Attempting to evaluate the Riemann Zeta Function at odd integers

Ajouté : 2026-05-10

[00:00:03]I'll give it a few seconds as well to actually make sure it's all right.

[00:00:15]Let me know when it's when it is as well by the way. Yeah, yeah, I can go for it.

[00:00:19]Oh, okay, perfect.

[00:00:21]I just want to know how well can you hear me?

[00:00:24]You're good.

[00:00:25]Speaking a little too loud.

[00:00:28]Um, okay. I can hear you fine at least.

[00:00:32]Um, as this talk is titled, I'm going to be discussing the the evaluation of the Riemann zeta function at integers.

[00:00:44]Um, sorry, I was distracted for a second. Um, evaluated, yeah, of course.

[00:00:50]Um, surely everyone here has heard of the Basel problem.

[00:00:55]Um, you know, the sum of reciprocals squares um, gives you pi squared over six.

[00:01:04]Um, in fact, has anyone here not heard of this at all cuz uh, that would kind of uh, be a little important to start stuff.

[00:01:22]I'll take it.

[00:01:23]No, so I'll get this though.

[00:01:25]Um, yeah.

[00:01:28]pi squared over six.

[00:01:30]I almost said it.

[00:01:32]Um, everyone's heard of that. Um, it's like, you know, one of the most famous things that Euler actually did because it's spoken about so often.

[00:01:44]Um, 3Blue1Brown has a great video on it and I actually uh, want to present Euler's proof to him.

[00:01:52]Uh, I just realized I'm not streaming at the correct I'm streaming both of them.

[00:01:59]Huh.

[00:02:00]I don't have it saved on me.

[00:02:02]I was going to just copy it over.

[00:02:07]I kind of drag and drop. Does that work?

[00:02:09]It does.

[00:02:12]Oh, it doesn't. No, devastating.

[00:02:15]Okay, hold on. Let me just switch files, it's fine.

[00:02:19]I changed stream.

[00:02:25]Okay.

[00:02:26]Uh can you can you see the notes for precalculus, I think?

[00:02:29]Yeah.

[00:02:32]Good.

[00:02:33]Okay.

[00:02:34]Um All right.

[00:02:37]Uh this this will come in handy as information later on.

[00:02:42]Um to my knowledge, this is how Euler himself did it.

[00:02:49]You assume that sin of x can be written as uh an infinite polynomial.

[00:02:56]Would you mind zooming in a little?

[00:02:59]Uh sure, sure.

[00:03:01]Uh there you go. Does that help? Uh a little bit less.

[00:03:05]Oh, yeah.

[00:03:06]Uh you assume that sin of x can be written as an infinite polynomial.

[00:03:11]You can express uh a polynomial in terms of factors of its roots, kind of like, you know, I've got x - a, x - b, et cetera, where a, b, and so on are the roots of the polynomial.

[00:03:27]Uh and then it's equal to zero, of course.

[00:03:30]So, that is what is being done here.

[00:03:33]We're saying that sin of x is defined as the product of all of its roots.

[00:03:40]Um and if you treat sin this way, you get the um this infinite product here of a sub n is uh given by this Euler identity.

[00:03:56]Uh trust me, this gets much more interesting later on. Please give it time.

[00:04:00]Um and yeah, of course, this is um sure most of you here have seen the proof. Um if you want to ask questions, you can.

[00:04:11]Uh but I don't I don't think I need to really go over this too much more.

[00:04:17]Uh you can I saw it quite easily.

[00:04:21]I was just going to make sure there's nothing in chat.

[00:04:23]Okay.

[00:04:25]Uh yeah, you cancel out the x squared, times it by the pi squared, and cancel negatives. Pi squared by 6. Very nice.

[00:04:34]Uh another thing lesser known that Euler did is he actually proved the general even case for not just n equals 2, but n equals 2n. Uh sorry, n equals 2k.

[00:04:48]Um I haven't read over these notes for a while. Yeah, okay, I thought so.

[00:04:55]So, yeah, you begin with the product formula for sine that was derived in this earlier section up here.

[00:05:04]Um you begin with it, multiply by pi x.

[00:05:09]Um so, you multiply by pi inside the x.

[00:05:16]So, you replace every x here with pi x, and it cancels out the pi squared down here, and they add the pi out front.

[00:05:23]No, am I lagging? Someone just left and rejoined. What does that mean?

[00:05:31]Everything seems fine. I didn't miss anything.

[00:05:34]Okay.

[00:05:35]I just wanted to make sure.

[00:05:39]Uh yeah, I just wanted to make sure cuz I I heard something strange. Um maybe it's just Discord.

[00:05:45]Yeah, uh then the logarithm.

[00:05:47]Uh and then differentiate and you get this horrible horrible formula here.

[00:05:54]Uh yeah, you get this horrible horrible formula.

[00:05:56]Um and after a little simplification, you can bring it down to this and rearrange to this.

[00:06:03]Um and you use on the base It's done it again.

[00:06:09]I'm going to assume nothing's changed.

[00:06:11]Yeah, and uh there is uh And here, this is the definition here of Sorry, this is the definition here um of the zeta function.

[00:06:26]Uh well, the Riemann zeta function in its summation form, that is.

[00:06:30]Uh which is perfectly valid to use for the zeta.

[00:06:34]Um you just change the index, you know, all that. I'll go to n minus one. The index increases, so you've got no division by zero.

[00:06:42]Um and rearrange. Yeah, you can pull it out.

[00:06:50]Blah blah blah, he set it equal What was I thinking when I was writing this? This is This is really This is not as boring as I thought it uh not as interesting as I thought it would be.

[00:07:00]Um In fact, I'll I'll give some motivation in a second for why we're doing this. Yeah, and you can bring out this.

[00:07:08]So, uh yeah, now I'll I'll talk motivation because motivation is actually quite important here.

[00:07:15]Um it's not I wouldn't say it's the best motivation, but I I personally like it.

[00:07:22]When we study mathematics, we tend not to study it for the math itself, if that makes sense.

[00:07:31]If we study math for the sake of studying math as a society, as a collective, we wouldn't actually do anything because math you know, mathematics is uh to an extent only as good as it is useful.

[00:07:49]So, of course we need uh we need to send justify why we should care about the zeta 2n or zeta of something in general.

[00:08:03]Uh and a few of a few of a few of my favorite reasons include its appearance in physics.

[00:08:11]When you're doing quantum mechanics in some quantum field theory cases, you do end up dealing with an infinite sum of reciprocals.

[00:08:20]And that infinite sum uh is not weighted and you end up with a direct zeta.

[00:08:29]So, being able to evaluate this for the precision of physics is always important uh because physics leads to engineering and can lead to uh just it it chains all the way up.

[00:08:44]Engineering, computer science, philosophy, and all all leads to other various various things which are incredibly important to society.

[00:08:53]Another reason, I think this is the better reason, is when we ask a question in maths, we tend not to ask like we tend not to just ask the question and immediately solve the problem.

[00:09:10]Uh in fact, the talk later, as you can see down here, is the zeta odd work.

[00:09:16]This, the as I'll get to in a bit, the Riemann zeta function evaluated at positive odd integers is not uh very understood.

[00:09:31]We understand that infinitely many of them are irrational. And we understand that uh zeta of three is proven to be irrational.

[00:09:41]And then there's a couple of other specific things, and that's all we know.

[00:09:44]We know nothing more, almost.

[00:09:51]When when when we're studying maths, we end up developing new fields.

[00:09:57]A great example of this uh is the uh Fermat's Last Theorem. And I'm sure as computer scientists this for the most part in here you are computer scientists and maths linguists. Um I do both.

[00:10:14]>> Sorry. My brother I just thought of Very cool.

[00:10:17]Arguably Arguably better.

[00:10:19]Um but yeah.

[00:10:21]Um Computer science is of course the foundation of cryptography.

[00:10:27]Without cryptography our data wouldn't be secure. We wouldn't be able to trust anything. And this is where our sponsor for today's video I'm good at um I can't believe I just Anyway, back on track. We We end up developing new fields. Fermat's Last Theorem ended up developing the field of elliptic curves.

[00:10:49]Uh elliptic curves I'm personally and maybe someone in here knows more about this than me.

[00:10:55]Uh to my knowledge elliptic curves and particularly elliptic curves along p-adic numbers are very very important to cryptography.

[00:11:06]I know they have some special properties and those special properties are what led Andrew Wiles in I think 1988 or something like that, maybe '86.

[00:11:17]Um back then uh he reduced Fermat's Last Theorem to a an elliptical problem.

[00:11:24]A problem uh It's It's got It's made by two Japanese people.

[00:11:31]He found this problem and he spent a few years on it and solved it.

[00:11:35]And but he developed the cryptography by pursuing a completely unrelated question. And that is why I think we should study this.

[00:11:47]Um But yeah.

[00:11:51]Negative integers, cool cool.

[00:11:53]Odd integers, have no idea.

[00:11:57]Um and this is where it gets fun.

[00:12:01]I can move on to the stuff that I've actually written up.

[00:12:06]Uh me and my friend sometime last uh last January, so about a year and a half ago now.

[00:12:13]Uh 1,500 words. Didn't realize it was that big. It's not incredibly long, though.

[00:12:18]Um but yeah.

[00:12:20]Oh, uh sometime last January, me and my friend uh went to a whiteboard.

[00:12:25]And he showed me what I call here part one.

[00:12:30]Um I didn't really know what to name it when I wrote it, and I haven't bothered to change it since.

[00:12:35]So, he showed me what I call here as part one.

[00:12:39]Uh this was his own derivation.

[00:12:42]And he thought it was pretty interesting.

[00:12:45]It did the thing again.

[00:12:47]This is really annoying.

[00:12:49]Um So, this here is identity. And I don't think this itself is Ramanujan's master theorem.

[00:12:59]But it is a very uh It's called identity. I think I made a mistake here somewhere. Yes, um I was meant to put 2x dx. That doesn't look right.

[00:13:11]One sec.

[00:13:12]I I was going to say I I remember I made a mistake here.

[00:13:15]Yeah.

[00:13:17]Um So, we we let x equal 2y. We use the U substitution. And it gives us dx equals 2dy.

[00:13:24]Then we relabel y back into X.

[00:13:28]Um Yeah, it's just a substitution. Nothing changes.

[00:13:34]Um But, after expanding everything out, uh you get this.

[00:13:43]times Put 2 to the n out front.

[00:13:46]You get all of this.

[00:13:48]Uh and you can split this then up into two fractions of denominator e to the x minus 1 and e to the x plus 1 uh in a sort of partial fraction style.

[00:13:59]I didn't actually prove this here, but uh it is I I I can't imagine it's too difficult to show that these two integrals are the same.

[00:14:10]Um So, then we let uh We also notice that this integral here is exactly the same as our I integral above.

[00:14:22]Uh so, we let J equal the integral right.

[00:14:26]Uh and then we can reformulate.

[00:14:28]This is the case.

[00:14:30]Rearrange and we arrive at the fact that J is equal to X.

[00:14:37]Um Something I noticed not not too long ago now.

[00:14:41]Um you know, that might uh that might strike someone's interest in here.

[00:14:46]Uh the Dirichlet eta function is defined as the alternating version of the Riemann zeta function.

[00:14:54]Uh Did I define the Riemann zeta function in here anyway? I don't actually I didn't.

[00:15:00]That's so silly of me. I didn't define the Riemann zeta function. Um I need to I need to do that, but oh well. Oh well.

[00:15:11]We move on. But, yeah, the the Dirichlet eta function is defined as the alternating version. So, 1 minus uh 1 over 2 to the n plus 1 over 3 to the end and so on.

[00:15:24]Um and you can also write it as such.

[00:15:30]This here is another formulation of the Dirichlet eta function.

[00:15:35]Um so what we have here really isn't just uh J equals this, it's J equals gamma times eta, which I I only noticed recently and I found it uh I found it pretty cool.

[00:15:50]Interesting.

[00:15:51]Um This is the Mellin transform and this is actually the tool which allowed us to to this.

[00:16:01]Um once again, it wasn't actually us who derived this. This is a known formula.

[00:16:07]Um as you can see, this is x to the n minus one, x to the n minus one, f of x here is our one over e to the x plus one.

[00:16:16]e to the x minus one, sorry.

[00:16:18]Um property about the Mellin transform uh sorry, a property uh that the Mellin transform has is that for f of x for any f of x which you can apply the Mellin transform to you can always write it as such.

[00:16:39]Um let me take a drink.

[00:16:44]Yeah, you can always write it as such.

[00:16:47]Um but I thought this doesn't seem to have any justification.

[00:16:55]Um but yeah, you need to have your f of x continuous on the interval zero to infinity including the zero.

[00:17:08]Uh and since e to the x minus one uh one over e to the x plus one is discontinuous at zero, we couldn't use it.

[00:17:16]Now that we have this, we can use it.

[00:17:19]Uh so, we apply the non-transformed and we acknowledge that it's J.

[00:17:25]So, >> [clears throat] >> uh we can note uh this is the case.

[00:17:38]And this Sorry, I've got massive brain fog.

[00:17:40]Not a good day to do this.

[00:17:42]Um How did I derive this? I know.

[00:17:49]I'm sorry.

[00:17:51]Um gamma of n phi of negative n.

[00:17:54]What Where have I pulled that from? Oh, my I feel so silly right now. I'm sorry.

[00:18:00]Oh, here. Right, yeah, of course.

[00:18:02]Uh this is the same phi of n, by the way.

[00:18:07]Um Ah, I see. Okay.

[00:18:14]So, this left integral here is our J, if we insert this as defined here.

[00:18:21]And J equals gamma of n phi of negative n.

[00:18:26]So, gamma of n phi of negative n equals what we define what we derived earlier for J.

[00:18:32]We pull that out.

[00:18:34]Um we rearrange for zeta in terms of phi and we get this.

[00:18:42]Um simple enough.

[00:18:45]But yeah, we will prove later on that phi of 2n is zero.

[00:18:49]And that is that is pretty interesting because it actually ties perfectly back into uh what I skipped over earlier.

[00:18:58]That uh for all negative even integers, the Riemann zeta function evaluates identically to zero for every even integer.

[00:19:10]Um it sort of oscillates.

[00:19:13]Um, yeah.

[00:19:17]Uh, I I I found that interesting. And that actually comes in a bit later on.

[00:19:21]Uh, quite significantly.

[00:19:24]Um, yeah, this is the definition here of the Taylor uh Taylor expansion of Mhm.

[00:19:32]anything really, but specifically one over e to the x plus one.

[00:19:38]Um, so we can set these uh equal to each other.

[00:19:44]And we can evaluate this term by term.

[00:19:48]But setting that negative one to the r uh x to the r over r factorial on both sides cancel.

[00:19:55]So you get negative one to the r times phi r is this. As we equals this evaluated to the zero.

[00:20:03]Um, so now let's give that a name, this thing on the right, because it's very clunky and it's going to be very useful.

[00:20:14]We'll call it D.

[00:20:15]Uh, honest to be completely honest with you, I thought of D because differen- differentiation.

[00:20:23]It's kind of pointless and silly.

[00:20:26]But yeah.

[00:20:27]Um, uh yeah, to find this negative one to the r phi r.

[00:20:35]Um, these are three more completely known equations.

[00:20:43]Um, this is I I I actually don't know where this relation comes from.

[00:20:51]But you can derive this through right down to it.

[00:20:57]Uh, you derive through this.

[00:21:00]And uh a couple of other facts which you could probably pick up on uh uh yourself if you actually went through it.

[00:21:07]But, yeah.

[00:21:08]Uh this will be useful.

[00:21:10]This was derived in the other document, and this is a brand new fact.

[00:21:16]Zeta prime here, uh I know I notated it kind of sloppily, but it the prime means with derivative with respect to n, of course.

[00:21:27]Um But, yeah.

[00:21:29]That is Uh oh, yeah, sorry. And then we rearrange in zeta of 2n + 1, because that's what we're going to need later.

[00:21:40]Um Sorry.

[00:21:48]Uh yeah, okay.

[00:21:49]So, we'll start with uh this this equation here, and which we derived earlier, this here.

[00:22:00]Um Go back down to it. There we go.

[00:22:06]We can substitute in uh our definition of phi of n in terms of d, which I actually Oh my days, this is very poorly written.

[00:22:18]Oh, wow.

[00:22:20]Um phi of r is d sub r of x times -1 to the r of uh -1 to the r, sorry.

[00:22:29]Uh so, we can substitute that in, and I took out a negative on the bottom and just added one on the top.

[00:22:39]Um Ah.

[00:22:44]See, right.

[00:22:46]I didn't introduce the fact I was going to prove this here. Oh my word, this is so poorly written.

[00:22:52]Um But, yeah.

[00:22:55]We can, of course, define d sub zero of x as the uh well, this is the definition of d sub zero of x, right?

[00:23:02]Um We sub in negative X here.

[00:23:06]And we can then take derivatives.

[00:23:09]And by the definition of D sub n, and so yeah, D sub n of X taking another derivative of it with respect to X just adds one to the uh the subscript.

[00:23:22]So, the nth derivative is D sub n.

[00:23:26]Uh and if we if we let n be even and set X equal to zero because that is uh how we define it here.

[00:23:36]Uh we define this to be I sorry, we define D sub n to be D sub n of zero which is -1 to the n times 5 over n.

[00:23:47]Um Once again Yeah, here it is. X is zero.

[00:23:55]Uh we can take as many derivatives as we need.

[00:23:59]And go to 2n.

[00:24:01]This becomes just a one.

[00:24:03]And you can take this off to the other side and you get the below.

[00:24:08]Uh derivative of zero is and one's there.

[00:24:14]is more precise.

[00:24:16]Um So, yeah, then you can conclude that D sub 2n equals zero logically.

[00:24:25]Um Okay.

[00:24:28]Yeah, uh angle 2n at quotient rule to the uh both stated formula, but this one here.

[00:24:39]Yeah, uh Uh this is the case for the first term of the nth derivative of everything that isn't D.

[00:24:49]Now, this is the case for the rest of the term.

[00:24:53]D sub 2n is zero as we've established, so we can ignore that.

[00:24:57]And I will not take the prime sub 2n as uh no, sorry. I will not not take derivative with respect to n as prime.

[00:25:08]So, now we have the final big formula.

[00:25:13]Kind of nasty, and it's in terms of this weird function that we just kind of plucked out of thin air, essentially.

[00:25:20]Um but I mean, it's a logical chain, and it is uh complete Yeah, it makes sense.

[00:25:28]I want to ask, has anyone here spotted any errors other than incredibly poor wording on the Apple?

[00:25:38]Um I'm just going to send a message in chat. React to it if you haven't seen an error, so I can gauge a simple regular dot.

[00:25:59]All right.

[00:26:02]Um seems legit.

[00:26:06]So, I will continue.

[00:26:08]Um this actually seems arbitrary without having been well-defined anymore.

[00:26:15]Uh my friend the my friend who worked on this with me was called Z.

[00:26:19]Um he actually wanted me to send a picture of him at the end of this, so uh to honor him, I will show you one at the end.

[00:26:29]That was weird.

[00:26:31]Um yeah, we let D sub n of x equal this.

[00:26:36]And why is D sub 0 of x?

[00:26:39]This was something that we tried right at the start. Remember um our I was trying to crack D sub n of x and figure out if there a way I could write this naturally without it being defined in certain terms of some strange things.

[00:26:59]I need to take the the derivative uh the derivative with respect to n of sorry uh n being discrete.

[00:27:08]Um n being discrete makes this hard, so we technically need an analytic continuation, and all of this has assumed that we are working with an analytic continuation.

[00:27:20]Uh more example of poor wording.

[00:27:22]Um But yeah, we the main reason for assumption of analytic continuation as well is that it did not exist an analytic continuation of uh y sub n of x with respect to n.

[00:27:41]We would be working under the assumption that there exists uh no closed form.

[00:27:52]Because any closed form, if you can write it, also must then therefore be equivalent to the closed form that we have generated.

[00:28:02]So, if there exists a closed form, y sub n has an analytic continuation, and we assumed that it does for the sake of argument.

[00:28:12]Um this is why we get y squared minus y from.

[00:28:16]If you take the derivative of if you take the yeah, if derivative of uh y zero with respect to x, you get the one derived.

[00:28:27]Um And then we can take the derivative of uh yeah, you can take the derivative of y with respect to x, get this.

[00:28:36]So, then we take derivative with respect to x of this thing here on the right.

[00:28:45]Um if I find what we're doing that, yeah.

[00:28:55]Oh, sorry. I take derivative with respect to Y, not with respect to with respect to X.

[00:29:01]So, yeah.

[00:29:12]Just double checking. Yeah, yeah, this is quite logical if you actually follow it through.

[00:29:18]Um, we can take Well, let me So, of course, d p by d y equals n d p d x times uh d x d y uh which is one over this.

[00:29:39]Uh, and that ends up canceling with this. So, you just end up with d d x, which is d n plus one.

[00:29:47]Uh, so it is.

[00:29:57]Um, so, yeah.

[00:30:00]If you move this to the side uh with a bit of tiny bit of work, you can show that this is the case. Derivative with respect to Y d of n is p sub n plus one.

[00:30:12]And we also recall this p because we hoped it would be a polynomial.

[00:30:17]And uh we were partially correct, if you want to call it that.

[00:30:26]Yeah, we can take we can take this uh where we assume that there are coefficients f n a also n and uh some sum over k.

[00:30:40]We write p sub n of y as a polynomial y because almost by definition it is a polynomial in y. It reduces respect to x of it and you get just a larger uh the next degree polynomial.

[00:30:59]So yeah, um this is pretty much by definition.

[00:31:05]Uh this one here and this one here is just you add one to all of the ends.

[00:31:10]Simple.

[00:31:13]So then we take x squared minus y on this side and then by definition uh y squared minus y times piece of n of y is equal to piece of n of y.

[00:31:30]Uh and then we take the derivative with respect to y of piece of n of y as shown here, you get piece of n plus one of y.

[00:31:40]Uh which is what allows us to set these two equal.

[00:31:43]Um once again, I think this is poorly written.

[00:31:49]But I'm not too sure actually. Doesn't look that bad.

[00:31:52]Uh it was poorly derived, I can say that much.

[00:31:56]But this formula is correct and from it we can set uh oh yeah, from it we can set this equal to this and derive this functional equation.

[00:32:12]We tried to solve this. We spent a lot. Personally, I spent probably in the realm of 20 to 30 hours um deriving results for this. I ended up uh I think I got like four or five results in the end because uh this is not my forte, I can say that much.

[00:32:36]Let's go I posted it to Math Overflow in I want to say in July.

[00:32:45]Um No, it must have been later in July. I think it might have been around October.

[00:32:50]Um and thank you very much Theodore Petroff.

[00:32:54]You solved the equation within 10 minutes of me posting it, nullifying my hours and hours of work.

[00:33:03]Um and it these are Stirling numbers of the second kind.

[00:33:07]Um and they have a definition, which I don't actually have here.

[00:33:13]Um and it is Stirling numbers of the second kind and please don't quote me.

[00:33:21]Um I'm going to write this K cuz there are two n.

[00:33:27]No, I wrote n.

[00:33:30]Um I choose I choose K times um I think it's like K factorial or something. Uh K factorial then factorial maybe.

[00:33:43]This looks around about right.

[00:33:46]Um it's very it works. It's very nice.

[00:33:51]This is where we are stuck.

[00:33:54]We still need to differentiate either P or D with respect to n.

[00:33:59]Or n is in the upper bound.

[00:34:04]And we yes, we've tried generating functions, I've tried ordinary and exponential generating functions and there's definition of Stirling numbers as a contour integral as well, but that is derived from its generating function.

[00:34:18]And this is where we're stuck.

[00:34:22]The main reason I am even in this server is because I want to know if anyone here is able to help us in any way to solve this.

[00:34:36]I think it would be interesting and we think that our method may have merit, but probably leads to a circular end as we derived uh as we've gone through this hundreds of different ways. Hund- maybe not hundreds, but like at least eight.

[00:34:52]Um I remember I tried taking the definition of B sub n and I remember I uh applied 4A and 4A inverse applied.

[00:35:05]Uh I tried Laplace just for the sake of it.

[00:35:09]And uh we we never got anywhere.

[00:35:13]Um but of course Dirac came along and solved the functional equation for us.

[00:35:20]And now we're here.

[00:35:23]Um just unmute, say whatever. I am wondering if anyone here has anything they could add to it.

[00:35:35]Any ideas we could try.

[00:35:37]And we we can even try and work through it now.

[00:35:40]If you uh you think you know how to Swab, I feel like you of all people would have something to put in here.

[00:35:52]Why me? I got nothing.

[00:35:54]>> [laughter] >> I don't know.

[00:35:57]Incredible.

[00:35:58]Um yeah.

[00:36:02]It I we don't really know what to do.

[00:36:06]Cuz I tried rearranging this amazing like uh adding in another sum here that you know, by this definition here.

[00:36:15]Uh and then changing uh order of summation and it switches the indices and then instead of getting an n minus one in the top, I get n N And it just I can't seem to make it not discrete.

[00:36:31]Um I I did think about the Euler-Maclaurin expansion.

[00:36:38]Don't know how long that would take me to derive.

[00:36:42]Um does anyone at all really have ideas as to what to do?

[00:36:53]No?

[00:36:55]Incredible.

[00:36:57]Um fine.

[00:36:59]I'll just send the picture he wanted me to send.

[00:37:02]He wanted me to show everyone.

[00:37:04]Um It's now in the It's now in the general chat.

[00:37:09]Uh is it general? Uh it's an announcement, sorry. There we go. That's what he wanted me to show you guys.

[00:37:15]Um Uh that's all I've got, honestly. No, this seems hard. Yeah, that's why uh professional mathematicians have been cracking at it for 40 years, and nobody's gotten anywhere with it.

[00:37:29]I don't know what to do.

[00:37:33]Um 120 cell, maybe? Maybe you'll know something about this?

[00:37:43]Anything?

[00:37:46]No, sorry.

[00:37:49]Bro's desperate. It's okay.

[00:37:51]Yeah, I am, to be honest.

[00:37:54]Um I think the last bit of work we made on uh last bit progress, sorry, we made on this was maybe November.

[00:38:02]Um and we got a call we got on call the same night that I saw the uh the looking for math gods video.

[00:38:11]Same night I saw that, I called him, and I was like, bro, maybe maybe we should work on it a bit more.

[00:38:18]But we just don't know where to go.

[00:38:20]Um Yeah.

[00:38:29]I think that's everything I've got, to be honest with you.

[00:38:32]Um I I can go through the the derivations of I well, I can uh you know watch through a little bit more of the derivations here, but I don't feel like that's as that's as interesting as this work here.

[00:38:49]But, of course uh if you're interested, I can just send these notes uh into general or into the event announcements channel or something.

[00:39:00]Yeah, sure. Post them. Somebody might be interested.

[00:39:03]Yeah.

[00:39:13]And then there we go.

[00:39:18]Yes, the big buttons.

[00:39:20]All right, then.

[00:39:22]Uh I do think that is all I have for today, then.

[00:39:25]Uh I don't I don't think I can give you anything more. Cool.

[00:39:29]Thank you.

[00:39:30]Thanks for the talk.

[00:39:33]That was it.

[00:39:34]Okay.