e^u du, like you could if you started with the function e^(2*x). When the derivative of the inside is a constant, it's a trivial matter of multiplying by 1 in a fancy way, to generate the derivative outside, and pulling the denominator of that fraction out in front. But when the derivative is another function, you have to find that function outside, in order to use it for generating du. By the fundamental theorem of calculus, the integral of the function, when differentiated, has to give us back the original integrand. Provided that it is a continuous function.

[@thatdnahero]

When I clicked on this video i thought it was about functions for elementary schoolers

[@The_Shrike]

So what would the indefinite integral be? If I handed it to a math wizard would he just spit it back at me and say “non-elementary” or is there some “non-elementary” answer?

[@johnathanhaw8968]

that was beautiful

[@SuperWaffleTime]

nice rungne sweater. tryna catch magnus' attention?

[@sinecurve9999]

Pastel green on pastel pink is not good for legibility.

[@craig4320]

With a certain mild color vision impairment, it's almost completely illegible.

[@danielmilyutin9914]

I didn't get how p(x)/q(x) became r(x)/(x-a)^m. Probably it should become finite sum of such terms with different r, a and m? And then we may use similar argument as well.

[@ingiford175]

Yes, i was expecting the p(x)/q(x) to become r(x)/ ((x-a)g(x)), how do you know that the root is a multiple root and only root?

[@jacemandt]

I had a little trouble with this, too. I think it goes like this: He is assuming that r(x) is some rational function, so it's of the form a(x)/d(x). Since d(x) must have a root, we can express d(x) as b(x)*(x-a)^m, where m is the multiplicity of the root. Then he renames a(x)/b(x) as p(x). Note that a minute after introducing this rational r(x) he says that p(x) is a rational function a(x)/b(x), and this is how he constructed it.

[@danielmilyutin9914]

@@jacemandt So r(x) is rational. I see it now. Thanks.

[@stephenbeck7222]

r(x) was always rational from the theorem. I think the confusion lies in that we normally write p(x) as a polynomial function as the numerator of a rational function but he is using p(x) as another rational function that has no more factors of (x - alpha).

[@ingiford175]

I thought p(x), q(x), r(x) as all polynomial functions. Only the first two are, r(x) is a rational function. Thanks for the comments (Stephen and Данила).

[@redwanekhyaoui7232]

We need a proof for this theorem

[@JGHFunRun]

It feels so weird that the r doesn't exist. Like It seems like it really should exist but nope no cake for you or me I guess

[@abdelmoulamsaddaq8240]

I am giving up!

[@juanlatorre9390]

Why did not you just solve the differential equation? It seems to be easier than all that stuff you did.

[@erikstanton3908]

dx! You forgot the dx!

[@deanstanleynapawutut2842]

Because you can learn it n college

[@direstorm7324]

Fedex?

[@polychromaa]

:)

[@joryjones6808]

A function being non-elementary is not really an intrinsic property of that function but rather that no one has been clever enough to find a closed form anti-derivative yet. In short, a function's non-elementary-iness says more about us than the function itself.

[@reeeeeplease1178]

Or that none exists in terms of our defined functions?

[@uriaviad9617]

You are actually just totally wrong. The term elementary here is not subjective, the definition of elementary function is just kind of arbitrary and only includes rational functions, exponential functions and their inverses. With this definition and the theorem he used in this video you can prove that the anti derivative of e^x^2 has no close form only containing these type of functions. It's not that no one has been able to find a closed form, one just doesn't exist. The term closed form is not well defined and changes based on context, in the context of finite operations of elementary functions this anti derivative provably has no closed theorem, in other contexts people have found represntions in terms of other non elementary functions.

[@japanada11]

A number being irrational is not really an intrinsic property of that number but rather that no one has been clever enough to find a fraction representation for the number yet. In short, a function's irrationality says more about us than the number itself.

[@carultch]

@@japanada11 Not really. For irrational numbers like pi, e, and sqrt(2), it isn't just that we aren't clever enough to find a ratio of integers to make them rational numbers. There are proofs that it is impossible for these numbers to be exactly a ratio of integers. There is an objective definition of irrational numbers, and it isn't a consequence of our number system or our lack of success at trying to find the ratio of integers that makes these numbers.

[@japanada11]

@@carultch Yes, you're right, and that's the point: I wrote that comment specifically to highlight the flaw in the logic of the original comment I was replying to. (Perhaps I should have added a /s tag to be perfectly clear)