I'd never heard of this before. Very interesting and an easily followed derivation. I'm not sure the video zooms are worth the extra editing effort. The choice of colors, consistent & legible handwriting, generally good parallel lines of writing, and clean arrows and boxes really carries these videos. They are easier to read and follow than many computer text videos and much easier to follow than most handwritten math channels. These videos are hard to improve on with regard to quality. Clean, elegant, easy to follow, and interesting variety of topics; I personally cannot ask for more. The only thing I've not seen on the chalkboard is some of the tricks us old-timers used to do for emphasis and variety like edge of the chalk shading and getting dotted lines by "pushing" on the chalk at the correct angle with the chalk pointed slightly forward so it "skips" along.
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.
I'm a physicist and even I had a little pearl clutching moment when you set u = sigma*x Like sure, I know this integral should obviously be from 0 to infty, what else would it be, but this is incredibly sketchy lol
Sigma is an operator where exponential notation is used for repeated composition, but still x^n * sigma^n becomes (x*sigma)^n, not to mention the rest. It could make sense if sigma is a linear operator, since matrix multiplication corresponds to composition, but that's not our case. What kind of umbral calculus is happening here?
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.
I'm just noting that that integral is the Fourier transform of f(x) on the multiplicative Lie group of positive real numbers: the Haar measure is dx/x and the character is x^t, so integral of x^tf(x) against the measure dx/x.
Ooh, this reminds me of the "insanity with a method in it" us particle physicists routinely pull with quantum / thermal field theory 😛 As the saying goes, the only math you need for field theory is the gaussian integral (but with all stops pulled).
The "function" sigma is not a function and is not even well defined. if a_n is equal to a_m but a_n+1 does not equal a_m+1 then sigma maps the value a_n to two different values. Similarly with the inverse of sigma.
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.
Thank you Dr. Penn. For me this was a good companion video to the one by Maths 505 where he uses Ramanujan's Master Theorem to evaluate the first two Fresnel Integrals.
I see that sigma is a linear shift operator, but when making the change of variables u=x*sigma, u should be operator valued, so how can u be treated as a real number with limits 0 and infinity?
@@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.
For the gamma integral to converge even for complex numbers, the u-substitution must approach infinity (so no -∞, +-i∞). So we extend that reasoning to the operator
Perhaps I really didn't understand the argument well, but around 8:00 it didn't make any sense for me to pass from "x^n . sigma^n" to "(x . sigma)^n". To my eyes it was totally cheating with notations: x^n denotes the nth power of x while sigma^n denotes applying sigma n times... I can't understand it works when the sigma operator is not simply "multiply by a constant" in other terms a_i is a geometric progression.
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.
That was so cool! I normally can see where things are going in your videos a decent ways ahead of time (which just indicates that you teach well - I am far from bright), and that happened in this video too, but the final result nonetheless blew me away. It was like a thunderclap in a clear sky. I literally just exclaimed "That is so cool! That is SO cool!" in public because I was so shocked. I followed and predicted everything until the actual numerical part, and that blew my socks off. (There are a few theoretical details which I would need addressed in a formal mathematical paper/proof, but the pedagogical value of forgoing them was sufficient justification of the choice to do so)
@@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.
It seems that something is quite wrong in this video. At 6:21, the integral depends on f(x), which depends on a_n for n >= 0. But the end result at 12:00 depends on a single value of a_{-t}. So if this is correct, there must be some relationship between the a_n, which has not been clarified.
There is one minor quibble here. The reversal of the integration and summation requires that the a x is convergent. From the alternating.series theorem we know this is the case if ax tends to zero. In the worked example the ax term tends to zero so the theorem applies. In other cases it may not hold. The worked example is nice as the result can be computed by hand and the result checked.
the sigma function reminds me of another function i used when i was playing around with a "generating vector" (a vector in an infinite-dimensional vector space where the e_n component is the a_n term in the sequence) called the left shift. the left shift is a linear transformation that brings e_n to e_(n+1)
@@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
I get that this is an entertainment channel but why spending 15 minutes on "trivial" analysis and not talk about the elephant in the room? My functional analysis background may be too weak for this but there's an unjustified abuse of notation here: why sigma_n(a_k) becomes similiar to a (linear) operator that can be "exponentiated" and later on (substitution) even treated as a variable (apparently)? If we just play with it, it could have also become something like n*sigma(a_k) or sigma(a_k^n) or whatever: why it's not the case? Maybe great math has always come from these surprising abuses of notation (other than conceptual) but this is closer to magic than to rigorous reasoning IMHO and isn't very educative/informative for people curious in math.
Nice, the final theorem. I don't understand how you treat the limit after the change of variable of the integral at 11:30. The upper integration limit is Inf*sigma but you use just Inf.
Love this channel, but geez this video is an absolute mess. Sigma is a function on sequences, but then we treat it like a real number in the integral? exp(-x sigma) is just suddenly fine? Then the u sub, so dx is du times sigma? That's definitely not gonna be integration over the reals. Then at the end, we only proved that a_n=(n+4)!/4! for n>=0, we could define a_n for n
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
I am an electrical engineer like Nahin and not a mathematician. But I kept an eye open for Ramanujan's MASTER THEOREM and was of the opinion that his mastery of infinite q-series raised to 24th power, with witch he solved the partition function (a method claimed by Ramanujan that was taught to him by Goddess Swaraswati [consort of Vishnu} and which string theorists employed to make sense. I never heard of the Master Theorem you just solved in your usual style (I notice 1/12 at the end of your deduction, implying a connection with his q-series). Previously I requested you to explain the q-series and I wonder if your current video is a reply to my request?
While you start with the function f(x) rather than the sequence a_n, how can you extend a_0, a_1, a_2 ... to a_-1, a_-2, a_-3? Is there any condition here?
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.
So many questions. So, sigma is not a function? (sigma)^n is not a number. It's the function being applied n-times. Also, in your example the n=0 term doesn't work. Like 5*6*7*... (n+4) But when n=0, 5*6*...*4? 😢
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.
Just some feedback; for me, the zooming is a bit jarring and distracts from the content. Not sure if anyone else feels the same. Thank you as always for the great videos.
But... am I missing something or the theorem was never clearly _stated_ (after supposedly having been derived) in the video? Also, maybe I missed some details, but the way the theorem is "stated" doesn't make any sense: the values of the sequence a(n) do not appear for negative n in the definition of the generating function. But they do appear in the final expression a(-n)Gamma(n). This cannot be true, cause I could take a(m)=0 by definition for every m
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.
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 don't see why these formal manipulations make sense. Particularly, I don't see you state any restrictions or regularity assumptions (except for a convergence criterion, which won't matter for what I'm about to say) on the Z-indexed sequence a_n. So suppose we take two such sequences a = (a_n), b = (b_n) which differ only at say n = -3. The functions f_a, f_b that you begin with depend only on the nonnegatively-indexed a_n, b_n, so f_a = f_b. According to the end result around the 12-minute mark, we should have for example that the Mellin transform of f_a = f_b evaluated at t = 3 yields Gamma(3) a_{-3} = Gamma(3) b_{-3}, which is obviously nonsense.
sigma^k (a_n) = a_{n+k}. WOW !!! This is a generalized residue theorem, a generalized gamma function, and a way to integrate entire classes of functions...
Doc, why it doesn't work when I move the sum n relative constants outside the integral then do the direct integration on x^(n+t-1)? Cos the intergral limit exceeded the "taylor" expansion limits?
Sigma doesn't make any sense at all to me. In the definition of sigma, it has sigma^n(a_0) = a_n where sigma is the recursion and n is the number of recursions. Okay, but sigma^n is treated as an ordinary power later on and not the number of recursions. You can't have it both ways. It would actually be sigma^n(a_0) = (a_0)^n. However, the answer of the example integral still manages to be correct. It's the most ridiculous abuse of notation I've ever seen.
Mixing notation for inverse function (-1) with multiplication indices? Dunno how that works.😬😱 Might be helpful to use notation sigma^(-1) for inverse function and sigma^(n) for n-composition?
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) 🤔
Secretive derivations are at least almost isomorphic to proofs by intimidation lol please, take as many liberties as possible, but give it a narrative 🙏
@@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?
Wait, if t=-1 this seems to imply that the integral of x^-2 f(x) is f(0), which seems weird. Is it crazy crap like assigning finite values to divergent integrals? Where can I read more about this?
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
This video is doing the manipulations with the wrong philosophy. The operator "sigma" shouldn't be the "composition operator" (whatever that means), it should be a translation operator. Translation operators have an easy-to-understand meaning as differential operators. BUT, the idea of a composition operator, while based on a misinterpretation, is utterly intruiging, and certainly this is what Ramanujan had in mind, so I shouldn't complain too loudly. To define a composition iteration operator, there needs to be a function f(x,s) analytic in s, such that f(x,n) = g^(n)(x), and a further condition on imaginary s to prevent us from adding cos(\pi s) to the thing (usually such a condition is a condition on the rate of exponential growth in s, see Carleson's theorem, but since g is general, just leave the condition vague). Then the 'composition operator' becomes a translation operator in s, and your derivation makes sense. But I am sure your source was using sigma as just a simple-minded translation operator, not as this sophisticated 'composition operator'. Considering how singularities multiply under functional composition, I am not sure this composition operator even exists for any function with a branch cut, for example, sqrt(x-1) (compose this with itself n times!). The theorem itself doesn't require such sophistication.
hey Penn, awesome video, but I don't get at 13:02 I don't get the binomial expansion; wouldn't it be valid only for x between -1 and 1? can u still apply it? 🥸
I'd never heard of this before. Very interesting and an easily followed derivation. I'm not sure the video zooms are worth the extra editing effort. The choice of colors, consistent & legible handwriting, generally good parallel lines of writing, and clean arrows and boxes really carries these videos. They are easier to read and follow than many computer text videos and much easier to follow than most handwritten math channels. These videos are hard to improve on with regard to quality. Clean, elegant, easy to follow, and interesting variety of topics; I personally cannot ask for more. The only thing I've not seen on the chalkboard is some of the tricks us old-timers used to do for emphasis and variety like edge of the chalk shading and getting dotted lines by "pushing" on the chalk at the correct angle with the chalk pointed slightly forward so it "skips" along.
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.
The editor is having fun with these lately and I wholeheartedly approve.
16:43
The editing of this channel has got a lot better lately. Well done! Nice little poem in the description today.😊
Thank you very much! I really appreciate comments like this.
I'm a physicist and even I had a little pearl clutching moment when you set u = sigma*x
Like sure, I know this integral should obviously be from 0 to infty, what else would it be, but this is incredibly sketchy lol
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
Sigma is an operator where exponential notation is used for repeated composition, but still x^n * sigma^n becomes (x*sigma)^n, not to mention the rest.
It could make sense if sigma is a linear operator, since matrix multiplication corresponds to composition, but that's not our case.
What kind of umbral calculus is happening here?
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.
I'm just noting that that integral is the Fourier transform of f(x) on the multiplicative Lie group of positive real numbers: the Haar measure is dx/x and the character is x^t, so integral of x^tf(x) against the measure dx/x.
Ooh, this reminds me of the "insanity with a method in it" us particle physicists routinely pull with quantum / thermal field theory 😛
As the saying goes, the only math you need for field theory is the gaussian integral (but with all stops pulled).
Apparently this technique is used to compute integrals related to Feynman diagrams, so the connection is even more direct lol
The "function" sigma is not a function and is not even well defined. if a_n is equal to a_m but a_n+1 does not equal a_m+1 then sigma maps the value a_n to two different values. Similarly with the inverse of sigma.
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.
Thank you Dr. Penn. For me this was a good companion video to the one by Maths 505 where he uses Ramanujan's Master Theorem to evaluate the first two Fresnel Integrals.
11:30 How do you integrate over the operator u? Spectral integration?
I see that sigma is a linear shift operator, but when making the change of variables u=x*sigma, u should be operator valued, so how can u be treated as a real number with limits 0 and infinity?
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
For the gamma integral to converge even for complex numbers, the u-substitution must approach infinity (so no -∞, +-i∞). So we extend that reasoning to the operator
Michael Penn is slowly becoming my favourite math channel
Have you seen the channel called Gamma Digamma?
Perhaps I really didn't understand the argument well, but around 8:00 it didn't make any sense for me to pass from "x^n . sigma^n" to "(x . sigma)^n". To my eyes it was totally cheating with notations: x^n denotes the nth power of x while sigma^n denotes applying sigma n times... I can't understand it works when the sigma operator is not simply "multiply by a constant" in other terms a_i is a geometric progression.
Very cool stuff! I have the feeling there are even more tricks in that area of Ramanujan.
Thank you so much for this Professor and Team Penn!!
This feels very fast and loose! And I'm a physicist!
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.
Can you explain why it doesn't cause problems to replace x^n sigma^n with (x sigma)^n? It seems like sigma and x should certainly not commute
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.
That was so cool! I normally can see where things are going in your videos a decent ways ahead of time (which just indicates that you teach well - I am far from bright), and that happened in this video too, but the final result nonetheless blew me away. It was like a thunderclap in a clear sky. I literally just exclaimed "That is so cool! That is SO cool!" in public because I was so shocked. I followed and predicted everything until the actual numerical part, and that blew my socks off.
(There are a few theoretical details which I would need addressed in a formal mathematical paper/proof, but the pedagogical value of forgoing them was sufficient justification of the choice to do so)
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.
It seems that something is quite wrong in this video. At 6:21, the integral depends on f(x), which depends on a_n for n >= 0. But the end result at 12:00 depends on a single value of a_{-t}. So if this is correct, there must be some relationship between the a_n, which has not been clarified.
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
VERY COOL. I've never heard of this concept and its pretty cool how it can be applied to solve otherwise complicated integrals.
There is one minor quibble here. The reversal of the integration and summation requires that the a x is convergent. From the alternating.series theorem we know this is the case if ax tends to zero. In the worked example the ax term tends to zero so the theorem applies. In other cases it may not hold.
The worked example is nice as the result can be computed by hand and the result checked.
the sigma function reminds me of another function i used when i was playing around with a "generating vector" (a vector in an infinite-dimensional vector space where the e_n component is the a_n term in the sequence) called the left shift. the left shift is a linear transformation that brings e_n to e_(n+1)
question: I didn't get the substitution u=x*sigma, It seems to me u becomes an operator itself, how can you integrate it from 0 to +inf? thanks
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.
I don't get how it's justified to pull out a_0 from inside sigma that is inside an integral to the outside of the integral
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
I get that this is an entertainment channel but why spending 15 minutes on "trivial" analysis and not talk about the elephant in the room?
My functional analysis background may be too weak for this but there's an unjustified abuse of notation here:
why sigma_n(a_k) becomes similiar to a (linear) operator that can be "exponentiated" and later on (substitution) even treated as a variable (apparently)?
If we just play with it, it could have also become something like n*sigma(a_k) or sigma(a_k^n) or whatever: why it's not the case?
Maybe great math has always come from these surprising abuses of notation (other than conceptual) but this is closer to magic than to rigorous reasoning IMHO and isn't very educative/informative for people curious in math.
Every mathematician should know this!!
giving us some insights into the thoughts of a true genius. thank you so much:)
That is very kind of umbral like...
Mellin transform! One of my favorite integral transforms.
Nice, the final theorem. I don't understand how you treat the limit after the change of variable of the integral at 11:30. The upper integration limit is Inf*sigma but you use just Inf.
for f(x) in the example the maclaurin series converges for -1/5
sorry -1
Love this channel, but geez this video is an absolute mess. Sigma is a function on sequences, but then we treat it like a real number in the integral? exp(-x sigma) is just suddenly fine? Then the u sub, so dx is du times sigma? That's definitely not gonna be integration over the reals. Then at the end, we only proved that a_n=(n+4)!/4! for n>=0, we could define a_n for n
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 TH-camr.
Since σ exists and is invertible, either a is periodic, or all a[n] are different. If a[n]=sin(n), what is σ?
11:32 - The Gamma Function strikes again! ^.^
The pan-and-zoom stuff is a bit disorienting, and removes context from what's being written.
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
Why is a-3 equal to (n+4)!/4! and not 0? The sum written at the start of the example starts at 0, and we could consider that a-1=a-2=a-3...=0.
No
I am an electrical engineer like Nahin and not a mathematician. But I kept an eye open for Ramanujan's MASTER THEOREM and was of the opinion that his mastery of infinite q-series raised to 24th power, with witch he solved the partition function (a method claimed by Ramanujan that was taught to him by Goddess Swaraswati [consort of Vishnu} and which string theorists employed to make sense. I never heard of the Master Theorem you just solved in your usual style (I notice 1/12 at the end of your deduction, implying a connection with his q-series). Previously I requested you to explain the q-series and I wonder if your current video is a reply to my request?
While you start with the function f(x) rather than the sequence a_n, how can you extend a_0, a_1, a_2 ... to a_-1, a_-2, a_-3? Is there any condition here?
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.
Nahin's book is really great
I liked this one! Great vid! I also liked the zoom effects. Just don't over do them!
So many questions. So, sigma is not a function? (sigma)^n is not a number. It's the function being applied n-times.
Also, in your example the n=0 term doesn't work. Like
5*6*7*... (n+4)
But when n=0,
5*6*...*4? 😢
I'm pretty sure the book is not inside interesting integrals, but irresistible integrals.
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.
That’s fucking insane. Ramanujan gd dm it
Ramanujan: gods damn not god(he was a hindu)!
Thank you!
This was really exciting!
I don't quite understand what the "theorem" is supposed to be. It just feels like a derived result. What's the application?
What happened below your left ear?
I can't figure out where the row for 1/(1+x)^5
Just some feedback; for me, the zooming is a bit jarring and distracts from the content. Not sure if anyone else feels the same. Thank you as always for the great videos.
But... am I missing something or the theorem was never clearly _stated_ (after supposedly having been derived) in the video?
Also, maybe I missed some details, but the way the theorem is "stated" doesn't make any sense: the values of the sequence a(n) do not appear for negative n in the definition of the generating function. But they do appear in the final expression a(-n)Gamma(n). This cannot be true, cause I could take a(m)=0 by definition for every m
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.
Stunning result in the end, but I find confusing why it is assumed, or seems to be to me, that sigma applied n times equals sigma to the n-th power.
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 don't see why these formal manipulations make sense. Particularly, I don't see you state any restrictions or regularity assumptions (except for a convergence criterion, which won't matter for what I'm about to say) on the Z-indexed sequence a_n. So suppose we take two such sequences a = (a_n), b = (b_n) which differ only at say n = -3. The functions f_a, f_b that you begin with depend only on the nonnegatively-indexed a_n, b_n, so f_a = f_b. According to the end result around the 12-minute mark, we should have for example that the Mellin transform of f_a = f_b evaluated at t = 3 yields Gamma(3) a_{-3} = Gamma(3) b_{-3}, which is obviously nonsense.
This is an excellent lecture. Could you comment on convergence .
sigma^k (a_n) = a_{n+k}.
WOW !!! This is a generalized residue theorem, a generalized gamma function, and a way to integrate entire classes of functions...
thats really nice!
That was a great video
Can't tell if that's a hole in his ear or an earring.
Michael Penn math presenter rules all the others!
Doc, why it doesn't work when I move the sum n relative constants outside the integral then do the direct integration on x^(n+t-1)?
Cos the intergral limit exceeded the "taylor" expansion limits?
Might want to consider audio-balancing the sponsor against the rest of the video. Can be somewhat jarring.
can anyone explain to me what would happen when t is not an integer?How would we compute a_t
This feels very OP and like a magic trick. Amazing but for sure it can't be real.
Sigma doesn't make any sense at all to me. In the definition of sigma, it has sigma^n(a_0) = a_n where sigma is the recursion and n is the number of recursions. Okay, but sigma^n is treated as an ordinary power later on and not the number of recursions. You can't have it both ways. It would actually be sigma^n(a_0) = (a_0)^n. However, the answer of the example integral still manages to be correct. It's the most ridiculous abuse of notation I've ever seen.
Appreciate.
why does u =x sigma not act on a0 in the last step?
Cool
Where can I get the notes ?
Can any one please help me to find out???
Mixing notation for inverse function (-1) with multiplication indices? Dunno how that works.😬😱 Might be helpful to use notation sigma^(-1) for inverse function and sigma^(n) for n-composition?
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) 🤔
so cool
Mellin Transform OP
Secretive derivations are at least almost isomorphic to proofs by intimidation lol
please, take as many liberties as possible, but give it a narrative 🙏
To me it almost feels like an April Fool's Joke, but a couple of months too early.
I like his notebooks; the darkest side of 20s to 30s mathematicians they stole many of his theory and formulas. It’s very bad but none speaks.
Can someone explain why the t=3 in the example?
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?
kinda severe zooms --- guess if you're gonna zoom then ZOOM
Wait, if t=-1 this seems to imply that the integral of x^-2 f(x) is f(0), which seems weird. Is it crazy crap like assigning finite values to divergent integrals? Where can I read more about this?
Gamma of negative integers diverges.
@@byronwatkins2565 right, I forgor 💀 about that.
Why is it called master?
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.
This video is doing the manipulations with the wrong philosophy. The operator "sigma" shouldn't be the "composition operator" (whatever that means), it should be a translation operator. Translation operators have an easy-to-understand meaning as differential operators. BUT, the idea of a composition operator, while based on a misinterpretation, is utterly intruiging, and certainly this is what Ramanujan had in mind, so I shouldn't complain too loudly. To define a composition iteration operator, there needs to be a function f(x,s) analytic in s, such that f(x,n) = g^(n)(x), and a further condition on imaginary s to prevent us from adding cos(\pi s) to the thing (usually such a condition is a condition on the rate of exponential growth in s, see Carleson's theorem, but since g is general, just leave the condition vague). Then the 'composition operator' becomes a translation operator in s, and your derivation makes sense. But I am sure your source was using sigma as just a simple-minded translation operator, not as this sophisticated 'composition operator'. Considering how singularities multiply under functional composition, I am not sure this composition operator even exists for any function with a branch cut, for example, sqrt(x-1) (compose this with itself n times!). The theorem itself doesn't require such sophistication.
oha
Michael why do you start every single sentence with the word "so"? Mix it up a bit, it's quite tiring to listen to
Stephanie face reveal when?
Lol nah 😊😊
hey Penn, awesome video, but I don't get at 13:02 I don't get the binomial expansion; wouldn't it be valid only for x between -1 and 1? can u still apply it? 🥸