 Yes, yes, yes Okay, so I don't know if people are coming late or just not coming Several people who usually have been coming aren't Today so today I wanted to Finish the discussion of the Maybe I will take off my jacket. It's a little too warm Finish the discussion of what I called the asymptotic trick, which is I told you is not new it's just a different way of Presenting it's apparently equivalent to something called the Richardson transform But I showed you various examples and last time I started With the examples and then showed the method and I want to give a little more details First about the method and then say that there's a more general there are more general classes So when I talked about the trick last time I started with the simplest situation and then I slowly Added variables. Well, I started by saying them intermediate one, but let me start at the very beginning So you have a sequence a n You you it's a sequence who is asymptotics You want to understand so I don't really need a mass that's very silly asymptotics You want to understand and remember that the typical situation Was that each a n can be calculated to as essentially unlimited precision Maybe it's an exact number like an integer or rational number or an algebraic number Maybe it's given by some process which you can compute 200 digits if you want maybe some combination of gamma functions Or the limit of some other process, but you know the a n's to very high precision But you know, you know a certain number are known Maybe n of them are known and then as I told you don't really know it need to know All of them from the beginning you might have a sequence of n successive values Starting at some large integer n let typically maybe you know the numbers Or you can you know all of them, but you can compute in a reasonable amount of time Let's say the numbers 1,000 up to a thousand and fifty so then you do fifty fifty one numbers But that could be much faster much cheaper than calculating all the thousand and will still work very well for the method If you try to use the first fifty numbers, of course You won't if again as good precision because obviously the asymptotics become much more exact when n is large But you don't need n to be huge so the simplest situation was that you expect or you know or whichever it is that This is given by an asymptotic series. So last time I put little a n, but now I'm putting Ck let's say Ck over n to the k. So we expect an asymptotics of the simplest possible form to simply a power series in one over n and Then we want to get C0. Okay But that was the basic situation the one I called last time the simpler problem. So let's call that one in Circle, I'm just repeating now. So I remind you that the numbers you pick an age Which last time just to be to make the mnemonic easier was was eight That's not eight factorial. It's eight. So you pick as smallest numbers. Sometimes you can take 30 Sometimes you can take eight it depends a little how much information you have and how much precision you have and how many numbers You have and then the precede the idea is very very simple you take You multiply your sequence remember that was the mnemonic Multiply a n by n to the 8 which is now n to the h sounds almost the same Different take the h difference Then the series starts with a zero. So when you multiply band to the h is a zero n to the 8 Plus the next few terms. That's a polynomial when you apply the h difference to a polynomial of degree h Then you just get a constant and that constant is h factorial times the leading term So if you divide by h factorial, but you'll get is a new asymptotic series in one over and it starts with the same n is before it starts with the same Sorry not the same It starts with the terms as k goes from one to h the new coefficient which used to be ck has now become zero Okay, and then the next term was minus one to the h which you didn't see last time because it was plus one Times the exactly the previous coefficient So we see nine last time over n to the h plus one and the next one is minus one to the h times h plus one Times c h plus two over n to the h plus two. We no longer care so much. So this is therefore a zero Plus oh with you know, assuming that the C's are of reasonable size This is still a reasonable size of constant in the o constant And so the error is now one over n to the h plus one Whereas if you just took the original series then of course this one and you just want C zero of course This is already C zero plus o of one of rent so it does converge But very slowly if you take a thousand terms then you'll be off in the third digit But now if you take a thousand terms you'll be off In the ninth digit and you see here that I don't need this uses only I Was doing the upward difference who use only a n a n plus one up to a n plus h That's what I was saying before if h is going to be eight and n is going to be a thousand so that the error So a typical values if n is a thousand and H is eight then the then you have only eight terms are needed only nine terms are needed So if it's very expensive to compute Then you only have to do nine terms from a thousand up to a thousand and eight And then you and you get the precision The error is at the order already of ten to the minus twenty seven But here you see that if you take the same thousand bit if h is seventy Which is quite often the case so you calculate a group of seventy numbers starting at a thousand then the error will be ten to the minus seventy one times three so about two hundred and ten so now you have two hundred digits and You still only need seventy one terms so you don't need all thousand So the method is in that sense very inexpensive. You don't need a huge number of values You do need them to high precision because this difference will magnify any mistake you make So if you did this calculation typically in Paris with the default precision you just switch the dongle of 38 digits You'll get complete nonsense. You have to set the precision to 500 or a thousand or 200 depending on the situation you try it blows up you increase it And then you have to remember so just if you do want to use that he you said backslash P Which means precision equal for instance 500 so that means compute Everything to 500 decibels that's all you have to do it'll do everything from then on the 500 decibels But then if you print out a number like the square root of two or or you know pie You'll see 500 digits, but you don't really want to so here's a command which is not based on mathematics like all the others It's there are many defaults and one of them is the format in which you print out a number and you have to change that default to for instance g 0.0 3050 let's say I Mean this you just have to look up the manual if you don't know by heart and this means But print display only on the screen or display only 50 digits So you compute the 500 digits, but you only show a number the first 50 so you can of course print in any format you want Okay, so that's the basic method and you see that with a very limited number of terms like eight or seven Do you increase the accuracy from ten to the minus three to ten to the minus thirty or ten to the minus two hundred? So that's very pleasing then more generally same assumption but we want also C1 C2 etc. Let's up to see something of course you can't go up to a million typically you can get 10 coefficients sometimes 50 I mean it depends again how accurate how many numbers you're using and then of course It's really clear you use one gives you a zero because that's what we just did it gives you a sort of high precision And now you take a n star to be a n you subtract sorry a zero is C zero excuse me because that's Here I made a mistake nobody stop and you're supposed to say when I say nonsense happens to be all the time It's a power series. Of course. It's here, too Of course, I don't I'm doing it on purpose to test you and you failed But actually I wasn't doing it on purpose and I keep misspeaking and miswriting these days So please tell me because even if it's obvious to use somebody else may be confused and you're doing your fellow listeners Service so let's say you want to compute you know 20 terms well, then it's clear you subtract this constant term Which is a infinity from this and you multiply by n and then method that one of course now starts with c1 plus c2 over n plus other dots so with method 1 you still get c1 and then of course we take the next one a and your Computer you don't have to give them different names. You just keep renaming a n you subtract Well, you do it again and again So you can since you already have a table so you if you had a table of 70 values starting from 1,000 to 70 Then once you subtract c0 you still have a thousand values when you multiply ban You still have a thousand so you've a new vector of the same length you do exactly the same thing 70 times if the accuracy is good enough and then you don't have to read because subtract c0 You can just take a n star and subtract c1 or of course if it makes you happy Paris won't care that takes Microseconds or something you can subtract both terms c1 over n and multiply by n squared so this one will start c2 plus c3 over n and Then of course method one again will give you c2 etc so by induction that was how we got a power series and then I'll remind you very briefly of the next three levels if n had a term n to the gamma times c0 plus c1 Over n so if it was a little more general you could of course also write that as c0 Well, if I called it minus gamma it would look better. It would be cn And the gamma plus c1 and the gamma minus 1 so it's still an expansion in powers of n But they're now shifted powers and that's extremely frequent that happens all the time in actual practice Gamma is often half of course if you guess what gamma is If you think it's likely to be a half or a third or a multiple of a third you just try different ends Different end of the gamma once you've guessed end of the gamma It's very true if you divide a n by into the gamma apply the first method And if it converges to a constant to a hundred digits then then that was the right gamma Because if if you were off by n to the one hundredth it would slowly diverge But if you don't know at all then gamma is sometimes some weird number then remember so this is the continuation of three continued You state now this time a n star as the quotient a n over a n minus 1 or a n plus 1 It doesn't matter Let's say this and that this will start End of the gamma over n minus 1 to the gamma will be 1 minus 1 over gamma to the n Times c0 plus c1 over n plus dot dot dot over c0 plus c1 over n minus 1 Plus dot dot dot so this if you work it out starts 1 minus n over gamma Sorry one minus. I'm really tired. I told you you have to watch me because I'm writing nonsense So this would be 1 minus gamma over And it's still wrong. It's to the minus gamma So it's 1 plus gamma over n and the next term would be gamma plus 1 by no mil two coefficient over 2 This one I think is c1 over c0 over n squared. So if you want it you could apply Using method 2 which gave you the cofist where you don't we know that the one is one So I don't even have to search for it. I can subtract it So if you want I could I could subtract from one subtract one right away Malt bar by n then it would start with gamma and I'd use two or I use I'd use method one So this gives me gamma It also once I know gamma gives me that but there's no point and then it implies once you have gamma if I make a N in a double star is simply the original a n times n to the minus gamma then of course this one now does start C1 over n and so on and so we use Method 2 I mean steps 1 and 2 to get all of the C C case So then we get all of them and then the next level was even simpler because if you had n to the gamma I mean the idea was even simpler. Of course, it's one more step If it's this if a n is this then again you take a n star To be a n over a n minus one So you if you had 70 terms of this you'll have 69 of that you'll have one fewer because you you know You're taking the quotient, but this will then be beta times 1 minus gamma over n plus Exactly what I wrote sorry plus gamma for n plus exactly what I wrote here, but we don't care So this now if you apply method 1 will give you beta to the very high precision and then of course once you beta You divide by beta to the end and then you're back in the previous case And then you get gamma divide by that and finally in the last case if a n has a power n factorial to the alpha and later in the course I'll talk about the two cases that are of most frequent occurrence when alpha is one or two So when it's factorial the divergent like to n factorial or well not really two That's the most common actually one or bigger than one So that's rapid factorial the divergent and super factorial the divergent super rapidly divergent But here if it's n factorial to the end times all the same stuff made at the end end to the gamma C 0 plus C 1 over n then once again You take I don't have to keep giving a name a n over a n minus 1 Will now be n factorial of n to the 1 is n to the alpha Times beta times 1 plus gamma over n and so on and so this is exactly the form that we had in 3 So now if you use method 3 or you could take you could also do it again take the next difference But that is not 3 that will give you gamma and then sorry alpha Here it was called gamma, but here it's become alpha and then of course once I've alpha I just divide the original a n by n factorial to the alpha use this to get beta divide by beta Then use the next one to get gamma divide by n to the gamma and I'm back in business So that's what I told last time I gave several Examples and so this was already a reasonably wide class of functions that covers a very very large number of sequences You find in real life, and I mentioned many examples But the problem is the problem I mean the fact is that often you encounter more tricky kinds of things. So for instance a problem at t-share Director here asked me a couple of weeks ago. There were some numbers and the numeric suggested Well, it didn't grow exponentially or factorial or even like a power It was almost bounded But there was a log term and then there was a c0 plus c1 over n and so on actually you can prove that in that case But you don't get at all the Explicit formula for c or you would if you did a lot of work, but it's quite tricky So let's say you have this well then if I make a n star just be the first difference Delta a n so this is a you know remember a n plus one minus a n well to leading order I mean here. I'll get the To leading order the first difference is like the first derivative the derivative log x is 1 over x So you'll get c over n This term of course the difference goes away. This we see 1 over n squared But actually it's not exactly n because it's c times the log of n plus 1 over n Which is log 1 plus 1 over n So the next term would be minus a half c over n squared and then here when you take the difference You'll get minus c1 Etc, but I don't care again because this then by the previous method just multiple by n So by the previous Will give me c and then I just divide a n minus c log n and and apply the method to that So that's obvious and certainly you know one things for immediately But unfortunately very very often first of all you don't know what form you have but to be honest if you don't know at all You can't do anything you have to study the little and make a guess you have to make an ansatz But it's I mean this does happen I've seen quite a few examples. I said two weeks got each as miss specific question coming from a You know physical calculation that there was a sequence of numbers. They're very slow to compute So I computed to 500 digits the first thing 200 and took even Paris which is very fast all night Because this was to very high precision and they're complicated sums of one over derivatives of virus trust p functions I mean it was a messy thing So you could calculate 200, but you couldn't calculate 20,000 of them And so you wanted something that would be accurate and this worked very well and gave c to lots and lots of digits But that is I say it's an easy case that the reason is there's only one free parameter at c and you can eliminate it by taking The first difference you get c over n, but let me give some other What about so I'll give Three examples of things that might come up For examples, maybe so a n looks like log n again, but instead of a single c It's the sum of a lot of C's you know So again a power series and this is quite common because after all the derivative of n to the minus alpha And to the minus lambda the lambda derivative up to sign is n to the minus lambda log n So the function I've mentioned that before that was the thing about g multiplicative When you have powers, it's it's very natural often to powers times log n or even some as log squared But once you've understood what to do here then It's easy enough to put in a second term But it's not so obvious what you do here because you see here if you take just If you take the difference then the first log n term will just become one of rents You're back to power series But the next log 10 n term is log n over n and that doesn't do that nice thing So if you think about it five minutes, you'll probably see the trick but I'm you know, it's supposed to be telling you the tricks. I'll tell you what you do So this is one then another one which is very common is you expect or you have maybe they're the powers in front in factorial But that you could do is before but you have n to the k over two or maybe I'll show you an example that came up in real life my real life I mean it was some the conjecture of George Andrews is a very famous and very good Governors India certain funds had made a conjecture on their growth it matched the data rather poorly and I studied it And that's when I sort of but I didn't find the right method and so I'll get that an example very sure But there it turned out in the end that it was a power series not in one over n But in one over the cube root of n so it was even worse So here it's like a power series plus one over square root of n times the second power series, but here it's like three power series so that's a worse situation and then another bad situation is There could be a term beta to the n times something nice. Sorry. You're a beta to the n plus a term, you know beta Some other term plus let's say there are three of them And you kind of expect it to be three or you've expected small numbers So you try the method with one it doesn't work you try the method two, but I haven't yet told you method in two And then you try with three left There's no point Generalizing if you can do two you can do three so but now you see if beta one Has the same has bigger absolute value than beta two So maybe one of them is a third and one of them is a quarter in absolute value They might become complex numbers then you've no problem at all if n is a thousand the this is exponentially much smaller than that So it's completely dominated So if these are all real numbers the real numbers are either positive or negative the worst you could have is a to the n plus minus one to the n Times some other a to the n and then of course you see immediately what you do You just take half of the coffees you take the evens and then you do the odds you do them separately But the problem is very often even if the a n's are real beta one and beta two might be the complex numbers of the same Out of the same value and then what you have and but you don't know what it is And so you have some absolute values to some beta to the n multiplied by some cosine thing And then it's oscillatory and so if you make a graph it runs all over the place And it's not obviously put to do and to be very honest I don't know if I don't know what the betas are at all Then I haven't found a good method It's something I'd be very happy if some of the people in the course here or virtually a thing can tell me either something for the Literature that they they saw the trick. Let's assume that we have exactly this So let's not worry about the power the shifted powers. That's now a detail Let's say you have one power series and one of rent here and another one here and beta one and beta two And somebody tells you it is like that and the problem is to find beta one and beta two well very roughly you can find them quite easily of course by By just you know taking n fairly large and then looking at the Oscillations and I've actually done that sometimes and finding the frequency of the oscillators you look at the graph and goes up and down And then you figure out what's the rough? Angle amplitude Whatever it's called phase of beta one over beta two and then you correct it by that and then it oscillates much Less and you can correct again and at the end you can find the ratio beta one and beta two fairly accurately And then you can put that in but it's a mess However, if you do know what beta one and beta two are even if there are several then you can make the method work So that that's an open question to me I don't think it's a very deep research problem But I've never seen how to do it and it's come up once or twice in my life that I had a series like that with unknown constants So before I show you the method Let me show you two examples Just to give you an idea of what we're up against Both are examples from from real life another sequence that actually you know came up in both cases things that I'd seen or that colleagues asked me about and that I want to find in the method, you know They said can you you know how to find asymptotes? Can you tell me the asymptotes of these numbers and in each case the numbers themselves are given in an explicit way They're rational numbers. You can compute them with my shoes untied. I don't want to trip You know you want to You actually want to find them. So let me tell you what the two examples were one came from Two friends of mine who are both apologists. One is the colleague with him I've been doing the work that I lectured about last year. So then if you know the name Greek starboard scarif alitas for Greek American by now he lived in America for half his life and Roland van der Reen then was also Friend I mean I know much less well and they wrote a piece I'll just put what they wanted the asymptotic expansion. It's words that even I don't know they explained to me with the words mean But just you see it's something else of the standard evaluation So I'm quoting from from the paper that this is a number. Let's call it a n Of the one skeleton. I'll just leave that out of the three-dimensional cube So this is something called from a field called quantum spin networks and you take Here you take a three-dimensional cube so an ordinary cube not a square and you take its one skeleton which looks like this and So that's a graph and that's the particular graph well So I guess I should say it of the one skeleton of and then to any such a graph you make a thing and they wanted to know It has for every end There's a number which is a rational number called the end evaluation and they they knew how to compute it And so here's the form but you'll see it's it's completely computable. It's an explicit integer Well, I can even first give you a little table and and an So it starts very harmlessly with one But it grows quite quickly the second one is already over two over six thousand the third one is already 505 million and the next one three. I'm not going to give you any more is seventy seven four one four four Oh, oh, oh Oh, oh, oh So this is a sequence of numbers that grows very quickly. They're integers I can give you the exact formula not that it makes any difference from what I'm saying but just so you see it's something and Then we're going to take the sum over all J for which the Sum end is not zero which is some finite set that you can see as soon as I finish writing it You take the binomial coefficient K over J minus 3n expression squared You multiply by the binomial coefficient 2n minus K over 4n minus J Multi that by the binomial coefficient J plus 1 over 2n plus K plus 1 You take the alternating sum that over all J So J has to be less than bigger than 3 and less than 4n and J plus 1 has to be bigger than this So you and also here they're you know further inequality So it's at most between 3n and 4n but only some of them contribute and then you take that whole expression to the fourth power So to compute it to compute the nth term is only n squared terms Because here if you multiply this out would be a five-fold sum But of course for each K you just compute the sum It's called O of K term O of n terms as you see n terms and then there are O of n terms here So altogether you have to compute, you know n squared steps So if you go up to a thousand it's a million steps Which the computer can still handle for a five-fold set you maybe couldn't go that five-fold sum an honest one So you can compute a lot of these numbers, but you have to use the method and here I'll just show you the bottom line as I say to show you what you're up against and also to bring out a point that I Won't come back to at the end Which I've already mentioned also in the first day Which is if you do find these numbers numerically you want to know them exactly and how can you recognize them? So here the unsets are not the unsets the results on numerically whether it's finally got proved I don't know there's a constant which is a real number and Then there's a big exponential, but this one we know from the theory. It's three to the 12th So it's three to the 12 n and the power we also know of course you could use the method to find but but here And you just try some integers and it works with n to the fourth, but then There's a power series Just like what I had before times I'll just call it m1 of n the m1 of 1 over n so m1 of x is a power series actually with rational coefficients and Then there's a second power series. So this is just like what I said before There are two power series Both with rational coefficients Except that when you do it numerically you find that this is rational coefficients But this one doesn't have rational coefficients the second one But if you look at the actual coefficients You find that the difference of two of them or a linear common a very sort of linear combination The beginning is a fixed is a number and so what that means is you get a constant and the constant is independent of log And that's very common because you should think of log n in all of mathematics when you see a log It's always up to a constant because log is the thing whose derivative is one of Rex So sometimes you do get log in mathematics, but sometimes you get log n plus gamma for instance or you'll get something else So when you see a log you should always imagine remember to eliminate it before I took the first difference and killed the log But that means that there's an implicit constant which however also has to be a real number But it's not a rational number So here one of the series the log n series is rational But the second series is rational plus multiple to c but that part is the same power series So once you found this you know, okay So this would be completely in the limit of what I told here the case that I put here to power series Because since I know all the three to the 12 and in the end of the fourth of course got removed that and the s zero I could even put into these power series That's what I did before some ck over n to the k But I don't want to because as I mentioned very frequently the power series are a fixed Real number complex number times the power series with rational coefficient They're much easier to recognize much easier to write down. So here indeed. I want to find s zero So if it were just that it would be easy, but there's there are more terms exactly like what I said before there are three Exponential terms so the remaining thing it looked like only one, but it's actually two of you think about it There's another constant, but this time it's a complex constant and then there's another exponential and another power The power is now not the fourth power, but the fourth Again, I don't expect you to be interested in this example. I haven't even told you what this definition is That's the formula, but who cares? But the point is in real life like you have very good topologists studying quantum invariance and suddenly they did this sequence of numbers And they want the asymptotic. It's given by this formula It's also there's a recursion and then you won't find the asymptotic and it's extremely complicated But what makes it hard here is that there's a new Exponential term and it's now like one plus the squared of minus two to the 24th But one plus squared of minus two absolute value squared is of course one plus two which is three So the absolute value of one plus squared of minus two is the squared of three you take the 12 24 and it's the same three to the 12th n is this and because it's a real number and this is a complex number You have two oscillatory things so we actually hear of three terms We have well it if I replace n by n over 12 we have three One plus the square root of two or maybe square root of three I can say to the 24th Then it's right one plus squared of minus two and one minus squared of minus two These are the three numbers, but they all have the same absolute value squared of three So they all contribute and it's heavily oscillatory. So you don't see anything when you make a table It's a huge mess and and all of these terms are the same size and then of course there's still that's just the leading term There's of course still a power series and as you could immediately guess the m3 of x no longer has rational coefficients, but It's a power series with coefficients, you know in q plus q is squared of minus two So altogether if you have n terms you have two complex constants and a real ones That's four complex cons of four real numbers to find and then if you want the first n terms of the series you've one two three four times n rational numbers so there's a lot of finding and This was partly to here We knew some information with a recursion as we saw that the possible exponents like a differential equation You can see the possible growth has to be from the singularity the only possibilities were three to the 12th And one plus squared of minus two to the 24 and the one minus So that's what I said if I've even three betas as I do here But I know what they are then I know what to do and so here one could find it So just for fun and to emphasize the other point about recognizing here was the answer So explicitly s zero That's a very easy number to recognize because you could easily imagine You know anything in mathematics contains the primes two and three practically and it tends to contain pi and so is s It's a multiplicative thing. It's not an additive thing. It's a multiplier and therefore you shouldn't if you want to recognize it You want to write it multiplicatively you don't want to write is 2 plus pi you want to write is 2 over pi So for that you use you take the log of this number once you have too many digits You take the logs of two three pi you could try if it doesn't work You put in some more things you throw in and then you use the standard thing called linear dependence, which Paris has So here you would just take you take the vector log once you've computed log s zero in Paris You would take log. I don't even need that you would take the vector s zero two Three and pi which in Paris is a capital pi because they like to use Capitals for the there are only three constants in pi everything else you have to everything else you have to compute i pi and the orders constant So you take this vector of four numbers you take its log And then you apply linear dependence to that to some precision and then it'll tell you immediately that it's you know The output will be exactly one four minus five and six I think so and then that tells you that that combination of the logs is zero That's the combination that is small and so you find as they are you can recognize it very easily S1 is already a little uglier but Since one plus squared of minus two up to three It's I mean it's just this prime over three you might well imagine that it plays a role And so you might try in your unknown things with the logs the log of one plus squared of minus two But you also need a one plus i that you might not think of However, since it's multiplicative, and I'm doing the logs you could take s one squared And then this would become an i and then s one to the fourth this would become minus one and s one to the Eighth would would have just no sign so you know you you can easily Try different things anyway. It's not terribly hard to recognize this number, but it's already a little trickier So I'm also making the point that when you do this kind of thing The answer comes out as a real or in this case a complex number to 50 digits You have to make a guess what kind of a number it is then using this kind of trick You can easily find out if your guests right find the numbers and once you found them Then you type this number in a parry and you find that it agrees to your number to 50 digits Then you know it's correct because the chance of that being true at random is very small see it's already trickier But I told you logs tend to have a gamma so you so when you're doing this now we do it additively see s One and as there were multiplicative concepts We have to look at their logs, but C is an additive constant, but it's a logarithmic additive constant So you don't use E. You might use one. That's the log of E. It sometimes comes up. So in practice I would probably take linear dependence of C I might try one so you know there might be a rational number as part of it Why not then I would try again maybe two and three are Plausible numbers and then I would try orders constant, which is called Euler This is as I said to parry as uses capital names For standard cons, but there are only three in Paris pie I and Euler you might say what happened to E Isn't that the most important constant mathematics? So here's something I learned from parry Sorry from parry, but from Archie Cohen, but I've heard it said by other people E people if you ask any random mathematician name the two most important constant mathematics Everybody will say E and pie or pie and E. You never get any other answer But the correct answer is named most The 20 most important constant mathematics E should not be on the list He is not at all in important constant In fact, I only know one or two results in all of mathematics where the number E occurs So Zeta 3 occurs way fast even Zeta of 11 occurs more often than E And gamma even gamma squared occurs. I mean, I'm not even talking about algebraic numbers And so you might think but everybody knows for instance, I bet you could say that you know E by heart And indeed I do know for my childhood a few digits think it's four five nine zero four five But I won't write them down in case I'm wrong But that's because I learned it as a childhood. I didn't know that it's not interesting It's just like remember when I told you that benouin numbers are benouin numbers are usually defined in every book by generating function then the benouin polynomials are defined in terms of them and then you specialize Sorry, that's the benouin polynomial, but that's actually wrong Benouin polynomials are simply you define them in one fell swoop and then their constant term is bn And it's the same here if you know E you've e to the x the e to the x is important You'll never hear me say that that's not an important function And the proof is as you all know that of course L all of this is due to Euler It both is about the simplest to well factorial generating function You can imagine and it's the limit of 1 plus x over n to the n So it's an incredibly important function But there's no particular reason to care about its value at x equals 1 The only reason we've been learning this school at all is because as a function It's called the exponential function So let's not call it e to the x so that if I if Piree did at that constant You would always have to write e to the power x But in fact, there's a function in Paris called x with x and that's the good function of its equivalent But you don't want to compute it as e to the x you want to compute it depending whether x is large response of other way It's a function and of course it has the property that x of x plus y I'm not going to write it out because we all know is the is the product of the two and in particular X of n is the nth power of e so x with n is e to the n and by interpolation indeed x with x is the Xth power of the value of that function at 1 but the value of that function at 1 doesn't play a role anywhere So any special function has interesting values for instance the gamma function has interesting values of positive interest at the n factorial negative interest about their infinite and at a half because that's the square root of 2 pi which plays a role in you know physics in many places and in mathematics and then of course at 3 halves 5 halves minus the half by the function equation It is somewhat interesting values at a third two thirds and so on but already a little less But we'll see it in a bit But it's not reasonable to ask what is the value of gamma of e I mean nobody cares what gamma fees There's just a random number and the same way if you take the sine function sine of 2 pi over 5 is a wonderful number It's algebraic but sine of 1 is a completely idiotic function It doesn't doesn't tell anybody anything that did the imaginary part of e to the I and it's not the natural and in the same way X 1 is not a natural parameter at all So the number e but everybody because we learned it in high school We know the number e and because it's faster if you're riding by hand I mean everybody does it's faster to write e to the X then X with X It's you know several so it's less chalk and less pen But it's not really the number e to the X power It's a function called X with X. Well, I could say the most important definition Of course as we all know is it's the solution of the differential equation y prime equals y There's nothing to do with taking a specific real number to the X power. Okay, that was the digression But I warned you several times before and if I hadn't warned you you would have noticed Now let me finish the sentence and then you can ask because it's not even a sentence It's a formula and I've been digressing a lot So it is a combination in fact of log 2 log 3 and gamma and before I answer the question Let me give you the beginning I have four coefficients here. I'll only give you the first two M2 I'll only give the here. I get the first three here I'll only give the first well It's also three because it starts with zero and then 689 over 864 X squared and M3. I'll just give 2 Again, it starts 1 minus 2080 minus 43 times the squared of minus 2 over 11 52 times x plus dot dot dots you see it's really tricky numbers and All of this you get by pure interpolation as it says it's easy with the small variant that I'll show you in five minutes Which takes two minutes to explain and then you understand it's and it's even not very hard to find So once you know but these here in this case there were three numbers once you know what they are You can apply the method but this is to show you that in in real life these were numbers They came up you can sometimes expect to get extremely complicated Asymptotics this asymptotics I wrote it here You know has all of these constants several two complex of one complex to real constants two power series with rational coefficients One power series with q squared of minus two coefficients and you can find all of that and you can easily get like ten terms of this Coefficient so now questions Yeah, of course. Yeah, I mean I would put on the first line equals x for one I mean naturally it's capable of computing that function at one But if you I mean just for fun, I can ask you you're certainly the most serious analyst here But anybody does does any of you know place in mathematics where he occurs? Because everybody laughs when you say that it sounds very provocative perverse, but it's simply true They you I mean I actually do know one because I'm a number theory students I do see numbers there is one thing called the Bell numbers You know the Bell numbers actually right now. I forgot their definition. I Know it very very well Yeah, here if I take the sum One over n factorial times. Let's say n to the fifth Then this will be what I don't remember what it is Of course, it converges and this will be 17 may be wrong because I didn't do it on my computer If you put any integer power over n factorial, then you always get an integer times e But that doesn't mean that he is a natural number. It just means this is in the literature because naturally if I put here You know a third to the a third to the n Then of course I'll get a rational number times e to the one-third I mean it's still x the n over n factorial the only reason it it's ease because I put one but they happen to be famous numbers Called the Bell numbers. So here if I put n to the k Then that's it strictly you can see in your head that it's true You'll get an integer plus div integer times e so there e comes up He's no longer asking any question You can give it to me and I can give it or you can give it You can be the person gives it whoever needs it or maybe you can anyway, it doesn't matter If people need the microphone, so sorry, it's true when there's a sorry. You're really just If somebody you sort of reminded me, please when somebody asked the question here You're supposed to remind me that I repeat the question then out of the mic Those so the question was in but since I answered it I think people could guess if you did leave e and party what you do you would write e equals x But then you pretty semi-colon so you don't have to look at it and then you would have e But it will never happen. I've been using Paris for 35 years or something I don't think I've ever had to write exp of one for any reason at all So that was one of the things I learned from Paris that it's a completely irrelevant Constant okay, so that was a non-trivial example, and I wanted to give another one But maybe I'll first give the method so that if I run out of time I think the others takes a while to tell and if I don't finish I actually prepared a whole second topic And I was desperate and very nervous my because I hadn't finished preparing but nice I won't get to it anyway because of all the digressions, but That's meant to be Relaxing and you're not supposed to mind if you do mind. I'm sorry so I'll give you another example That as I mentioned already very very soon as soon as I finish explaining the method what you do So let me say what you do for instance in this case Okay So what do you do? Well, you apply the method against remember with the method was you choose age equals eight or some other reasonable number you multiply by A n you take the h difference and then you divide by eight factorial remember why we did it because A n we were assuming then was a power series and one or friend it starts C 0 plus C 1 or friend when you multiply by n to the 8 It becomes a polynomial of degree exactly 8 Plus the next term is o of 1 over n when you take the 8th difference You get a constant plus o of 1 over n to the 9th and you can see the constant so the trick is really pretty simple You take the h plus first difference who differentiates once more well Then the power series part will go away because the power series when I multiply by n to the 8th This term will become under multiplying by n to the 8th C 0 n to the 8th plus dot dot dot plus C 8 Plus C 9 over n and then when I take the 9th difference and divide by 8 factorial or 9 factorial It doesn't matter because they're both 0. I'll get 0 Well, I'll just get 0 and then I'll get C 9 It's the 9th difference so I actually will I could even divide by 9 factorial Maybe I should divide by 9 factor. It doesn't really matter very much You'll get roughly C 9 over n to the 10th and so you've but remember my n is a thousand so thousand Zero to 30 digits. However, I've now killed this term But I haven't killed this term But what happens when you apply the 8th difference to this remember that the 8th difference is very nearly the same as the 8th derivative What happens if you differentiate a product of two things when you apply Leibniz? You either differentiate this or you differentiate this the next time also So when you do it eight times again like this you look at some terms who differentiate this some terms and this 8 minus that number Well, if you haven't differentiated this at all you've differentiated this eight times just as we did here It will be over one over n to the nine and when you applied the it's the same as here So we're treating it log n as a constant it will go away So the log is simply gone and so what will happen is when you differentiate you have to differentiate log n once Now it's one over n and so this will start after one differentiation of one difference But it's the same you'll have roughly C zero over n Squared oh sorry over n because it was C zero times one and then when we do another eight Which is what I did I live eight factorial C zero over n to the nine and so when I multiply the whole thing So this one so here I'm multiplying by n to the eighth and Then it's whatever it is and then I take eighth different the ninth difference again over eight factorial That means the first difference and then another eight difference and then you divide by eight factorial what you'll get is Again o of one over n to the ninth or eight or tenors anyway something except that's only if you did differentiate the log n But if you didn't differentiate the log n Then sorry this time. I of course don't do it having done that. Sorry. I did this stupidly the log n will still No, no, it's okay. Since I've differentiated nine times I'll now have a new series and it will start with C zero over over n and So now you can apply the old method and you'll get C zero and then you subtract C zero and you get the next one and so on So in other words that that's the trick. So we've split it into two parts and Here you can essentially do the same what you say is okay this let's say in the simplest case This is the sum C k over n to the k and this is the sum, you know some D DK over n to the k So, but you know beta one and beta two well, then you do exactly the same thing. You simply divide By the known beta 2 to the n a n and this would now be some new concepts Which is beta one over beta 2 to the n times of power series m of 1 over n plus another power series I'll use the same notation as before But now you do the same trick this power series you multiply by n to the 8th But you differentiate nine different take the difference nine times that will kill it completely It won't kill this but it'll turn into some it'll still be a multiple beta one or for beta 2 to the n times of power series Now you take that new sequence Divided by beta 1 over beta 2 to the n it's now power series You read off the leading term and then you continuous you get one term at a time So it's it's more of a nuisance, but you see that the ideas is easy and the same here because let's say we have C Zero of this then what I'll have is a power series In one over n called M zero plus another power series times one over the square root of n So now I can choose which one I kill I either take the n to the 8 8 n and take the n plus first difference That's like what I did before that will kill this one But this one will survive and have a square root of n at the end I have to multiply by power of square root and put it back and unravel it So this will give me something that will give a new sequence and then you apply the method again And that will give you essentially the leading coefficient C zero Sorry D zero because this one kills this so I'll be left with in the notation. I just wrote I don't know what I called it. Didn't I just write it somewhere. Oh here It will give me you essentially M one of zero It'll give me the constant term of this series or I could do it maybe even more intelligently. I could take 8 plus a half An and then again Apply the h plus first difference So if I multiply a n not by n to the 8 but n to the 8 plus a half Then this will start the polynomial degree 8 if I take the ninth difference I've killed it up to very small terms But this one won't go away and so I'll get a new series which have a square root of n I divide that by squared of n and find the leading coefficient and then I have You know the constant with M zero and of course exactly the same if it's n to the two-thirds But now it's a three-dimensional so it would now be M zero. It's a power series Plus one over n to the one-third times m one of one over n Plus one over n to the two-thirds times the power series So you don't think that it's a power series in one over the cupid of n But it's three power series in one over n. Well, if you remember that differencing is like differentiation This would be like splitting a multi into three pieces This is actually the beginning of the theory of I'm an example of the theory of d modules You have a differential module you can differentiate power series But you've split into something finite I mentioned you look at the different pieces and how they interact So I'm not going to say if you want I can write out the details what you do here Well, actually I did do it in this case. So I can say it so let's say that I have a sequence and I think that it's I'll go back to the C's Notation C zero one-third plus C two over n to the two-thirds So justice in this example square root. I could multiply by n to the eighth take the ninth difference and kill this and That's what I'm going to do and then I'll kill that and then I'll get C two first It's kind of silly you'd like to get C zero first it would be more intelligent multiply by n to the two-thirds Plus eight take the ninth difference kill that then kill this and then you left with that So as it happens just this morning. I quickly when I did this problem So I'll show you the problem actually had and then From Andrews work So I have some numbers Yes, please now you have the mic so so in this case is for example usually of course I mean if you had the if you could compute to every single a n to high precision Even if n is large you could just consider a new sequence a n star by being a subindex n square You know naturally that's that's that's that's just what I'm about to say this idea would work. I mean practice Or it's extremely poor. I mean that's what I did for years So it's well, I was gonna say so it's not stupid because I did it is stupid But I did it for years. It's the obvious thing to do But I'm just about to give you an example So let me finish the example and then or rather give you the example haven't started it So this was a sequence of numbers found by Andrews as I say Absolutely leading Covenators and Q series person and let me actually tell you the story because they're very interesting So in the end I made it he had made a conjecture Which turned out to be wrong well, he stated a little vaguely so he was vague enough that it wasn't wrong He stated we said because there were some two terms He said because we expect the first term to dominate we expect an asymptotic to roughly the form But if you assume that the first term dominates you get an exact asymptotic with a constant that constant turns out to be wrong And the true answer it's two-thirds of the constantly had the rest is asymptotic for correct So and you know he has a very good intuition, but it's because of this tricky Asymptotics, so I'm going to make a power series G3 of Q. So this the Andrews Q series Well, he had the Gk for every K But G2 he could write based in terms of a Multireform and times a mock multireform those are the things we understand so what you get exact asymptotics But he wanted friends G3 you could also look at G4, but it gets more complicated So it's a power series which starts in it's just some power series with integer coefficient You can easily compute a thousand coefficients And the question just like in what we were doing before let's say you want to apply the circle method Then you would want to know the asymptotic said this is Q tends to one So I'll give you the definition. They're actually two one of them Let me write Qn as One minus Q up to one minus Q to the end I think I've used that notation last time and then you know Q infinity is of course the the infinite product one minus Q One minus Q2 squared to infinity. So it's essentially the dedicated a to function So here it's a double sum which means a bit of a nuisance to compute. It's the arth Q factorial, but with argument Q cubed and then the S I did our first and then S for some reason. So this one is R So it's the arth and the S and the numerator. So it's a little like what's called a num sum It's an alternating sum which makes that's why you know The growth is hard to get up because if the terms were all positive You could just use a stationary phase and get it immediately So the numerator you have a Q to a quadratic function of r and s So it's a perfectly explicit thing But that's a little slow if you want a thousand coefficients You have to go up to roughly a thousand well the thousand over three and thousand over four So you have roughly for each term n squared and you're going up to n so it's n cubed It's a thousand cubed. So it turns out that there's let me call this p of Q Is you know the infinite Pochamer symbol so this product which is very easy to compute Then it's p of Q to the fourth and he proved this this is not at all the novice identity So there are two terms and the first one is very easy It's just a product of p's so a product of eight of hunks with arguments four six three and 12 except I miss wrote no, I didn't miss right It's in fact correct Okay, and then you have a second term which again has several of these infinite products So here it's p of Q times p of Q to the 12th over p of Q squared Q to the 12th and now it's again a sum But now it's just odd numbers and it's minus four over n if you're not a number theorist You might not know it's zero of n is even and otherwise it's plus or minus one alternating and then it's again Like before a product of Pochamer symbols. Oh, I didn't tell you the more general Pochamer symbol So the more general Pochamer symbol X Q n Well here I actually don't need X Q and I only need X Q infinity is the product one minus X times one minus Q X So it's the shift of the thing Okay, so that's called X Q and it doesn't matter at all I'm just writing the formula for completeness and said you can see a very non-trivial identity approved by one of the Absolute masters so Q cubed n and Q to the 3n Q to the 4th infinity and in the numeric you've Q to the 3n So this one convert as much less well because if you want to get up to a thousand you have to take a thousand terms here here This actually wasn't true But I said that this inefficient because this is quadratic if I want to go to a thousand arenas are only the order of the Squared of a thousand and I'm multiplying so it's actually not so bad name. We've both formed this work But then the question is what is the asymptotics as Q approaches one? And so you write, you know Q as always is e to the minus t as we've been doing it's really Supposed to be e to the 2 pi i tau, but you might as well do it on the imaginary axis So it'll be simpler and so he made the conjecture This was Andrew's question mark that it's asymptotically equal to a constant over the squared of s I'll put the constant a second times e to the minus pi squared over 36 s So it blows sorry if he called it s and I'm calling it t so it's t so it blows up exponentially in well actually goes to zero exponentially in One over t as t goes to zero But the constant and he said if you he said, you know some numerical experiments But he's not a great computer also this was quite a few years ago computers were slower He said the experiments suggest that when Q is very close to one that this term is more important than that term This term you can find the astronauts equally easily using modularity and what you find from that is pi over 2 t So he just said that he just made the conjecture c over the squared of t times that and that is actually correct But his argument that he gave in the text is it should be like that because this term should be negligible compared to the first But actually it turns out that the correct answer So this is without a question mark. It is exactly of the right of the form. He said but there's a two-thirds and so the reason that it's Hard is the asymptotics when when you find it I'll just give give the answer the bottom line So gee it well when I wrote this this it's not published. It was notes for myself but I shared it with various friends and then It got further shared and there was a paper But I was very happy because I wasn't working on it the final paper. It's I've forgotten the third author So it's dream on the rolling Gryffin I don't membranes. Oh joy, maybe I'm sorry I didn't I have it on my computer, but I didn't print out the page of that But the conjecture that I made which is now no longer a conjecture But I found everything numerically how they're following form. It is indeed pi over squared of 2 t times e to the minus pi square over 36 t But now it's a power series in It's a power series in so we call it m0 of some power series of one over the cube root of T Now you might say this is different. It's not a sequence, but all I have to do is specialize to t is sorry This is a power series in t. T is very small. It's a power series in cube root of t But if I just compute this Numerically, which you can do quite quickly in various ways Then you you can compute if t is one over n and then you get a sequence and you apply the previous method But the actual form there were three power series and so to do it I didn't have any idea did exactly what what a Manuel wrote. I just replaced n by n cubed So let's so if I take g3 and I evaluate t should be very small So I take I took one over n cubed But this thing although you can compute it quite well, but still if q is very very near one It's very small. So I only computed using 50 terms so I used This where n goes from 1 to 50 And then I use the interpolation method indeed you get all the coefficients and then you have to recognize them separately But once you've seen it's in the cube root of n Then you can expect that the coefficient of one over the k is some constant times a rational number if k is zero 1 mod 3 a different real number times rational if it's 1 mod 3 and a third and Indeed I found that I'll give you the numbers in a second. So what I actually found so indeed So what I'm saying is you would expect a constant times a power series with coefficients in T cubed I mean every constant is of course every power series and t well It's now it's going to be in one over n because I've cubed it So now it'll be one over n cubed and then and so on and let me not write it out There are three constants now when you looked at them the leading term turned out really easy to get and it was two-thirds as I told you but in fact if you looked at the next coefficient that was the the cube the cube of the cube root of n it was 124th and the next was one over 24 squared times a half and the next was one over 24 cubed times the sixth So it was simply the exponential series So that when you could simply recognize after a couple of terms and I'm not going to bother to give all the first coefficients Because we all know the exponential function, but the others Turned out to be Well, as I said, there's there are constants and then there's a power series Let's call it m1 of t and then another constant t to the two-thirds times another constant in t so this is a power series in in Oh, sorry This is a power series in t because this is a part in t to the one-third But you have to do it separately and then what I found with my computer using exactly what Emmanuel said So it was only 50 terms, you know, and 50 terms you don't expect to get a lot It's only 50 terms because I'm computing up to 50 cubed. It's healthishly expensive And so what I needed to make the method work at all actually got quite a bit of the power series I'll put m1 m2 is similar and So it's t one is I've told you often these are rational numbers And so once you have them too many decibels you easily recognize you'd expect small denominal So you multiply by powers of 2 and The next one is 203 over 2 to the 9th times I can't even read it 3 cubed times t squared plus and the coefficient of t to the 6th is already You know a very big number in a big denominator, but just powers of in fact, it's only of 2 and 3 So they're actually this is actually a power series with integer coefficients up to a rescaling So and I found several and similarly m2 is whatever it is. It starts with 5 okay, and The constants and this I want to say because this brings me back to the thing about recognizing numbers the constants were Well, if you can see from where you're sitting You won't be able to read it But you can see it on the page it goes from the left of the page to the right of the page and c2 is some other number 0.032352 and both of them I had with this to I think 50 digits Okay, all of which turned out to be correct when I recognized now you have to recognize those numbers Then then you have a conjecture and so you recognize and I just wrote in my note We recognized in me actually the first one is three to the minus two-thirds over two gamma of two-thirds and the second one is Three to the minus a third over eight gamma one-third and again if you just look at a random real number You can't see that but here. We're not doing random We know it has to do with three there's a period of three there are threes here It's there's a three even in the name of the function This is the third one we sort of expect something with three and if there are pi powers of pi like squared of pi Is gamma of a half but here we should expect gamma so actually what I tried when I did it is linear dependence with gamma of a half which is squared of pi gamma of a third gamma of Sixth you don't need the others because of in duplication where you maybe don't even need both of those And then they both turned out to be an instantly Recognizable and later as I said this has been proved But just to finish the numerical part in a manuals question I used only 50 terms to get all of this precision Which is really surprised when you think that I'm now taking 50 terms to get asymptotic series, but I Needed to make it work. I Needed 20,000 Digits of decimal digits of precision and that's really painful. And so this took me about took the computer about four hours Now with the method that I explained here, which that is your question Is it worth doing that couldn't you just do that if you have enough? Yes, but to have enough you take 50 it's a very lousy small number and then you were 50 cubed It's a lot of work and so it took four hours this morning. I checked my thing. Unfortunately I didn't print out the now I did put my computer at home But the cloud can actually the institute didn't work so I couldn't I forgot to send myself by email So I can't give you the numbers, but I know that the computing time now I got 300 digits instead of 50 and that the computing time now is five seconds Instead of four hours, so in other words, it is worth knowing these these tricks So I've told you the trick wait a second. I started four. I'm supposed to end at 530 So I think that's all I wanted to say about the I wanted to give the two very non-trivial examples This is spin network of the One skeleton of a cube that you know that these wildly complicated numbers where there were like seven different terms in the asymptotics involving exponentials of Algebra numbers, but luckily known algebraic numbers and once again I do not know how to do this if I've two or three or several terms And I don't know what they are and then this one where when I did it at the time it took me several hours and Also with a lot of trouble. I got six coefficients of these now Of course, I could get 50 coefficients of that in in the same few seconds So it is worth, you know having these Small things and using this D multiple idea. So the D multiple idea once again Is you imagine that your thing is a sum of several terms and then each one you stare at that one has the form some fixed function Times the power series. So if you divide by that function, it's a power series So if you multiply by n to the eighth and take the ninth difference, it's gone So therefore you take your whole expression multiply by that free So now it's what we expect the sum j from one to three of some unknown function Sorry known function like for instance n to the you know j over three there's something like that and then so this is known Like n to the one over three or beta to the n But we have to know the beta and then some mj of one over n which is an unknown power series Just like before so if that's what a and is and of course I put three but it could be any other number then what you do is you divide a n by f3 of n multiplied by n to the h and Take the h plus first difference and again divide by h factorial. Actually. It's not quite I'll show you that I do want to say one thing about that and then you get something you where you've eliminated the first term But now the next one will start f To divided by f1 times an unknown m1 you've taken the difference It becomes a mess if it's not easy functions, but you multiply back and you you played the game But once you've eliminated once you can imagine that you can do it a little again and eliminate twice and get the last one So that's all you do so the basic idea is break up your your ansatz your guess in to find that the many power series of one over n times known functions and then you're kind of cooking with gas and Even if they're slightly unknown functions like we haven't quite known all the exponents you could you can do the same tricks I did last time at the beginning this time Let me just say one thing because this today. Unfortunately. I don't well I do have the numbers because I worked them out again in my office at the one on the computer program So let's actually do this in this case. So I have this a and I'd written it before It's the sum ck over n to the k over 3 So I'll start now. I do want to look at my notes. I get the Things right. I don't know in fact where I put my notes here So I have my sequence which with the ansatz it's a power series and one over n like here And I want let's say c2 actually It's a good example because I didn't want Caesar. I knew what it was Of course in that case I could have subtracted it off and just done one of the others But let's say it's this then what I'm going to do is Is Let me I just defined the q factor factorial symbol, but there's another pochamber symbol Unfortunately the same notation xn when you're not doing q series This is just what's called the rising factorial x times x plus one up to x plus n minus one So it's it's called the shifted factorial or the ascending pochamber symbol So that's an easy thing. So now what will I do? I'll take a n and I'm first going to eliminate So again, this is the sum c3k Over well c3j over n to the j Plus n to the minus a third times the sum c3j plus one over n to the j Plus n to the minus two thirds times the sum c3j plus two Over n to the j So again, we have three power series and I first eliminated the first one by the same trick that I've told you so you Except don't if your cop take notes don't write that because it's a poor idea You multiply your series by n to the h And so then you differentiate h plus one times because that will kill this thing kills The three the c3j terms, of course not all of them only up to h But then you're left with an arrow, which is you know, n to the minus nine and this can be big and we can ignore it So we do that but I shouldn't divide by h to the n Why did that divide before because when I had this c0, I multiplied by n to the eighth I got c0 n to the eighth. Let's just call it c and then when I took the eighth derivative Eighth difference or derivative. It doesn't matter. It's the same to leading order I got eight factorial times c so I should divide by h But now I don't want to divide by h Because I don't care about this term whatever I double divide by whether it's h factorial or h plus one factorial It's going to kill that term But now of the next term the c1 which I now want to tell so the next term Is going to have a one over n to the one third So But actually I don't care about that. I'm going to kill it The one I care about is going to be the the leading term because I'm only going to get one coefficient c2 Then I'll do the same to get c1 c0 Probably the output and then I do it again So it's always one at a time one step at a time like I'll call it synonymous So what I imagine that my thing is actually c n to the minus two thirds So when I multiply by n to the h it becomes c times n to the h minus two thirds When I take the h plus first difference Then I get the h plus first derivatives. It'll be h minus two thirds times h minus five thirds All the way down to h minus two thirds minus h plus one which is therefore minus a third And then of course the power has gone down by h plus one So it's also now become minus five thirds. I guess if I made a mistake if I made a mistake I made a mistake. Let's not worry too much So I'm all planned by n to the h so this number you see is the Rising pochhammer symbol starting at minus a third and of length h So I should not divide by h by h factorial. I should divide by the shifted Well, it's h plus one factorial because I'm taking the h plus first difference It's one third h plus one or of course I could put two thirds index h That's the same up to a factor of minus three So now if I do this that still kills this one But now I have to look what happens to the next term n to the minus a third Well here I've n to the h minus a third and I differentiate minus h minus one So that's n to the minus four thirds here And so I call this one a n star as we've always done And now the next stage is I'll take a n double star And now I'm going to multiply since I expected to start with h to the fourth terms I'm going to multiply by h plus four thirds Times a n star and that should start like n to the h So now I again take the h plus first difference, but I divide again And if I remember correctly, it's simply minus two thirds this time times h plus one And that will now start c n I think was minus a third And now I'm done. I just multiply by n to the third and apply the old method So I do it in three successive things and that worked. I did it as I said this morning It took half an hour to program because I kept making mistakes with the Prefactor until I did it correctly But indeed it gave the number immediately to a huge number of places and all of them, of course, were correct So that's the asymptotic trick I wanted to start a new topic And it's kind of fun, but on the other hand Then I would talk now for 10 minutes and then talk next time So I suggest that I stop and I take questions and you try to make as many questions You have also people in the audience and otherwise we just go home So please a quick question. So when you apply this method, you're going to have to apply it two times, right? I mean, no, maybe even three times with the age. So three times the original one So three times, but if it's a d It seems to me that you want to let me answer If it's a d-dimensional mode to remember the general so here it was three if it's If it's dimension d you have to apply d times indeed Yeah, so please it seems to me that the first time that you apply you want to get a somewhat larger age and then a somewhat smaller age And then at the end you want to get absolutely not no because the turn that you're left with No, no, I understand what you're saying. I thought that too, but it's not true It's not true because you see when you do this Remember that's a very good question and it's an easy miss deception I thought that too at the time I started with 2h and 3h. It's it's simply not right Remember the original method if I'd see one end Then I even wrote it out painfully and everybody wondered why I'm bothering to to do it, but just to make it Visual what's happening when you multiply by n to the eighth Then you get c zero n to the eighth plus dot dot dot plus c eight plus c nine over n Now when you take the eighth derivative, which is what we did in the original thing and divide by one over eight factorial You get c zero you've killed everything else, but you simply get c nine over n to the ninth There's no increase or loss of accuracy So the process I did and if this had been h plus one that maybe this would have been n to the 10th or eight n to the eighth So if I'm taking eight that error is still n to the eight and it doesn't bother me at all because this method So long as the number of times you differ differentiate or difference and divide by h factorial and the n to the h Are roughly the same within a bounded distance. Then you don't change the power At all the n the n to the ninth became n to the ninth You change the constants on us by zero and you're happy and otherwise remember the next one here was was nine C 10 it's not such a big deal and here it might be a third and this might be a third times five thirds But there are the order of one So you don't change the so each time you do it the remaining terms are no bigger than they were just all the immediate terms have been killed And so now the new term still has Is exactly the format had before but a bunch of of the c three i are now zero And now you do it again in the next and so it's it's actually very surprising I also assumed you would need of course morally you have tripled h Because what is true when I take the difference of sequence of 50 numbers the eighth difference I'm not with the sequence of 42 numbers If I take the first difference a a 1 a 2 minus a 1 a 50 minus a 49 only 49 numbers left So indeed if I had 50 numbers from a thousand to a thousand 50 Then before I could take up to the 49th difference, but now I couldn't If I take the eighth difference after one step only of 42 numbers then i've 34 and then i've 26 So I can't have aged bigger than a third of the number I have and at the end of my vector I'd like to vector to the least length two or three so I can see that the numbers are very close So that and it means that indeed you've done three h steps So it's as if you'd started with three h morally And you saw that we had these shifted factorials the shifted factorial Was just something like a gamma of h plus a third over gamma of a third And so roughly I divided if you do the whole division By you know h factorial times this and this of course is roughly I mean whatever it is up to power It's roughly gamma of three h so indeed morally It's exactly as if I take the original thing and done it with an h which was three times as big But the three if it's 25 has been split into three steps of eight each So you don't change the age at each stage But you do so you end up however the number of Calculates you have to do will be three h if there are three terms And the length of the vector you need has to be a little bigger than three h or or you want of anything left You'll have killed everything so in that sense there there is of course a lot naturally You can't expect to get as good precision when there are three terms of different sorts as if there were just one and the others were up Zero So thanks. I was very good. Actually, there are several very good questions There's only good questions, but It's very helpful Does anyone else has a question a suggestion or for that matter a sequence? Oh, yeah, you should have the microphone especially your voices Not as loud as Emmanuel's even when without the microphone So when you say you use computers, you mean you use your like home laptop or like oh, yeah It's not about the speed of the computer if you use a bigger computer smaller when I say it's a tenth of a second Maybe another computer. It's a 20th of a second. Maybe it's a fifth of a second. That's not important Let me answer that in more generalty many people Including Emmanuel who don't do a lot of numerical computation say oh, well, I don't do that. I'm not very good with With computers. I don't know. You know computer languages all of this isn't about I could write some programs in Paris They're one line and you just do what I say you just take the difference. So you write, you know, bn So typically you write You know bn is Say in paris you would put b of n. Maybe it's a vector b of n is a of n plus one minus a of n You'd only be a computer actor to be able to read that line You just write something that looks what you're doing and the question whether it's quick or not is not if you're a good Program, but whether your algorithm is good. The algorithm is good It's very short and then paris or any other program will do it You know extremely quickly even mathematical will do it quickly But when you say you try something like it takes hours then it's it means that you you're on your I don't remember which computer this some notes. I wrote myself 10 years I looked down my computer and there was a footnote It took four hours, but it was a I don't know what you're asking I certainly didn't do it in my head. I don't have a portable telephone So it was on some on some computer at home Whether it was you know laptop or but I mean it was not a mainframe computer It was whatever I have at home and it uses paris and and to you know So this was 10 years ago So actually I read did the ones at some competition that it said it took 1.3 seconds and I did that one today and it took 0.3 seconds because my computer now is Four times faster than it was but that doesn't matter. I'm comparing five seconds and four hours And so five seconds on a faster computer is one second and four hours on a faster computer is two hours But the point is that one out of them is much better than the other the whole thing is not About computer power, but it is about software because if you try to do this in Fortran Then it is you know even Multiple precision it goes up to 32 digits. None of this will work with 32 digits Because the method is very unstable as I said here for this terrible thing That was why it was so slow. It wasn't computing 50. Actually, that took some time too because it's 50 cubed But the real problem was each one had to compute to 20 000 digits And it's given myself and that you know, it's it's an infinite sum But to get that much accuracy you have to take a huge number of terms And only because pari is so fast the program is quite short. It's two lines But it's very very inefficient So to compute, you know up to 50 cubed and to and to 20 000 digits that took hours And we'd still take hours even on a faster computer today might take, you know, one and a half hours But once I use this method first of all, I'll need 300 code to see here I was calculating up to and up to 50 cubed which is after all 125 000 Today since I was using the better method, of course, I didn't want just 50 terms I wanted more because I'm doing it three times, but I actually used 300 terms But here I needed 125 000 terms and here I needed these two 20 000 digits But today I only compute and I think I took 500 terms to get 50 digit accuracy yet But I only needed 300 digits and this one took five seconds So I need way way way fewer terms and much less precision I didn't need this incredible precision because that magnified the bigger end is the more you're magnifying these small effects So it's uh, I mean it's it's reasonable But it's not a question of, you know, putting it on parallel computers, you know 10 clever, you know craze and different Institutes all over the world. It's nothing about that And my experience of many years of scientific computing. It's almost never about that About once every two years have a problem that I ask our computer specialist Can we do this, you know, that it's running some taste on a little network because it's it's still taking hours But usually making things go fast is not about a faster computer more powerful Parallel processing any of those computer things. It's not we're not doing information science It's about thinking of an algorithm that will be That's getting to the heart of the mathematics of the problem more and here the heart of the mathematics of the function a sequence which looks Like ck over n to the k thirds What it looks like is you have to replace n by n cubed and then you do it and that's very inefficient But if you realize that this is a sum of three functions Which are simply power series and one over n which after rescaling our polynomials, which you can kill by differencing Then you've seen something nice And actually to me all the program is about that. It's not even getting the answers In order to make the program run well, you have to understand better What the numbers are telling you and what they're asking you and so to understand numerics well It's essential to program Not to get the answer but the computer will help you to understand And also if you make a mistake like this morning at the wrong factor Well, first I just put h factorial and then it didn't converge to I mean it's still converged But it converged somehow to the wrong thing Because I was getting a product of factorials and I couldn't recognize it because actually it are product of these factorials Which is many many primes and it's uh Because you have a third times four thirds times seven thirds up to eight plus a third So it does all the primes up to 27 So if I enough digits I could recognize of it It was and then I realized I was being stupid because I shouldn't divide by h factorial But if I shifted it's the computer that told me that because when I made a mistake The computer immediately said it's not working on you the numbers didn't converge well and so It's it's to me all of us mathematics I'm in number theory the computing is about understanding how these numbers they're internal structure, but understanding them concretely So and it's well, of course sometimes you still need a faster computer Obviously if I programmed that on 20 parallel craze then it would have taken only 10 minutes instead of four hours But it was much more efficient to wait for 10 years and then understand the right way and then it took five seconds So whether that's actually faster is a quest up to matter of discussion Anyone else wants to ask you a question now we aren't oh, it's still one minute left even officially So we've got to have one more question It's probably about to turn to 30 I don't know. Oh, hi. I can't see you, but I can hear you Campbell. Yeah Well, no, I'm looking at a screen, but he's not on the screen. It's just his name Anyway, it doesn't matter. I know what you look like So now I can see you but you're very small. So what's your question? Did you have a question now? We can't hear you. I can see you Un can you unmute yourself? Because in principle everybody is not muted. I was assured by Marco So, but I'd heard him before he said hi. I heard his voice Oh He's been switched off, but we haven't got left right so they can still hear me So then I can apologize. I'm sorry Campbell. They'll have to ask next time or send me in. Can you hear me now? He came back. Sorry. Sorry. That was my issue. Um, no, so this is more of a comment. Um Yeah, so basically that open problem you mentioned with the asymptotics. I think I've got a method I'll send it through to you Which problem about which asymptotics? So the if you have two different bees Uh, beta, sorry for the question here about beta one and beta two Then I'm so if you don't know, yes, okay. So what do you do if you don't know beta one and beta two? Yeah, so you construct a new sequence. It's a bit more costly than um, you need a little more data than you would When you know beta one and beta two, but you can basically You basically construct a new sequence Um With a bunch of binomials, uh, I can I'll send you through the well wait I think it's a new sequence Campbell. Yeah, can I ask you not to send it to me? Because now you've piqued my curiosity So let me think about it again knowing that it can be done If I if I can think of a cute way to do it because actually the second thing I was going to tell today but didn't didn't get to is Was also about recognizing a sequence of numbers that had to be at the end And there was a trick there and I wonder if you can be combined. So please Keep it and send it to me in a couple of days and next Tuesday I'll present what you said today is thursday, right? So the next course is tuesday Same time but tuesday and thursday and then it changes to monday and wednesday and hour earlier So I will if you send it to me by email, but you can send I just want to look for a day or two Now I want to think if I can think of anything So So now I'm I'm curious because I told you I failed so And maybe if if I find a different one then we'll have two ways and see Yeah, yeah, and as I said mine's a little costly. So if there's sort of a more efficient version, that would be better So thank you very much. Anyway, sorry Okay, yeah, I'll send it through. Well, that's way beyond the call. Sorry comment not a question I ask people to ask questions not to answer my questions, but of course that's even better if you can do it Can't look by the way is my doctoral student and he's extremely good. So he he often can do things that that I can't do I mean, it's kind of the it's the norm Okay, so Then till next tuesday and thank you all of you for still coming there now only There are eight people today that were like 20 at the beginning Sorry, well, there's No, well, that's that's good And I think some people watch it later like the people in China