The Basel Problem Part 1: Euler-Maclaurin Approximation
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- Опубліковано 2 чер 2024
- This is the first video in a two part series explaining how Euler discovered that the sum of the reciprocals of the square numbers is π^2/6, leading him to define the zeta function, and how Riemann discovered the surprising connection between the zeroes of the zeta function and the distribution of the primes, leading ultimately to his statement of the Riemann Hypothesis. This video focuses on how Euler developed a method to approximate this sum to 17 decimal places, as well as how the Bernoulli numbers naturally appear as part of this problem.
This mathologer video touches on many of the same topics we do in this video, including Euler-Maclaurin and sums of powers. It is highly recommended! • Power sum MASTER CLASS...
Theme music by Keith Welker.
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32:39 the Bernouilli coefficients Bk(0) and Bk(1) are wrong for k = 1 (signs are inverted).
Thanks for the great contents.
Oh you're totally right, and I think you're the first one I've seen point that out! Sadly there is no way to fix these things on UA-cam!
@@zetamath Yes... You're not alone, a lot of youtubers doing math videos have an error here and there... that's life I guess...
@@zetamath Pin this comment
@@columbus8myhw Done!
These small mistakes may act as a litmus test, indicating that the community is really paying attention.
It's hard to understand how smart Euler was. Always breathtaking when his derivations are worked through
yeah like, his intuition was insane, the greatest thing about it isn't even the proof itself but the pattern matching used to discover them
Yeah with all his infinite canceling n shrt
Yeah sometimes you can sniff out the Euler in some derivations
I remember that a year or so ago I stumbled onto this video and, although it was amazingly put together, I was completely lost. Returning to this after completing an introductory calculus course and understanding the arguments is a positive experience like no other. Thank you so much for this resource!
5:40 Why did they calculate the decimal expansion of pi^2/6? Why did Euler know that decimal expansion so well as to recognize it?
ikr, how tf do you just go OH YEAH...THAT? man that just looks exactly like pi squared divided by 6.
my guess is that he looked at that number and started multiplying by random shit in his head like 2, 3 and 6 and realized it was something like 9.86.. and remembered the first digit of pi is 3 and probably remembered somewhere where he had to use pi² and saw its decimal approximation (he worked in a lot of applied math so it'd make sense that he knew a lot of decimal approximations of constants and operations on them, specially pi)
The triangle area proof is pretty simple with a telescoping series, but that visual demonstration is so much more elegant.
@6:35 As the triangles slide to the left against the y-axis it becomes clear that all triangles have a base of 1, and therefore the sum of the area of all infinite triangles is 1/2 times 1 times the sum of the heights of each one of the infinite triangles. Now the heights of all infinite triangles add up to 1, as geometrically shown on @6:45.
Very interesting video: the length is not at all a deterrent considering how well explained and entertaining this is.
Thank you very much for an interesting video digging deeper than the others!
48:11 : You " _can in no way imagine this sum to be calculated by hand to 20 decimal places_ "? I just timed how much it would take me to calculate it (using a pen and paper). Took 1 hour 13 minutes, including going to the kitchen to heat my tea in a microwave 3 times. I have never been fast, and I stopped calculating on paper some 30 years ago when calculators became ubiquitous. My point is, I would not be surprised if this can be done in 20 minutes if you practice calculations by hand all the time. (You could miss the fact that all the fractions involved have relatively "easy" denominators, that is, the decimal fractions have a relatively short period:
1/6 = 0.1(6)
1/30 = 0.0(3)
1/42 = 0.0(238095)
1/30 = already calculated above
5/66 = 0.0(75)
691/2730 = 0.2(531135)
7/6 = 1 + already calculated above
3617/510 = 7.0(9215686274509803) -- the most difficult division here, as 1/51=1/(3*17), and 1/17 has a period of whole 16 digits
43867/798 ~= 54.971... -- I never figured out the period, as this is divided by 10^19, and you only need 4 digits (54.97) to have 21 digits in the result.)
This is real great stuff. Many thanks for it from Germany!!
How the heck Euler figured out that 1.644... is pi^2/6?
If you have no distractions, a lot of free time and an infinite passion for mathematics, you might have made the connection too.
Most underrated youtube channel. Math does not have to be difficult.
That was wonderfully entertaining! Thank you, sir!!
wow... that was really fascinating. I'm not a professional mathematician but like maths so actually did an investigation into sum of 1/x^n .. a few years ago and that was indeed the first time I came across the Bernoulli numbers. (they popped out when I inverted this matrix ( very closely related to pascals triangle)) . So then I found them on wikipedia but didn't make much sense of it. Really this video makes it much more clear what these things are (than any other i've seen). Thank you.
Boy, incredible content!
Thanks for the excellent explanation!
Way to go, thanks, you create access to “next level” ideas
Nice work.
I love your vids!!
Rather be lucky than good. Good job on the video!
I just happened to see this today. I have wanted to give a talk on this for over ten years but could not make it clear enough for a math circle. Now I don't have to. Thanks
great video !
great video!
25:18 the "special A" was written identical as "area A", be careful to distinguish between them.
Thank you very much
Thanks!
pleaaaasee make more of these or anything that has nice. mathematical objects like infinite series and stuff like that
Excellent
Been enjoying your videos. So how did Euler previously know the value of pi squared over six so that he could match it to his approximation?
My understanding is that it wasn't that he had calculated it previously, but rather that due to spending a lot of time doing these calculations by hand (as mathematicians of that era did) he was able to eyeball it and think "hey that's pretty close."
You sure he didnt just keep along list of random decimal expansions and then recognise it froom the list. If i were euler, i would have had a list of hundreds of radicals, and other irrationals.
Perhaps he saw that the value he got wasn't far off 10/6 (which is 1.666...), so that might have led him to search for a particular numerical value close to 10. From that to pi squared it isn't too far a step, since anyone familiar with doing calculations by hand knows pi has that nice property that its square is almost ten.
@@simonwillover4175 Euler had an incredible memory, he likely would have a vast bank of common decimal expansions in his head without needing them written down
2:40 - There is some missing logic here. By the same argument you get 1 + 1 + 1 + 1 + ... = (-1 + 2) + (-2 +3) + (-3 + 4) + ... = -1 + (2 - 2) + (3 - 3) + ... = -1. Telescoping sums only work if the final term approaches zero. Nowadays you could compute the sum in a spreadsheet, look up the the first few digits in OEIS (A013661 btw) and boom, pi^2/6. Sometime I wonder how much math is missed because we have computers for computations. But I guess without computers no one would have ever heard of the Mandelbrot set so it more than balances out.
Thank you.
Please can we get a lot, lot more videos on this channel?
When I was trying a different method, i discovered a nice little identity, which is undoubtedly known, but I could not find it on the wiki page and to me it was new.
Sum (m>=1) ζ(2m) - ζ(2m+1) = 1/2
I wanted to approximate the integral by a discrete sum:
set f(x)=1/x^2, with n-th order derivative (-1)^n(n+1)!/x^(n+2)
The taylor series of f at a is given by Sum(n>=0) (-1)^n (n+1)(x-a)^n/a^(n+2).
Integrating f(x)dx from x=a to a+1gives F(a+1)-F(a) = Sum(n>=0) (-1)^n/a^(n+2)
This is valid when a>1
Integrating f(x)dx from x=2 to infinity gives 1/2 = Sum(a>=2, n>=2) (-1)^n/a^(n) .
It seemed okay to swap summation order to arrive at the identity.
43:50 so next you notice that k is always odd here and the terms are being divided by (k+1)!, leaving a factor of -1, allowing the formula at 43:36 to be simplified to 1 + 1/2 + b2 + b4 + .. + b(2k), right?
edit: apparently not, that's pretty interesting
That doesn't converge. (In fact neither does the version with 10, but it gets close before it goes far away. Doing it with 10 does this better and for longer.)
17:27 ik I’m probably being dumb, but how can you assume that f is linear, and then proceed to assume it’s quadratic?
My guess is that if f(x) behaves nicely in some way (i. e. looks sorta like some polynomial with degree n), you can assume that for the nth derivative of f(x) that remainder integral term is sufficiently close to zero and you end up with a nice approximation, which would also make sense considering that the Euler Maclaurin Formula is an asymptotic expansion, meaning after some term the series diverges. So this is just a kinda esoteric derivation of those coefficients by assuming that any derivative of f(x) is constant meaning that integral term is zero, but you could probably add any arbitrary coefficient each time you integrate by parts and you would get some other formula, except it wouldn't be as useful for approximations. That's my guess.
We're assuming with different F functions. It might have been more clear to say "consider if F was linear" then "consider if F was quadratic" and so on
@@romajimamulo yea thanks I was a bit confused :p
Hi sorry i've seen a proof of this lemma before and it invovled exact formula for bernoullis' numbers by considering function (x)/((e^x)-1) is there any other proof possible just by the given defs and lemmas in the video ? in that case may you write it for me ?
As with most things in math, there are many different possible proofs. When it comes to the Bernoulli numbers specifically, I generally find them mysterious and unmotivated, so I did my best in this video to give a proof that suggests they are discoverable.
@@zetamath that's deeply respectful. i've faced almost same situation about binomial theorem but this always raises important questions for me if there is two definition of something how can we derive one from the other in this case by given info in the video i was able to write expression (n=0 to infinty)ΣBn(x^n)/n! how can we derive the other definition of bernoulli polynomials involving (x)/((e^x)-1) ?
Maybe I'm not understanding correctly the procedure. But at 51:47, couldn't this be attributed to precision errors in a 64 bit floating point representation? I wasn't expecting it to be flat like that 🤔
They don't use 64 bit floating point representation for problems like this.
55:06 that integral at the top, did you mean to integrate along x instead of t?
I didn't get this: at 17:05 we integrate by parts again over the expression obtained by letting f(x) be linear. So, how can we proceed in the integration assuming that f(x) is quadratic, if the initial expression was derived on the basis of f(x) being linear?
The initial expression was not derived on the basis of f(x) being linear.
NITPICK: At 2;46 your graphic is misleading. When you introduced the parentheses you should have taken the leading minus out of the parentheses I assume you are not equating an infinite product to an infinite sum, which is often ok.
That's a fair point! We strive to have our graphics be as clear and correct as possible but that one escaped us. The parentheses are meant only to group the terms of the sum, not to imply multiplication. Hopefully it is still clear to most people what is going on there!
That process for finding A reminds me of the expectation and variance of a uniform distribution. What I mean is, ((f(0)+f(1))/2 looks like a mean and (f’(1)-f’(0))/12 looks like a variance. Does that make sense?
For the accuracy graph, I didn't undersstand if the error comes from the method of approximation or from imprecise numerical calculations
Because it looked like what my numerical analysis teacher showed us when he demonstrated that approximating pi using regular polygons quickly gave absurd answers even though the method was correct.
Oh and I forgot to say the video is really cool, loved learning more about the Bernoulli polynomials and numbers
Is the worsening of the approximation of the sum after including lots of terms do to numerical error in representing tiny values of high order derivatives? Or does the sequence formed using the Bernoulli numbers actually form a divergent series?
It is actually divergent, as weird as that is!
Anxious for part 2. Coming soon?
Hopefully. I got 90% of it done before teaching started up again, and then have since been consumed by making videos for that! I look forward to getting back to this though, there is much more to come!
why the accuracy graph diverges for euler maclurin series at k=100, pleasee explain
at 2.44 shouldn't the terms be added not multiplied?
Hi! Do you (or anyone in the comments) have a source for the material in this video? As in, how you know Euler approached this problem in this way.
this may be a really stupid question as im in no way smart enough to figure out if these two things are related in any way but does the 1.6449 in pi^2/6 have anything to do with the 1.6449 in normal distribution tables? again it might be a complete coincidence but i remember pi being involved in the normal distribution somehow
If there is a connection between these, I do not know it. But that doesn't mean there isn't one! Anyone reading this who knows? I'd love to hear it if there is a reason for it!
Could you tell me please where exactly are you finding 1.6449 ? In what normal table? How is the table labeled?
Probably the 90% quantile of the standard normal distribution is meant, which is 1.64485362... as opposed to π²/6 =
1,64493406...
Doesn't seem to have a specific meaning.
The only problem with the zeta math videos is that there are not more of them. It is really sad because they are of great quality
On 35:50.
All of the Vk-1 are polynomials with grade k-1, which means that Bk=AVk-1 has degree k.
Then, if k is odd, the function Bk, an odd degree polynomial, is also odd (meaning Bk(x)=-Bk(-x) for all number x). With that, knowing that for any odd function f we have f(0)=0, bk must be 0
Hi, I was wondering how you proved the odd Bernoulli numbers vanish? I've found proofs online but they rely on an alternate Taylor series definition of the numbers. I also saw a proof in the comments that claimed that odd index implied odd degree, which in turn implied that Bk(x) was an odd function and therefore must vanish at zero, but not all odd degree polynomials are odd functions (e.g. (x+1)^3 is neither odd nor even). Please let me know what you did!
These Bernoulli polynomials (see 32:45) would be either odd or even functions if you shift them by 1/2 to the left. So perhaps it would be more straight forward to define them on the interval [-1/2, 1/2]. However, I suppose that this would result in clumsy formulas in the applications. So it was decided to define them on [0, 1], at the cost of slightly more complicated coefficients.
i see what you are doing, instead getting a million from proofing reimans hypothesis, you publish content about it to make the million 😂❤
why are the odd Bernoulli numbers zero? can someone nudge me in the right direction?
why can you just choose any C? i thought you don’t use the +C when doing IBP
2:15 aww you're so kind :,)
To say that the first series = 1 implies that infinity must be odd. Is this true?
What? No.
Euler probably knew pi^2 really well, and then /6 is trivial
49:20 that is fkn sick
But how did he know that he could just approximate that integral term as 0?
Perhaps you mean that integral term at 32:23. Unfortunately, he simply drops this error term in the following. Doing it correctly, you can use this error term to calculate your accuracy. Peculiarly, this integral at 32:23 can be estimated by
Bₘ* max[x=0..1] f⁽ᵐ⁾(x)/m!
where m = k+1 if k is uneven. This formula relies on the fact that |Bₘ| ≥ |Bₘ(x)| for all even Bernoulli polynomials.
To get an estimate for the error of the result of 48:22 of the Basel problem, we have to sum all these error terms up for all the intervals between the integers from 10 to ∞, for which we could use an integral once again. As 48:22 sums up to the 18th term, using B₁₈ as the last Bernoulli number, we use the 20th derivative of 1/x² for the error term:
(1/x²)⁽²⁰⁾ = 21! / x²²
The maximum of this function within an interval of ℝ⁺ is always on the lower bound of this interval. So we get:
Error ≤ | B₂₀ / 20! * ∑ [k=10..∞] 21! / k²² |
≤ |B₂₀| * ( ∫ [x=10..∞] 21/ x²² dx + 1/2*21/10²² )
= |B₂₀| * 1/10²¹ * (1+21/20) = 174611/330 * 41/20 * 1/10²¹ < 1.085E-18
using the value of B₂₀ = -174611/330 of 35:30 in the video.
Now the calculation of 48:22 gives Euler's result at 48:38:
1.644 934 066 848 226 436 95 ...
while the true value of π²/6 is
1.644 934 066 848 226 436 47 ...
so the true error is smaller than 5E-19. Which means that the above estimation of 1.085E-18 is quite a good one.
Give us more
But how did Euler recognize the decimal value as being pi^2/6 ? That part seems highly non-trivial.
that’s what i’m thinking
He probably knew pi^2 really well.
The equality at 2:49 can't be right, as infinite sums are not associative. However, finite sums are associative, and thus the finite sums must be 1+1/n, which does converge to 1.
_"infinite sums are not associative"_
Not quite right. In _convergent_ sums (series), you can set as many brackets as you like without changing the limit. You only have to take care when removing brackets, as the result might not be convergent any more. So the equation at 2:49 is totally correct.
@@miloszforman6270 Not quite, only unconditionally convergent infinite sums are associative.
@@codetoil
I think that you are confusing "associative" and "commutative". Re-ordering of a series has not much to do with setting brackets. E. g. you can change the limit of the alternating series ∑ [n=1..oo] ((-1)^n)/n = -ln(2) to any real value, or even to positive or negative infinity simply by re-ordering. Setting brackets, on the other hand, won't change anything.
@@miloszforman6270 You sure it works like that for infinite series? When you change the perenthesis (associativity) in an infinite series, you are effectively changing the order in which you add the terms (if you reorder individual pairs of terms, which does commute, to compensate)...
@@codetoil
I can't see what you mean. Perhaps you can try to give an example where setting brackets within a convergent series does change the limit value. But I'm convinced that you can't give such an example, as none does exist.
_Removing_ brackets might result in a non-convergent series. E. g. (1-1) + (2-2) + (3-3) + (4-4) + ... clearly converges to zero - but only if these brackets are still there.
Oh god, anything but an infinite lake of infinite infinitely far lighthouses...
But how did he know it had the same digits as π^2/6? Did he just happen to know those digits?
This part is also mysterious to me. I've always justified it in my head as "Back then people did a lot more of those kinds of calculations by hand, so their intuition was a little better" but that doesn't really wholly get me there. If anyone has any insight I'd love to hear it.
_"But how did he know it had the same digits as π^2/6? Did he just happen to know those digits?"_
Of course a guy like Euler knew a few digits of π² by heart. Now what could ∑ 1/n² be? Let's try the first 10 terms, and improve the sum by the integral, that is, 1/n: We get 1.64977, which is 99.7% of π²/6. Such a calculation took Euler a minute or two - and now he had a hint.
Calculate up to 20: oh, we have 99.9% of π^2/6. Peculiar, isn't it? Continue with 50: We have 99.99%. That can't be wrong, doesn't it? We must keep in mind that Euler was _really quick_ with numerical calculations.
In fact, the first series cannot be equal to 1 because for any finite k, there is always a residual term of -1/(k+1). If you mean that the limit of the sum as k->∞ = 1 then OK but I don't think your logic is correct because no matter what infinity is, when you evaluate the sum, you have to use an actual number and then you have the residual - hence the limit.
_"when you evaluate the sum, you have to use an actual number"_
The partial sums form a sequence, and this sequence converges to a limit. So what are you talking about? The sum of a series is usually defined as the limit of the sequence of the partial sums.
3:00 That is how you can prove that (1 − 2 + 2 − 3 + 3 + …) = (1). Nice.
Those were reciprocals, not the integers: 1 - 1/2 + 1/2 - 1/3 + 1/3 + ...
@@vocnus Still, such grouping in infinite series is very dangerous and can lead to absurdities like the one with integers. I say absurdity, because if you group otherwise, you find minus infinity. Nice.
The thing to check is that the limit of the last terms of partial sums is finite. Grouping adjacent terms isn't a problem, but you have to remember the unpaired last term which needs to approach something (ideally 0).
@@iabervon the "last" term of an infinite series of numbers ??? What's that?
Also, if "Grouping adjacent terms is not a problem":
(1 - 2) + (2 - 3) + (3 - 4) + (4 - 5) = -1 -1 -1 -1... = - infinity
1 + (-2 + 2) + (-3 + 3) + (-4 + 4) = 1 + 0 + 0 + 0 ... = 1
@@ericbischoff9444 An infinite sum is formally the limit of partial sums (n going from 1 to k) as k goes to infinity. Each of these partial sums is finite and has a last term. For example, (1-2)+(2-3)+... is formally the limit as k goes to infinity of (1-2)+(2-3)+...+((k-1)-k), which does simplify to 1-k (regardless of grouping) but that limit doesn't converge. The danger comes from looking at a telescoping sum and assuming that it simplifies to something where k doesn't contribute. The video is careless in this way; it just says (1-1/2)+(1/2-1/3)+... is 1 by canceling everything, when it's really the limit of 1-1/k as k goes to infinity. There's also an issue if your individual terms don't go to 0 that the informal expression with ... may not clearly distinguish between sequences of partial sums that are different and don't have the same limit. E.g., 1+1-1+1-1+... might mean partial sums that simplify to alternating 1 and 0 or always 1 or always 0, depending on what you pick for adding one more term to the partial sum.
Using lower case "d" in TWO different ways in the same equation is REALLY confusing.
WHY DID YOU USE LAGRANGE NOTATION :skull:
That’s wrong he didn’t know Its Pi squared over 6 because he knows that number is equal to that , He was used to series particularly product series and Taylor series for functions and he knows that 1/n squared pops there
@@angelmendez-rivera351 ? why
PROBABLY I ENTERD A ROOM WHICH IS DARK SOME LIGHT IS THROWN ANALATIC SHOULD BE SHOUD BE EMBEDDED WITH PLENTY OF EXAMPLES IE IT SHOULD BE BROKEN INTO MANY VIDEOS WITH EXAMPLES THAN VIDEO OF AN HOUR LENGTH
An infinite summation, surely, can only be approximate...since it never ends.
Hahaha, really miss the smiley on UA-cam
? no the """""infinite""""" sum is just really the limit, ergo, what the sums approach
Imagine if euler had access to computers and AI 🤔
Eskerrik asko!
Thank you so much for your support!
zeta :ahi:
I am sorry but when I see the approximation of the sum with the 17 decimals I cannot guess this is close to pi square over 6 😂😂
Wow, that sigma is probaly infinity
Written in arithmetic:
Inf·1/n²
n=1
So inf ·1/1 = inf
what?
@@DOROnoDORO *insert highpy grapgic spongebob picture here*
Candles
I computed the first 100 values of the harmonic series, years ago. I think I found it approached 7.5, but never got there. The 1/n terms became increasingly infinitesimal as n becomes larger, but never reach an endpoint. I found the Harmonic series must diverge because it does not converge to a number. I reported my findings to a youtube presence and he trashed me for questioning Nicole Oresme's (incorrect) explanation, but could not substantiate his attack on me with any facts. Use of the integral test shows divergence.
Anyway, thank you for explaining stuff.
That's quite a story, your ignoring what was *correctly* found by a bishop *in the 14th century* .
I always wondered why he didn't multiply by 6 to get pi squared
•ꔹ
Tired of seeing this problem. Have seen dozens of videos where it has been explained. It's been overdone.Would like to see how mathematicians are trying to tackle the sum of reciprocal of consecutive cubes. That problem is still unsolved. What results have been found out?
Its been found to be irrational and is called aperys constant in honor of the guy proved its irrationality, i dont know what the proof looks like but i bet its way to complicated for a simple yt video adressed to a general audience. Also the Euler Maclaurin Formula hasnt been covered that much on youtube yet so this is a welcome contribution to yt math.
@@angelmendez-rivera351 why is your reply so mean? I was only asking a question. You did not have to be so cruel. You also did not read my question. I know the reciprocal of the squares is known. I asked about the reciprocal of CUBES. I have not seen results on that problem.
Bela's point about the Euler-Maclaurin formula not being covered in UA-cam videos is really important; for me, this video was extremely helpful to understand where the formula comes from. I have not seen any video like this one yet. With respect to the Basel problem, the video used it just as an "excuse" to explain the formula. In fact, the video didn't explain the Basel problem like most other videos, which usually focus on demonstrating the equality with pi²/6, it focused on how to approximate the infinite sum, which most people don't think about.
Sorry. The premise is entirely ridiculous. Euler didn’t think “Aha! This series looks like pi^2/6!” and worked backwards! Using that premise, Euler could’ve just have equally said (using the same premise as the video proffers) “Aha! This series looks like pi^34/76231174” and Euler worked back from that. Poor mathematics and not worth wasting time watching.
Nonsense comment. Where did you get that pi^34/76231174 from? That's pure nonsense.
Thanks!
Oh wow, thank you so much for your generosity!