why some series are "regularizable"

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[@Calcprof]

Many of these divergent series (some of which are very useful for numerically computing special functions) can be had by integral transforms, moving the contour for the inverse transform the "wrong way" (whereas the residue theorem gives you the convergent series.)

[@SyroccoModding]

This is interesting, do you have an example, link or ressource to learn more?

[@Calcprof]

@@SyroccoModding Here's a little about computing the exponential integral with a divergent series: en.wikipedia.org/wiki/Exponential_integral . The inverse transform method of getting the divergent series see Integral Transforms and Their Applications (Texts in Applied Mathematics, 41) 3rd Edition by Brian Davies (Author)

[@SyroccoModding]

very cool, thanks@@Calcprof

[@lavneetjanagal]

A quick calculation gives Ei(1)/e as a value assignment for the second series. Here is how: 1. Introduce the variable x as in the video. Take Borel transform of the series, you get BA(x)=1+x+x^2+... = 1/(1-x) 2. Now, to get the Borel sum we perform Laplace integral F(x)=\int_0^{\infty} e^{-t} BA(x t) dt We want to compute F(1) to get the original sum. The integral does not converge in the Riemann sense, but Cauchy principal value can be assigned to the integral , which is exactly Ei(1)/e ~ 0.697175

[@GeoffryGifari]

if we can assign a finite value to both series, what would that imply?

[@DrYankeeDoodle]

@@GeoffryGifari it seems like kinda ''domestication'' of the wild numbers 😁

[@xinpingdonohoe3978]

​@@GeoffryGifari I'd imagine it like "removing" the infinite part. A bit like what you do with a pole. You remove the infinite aspect, and are left with a residue. Maybe it's in the same vein as that.

[@aweebthatlovesmath4220]

15:00 with Ramanujan summation formula we can assign a value to the second sum! this formula is usually a good way to give values to these sums.

[@viliml2763]

I'm getting 0.697175 + 1.15573*i, is that correct? Odd that it's non-real.

[@Qermaq]

And... ?

[@udic01]

10:36 Wrong 1/t=1-ln(u) t=1/(1-ln(u)) dt=1/[((1-ln(u))^2)*u]du but the result is the same. I guess 2 wrongs do make a right...

[@janeknowakowski5732]

Thanks. I wasn't sure if i spotted a mistake, but you reassured me.

[@emanuellandeholm5657]

Two wrongs did make a right. I was cringing so hard when Penn wrote dt = 1 /u sans the du

[@shruggzdastr8-facedclown]

For someone who strives to be an effective math(s) communicator, Prof. Penn has a disturbingly regular tendency to have calculation/notation/transcription errors in the working thru segments of his videos

[@MathFromAlphaToOmega]

By regularizing the first sum, one can actually approximate the integral of e^(-x)/(1+x) by hand pretty quickly. For example, one can write it as a continued fraction with a nice pattern and then take the convergents of that, or use the Shanks transformation on the original series. Carl Bender has a very nice lecture series on techniques like this.

[@davidblauyoutube]

I was looking into the first series just recently, this is perfectly timed. Thank you!

[@tomkerruish2982]

I worked on the second series a while ago. I used Borel summation and analytic continuation. I got a result that involved the exponential integral function, but the answer was smack dab on a branch cut involving the natural log of -1.

[@subrabalanmurugesan7159]

I too was thinking you could use Borel resummation to get some result for the second power series.

[@maximusideal]

At 9:18 I think you meant to say "delta" instead of "gamma."

[@landsgevaer]

I also thought that was a helluva doctor's prescription type of gamma if I ever saw one. Turned out to be a only moderatly scribbly delta...

[@goodplacetostop2973]

15:06

[@TechToppers]

Man... You are still here... Amazing!

[@trueriver1950]

10:03 "I would speculate that we probably believe that..." Love the multiple uncertainties in that phrase

[@peterhall6656]

Hardy's book "Divergent Series" is the gold standard. Norlund, Cesaro, Voronoi etc are all major players in this space. Hardy level analysis for these types of series is generally not covered at undergrad level except perhaps for the Tripos at Cambridge and maybe some courses at Princeton etc. I recommend Hardy's treatment - you learn a lot about what is "regular"

[@egoreremeev9969]

Even if the zero is the problem in case of second sum, you can "kinda" see that you can use the previous series if you plug in x=-1. But the problem with integration stays, there would be no "nice" way to set the value of the sum via this method. Although, if the point x=0 is just a pole and not a branching point there would be no problems. It's like with Riemann-Zeta function, because sum 1/n is just a pole for it (but I don't know how it is proven) there are unique values for the values Re s < 1.

[@MichaelMaths_]

You can technically assign Zeta of 1 to be the Euler-Mascheroni constant under a modified form of Ramanujan summation, but I guess it’s not nice with the fact that other values are consistent across many other forms of weaker regularization or smthn. Or making it have a weird discontinuity if taken

[@egoreremeev9969]

@@MichaelMaths_ oh, I thought about a different kind of assigning value - as a coefficient of 0-th term in Laurent series at the point. It's more of a definition than an actual way of calculating it, but I wonder if modified form of Ramanujan summation yields the same result. Or would it yield the -1 coefficient? EDIT: Looked it up - that's actually the case, they are the same!

[@MichaelMaths_]

@@egoreremeev9969 Yeah that is another way that gives the same value

[@GeoffryGifari]

Note to self: to solve math, make it complicated at first then simplify or take it to a limit later

[@qhrynxx1306]

0!=int[0,inf](e^-x) 1!= int[0,inf](xe^-x) 2!=int[0,inf](x^2e^-x) ... so when you take the sum= int[0,inf]((1+x+x^2...)e^-x) and if we use power series (i know it ll not converge [1,inf) range but lets forget it just to assign a number, the original sum is diverge anyway) itll be int[0,inf](e^-x/(1-x))=Ei(1)/e

[@toudeville]

Hi Michael! I’ve watched your video on the Virasoro algebra, and you really teased me about explaining the ramanujan formula 1+2+3+…=-1/12 using vertex operator algebra. Will you do a follow up video about this ? Have a great day!

[@AriosJentu]

I think integrating constant sounds a little bit incorrect, because it's a function. The correct way is the integrating factor, and you corrected it on the next board :) Awesome video

[@Kapomafioso]

14:27 "the only reliable value" - that's not true. You can formally plug in x = -1 and retrieve the first series, so your initial condition might be y(-1) = - delta instead. Not sure why this wasn't used, it seems kind of obvious, especially after just having done the first regularization.

[@mathaleph8501]

In fact, the function he defines is a power series which is not defined outside of 0. So evaluating it at -1 or 1 does not makes sense until someone define it (which was not done in the video).

[@Happy_Abe]

@10:50 shouldn’t dt be -1/u?

[@samagraarohan2513]

Yes. The u substitution was sketchy as hell

[@thomashoffmann8857]

The error above compensates that glitch 😉

[@Happy_Abe]

@@thomashoffmann8857 which error

[@EtienneSturm1]

what rule are you breaking exacly when you "assign" a value to the 1st serie?

[@ConManAU]

The rule that the value of an infinite series is the limit of the partial sums of its terms - because the series diverges, the partial sums don't approach a limit and so the series has no finite value.

[@michaelgolub2019]

Strange, the alternating series comprises only integers, but the result is fractional. Something like the sum of all naturals is -1/12 (Ramanujan).

[@untitled6400]

For me, this is another example of writing closed form of a infinite geometric seies without checking checking the limit of their common ratio. For context, write the factorials as integral representations of gamma function, then try to bring the summation inside the integration, then write closed form of geometric series.

[@Monkieteam]

What if you assign a lower bound of -1 instead of 0 and use the value assigned to the alternating series instead?

[@guerom00]

So, what's the loophole? Is it that y(1) is not defined in the first place?

[@PeterAuto1]

power series have a radius, within this radius the series converges, outside it diverges. But if you find a formula that describes the inside, you can calculate the value of the formula for points outside the radius. So you are basically pretending that the series is converging and look what value it would have

[@guerom00]

​@@PeterAuto1 so what's the radius of convergence of the power series defining y(x)? It seems to be 0?

[@burk314]

@@guerom00 Yes, the radius of that power series is definitely 0.

[@guerom00]

​@@burk314 OK, I'm confused... We define y(x) by that power series. It is defined for the single point x=0. So, what does the differential equation we arrive at even mean?! I mean, can we even define a y'(x) for a function defined at a single point?

[@pavlopanasiuk7297]

@@guerom00 technically there are no loopholes. The way we turned infinite sums into "something different" is by collapsing that whole power series into a well-defined diff. eq., which only relates to the original series at the convergence radius. You cannot predict the existence domain of that newly obtained function (the one that you have in diff. eq.). So if you had a power series with zero convergence radius, that diff. eq. may not have anything to do with that power series at all. Since these are all "shenanigans", there are no loopholes, we make some domain continuations and see that they behave differently

[@tomholroyd7519]

that's pretty awesome

[@nahbro1321]

Wait but this divergent series i am sure was sent by Ramanujan to Hardy in his first letter ...... y Euler is mentioned??

[@iabervon]

Euler also solved it, using the method in the video, and lived a hundred years before Ramanujan. Part of the reason Hardy knew Ramanujan was onto something was that some of the results agreed with obscure works in his library, while others were new, and this is one of the ones he could check.

[@nahbro1321]

​​​@@iabervonunderstood👍 Ramanujan was truly a genius.. I mean he extracted maths ...and re-invented the formulas from Euler's mind at such young age...like nothing happened...

[@keithmasumoto9698]

So how are these values like -1/12 and 0.5963 useful?

[@pavlopanasiuk7297]

In quantum field theory you occasionally end up with regularizing infinite integrals, which is essentially subtracting infinite terms, causing divergence, and it brings you to a physical result. The values of those regularizations have a direct link to the discussed in video

[@kevinmartin7760]

At 8:00 he seems to be talking about a constant of integration for the RHS of the equation on the board, but the integral is a definite integral and so no constant of integration is involved at all. However, there is still the constant of integration hiding in alpha, and this should appear multiplied by y in the LHS and within the integral, divided by t, in the RHS, Although the RHS is the integral of a sum, it can't be split into the sum of two integrals because integral{0 to x}(C/t dt) is infinite IIRC. The Force is weak in this hand-waving... or at least the explanation is incomplete.

[@pierreabbat6157]

Both these series converge in all p-adic fields.

[@Alan-zf2tt]

what is a p-adic field and convergent values?

[@FLCoeur]

Does this mean we can find the coresponding value of the constant in p-adic fields, like we have done for e and ln?

[@user-gs6lp9ko1c]

"Monkey business"? 🙂

[@peasant12345]

Why you can differentiate an infinite series term by term in the first place. You might need to use dominant convergence argument.

[@davidblauyoutube]

For the second sum, I took a similar approach with y(x) = sum_{n=0}^{\infty} n! e^{ -(n+1)x }. Then the goal sum is y(0) as before, but now y' = 1 - y e^x. This equation is solved as y(x) = e^{e^x} ( c + Ei(-e^x) ) where c is a constant and Ei is the exponential integral. Note that in the limit x->\infty, one has y->0, and both Ei(-e^x) and e^{e^x} Ei(-e^x) both ->0 so we can take c = 0. Then the goal sum is y(0) = e Ei(-1) = -δ.

[@looney1023]

Is there a special function that corresponds to families of that first series the same way the Riemann Zeta function corresponds to 1+2+3+... and it's "siblings"?

[@alielhajj7769]

Interesting hypothesis please make a video to test it if true: this representation is actually the limit of the average of the partial sums

[@viliml2763]

The largest value dominates the average so nope. However, there's a closed formula for the partial sums and it is of the form `c + (-1)^n * f(n)`. In regularizations like this it usually makes sense to say that limit of a^n as n->infinity is 0 (see p-adic numbers) so the second term vanishes and you're left with a constant c and that is exactly the Gompertz constant.

[@krisbrandenberger544]

@ 14:00 The first term of the final answer should be negated.

[@krisbrandenberger544]

@ 13:54 The rhs should be -x, not x.

[@TaladrisKpop]

The problem with this kind of video is that it never properly defines what kind of operations are allowed to assign a value to a divergent series. As such, it is wrong to claim that no value can be associated with the second series, maybe another chain of operations could assign a value to it (someone did in the comments actually). In the present case, I am not sure that "using a loosely related function" could be accepted as "assigning a value".

[@ruffifuffler8711]

Here is a guess: If you build the proxy as a complex function, if you then see that you are adding more smoke stacks to the core, then it does not converge absolutely and conveniently to symbolic logic. But, taking successive logs leaves you with a rose or some other space formation that partitions the complex plane, making it less and less analytical, but more and more as a construction of an object, figure, or ornament?

[@ruffifuffler8711]

And while it's on my mind, the 14.2° is relevant as the upwards waterfall cut line, because the induction operator cannot bend the real axis to itself any further without cascading, which it then does, ...and that acts as a lance for any further intrusions.

[@mathisnotforthefaintofheart]

Euler "bends" the rules more than once in his life time.....not always 100% correct, but his results are marvelous nonetheless. Thinking outside the box does help

[@zh84]

The solution of the Basel problem is my favourite example. His reasoning is now not considered rigorous, but it gave the right answer!

[@mathisnotforthefaintofheart]

@@zh84The way how he did the Basel problem was "flawed" not because of being wrong, he wasn't. But he did not support his reasoning with the necessary rigor to justify his manipulations. So his proof was lacking. But nothing in his proof was wrong otherwise.

[@theupson]

"it all depends on what your definition of 'is' is" -bill clinton, visionary mathematician

[@aomaik7639]

Hi can you make video about Euler-Compretz constant with continued fractions for it

[@dirkdiggler8403]

To establish that a sum of whole numbers can be something other than an integer, one must engage in quite a bit of circumlocution. 😅

[@minwithoutintroduction]

19:28 🎉

[@gary.h.turner]

... thus proving that the infinite is really finite, and that therefore the whole of mathematics is inconsistent within itself! Oh dear! 😢

[@olli3686]

Why not just show the initial 1 as zero factorial?

[@ChemicalVapors]

Do different methods of assigning vales to divergent series give the same result?

[@drdca8263]

In many cases, yes

[@emanuellandeholm5657]

I was like Ei from the get. Nice problem!

[@vetbaitednv]

love these videos, keep it up

[@nicolasalvez1999]

how a sum of integers gives you a non-integers number?

[@FaranAiki]

Welcome to divergent series!

[@mathaleph8501]

Hey, there is something I truly don't understand. You consider a series who's sum is not defined at the beginning, ok. To give it a sense you define a function and say your series is your function evaluated at 1. Now my question is : how do you define your function ? I mean it is a power series with radius of convergence 0. Hence your function does not exists, as a function, outside of 0 so you can not evaluate it at 1, right ? Now your power series exists as a formal series for which you can solve the differential equation, but my other question is : how do you define the evaluation of a formal power series ? I don't understand how you can justify the existence of an object by using other undefined objects...

[@thomaslangbein297]

Pretty sketchy…April 1st?

[@Arycke]

Nah it's April 2nd

[@tomholroyd7519]

Euler bending

[@writerightmathnation9481]

The claim that we cannot “assign” any value to the second series is ambiguous at best. What’s it really mean to say “we’re cannot assign a value” to some mathematical expression. When someone puts quotes around an equal sign, they’re admitting that something got hidden. Please, before the video ends (I’m not done watching it), clarify what you mean by “cannot assign”. Why can’t I just say it gets assigned the value 1? Ok, so at the end he does a very quick “retraction” of the “cannot assign” part.

[@Alan-zf2tt]

All this suggests to me is .... due diligence. I admit that the brilliance of some to show something with at face value a ludicrous result must be great fun. A bit like "Guess why in Math" I wish I could but it would take years n years of study to build up such a wide range of skills and awareness. Interim conclusion: should use function notation and explicitly state how many transformations were used? Observation: could a concatenation of transformations me used swinging things through a floor or ceiling or stepwise function arbitrarily defined to confuse be dreamed up? (Maybe cryptographers already have?) 🙂

[@Trixex]

What a great video. You could show this to a (smart) 5 year old and they would follow.

[@driksarkar6675]

That’s a very, very smart 5-year-old…

[@Trixex]

@@driksarkar6675 is that you on the pic? Did you follow? XD

[@driksarkar6675]

@@Trixex I followed, but I'm no 5-year-old.

[@charleyhoward4594]

this was confusing ...

[@veteatomarporculo100]

So summing and subtracting integers numbers gives a non integer value xd