Thank you for letting me know. It was an experiment to see if it would be helpful. Once I gather more feedback like this I'll be able to decide if I should continue it or save it for particular circumstances. Can you think of a situation where it WOULD be helpful to you? Just wondering. -Stephanie MP Editor

@@MichaelPennMath I tried to think of one when I was writing my comment. :) I do like the occasional "sidebar" or overlay with quick footnotes and extra info. They are a good addition, as are the quick corrections of a typo or misspoken phrase. They provide useful extra information with minimal disruption. I think the video cuts where a previous board is summarized at the top off screen was a great idea that's been around since the beginning of these videos. I pre-date computers in the classroom and got used to the chalkboard style of math and physics teaching. I personally like the consistent full view because I sometimes am looking at previous lines and processing other parts of the board as Michael is writing. Watching these videos is relaxing for me in large part because of the pacing. A different generation more used to PowerPoints and graphics as well as the ability to freeze and rewind a video might not process a chalkboard the same way . I suspect I am probably not within the main demographics of most watchers, so take my preferences with a grain of salt. :). EDIT[. Oh, and... nice job. These are good videos.]

@@MichaelPennMathI also think that the zooms aren’t really necessary. It’s just distracting. (Overall I think Michael's videos don’t need and never needed this kind of editing, like special effects or something of that sort.)

@@MichaelPennMath Unfortunately I think zooming in makes the argument harder to understand: you can't, while zoomed in, glance back and forth between the new formula and the old one, or elsewhere on the board, to see what is going on.

@Michael Penn On occasion Michael starts writing smaller when he realizes he's running out of room. I can see the zoom being useful potentially in those rare instances, but I agree with the others here that it is more distracting than useful.

Like if I wanted to be a tiny bit more rigorous I'd go through an eigenvector decomposition, but the shift operator has no eigenvectors

Thank you very much! I really appreciate comments like this.

It is a operator. Idk if that makes a difference though

The notation he uses hides the fact, but sigma isn't supposed to be a function from R to R. Its input is a sequence a_n, indexed by n, and its output is the shifted sequence n->a_n+1.

Maybe a better way to think about it is that the sequence a_n is a function from Z to R, and the operator o takes a function f: Z to R and produces the function g(n) := f(n+1). This is called a shift operator or a translation operator, you can read about them on Wikipedia. The linearity of the shift operator is what allows us to use it in a variable of integration without problems.

Apparently (based on other comments, including a heart’ed one), sigma is actually the shift operator acting on the space of sequences (a_n)_n (By “(a_n)_n” I mean the entire sequence as one object, as opposed to “a_n” which I would use to refer to a particular entry in the sequence. More specifically, I mean “the sequence whose entry at position n is a_n”) sigma( (a_n)_n ) = (a_(n+1))_n i.e. the sequence whose term at position n is a_(n+1). Then, basically everywhere that a_0 is written in the video, interpret it as if it said (a_n)_n , except that at the very last step of evaluating the expression containing it, we take the entry at index zero of the sequence that the expression would produce if (a_n)_n had been written where a_0 was. So, uh, where there is an expression that is written like H(a_0) in the video (where H involves sigma) read it instead as (H( (a_n)_n ))_0 ... I think.

While I believe that there is probably a way to justify the slight of hand here, it is awfully sketchy and should have been addressed in the video.

@@TheEricthefruitbat Yes, the way to go here is actually the other way around. Int(f(x)*x^{t-1),x=0..infinity) is the Mellin transform F(s)=M(f)(s) of f(x). So you can prove it by using the residue theorem on the inverse M^{-1)(F)(x)=f(x). You just need to justify the closing of the contour (over the imaginary line) along the half-circle on the left part of the complex plane to apply the residue theorem.

​@@TheEricthefruitbat"should" is a strong assertion

This is too loose even for me and I'm an engineer!!

It feels very loose because it *is* very loose. And quite flawed, as some others have noted.

Apparently this technique is used to compute integrals related to Feynman diagrams, so the connection is even more direct lol

I think (but maybe I'm wrong) that a_n it s not any sequence but a specific sequence that admit a value σ such that σ*a_n=a_(n+1) for all a_n so sigma it's a constant and not a function In this way the proof works fine too and also you explain the very sketchy dσ^(-1) = dx that doesn't effect the bounds of the integral

​@@simonecutrona6337if that was the case, a_n is a geometric series with common ratio sigma. Then it's power series would be easily computable and most of what he did would be unnecessary.

From what I understand it works because sigma is not some random function. It is an operator that only applies on the sequence a_n. Therefore, x*sigma*x = x^2*sigma since sigma doesn't apply on x. Technically, sigma doesn't have anything to do with the integral (doesn't depend on x, doesn't affect x), but you can't take it out because the summation is inside the integral.

sigma is not a function of x, it's just an operator on the terms of the sequence, so it's not affected by the integral.

@@xizar0rg ok thanks. But my question was about the integral in “u”

@@rtheben I’m making a guess, and this might not make sense.: Maybe we should see it as a parameterized family of integrals, parameterized by the combination of the sequence (a_n)_n and chosen index value, but where for every value for these parameters, the upper limit for the integral is still +infinity? But, that’s a guess, and I’m not sure of it.

Its being used as a operator here. While this can be justified its a bit of a pain. Umbral calculus uses a lot of tricks like this. This sort of thing is reminiscent of calculus pre Cauchy.

A_0 is a constant. The sigma is an operator that maps a_n onto its sucessor. All of the a_n are constants. The intergral is with respect to x so it does not affect the a_n.

I think it is better to keep a_0 after the integral while pull it outside of the integral. Then by interchanging the summation and the integral, pull out the operator outside the integral, we can get the final result as well

Tell us about that “relationship that seems to exist” by doing a video on your channel.

the integral also depends on t, so i don't see the problem here

I think you seek verification that you are indeed bright

@@kqnrqdtqqtttel1778: No, I definitely am not, nor am I seeking praise. The point of my comment was that I was amazed by the simplicity of the answer when put into actual practice.

In the book Michael referenced, there's a note that references a paper on this theorem (doi: 10.1007/s11139-011-9333-y). In the abstract, it says "Ramanujan’s Master Theorem... provides an explicit expression for the Mellin transform of a function in terms of the analytic continuation of its Taylor coefficients" which is pretty cool. Presumably, in a more rigorous proof, we would see why the negative a_n's are necessarily the analytic continuation, rather than something else. My guess would be something to do with this sigma operator being a linear operator and analytic continuations "play nice" with linearity given how derivatives work, but I'm just spitballing here.

@@Alex_Deam Thanks for your reply. To my knowledge, the analytic continuation can be only applied to a function of continuous variable. How can you apply the analytic continuation to a sequence whose domain is discrete?

@@yuan-jiafan9998 I presume it's similar to how it's done with the factorial, in that the analytic continuation is actually being applied to some interpolation function that gives the same values for positive integers

Presumably G.H. Hardy cleaned it all up to something respectable, but all such details are here swept under the rug. I have observed that M. Penn characteristically ignores any objections to his content, but this is a whopper.

Ramanujan: gods damn not god(he was a hindu)!

sorry -1

No

It's just the notation, in this notation sigma^n means sigma composed with itself n times. This is the same as the notation ^-1 for the inverse

It still feels like an abuse of notation. Sigma^2(a0) = sigma(sigma(a0) and not sigma(a0) × sigma(a0), so why is x^n × sigma^n(a0) = [x × sigma (a0)] ^n?

@@robertmauck4975 I believe in the video we don't see x^n * sigma^n (a0) = [x * sigma (a0) ]^n but x^n * sigma^n(a0) = [x * sigma]^n (a0) which is different

@@robertmauck4975: No. This is the natural notation. It is notations like log^2 or sin^2 that are unnatural and exist solely for historical reasons (probably cause they are typographically simpler to write down)

@@rv706 I agree that it is more natural, but it is also nonstandard, and may at times be worth mentioning up-front that one is using it this way, to avoid confusion. One might also use f^{\circ n} for the n-fold composition of f. I suppose if one wanted to refer to the function x \mapsto (f(x))^n while avoiding the possibility of it being interpreted as f^{\circ n} one could maybe denote it as f^{\cdot n} ?

I notice Michael Penn never replied to this valid criticism; +1. This surprises and saddens me. Also surprising is that none of the fawning adorers gushing praise on this video picked on the fishiness of the derivation. A lot of his vids have high quality, but this seems to hit a real low.

@@toddtrimble2555 He did “heart” another comment which says a little about how to make, the thing that this video is really meant to be getting at, somewhat more rigorous. I don’t think it is reasonable to expect him to respond to every comment that points out a flaw in the reasoning. Though, it would be nice to, say, pin a comment which clarifies the steps that aren’t exactly valid.

Well, he did say that the terms of (a_n)_n are all related by sigma. Though, seeing as I think it would be better to frame this as sigma acting on sequences as a whole, and specifically being the shift operator, then in that framing, it still isn’t clear why the values of the sequence for negative indices should be determined by this. Perhaps one could have sigma be replaced with two things: • An operator sigma_o that acts on bi-infinite sequences, and is the shift operator, so that simga_o( (a_n)_n ) = (a_(n+1))_n (“o” for “operator”) • a function sigma_r which satisfies some other properties (I’m not sure what properties are needed here), and for the particular sequence (a_n)_n that we are considering, is such that for all n, sigma_r(a_n) = a_(n+1) (“r” for “relationship between successive entries of (a_n)_n”) Probably need some hypotheses on the sequence (a_n)_n as well. Or, I guess those could be viewed as properties of sigma_r along with a_0 ? Then, all the places where in the video he writes sigma^n , read that as sigma_o^n . (sigma_o is a linear operator, so this is fine) well, when he does the substitution of u = x sigma, ... I guess that could kinda-sorta be viewed with sigma being sigma_o ? Would still need a lot of elaboration to justify doing that, but I imagine there’s some interpretation that makes sense? (It’s not like OP made this idea up himself. The name Ramanujan is in the title for a reason.)

@@drdca8263 Of course he didn't make this idea up himself. Ultimately, the a_n must be glued together by a function a_x which is analytic in a real or complex variable x (so that in fact the values a_n for nonnegative n would uniquely determine a_x for remaining values of x, by a uniqueness of analytic continuation argument), subject to some further assumptions that would make the theory of Mellin transforms work out nicely. At least I guess it's something along these lines. So yes, for sure there are suitable hypotheses so that the statement itself becomes correct, and from what I read, Hardy figured out a correct statement and proof. But since his video makes no mention whatsoever that strictly speaking you need such hypotheses, it's disinformation. And his "proof" cannot possibly be valid, even if he *had* more carefully stated the theorem at the outset -- if it were valid, then it would be valid even under the minimal hypotheses that he did in fact use in the proof, and as we saw, one then easily derives nonsense. If he had said at the beginning that the proof to come isn't really valid, but is a handwavy and formal (in the sense of "on the surface") manipulation meant to suggest that a more careful statement is perhaps plausible, then that would intellectually honest and I would have no complaint at all. But since many in the audience seem to be relatively "young in mathematics", and trusting in the instructor, I just think it's super-important that he behave like a mathematician and be very open and honest about what he's doing. There's that (in)famous Numberphile video where they have two or three guys tag-teaming and giving a "physicist-style argument" for why zeta(-1) = -1/12, and I see this video as "pulling the wool over people's eyes" to roughly the same degree as that one, and I honestly believe Michael Penn knows better -- cool physicist arguments aside.

agreed. doesnt seem, to be in Nahin's book. irresistble integrals uses it but does'nt develop it

It's in the 2nd edition of Inside Interesting Integrals; not the 1st. The "derivation" is the same as here, but Nahin notes that it is far from rigourous. However, this is how Ramanujan originally did it.

They are literally the same notation! Composition with negative "exponent" -n means composition of the inverse function n times, by definition. This of course is well-defined only when the function is invertible.

@@rv706 f^-1 does not mean 1/f. There is no index notation for 1/f. f^-1 means inverse function of f. I'm suggesting this is poor notation and that f^-1 should mean 1/f. We could use f^(2) to be an index notation for multiple composition fof (composition) and f^2 would mean f x f. We could use f^(-1) for inverse. f^(-2) could then also be an index notation for multiple composition of the inverse f^(-1)of^(-1) 🤔

Its because of the (arbitrary?) choice to do the integral with x^2. You could do any integral with a power on top this way I think.

@@MarkTillotson Oh okay that makes sense, thank you

I still am not clear on the choice of T=3. Is it truly arbitrary?

@@nateinhouston I think if t was chosen to be 5 or more, the integral would diverge, so, I guess not entirely arbitrary. But maybe one could still use some (unrelated) methods to assign a finite value to the integral anyway, and maybe it would agree with the value from this method?

yeah i don't really understand why is it ok to see sigma as a real number.

@@behzat8489 if there can be Sigma male, there can be Sigma number too

Yep. But don't hold your breath waiting for the answers from this KZfaqr.

Gamma of negative integers diverges.

@@byronwatkins2565 right, I forgor 💀 about that.

Thank you for the feedback. If you didn't see my reply to a similar comment, this was an experiment to see if it was helpful or not. -Stephanie MP Editor

Master theorem is a term used for.results that can be applied to a variety of functions. Its not always used. The Furlani integral is one that can be applied to a variety of integrals but doesnt not have the term master

I think it's because you need a master's degree in occult mathematics to be able to come up with things like this.

To me it almost feels like an April Fool's Joke, but a couple of months too early.

Lol nah 😊😊