One of my favorite identities.

[@PotatoImaginator]

He always knows the good place to stop :)

[@TheAmbientMage]

Best road trip copilot ever

[@HagenvonEitzen]

Well it's no tthat hard to tell - it alwys happens at the end of the video

[@dhampson545]

He always stops at the end for some reason.....

[@QuantumHistorian]

I love how at 10:00 he spent multiple lines rewriting the lemma to be as a sum over all integers (rather than all natural numbers as it already was), only to have to essentially undo all that manipulation again at 16:00.

[@yds6268]

This is crazy. I thought I saw all the interesting zeta identities, but this is the best one

[@srijanbhowmick9570]

Can you share the other interesting zeta identities that you have seen ?

[@wesleydeng71]

@1:50 zeta(10) = (pi^10)/93555

[@byronwatkins2565]

At 12:35, z=pi a NOT pi/a in the sum.

[@gerryiles3925]

At 12:26 the Pi/a outside the sum was correct but the ones inside the sum should have been Pi * a...

[@Happy_Abe]

How do the pi^2’s in the denominators cancel out then from just a pi on top on the numerator?

[@gerryiles3925]

@@Happy_Abe The Pi's are still in the same places, it is the a's that move and will actually reduce to the necessary form...

[@Happy_Abe]

@@gerryiles3925 thank you!

[@quickmath8290]

have been lookin for this comment 😌

[@iamrepairmanman]

12:42 you accidentally replaced z with pi/a instead of pi*a

[@megauser8512]

The non-alternative sum results in a similar solution: 1/2 * ( coth(i) / i + 1 ), which reduces to 1/2 * ( 1 - cot(1) )

[@krtschil]

Are you sure? 1/2*(1-coth(1)) is negative which is not possible. The limit must be positive

[@krtschil]

According to Wolfram the numerical value is approximately 0.178954

[@khoozu7802]

​@@krtschillook carefully he wrote cot(1) not coth(1) for final part

[@Radana95]

The merge of sums at around 11:20 was really nice 👍🏻

[@Nico2718_]

So nice! I'm 19 and I just finished high school, I'm planning to go to university and study maths there, also because I miss some key knowledge since I went to a music oriented high school; but your explanations are so clear! I managed to understand almost everything, I only took some derivative things I didn't recall and everything regarding the hyperbolic functions (how you evaluate them and derive them) as given (and also the results of the zeta function at even integers, like I already knew the results had this form but I still don't know why), but besides that it was so clear, which is what makes you appreciate so much the end result! I love your videos, thanks!

[@goodplacetostop2973]

20:10

[@dicksonchang6647]

12:18 there is typo on the last sum, it should be πa rather than π/a

[@Happy_Abe]

How do the pi^2’s in the denominators cancel out then from just a pi on top on the numerator?

[@qzhong]

Because there's another pi outside the sum

[@Happy_Abe]

@@qzhong ah thank you!

[@donach9]

Thanks, that was confusing me

[@anishkrishnan9698]

yeah and from there its simple cancellation of the pi^2 terms, no need for relabelling or re-indexing

[@The1RandomFool]

There is a closed form of even values of the Riemann zeta function, but it isn't easy to prove. I used complex analysis along with the Laurent series expansion of cotangent, which I've also proven using the generating function for Bernoulli numbers and complex numbers. So consequently, the closed form of the zeta function involves Bernoulli numbers. Putting it all together, the problem in this video could have been calculated from just the generating function of the Bernoulli numbers, as an alternative. The lemma in the video can also be proven with complex analysis.

[@GeoffryGifari]

that lemma we use reminds me of the integration formula ∫ 1/(x^2 + a^2) dx

[@vasilismisoulis1833]

He has a typing error at 12:30 but the video is amazing

[@ClaraDeLemon]

That cotangent identity is a classic complex analysis exercise: show these two functions have the same zeroes, and their difference is bounded (and holomorphic), so it is constant, and evaluated at zero it is also zero :p

[@wynautvideos4263]

Wow i always wanted to see a formal proof of why you can factor sin and sinh that way but if its really that simple thats beautiful

[@chalkchalkson5639]

Isn't that a bit tricky since coth diverges at 0?

[@gutentag1752]

I wonder if you could also optain the same result using the Bernoulli numbers. ζ(2n) can be expressed in terms of these so there might be a way. Their generating function for t=2 would also nearly lead to the result in the video differing by a factor of just 2. Now I am quite tempted to try it out

[@lexinwonderland5741]

please do, and let us know how it turns out!!

[@noahtaul]

I was wondering the same thing, that RHS looks very much like the result of a geometric series

[@RandyKing314]

“going around right now” yeah it’s a fad😆. seriously though, good discussion on the identity, Professor, thanks for the vid

[@jordanrutledge7943]

I looked it up and apparently the numerator of the rational number for zeta(10) is still 1, it’s 1/93555, but zeta(12) it finally breaks the pattern 691/638512875

[@MrCreeper20k]

Out of college for one year at a job where I don't really use math. This channel keeps my passion for math alive! I can't wait until I get the whole work life balance thing right and I can start dedicating some time to learning math again.

[@tiltltt]

what a quirky function, I wonder if there are any famous hypothesis that use it

[@stewartzayat7526]

Certainly not. I can't imagine this function being important in any way

[@HagenvonEitzen]

@@stewartzayat7526 Certainly not when it is only defined for Re(s)>1 :)

[@emanuellandeholm5657]

The power of pi over something breaks down for RiemannZeta(12).

[@andrej8875]

It doesn't?

[@emanuellandeholm5657]

@@andrej8875 According to Wikipedia, this is where it happens

[@RobsMiscellania]

@@emanuellandeholm5657 It is still a rational number multiplied by pi to the 12th power. In general that will always be the case for even exponents, as Euler showed. So it's true, in general the number Zeta(2s) will not have the form (1/n)*pi^(2s), where n is a natural number. But Zeta(2s) = (p/q)*pi^(2s) is in fact always true for p,q,s natural numbers.

[@emanuellandeholm5657]

@@RobsMiscellania I think Professor Penn meant that the pattern was 1/k times an even power of pi, which breaks down for 2n = 12. I could be wrong tho.

[@RobsMiscellania]

@@emanuellandeholm5657 Yes, that pattern of even power of pi divided by a whole number breaks down for s=12. I don't believe there ever is an instance of it being true ever again for any even number 12 or greater. I can see how it might be misleading for someone who doesn't know the connection between Bernoulli numbers and the solution of the "generalized" Basel problem, where one might think that it truly is 1/k times the even power of pi.

[@chahn987]

Didn't know about the Weierstrass product until googling it after it was mentioned. Was actually playing around with an alternative expansion of sine not long ago by using the roots and plotted and ended up looking pretty close to sine a couple weeks ago after using a normalization factor. Didn't know this actually had a name

[@Calcprof]

Euler works out (I belive) up to ζ(20), and works out the recurrence in Introduction to the Analysis of the Infinities, available for free (in Latin) at the Euler Archive. There is so much computation that the latin is not hard to read, -- just follow the computation. There is also an english translation available from the usual sources.

[@dragonmudd]

So excited to see some hyperbolic trig functions in your videos. More please!

[@АндрейДенькевич]

Sum of Zeta(i) for i:=2 to n equals Sum of (a- 1/a^n)/(a-1) for a:=2 to infinity - zeta(0). Where 1/a^n is sum of carry*(a-1) from n-th to infinity digit of (-a)-ary number system (i.e. a-ary number system with carry to m-th digit not a^(m-1) but 1/a^(m-1), a will be infinity of that number system, dividing by (a-1) is needed to receive carry in each of digits (from previose digit) , it will be numbers 1/a^m (m=0..n-1) in each of n digits)). (a - 1/a^n)/(a-1) = 1/a^0+1/a^1+..+1/a^n. For example. 1/2^0+1/2^1=(2-1/2^1)/(2-1)=1+1/2; 1/2^0+1/2^1+1/2^2=(2-1/2^2)/(2-1)=1+3/4; 1/2^0+1/2^1+1/2^2+1/2^3=(2-1/2^3)/(2-1)=1+7/8; 1/5^0+1/5^1+1/5^2+1/5^3=(1-1/5^3)/(5-1)=1+124/(125*4)=1+31/125; 1/6^0+1/6^1+1/6^2+1/6^3=(1-1/6^3)/(6-1)=1+215/(216*5)=1+43/216; If to take difference between 2 cosequent Sum of Zeta(i) above then: Zeta(2) equals Sum of (a-1/a^2)/(a-1) for a:=2 to infinity- Zeta(0); Zeta(2) = 1/(2-1)-1/(2^2*(2-1))+1/(3-1)-1/(3^2*(3-1)) +1/(4-1)-1/(4^2*(4-1))+1/(5-1)-1/(5^2*(5-1)) +...; Zeta(2) = 1-1/4+1/2-1/18 +1/3-1/48+1/4-1/100 +...; Zeta(2) = Zeta(1)-1/4-1/18 -1/48-1/100 -1/180 -1/276 Zeta(2)=Zeta(1)- Sum 1/(a^2*(a-1)) for a:=2 to infinity; Zeta(3)=Zeta(2)- Sum 1/(a^3*(a-1)) for a:=2 to infinity; Zeta(4)=Zeta(3)- Sum 1/(a^4*(a-1)) for a:=2 to infinity; Zeta(3) equals Sum of ((a-1/a^3) - (a-1/a^2))/(a-1) for a:=2 to infinity - Zeta(0); Zeta(4) equals Sum of ((a-1/a^4) - (a-1/a^3) + (a-1/a^2))/(a-1) for a:=2 to infinity - Zeta(0); Zeta(5) equals Sum of ((a-1/a^5) - (a-1/a^4) + (a-1/a^3) - (a-1/a^2))/(a-1) for a:=2 to infinity - Zeta(0). Sum of Zeta(i) for i:=2 to infinity equals Sum of (all ones row)*(it's a-convolution(take 1/(element of Pascal matrix))* (all ones column) for a:=2 to infinity. (1 1 1 ...)* ( 1 =2^0 -> 1/2^0 1 1 =2^1 -> 1/2^1 1 2 1 =2^2 -> 1/2^2 1 3 3 1 =2^3 -> 1/2^3 ... ) *(1 1 1 ..)^T =2/1=1+1/1 + (1 1 1 ...)* ( 1 =3^0 -> 1/3^0 1 1 1 =3^1 -> 1/3^1 1 2 3 2 1 =3^2 -> 1/3^2 1 3 6 7 6 3 1 =3^3 -> 1/3^3 ... ) *(1 1 1 ..)^T =3/2=1+1/2 + (1 1 1 ...)* ( 1 =4^0 -> 1/4^0 1 1 1 1 =4^1 -> 1/4^1 1 2 3 4 3 2 1 =4^2 -> 1/4^2 1 3 6 10 12 12 10 6 3 1 =4^3 -> 1/4^3 ... ) *(1 1 1 ..)^T =4/3=1+1/3 + (1 1 1 ...)* ( 1 =5^0 -> 1/5^0 1 1 1 1 1 =5^1 -> 1/5^1 1 2 3 4 5 4 3 2 1 =5^2 -> 1/5^2 1 3 6 10 15 18 19 18 15 10 6 3 1 =5^3 -> 1/5^3 ... ) *(1 1 1 ..)^T =5/4=1+1/4 + (1 1 1 ...)* ( 1 =6^0 -> 1/6^0 1 1 1 1 1 1 =6^1 -> 1/6^1 1 2 3 4 5 6 5 4 3 2 1 =6^2 -> 1/6^2 1 3 6 10 15 21 25 27 27 25 21 15 10 6 3 1 =6^3 -> 1/6^3 ... ) *(1 1 1 ..)^T =6/5=1+1/5 + S(a)= sphere * uncurving of a-sphere with taking zeroinfinity inversion * sphere^T= a/(a-1)=1+1/(a-1) + ..................... = zeta(1)+zeta(0). So we can define zeroinfinity inversion following way: I(a)=S(a+1)-1; I(b)=S(b+1)-1; I('+')= sum of indexes of above sequence. For example "a+b=c" became "I(a) I('+') I(b) = I(c)". "2+3=5" became "1/2 I('+') 1/3=1/5 (ie 5-th member of above sequence - 1). a/n + b/n = 2-convolution (a b) * (n n) / n^2 =(a*n+n*b)/n^2. a/n + b/n + c/n= 3-convolution (a b c) * (n n n) / n^3 =(a*n*n*n+n*n*b+c*n*n)/n^3. a/c + b/d = 2--convolution (a b)*(c d)/c*d. a/c + b/d + f/g = 3-convolution (a b f)*(c d g)/c*d*g. So division operator A/B is n-convolution A*B and after that applying zeroinfinity inversion multiplying by 1/convolution B*B. So inversion of + is convolution.

[@ForcesOfOdin]

Lots of cool techniques here. I love the conversion of m=1 to infinity to m over all Z by splitting in half, using the squares to exchange the second half with negative indices, and then adding in the 0. And in the end we get something like a crazy new definition of e😁

[@Anonymous-zp4hb]

Beautiful. Great when I can follow along with little knowledge. And like you said, it's only the equating sinh(z)/z to that infinite product that wasn't fully justified in the video. But I've seen a similar breakdown of the sin function in a simplified solution to, funnily enough, the Basel problem.

[@edmundwoolliams1240]

KZbin's trending function!

[@anon5976]

Michael I don't know if you read these comments, it's fine if you don't. I just wanted to say that I'm so grateful for these videos and I have learned so many cool math tricks from you. Whenever I swap the order of summation or rewrite a geometric series using its closed form I feel like it's a micro homage to you. These videos are changing the world.

[@tomholroyd7519]

11:31 that's a nice capital sigma

[@CM63_France]

Hi, 6:49 : multiplied by some constant, which is equal to 1, by doing z=0.

[@patrickhickey7673]

Wow, that’s a really nice identity!

[@General12th]

Hi Dr. Penn! Very cool!

[@Handelsbilanzdefizit]

Of course, this sum is the inverse of: lim_n->∞ (sum_i^n exp(2i/n)2/n) And it explains why it works for even indices i.

[@lexyeevee]

ah, yet another place where using π instead of τ obscures a beautiful pattern - ζ(2) = τ²/24 = ½ B₂ τ²/2! ζ(4) = τ²/1440 = -½ B₄ τ⁴/4! ζ(6) = τ²/60480 = ½ B₆ τ⁶/6! ζ(8) = τ²/2419200 = -½ B₈ τ⁸/8! where B_n are the Bernoulli numbers.

[@anishkrishnan9698]

Not really, just use tau = 2pi and the alternating 1/2's and -1/2's just become ascending in powers of 2, still beautiful

[@lexyeevee]

​@@anishkrishnan9698 it's only a factor of two, so it's never going to be _too_ different, but with π you necessarily end up with an extra term somewhere - either keeping the ½s, or having odd powers of two. which is the whole objection to π: everywhere it shows up, there's a missing factor of 2 somewhere that makes everything just a little bit clumsier

[@Jack_Callcott_AU]

Michael's lemma could probably produce something interesting if you set x = a and integrate both sides.

[@jacksonstarky8288]

Another interesting topic would be the connection between gamma (the Euler-Mascheroni constant) and the zeta function... just because any math that connects pi, e, and gamma is beautiful and a possible starting point for the eventual proof of the irrationality and transcendence of gamma.

[@jacksonstarky8288]

I'd love to see a video about the general formula for zeta(2n).

[@veselindimov307]

As far as I remember it's related to Bernoulli's numbers in some sense and it's quite a long topic of its own. But I think Mathologer had done a great video about it. Maybe try to search it.

[@ירוןעובדיה-ל4ה]

תודה!

[@hydra147147]

Using Taylor series of cotangent gives equivalen proof: x*cot(x)=1-2x^2/pi^2*zeta(2)-2x^4/pi^4*zeta(4)-2x^6/pi^6*zeta(6)-..., so substituting x=1 and x=I respectively we get the same identities

[@ZipplyZane]

This is twice today that I'm seeing summation notation written differently. Here you just specify that the index is all rational numbers, rather than starting at -inf and going to +inf. And earlier I saw someone write n≥0 as the index instead of having n=0 on bottom and infinity on the top.

[@tokajileo5928]

what if you put odd numbers into the series instead of even?

[@timmaths]

Zeta of 2n gives nice answers with (smth)*pi^2n. Zeta of 2n+1 doesn't have nice values. Just for a start, Zeta(3) is irrational, and at least one of Zeta(5), Zeta(7), Zeta(9) and Zeta(11) is also irrational. We haven't found nice ways of expressing the values, so good luck calculating things with Zeta(2n+1)

[@Jack_Callcott_AU]

In the infinite series that Michael investigates the division by Π^(2n) makes the expression a rational number, as he said. Those rational numbers are related to the Bernoulli numbers. Michael should do a video on Bernoulli numbers.

[@Noam_.Menashe]

Another way to get the coth sum is by using imaginary values in the digamma function.

[@conrad5342]

At some point in time one has been surprised of pi, i and e combining as e^(i pi) = -1 .... Zeta funktion: Hold my beer.

[@maxfriis]

Damn - he is lucky :P

[@wyattstevens8574]

If this isn't the exact value, it's very close: the reciprocal of the alternating sum is at least *close* to 0.8 above that of the absolute sum. If the thumbnail's anything to go by, that is. Even with just these 4 terms, it's already very close!

[@ZakiZaki-w6e]

1. Déterminer les racines carrées du nombre complexe 33+56i 2. Soient x et y deux reels tels que 33x-56y=x/x”2+y”2 Et 56x+33y=-y/x”2+y”2 Donner la valeur de |x|+|y| “2=squard

[@ZakiZaki-w6e]

Help

[@timmaths]

Je ne pense pas que écrire un commentaire en français sur une vidéo en anglais soit très utile, mais voici une réponse. 1) 7+4i et -7-4i

[@purplerpenguin]

I'm sure I'm missing something really obvious here, but why is a function equal to the infinite product of its zeros? I was expecting some other, multiplicative, factor.

[@jaimeduncan6167]

The patterns break for 12, not 10, according to mathematica.

[@GeoffryGifari]

guys, is there anything interesting about this one: if we have a ratio of riemann zeta function between successive even numbers ζ(2n)/ζ(2n+2) = Constant × 1/π^2 (result from the video) if we take the limit as n goes to ∞, does the ratio converge to something?

[@rsrudhro9664]

WolframAlpha says it converges to 1, which kind of makes sense, as n -> infinity zeta(2n) is roughly equally to zeta(2n+2)

[@GeoffryGifari]

@@rsrudhro9664 converges to one? so as long as the zeta function of an even argument has the form of rational number × π^2 , the "Constant" term in the ratio ζ(2n)/ζ(2n+2) = Constant × 1/π^2 will end up approaching π^2? that's pretty cool. didn't see that one coming

[@tien-chunghu3615]

Can one get the Zeta(2) through the Lemma (Coth(pia identity) by taking a approaching to 0 ?

[@integrando1847]

Its just the taylor serie of coth(x)

[@Facetime_Curvature]

I'm kinda confused and haven't gotten much into the zeta function... but didn't you say that the zeta functions nice expression of pi to the s over some number breaks down after 10 or so. But here we are making an expression that is treating the reducibility of the expression as reciprocating by 2n over pi to the 2n? Since you are assuming over infinity doesn't that not work when n>=5? I'm sorry if this is incorrect in some obvious way, but I'm not quite tracking. Is this only a form of continuation of an approximate function or something?

[@nasim09021975]

Well, I have never heard of of anyone vocalize mˢ as "m to the s" 😅 Hmmm🤔 Generally , it has always been (well, almost always) either "m to the sᵗʰ power" or "m to the power s"

[@CTJ2619]

Wow nice video

[@GreenMeansGOF]

12:46 I don’t understand. The summand simplifies to 1/(1+(ak)^2). How does that reindex to 1/(m^2+a^2)?

[@TheAwesomeJordy]

Pi/a should have been Pi*a in the summand

[@GreenMeansGOF]

@@TheAwesomeJordy got it. Thanks.

[@chandranisahanone]

U r best ❤️🛐

[@gashtag9483]

Cool

[@shanathered5910]

cool

[@sinus_hiphop]

damn! 8)

[@bugzbunny109]

fist

[@Aditya_196]

🫠 despite knowing nothing I don't know why I watched this for 5 minutes

[@szymonraczkowski9690]

cool