Wonderful video. Finally an exposition on the zeta function that goes beyond merely saying "the zeros of the zeta function tell us something about prime numbers", but actually demonstrates it.
@@JamesWylde adblock exists, plus he works really hard on making these hour long videos which rival the quality of University level lectures so its totally justified i feel
When he discovered this proof for the sun equating to exactly pi^2/6 he must’ve been absolutely astounded and excited. Thank you so much for the amazing video.
That is certainly the niche I'm trying to fill. There is a lot of excellent youtube math content that is short form (which granted is what the algorithm loves) but here I'm trying to tell coherent multipart stories. I'm glad that appeals to you as well!
46:25 The numerator of the fraction in the summation should be (p^-s)*log(p), not log(p). It is corrected in 46:44 though as when the geometric series is written out in the third term, k is implied to start from 1 (as can be seen in 47:01). Otherwise after writing out the geometric series, k would start from 0, which doesn't make much sense in the context of 47:01. Not sure if I'm explaining myself clearly, but that's just what I observed.
I‘m just half way through another Zetamath re-run. Really amazing videos explaining quite high-level maths in an understandable way. I consider these videos to be all the way up there with 3b1b and mathologer. And this particular one is probably my favorite math video out there! Please make more of them!!
After watching so many videos on Riemann hypothesis I thought I will not find any new information from this topic...but this video has proved me wrong 👏
One of the best maths videos on youtube! Looking forward to watching more of your number theory videos, including the next one on analytic continuation.
Dude! Your videos are gold! The most precise and detailed derivations of the shown properties of the zeta function. I have seen no textbook that can cope with your explanations! Chapeau!
What an incredible, incredible series. It is currently my last summer holiday before going to university to study applied mathematics. I suffer from a lot of doubt about whether I've made the right choice, but videos like this one reassure me that I do want to study mathematics and all of its complexities and secrets. Thank you so much for making these videos.
Well done! At last I have found a channel that is interested in knowledge. Infinite series blew me away. The way this topic is presented makes me want to learn more.
What a wonderful presentation! You can't get such insight out of any book on the subject. For something like this alone, you just have to love TH-cam. Hopefully you can do more math presentation of such form, it would be highly appreciated. Thanks a lot, zeta!
Excellent videos, very clearly explained. You focus on explaining the essential concepts with great graphics but without of too many details that can sidetrack the listener. KEEP UP THE GOOD WORK!
I. POSTED A COMMENT ON THE PREVIOUS VIDEO. I'M A RETIRED EE, AND I HAVE BACKGROUND IN BASIC CALCULUS 1,2,3, DIFF EQU'S, SOME VECTOR , & TENSOR CALCULUS, SOME LINEAR ALGEBRA, SOME FUNCTIONS OF A COMPLEX VARIABLE. I'M WEAK ON THE FIELD THE PROBABILITY SINCE I NEVER TOOK A COURSE OR READ A BOOK ON PROBABLY AND STATISTICS. I'M JUST FAIR IN MATHEMATICS. I'M FAIRLY GOOD WITH MOST CONCEPTS. SO I DO WATCH AND CAN UNDERSTAND FOR THE MOST PART OF THESE MATH. IDEAS POSTED ONLINE. BUT I STRUGGLE AT TIMES. BUT YOU SIR, ----- YOU ARE A MATHEMATICS GENIUS. AND THESE LAST TWO VIDEOS I WATCHED ARE ABSOLUTELY AMAZING. I WAS NOT AWARE OF THESE CONCEPTS. I REPEAT THESE VIDEOS ARE ABSOLUTELY AMAZING. I AM ABSOLUTELY DUMBFOUNDED IN THE LAST TWO VIDEOS. I'M GOING TO CHECK OUT MANY MORE OF YOUR VIDEOS. YOU ARE A SUPER TEACHER. YOU TEACH MATHEMATICS VERY CLEARLY SIR. KEEP UP THE AMAZING WORK LET ME DO ON THESE VIDEOS. YOU HAVE BENEFITED MANY MANY THOUSANDS OR MORE SO MUCH WHAT'S THE IN-DEPTH DETAILED EXPLANATIONS YOU PUT FORTH IN THESE VIDEOS. YOU TEACH WITH EXTREME CARE AND CONCISENESS IN YOUR DEFINITIONS. IT IS SO WONDERFUL TO FIND A REALLY REALLY GOOD MATH TEACHER, REALLY KNOWS HIS STUFF. THANK YOU MUCH FOR THESE VIDEOS THAT TEACH SO MUCH ON ARE SO INFORMATIVE ON MATHEMATICS PRINCIPLES.
I TOTALLY AGREE WITH YOU. THIS MATH VIDEO IS COMPLETELY ALL C A P S MATERIAL: THE TYPE OF MATH VIDEO THAT MAKES YOU WANT TO GET UP OUT OF YOUR CHAIR AND SHOUT OUT IN APPROVAL!
Zetamath, you not only make great Sudoku puzzles, you make great videos. You have a really good handle on advanced mathematics and can explain it quite clearly. Thanks for sharing.
Gone through all the videos of the series (in the wrong order but w/e) and this gotta be one of the most detailed and sophisticated math series on TH-cam. Love it. Now I can only hope that in the final part you either prove or disprove the Riemann Hypothesis ;)
As was mentioned by other people here, the middle term at 46:45 is wrong (lacks a multiplicative p^-s term). So the sum at the right starts at k=1. This is correctly done in the following. Such minor errors could be mentioned in an "errata note" in the introductory text, or pinned at the start of the comment section.
Thank you for sharing this, like an accessible trail to a normally quite hard to reach location, amazing to see these sights … to gain a bit of insight
Your summing up at the end and equating it to the zeros was fantastic - and exactly what I've been looking for - will watch another 3 times to cement it into my brain. Now, if you could show a person like me, the step-by-step derivation of Riemann's product of zeros function (time index 48:51) in your next video, it would be much appreciated. Just an incredible video - thank you.
Getting to the step by step derivation of this formula is one of my goals for the series , but it will require a few videos to get there. Keep following, though, and we will make it there together!
Very interesting! In 46:30 we have the inverse of linear aproximation that can be used in Newton-Raphson method to find roots of zeta function from a initial condition s0.
Thank you for sharing these videos. One thing that seems clear to me, I hope I am not mistaken, is that Euler was both a doer and a thinker of mathematics. It is wonderful to see his persistence in trying to find a solution - and have we not experienced such things ourselves? The disappointment of something not working as intended seemed to prompt his creativity. At his level - I see math as a creative topic as much as any creative craft. Interim conclusion: degrees in mathematics should be arts and not science?
This is wonderful! I'm in awe of your work, you don't skip the hard part of a theorem and that's so nice. Thank you. I just have one question, I don't understand how adding the zeros of a complex function creates this kind of wave propagation in a function defined on the reals?
Very much appreciated your work... I want to understand more clearly that "'"how zeros of complex zeta function explains the distribution of primes??""" I hope your next video will address this....
Thanks! This is an amazing video. I'm trying to get the sum of 1/n^4 from 1 to infinity. Please help me get started on finding the formula for the partial sums of the coefficients of x^5. There's no obvious pattern that I'm seeing for 1/4, 7/18, 91/192...
At 50:26 I don't understand where the k comes from in p^k - but I suppose it will be explained in one of the next videos. Great contents, so enlightening.
The p^k is there to indicate that the sum is over all prime powers less than x, not just all primes less than x. The world would be a better place if that k wasn't there, but sadly, it is.
Phenomenal video I love the content, quality and length! thankyou! around min 23:00 you ask to find the sum of 1/n^4, it took me a few attempts and I was wondering if anyone has another solution; heres mine: take RHS and plug into x the value (ix) to get [ix*prod(1+x^2/n^2pi^2)], multiply this with original RHS to get L1: ix^2*prod(1 - (x^4)/(n^4pi^4)) next in LHS plug into x (ix) (to get -isinh(x)) to find i(x + x^3/3! + x^5/5! + x^7/7! + ...) multiply with original LHS and only look for coefficient of x^6 to get: something + -i/90 x^6 + something equating coefficient of the other product of the RHS's we get the answer.
The sine function being an infinite product and equivalently an infinite sum is similar to the Euler products being Dirichlet series. I think the odd values of the zeta function ζ(2n + 1) are transcendental because the logarithm may describe it in some. The logarithm is a transcendental function that switches multiplication to summation. The values ζ(2n + 1) possibly being transcendental may explain the difficulty in deriving them. The Riemann zeta function is the solution of no algebraic ordinary differential equation on its region of analyticity. See the abstract of “Does the Riemann zeta function satisfy a differential equation?” (2015) by Gorder. This makes the zeta a function a hypertranscendetal function. The zeros of the zeta function are found in its analytic continuation. Putting this together, the zeros of of the zeta function are no solutions to any algebraic differential equation.
_"The values ζ(2n + 1) possibly being transcendental may explain the difficulty in deriving them. "_ Now the values ζ(2n) are certainly transcendental - and they are well-known. For ζ(2n+1) it is not even known if they are transcendental at all, except for ζ(3).
Yeah - I get it. If you trip over a tree root - a square root even - and multiply that by a pie that is in the shape of a square then, X is important somehow and then another Pi (apple or steak & kidney is irrelevant) and add some binomials and then then a lot more squares and one more cube.
something is wrong with the formula reached at 46:28 because when p goes to infinty the terms inside the series goes to infinity , i think it should be p to the power of s not -s or we can add p to the -s in the numerator in fact that's what i got after calculating the derivative , idk if any one found another result plz let me know
There’s one small issue that is not explained in the video. At 53:30 the graph starts including the first few terms of the summation involving the zeros of the zeta function But it appears that any partial sum would be complex. To be exact, the terms are of the form (x^alpha)/alpha, where alpha is a zero of the zeta function and x is an integer or real number. In general, each term in the summation is a complex number because every zero of the zeta function is complex. In the limit, this sum may indeed converge to a real number but any partial sum is probably complex with a non-zero imaginary part. My question is whether you used the magnitude or the real part of the partial sum for your graph and does one give a better fit than the other.
Good eye, but sadly the answer is more boring than you might hope. Actually, the terms come in conjugate pairs, so when computing the partial sums, I included them in pairs to keep the thing being graphed real.
That's a good question, I'm not sure if they will get mentioned again in this story, but they come up so often, it wouldn't surprise me if we talk about them on the channel again!
I remember as a student being astounded by the fact that, no matter how large an integer n, there are stretches of at least n integers in a row with NONE of them prime! Sometime later (in grad school, I hope) it belatedly occurred to me that the same is true for the perfect squares ! 😲 Duh!
Hey, I just recently discovered you channel. Is there a possibility that you would make a video about automorphic forms? It seems that you mainly focus on analytic number theory. So it may serve as a rich and interesting topic to discuss on your channel in a more informal fashion.
At 47:00 , shouldn't the first term of the expansion be log(p)? [Expanding 1/(1 - p^-s) of 46:52 to a geometric series, the first term is 1. So the sum over all p^k (also in 46:52 ) must start with p^0.] Am i wrong?
The formula for the derivative at 46:25 is wrong. It must contain an extra p^-s factor. Nevertheless, the left hand side of 46:45 and the right hand side are equal once again, only the middle term is wrong. A double error, so to say, which cancels out. For the geometric series, we have 1 + q + q² + ... = 1/(1-q) and q + q² + q³ + ... = q/(1-q) The extra p^-s term changes it to the second version above, so everything is all right in the following. Had already been mentioned in these comments here, as I just noticed.
If all goes well it should be within the next couple of months, but I should be careful not to overpromise, as this one ended up coming out a full year later than I thought it would! Thanks for the compliment, I'm glad you're enjoying the content!
53:23 - 54:25 It would have been nice to state the number of zero pairs used for this animation. It's "the first handful", but what is a "handful of zeros"?
Can someone help me understand ? At 30:50 he says harmonic series of ζ(1) diverges I am no mathematician but i do not understand why does zeta of 1 diverges if you are constantly adding up smaller and smaller terms it should converge the only difference between ζ(1) and ζ(2) is that ζ(2) is doing it much^2 faster 😅
Why don't you look at Wikipedia: "Harmonic series", chapter "comparison test". It's a very easy proof that this series diverges. It is frequently given in schools. On the other hand, ζ(2) converges, as does ζ(1.1) or ζ(1.01).
This is amazing! Though I think the analogy you made between rarity of primes and integers or squares of integers and so on could be misleading. Still great work!
when you take the log of sin(x) you must remember that x must be greater than zero (sin(x) varies between -1 and 1) , otherwise log will be a complex number. How do you justify then the use of log(sin(x)) and the use of derivative of log(sin(x)) to obtain cot(x) ? (of course I know the results are correct.I just want a rational explnantion)
A more rigorous description here would talk about analytic continuations, using the fact that these things agree when x>0. it is a great detail to pick up on, but I didn't talk about analytic continuation until the next video, so I was a bit hand wavy here (as is the general theme of the series).
Then there would be complex roots and thus the statement of p having only 3 roots wouldn't apply. The narrator didn't say "3 real roots" but "these 3 zeroes" which makes the statement correct.
This proof is near and dear to my heart, and is certainly something I've contemplated how much I could say about. Unfortunately, the proof is usually the pot of gold at the end of the rainbow to motivate a year long course in abstract algebra, because it combines an awful lot of different math all in one place. If I think of a way to present it, I definitely well.
51:00 This "psi function" (ψ-function) is also known as the "second Chebyshev function" (besides ϑ(x), which is the first Chebyshev function). I tried to reproduce this calculation of 53:23 onwards using spreadsheet calculation. It's a bit clumsy at first if you're not that practiced with complex algebra, but it worked. I used the formula =SQRT(x) * ( COS(LN(x)*alpha) + 2*alpha*SIN(LN(x)*alpha) ) / (0,25 + alpha^2) and these have to be added up for all nontrivial zeta zeros ("alpha") 14,1347..., 21,022..., 25,010..., ..., (the imaginary parts! being used as reals here, the real parts of 0.5 only accounting for the leading square root) or better to say: for as many of them as appropriate. 100 oder 200 of these pairs give quite decent results at least for x
what you showed at 53:12 and at 54:30 was the clearer simplification of the Riemann Hypothesis idea on the internet I already know all of this but I just love to watch the Riemann Hypothesis every time I see a new video about it btw at the beginning of the video the Basel Problem explanation was a bit overkill which made it less good for me you should also check m.y. v.i.d.e.o.s.
Wonderful video. Finally an exposition on the zeta function that goes beyond merely saying "the zeros of the zeta function tell us something about prime numbers", but actually demonstrates it.
This channel needs more support, this is such amazing content! I feel like I learned so much in the past 3 videos, and I can't wait for the next one!
He just needs to drop the mid roll ads, if he did I would probably support it.
@@JamesWylde adblock exists, plus he works really hard on making these hour long videos which rival the quality of University level lectures so its totally justified i feel
Am I the only one that likes adds? They try to show you things that might interest you, and skip add helps them do that.
@@JamesWyldesuck it up and pay for yt premium
I agree it’s excellent
When he discovered this proof for the sun equating to exactly pi^2/6 he must’ve been absolutely astounded and excited. Thank you so much for the amazing video.
I wonder if he was more excited by figuring out the value was pi^2/6 or figuring out how to prove it!
You made this as clear as possible while still keeping a healthy dose of rigour and detail. Awesome work!
Where have you been? I am a math “hobbyist” and tutor. You have a marvelous way of explaining things.
Seeing what has happened to youtube math has been very incredible. I hope all of these channels get the recognition they deserve
This is amazing! You have a rigoros and fun way of describing :) We need more in depth math like you on TH-cam
That is certainly the niche I'm trying to fill. There is a lot of excellent youtube math content that is short form (which granted is what the algorithm loves) but here I'm trying to tell coherent multipart stories. I'm glad that appeals to you as well!
46:25 The numerator of the fraction in the summation should be (p^-s)*log(p), not log(p). It is corrected in 46:44 though as when the geometric series is written out in the third term, k is implied to start from 1 (as can be seen in 47:01). Otherwise after writing out the geometric series, k would start from 0, which doesn't make much sense in the context of 47:01.
Not sure if I'm explaining myself clearly, but that's just what I observed.
If not for you, I would have lost my confidence in math. Thank you so much!!
This should be one of the most famous math channels soon, because definitely it's one of best
I‘m just half way through another Zetamath re-run. Really amazing videos explaining quite high-level maths in an understandable way. I consider these videos to be all the way up there with 3b1b and mathologer. And this particular one is probably my favorite math video out there! Please make more of them!!
This is a superb explanation. My understanding increased tremendously. Thank you, imagine it takes tremendous work to put this together.
After watching so many videos on Riemann hypothesis I thought I will not find any new information from this topic...but this video has proved me wrong 👏
i do not why such a fantastic channel only got fans less than 10k, i hope more people can find this...
this video had more narrative tension than most movies i've seen. feel like i should cheer.
One of the best maths videos on youtube! Looking forward to watching more of your number theory videos, including the next one on analytic continuation.
Dude! Your videos are gold! The most precise and detailed derivations of the shown properties of the zeta function. I have seen no textbook that can cope with your explanations! Chapeau!
What an incredible, incredible series. It is currently my last summer holiday before going to university to study applied mathematics. I suffer from a lot of doubt about whether I've made the right choice, but videos like this one reassure me that I do want to study mathematics and all of its complexities and secrets. Thank you so much for making these videos.
Well done! At last I have found a channel that is interested in knowledge. Infinite series blew me away. The way this topic is presented makes me want to learn more.
this is byfar one of tte most amazing proof walkthroughs I've ever seen
That moment at 19:17...jaw-dropping! Such a good lesson in persistance - it's incredible that Euler came up with this!
Everything about your presentation style is AWESOME!
That was so enjoyable, and _crystal clear._
What a wonderful presentation! You can't get such insight out of any book on the subject. For something like this alone, you just have to love TH-cam.
Hopefully you can do more math presentation of such form, it would be highly appreciated.
Thanks a lot, zeta!
Excellent videos, very clearly explained. You focus on explaining the essential concepts with great graphics but without of too many details that can sidetrack the listener. KEEP UP THE GOOD WORK!
This video was so well-done that it motivated me to become a Patreon supporter. This channel deserves far more support!!
This video was so beautiful.....i got emotional by watching it.Very well done zeta math keep it up!!!!
I. POSTED A COMMENT ON THE PREVIOUS VIDEO. I'M A RETIRED EE, AND I HAVE BACKGROUND IN BASIC CALCULUS 1,2,3, DIFF EQU'S, SOME VECTOR , & TENSOR CALCULUS, SOME LINEAR ALGEBRA, SOME FUNCTIONS OF A COMPLEX VARIABLE. I'M WEAK ON THE FIELD THE PROBABILITY SINCE I NEVER TOOK A COURSE OR READ A BOOK ON PROBABLY AND STATISTICS. I'M JUST FAIR IN MATHEMATICS. I'M FAIRLY GOOD WITH MOST CONCEPTS. SO I DO WATCH AND CAN UNDERSTAND FOR THE MOST PART OF THESE MATH. IDEAS POSTED ONLINE. BUT I STRUGGLE AT TIMES. BUT YOU SIR, ----- YOU ARE A MATHEMATICS GENIUS. AND THESE LAST TWO VIDEOS I WATCHED ARE ABSOLUTELY AMAZING. I WAS NOT AWARE OF THESE CONCEPTS. I REPEAT THESE VIDEOS ARE ABSOLUTELY AMAZING. I AM ABSOLUTELY DUMBFOUNDED IN THE LAST TWO VIDEOS. I'M GOING TO CHECK OUT MANY MORE OF YOUR VIDEOS. YOU ARE A SUPER TEACHER. YOU TEACH MATHEMATICS VERY CLEARLY SIR. KEEP UP THE AMAZING WORK LET ME DO ON THESE VIDEOS. YOU HAVE BENEFITED MANY MANY THOUSANDS OR MORE SO MUCH WHAT'S THE IN-DEPTH DETAILED EXPLANATIONS YOU PUT FORTH IN THESE VIDEOS. YOU TEACH WITH EXTREME CARE AND CONCISENESS IN YOUR DEFINITIONS. IT IS SO WONDERFUL TO FIND A REALLY REALLY GOOD MATH TEACHER, REALLY KNOWS HIS STUFF. THANK YOU MUCH FOR THESE VIDEOS THAT TEACH SO MUCH ON ARE SO INFORMATIVE ON MATHEMATICS PRINCIPLES.
I TOTALLY AGREE WITH YOU. THIS MATH VIDEO IS COMPLETELY ALL C A P S MATERIAL: THE TYPE OF MATH VIDEO THAT MAKES YOU WANT TO GET UP OUT OF YOUR CHAIR AND SHOUT OUT IN APPROVAL!
Zetamath, you not only make great Sudoku puzzles, you make great videos. You have a really good handle on advanced mathematics and can explain it quite clearly. Thanks for sharing.
Excellent video. Provides illuminating depth on the relationship between Euler's and Riemann's work. I am surprised this does not have more views.
Gone through all the videos of the series (in the wrong order but w/e) and this gotta be one of the most detailed and sophisticated math series on TH-cam. Love it.
Now I can only hope that in the final part you either prove or disprove the Riemann Hypothesis ;)
Your long form deep-dive videos are gorgeous. Thanks!
As was mentioned by other people here, the middle term at 46:45 is wrong (lacks a multiplicative p^-s term). So the sum at the right starts at k=1. This is correctly done in the following.
Such minor errors could be mentioned in an "errata note" in the introductory text, or pinned at the start of the comment section.
Thank you for sharing this, like an accessible trail to a normally quite hard to reach location, amazing to see these sights … to gain a bit of insight
this channel is a hidden gem
Your summing up at the end and equating it to the zeros was fantastic - and exactly what I've been looking for - will watch another 3 times to cement it into my brain. Now, if you could show a person like me, the step-by-step derivation of Riemann's product of zeros function (time index 48:51) in your next video, it would be much appreciated. Just an incredible video - thank you.
Getting to the step by step derivation of this formula is one of my goals for the series , but it will require a few videos to get there. Keep following, though, and we will make it there together!
Thank you for your efforts to introduce the zeta function and its relation to the distribution of prime numbers.
Absolutely amazing how it all comes together in each one of these videos.
This is the most amazing math video I've seen in a while. And I've seen quite a bit!
I am getting goosebumps. This is my favorite channel ever. I love you
That makes me very happy to hear!
Just excellent
Thank you for creating this channel , at the first place..!!
Animations were superb well done. Though I did know most of this already the animations made it very enjoyable.
Very interesting!
In 46:30 we have the inverse of linear aproximation that can be used in Newton-Raphson method to find roots of zeta function from a initial condition s0.
Thank you for sharing these videos. One thing that seems clear to me, I hope I am not mistaken, is that Euler was both a doer and a thinker of mathematics. It is wonderful to see his persistence in trying to find a solution - and have we not experienced such things ourselves?
The disappointment of something not working as intended seemed to prompt his creativity.
At his level - I see math as a creative topic as much as any creative craft.
Interim conclusion: degrees in mathematics should be arts and not science?
I agree. Can't wait for another video
Ευχαριστούμε!
Simply great! Thank you.
Outstanding explanation.
This is wonderful! I'm in awe of your work, you don't skip the hard part of a theorem and that's so nice. Thank you.
I just have one question, I don't understand how adding the zeros of a complex function creates this kind of wave propagation in a function defined on the reals?
Thank you for the positive feedback! Your question will be the topic of a future video, so stay tuned.
Very much appreciated your work...
I want to understand more clearly that "'"how zeros of complex zeta function explains the distribution of primes??"""
I hope your next video will address this....
Thank you so much for this - it was amazing
Thanks!
Thank you so much! I think this is our first contribution!
I stayed up until 1am watching your damn videos.
good job.
Thanks! This is an amazing video. I'm trying to get the sum of 1/n^4 from 1 to infinity. Please help me get started on finding the formula for the partial sums of the coefficients of x^5. There's no obvious pattern that I'm seeing for 1/4, 7/18, 91/192...
What exactly is the question? Do you want to use the Euler-Lagrange method on ∑1/n^4?
One of the greatest videos ever !
Great video! Even better the second time.
At 50:26 I don't understand where the k comes from in p^k - but I suppose it will be explained in one of the next videos. Great contents, so enlightening.
The p^k is there to indicate that the sum is over all prime powers less than x, not just all primes less than x. The world would be a better place if that k wasn't there, but sadly, it is.
@@zetamath aaah it means that for example 1/ 7² = 1 / 49 is in the sum too if x is 100? thanks for the explanation!
32:35 -- sum of reciprocals of primes
Math is so beautiful
Eagerly waiting for your next video . Please upload fast :( .
Phenomenal video I love the content, quality and length! thankyou!
around min 23:00 you ask to find the sum of 1/n^4, it took me a few attempts and I was wondering if anyone has another solution; heres mine:
take RHS and plug into x the value (ix) to get [ix*prod(1+x^2/n^2pi^2)], multiply this with original RHS to get L1:
ix^2*prod(1 - (x^4)/(n^4pi^4))
next in LHS plug into x (ix) (to get -isinh(x)) to find i(x + x^3/3! + x^5/5! + x^7/7! + ...)
multiply with original LHS and only look for coefficient of x^6 to get:
something + -i/90 x^6 + something
equating coefficient of the other product of the RHS's we get the answer.
The sine function being an infinite product and equivalently an infinite sum is similar to the Euler products being Dirichlet series.
I think the odd values of the zeta function ζ(2n + 1) are transcendental because the logarithm may describe it in some. The logarithm is a transcendental function that switches multiplication to summation. The values ζ(2n + 1) possibly being transcendental may explain the difficulty in deriving them.
The Riemann zeta function is the solution of no algebraic ordinary differential equation on its region of analyticity. See the abstract of “Does the Riemann zeta function satisfy a differential equation?” (2015) by Gorder. This makes the zeta a function a hypertranscendetal function. The zeros of the zeta function are found in its analytic continuation. Putting this together, the zeros of of the zeta function are no solutions to any algebraic differential equation.
_"The values ζ(2n + 1) possibly being transcendental may explain the difficulty in deriving them. "_
Now the values ζ(2n) are certainly transcendental - and they are well-known. For ζ(2n+1) it is not even known if they are transcendental at all, except for ζ(3).
I would love to see a video that shows the analytic continuation of the Riemann zeta-function all the way to proving the -1/12 result.
Very beautiful work here
Excellent video
Yeah - I get it. If you trip over a tree root - a square root even - and multiply that by a pie that is in the shape of a square then, X is important somehow and then another Pi (apple or steak & kidney is irrelevant) and add some binomials and then then a lot more squares and one more cube.
something is wrong with the formula reached at 46:28 because when p goes to infinty the terms inside the series goes to infinity , i think it should be p to the power of s not -s or we can add p to the -s in the numerator in fact that's what i got after calculating the derivative , idk if any one found another result plz let me know
The derivative was miscalculated. There should be another factor of p^-s on the right. The correct sum simplifies to ∑ log(p) / (p^s - 1).
Best video ever!
At 46:49 it would be better to explan what does mean \sum_{p^k}\frac{\log(p)}{p^{ks}}=\sum_{p}\sum_{k}\frac{\log(p)}{p^{ks}}.
Awesome video. Thank you!
There’s one small issue that is not explained in the video. At 53:30 the graph starts including the first few terms of the summation involving the zeros of the zeta function But it appears that any partial sum would be complex. To be exact, the terms are of the form (x^alpha)/alpha, where alpha is a zero of the zeta function and x is an integer or real number. In general, each term in the summation is a complex number because every zero of the zeta function is complex. In the limit, this sum may indeed converge to a real number but any partial sum is probably complex with a non-zero imaginary part. My question is whether you used the magnitude or the real part of the partial sum for your graph and does one give a better fit than the other.
Good eye, but sadly the answer is more boring than you might hope. Actually, the terms come in conjugate pairs, so when computing the partial sums, I included them in pairs to keep the thing being graphed real.
@@zetamath Thanks! Would never have guessed the answer. Then, every other partial sum is real to infinity? That's amazing.
Amazing. Thank you so much.
Amazing! Will we get more on Bernoulli numbers?
That's a good question, I'm not sure if they will get mentioned again in this story, but they come up so often, it wouldn't surprise me if we talk about them on the channel again!
Great content! Just great!
Thank you 🎈
I remember as a student being astounded by the fact that, no matter how large an integer n, there are stretches of at least n integers in a row with NONE of them prime! Sometime later (in grad school, I hope) it belatedly occurred to me that the same is true for the perfect squares ! 😲 Duh!
Hey, I just recently discovered you channel. Is there a possibility that you would make a video about automorphic forms? It seems that you mainly focus on analytic number theory. So it may serve as a rich and interesting topic to discuss on your channel in a more informal fashion.
At 47:00 , shouldn't the first term of the expansion be log(p)? [Expanding 1/(1 - p^-s) of 46:52 to a geometric series, the first term is 1. So the sum over all p^k (also in 46:52 ) must start with p^0.] Am i wrong?
The formula for the derivative at 46:25 is wrong. It must contain an extra p^-s factor. Nevertheless, the left hand side of 46:45 and the right hand side are equal once again, only the middle term is wrong. A double error, so to say, which cancels out.
For the geometric series, we have
1 + q + q² + ... = 1/(1-q) and
q + q² + q³ + ... = q/(1-q)
The extra p^-s term changes it to the second version above, so everything is all right in the following.
Had already been mentioned in these comments here, as I just noticed.
when is the next video coming out
loved this one
If all goes well it should be within the next couple of months, but I should be careful not to overpromise, as this one ended up coming out a full year later than I thought it would! Thanks for the compliment, I'm glad you're enjoying the content!
53:23 - 54:25 It would have been nice to state the number of zero pairs used for this animation. It's "the first handful", but what is a "handful of zeros"?
This was me trying to avoid discussion the trivial zeroes and the non-trivial zeroes (discussed later).
Love your content
all those 3b1b references, i love it
great video!
True MatheMagic!
Can someone help me understand ? At 30:50 he says harmonic series of ζ(1) diverges
I am no mathematician but i do not understand why does zeta of 1 diverges if you are constantly adding up smaller and smaller terms it should converge the only difference between ζ(1) and ζ(2) is that ζ(2) is doing it much^2 faster 😅
Why don't you look at Wikipedia: "Harmonic series", chapter "comparison test". It's a very easy proof that this series diverges. It is frequently given in schools. On the other hand, ζ(2) converges, as does ζ(1.1) or ζ(1.01).
This is some seriously cool stuff.
Very Interesting Great Job
really interesting. keep going.
This is amazing! Though I think the analogy you made between rarity of primes and integers or squares of integers and so on could be misleading. Still great work!
Did anyone workout the proof for Σ 1/n^4? It took much longer than I anticipated
I'm here for this solution to. I'm having a hard time figuring out a formula for the partial sums. Do you have a formula?
Nice theme music :)
“Euler didn’t prove this either”
If it’s good enough for Euler it’s good enough for me
Particles possible paths through zero?
when you take the log of sin(x) you must remember that x must be greater than zero (sin(x) varies between -1 and 1) , otherwise log will be a complex number. How do you justify then the use of log(sin(x)) and the use of derivative of log(sin(x)) to obtain cot(x) ? (of course I know the results are correct.I just want a rational explnantion)
A more rigorous description here would talk about analytic continuations, using the fact that these things agree when x>0.
it is a great detail to pick up on, but I didn't talk about analytic continuation until the next video, so I was a bit hand wavy here (as is the general theme of the series).
@@zetamath ok
Is it just me? Or this is actually like seeing nature unraveling it's most intimate mysteries!
4:23 That's not true... If q(x)=x^2+1, for example, p still has only 3 real roots
Then there would be complex roots and thus the statement of p having only 3 roots wouldn't apply. The narrator didn't say "3 real roots" but "these 3 zeroes" which makes the statement correct.
53:50 how can you put zeta zeroes in real plane zeta zeroes are imaginary number.
unbelievable
Wow great, wouldn't it be great to see a vid about why there is no algabraic formula for x^5+...
This proof is near and dear to my heart, and is certainly something I've contemplated how much I could say about. Unfortunately, the proof is usually the pot of gold at the end of the rainbow to motivate a year long course in abstract algebra, because it combines an awful lot of different math all in one place. If I think of a way to present it, I definitely well.
51:00 This "psi function" (ψ-function) is also known as the "second Chebyshev function" (besides ϑ(x), which is the first Chebyshev function).
I tried to reproduce this calculation of 53:23 onwards using spreadsheet calculation. It's a bit clumsy at first if you're not that practiced with complex algebra, but it worked. I used the formula
=SQRT(x) * ( COS(LN(x)*alpha) + 2*alpha*SIN(LN(x)*alpha) ) / (0,25 + alpha^2)
and these have to be added up for all nontrivial zeta zeros ("alpha") 14,1347..., 21,022..., 25,010..., ..., (the imaginary parts! being used as reals here, the real parts of 0.5 only accounting for the leading square root) or better to say: for as many of them as appropriate. 100 oder 200 of these pairs give quite decent results at least for x
what you showed at 53:12 and at 54:30 was the clearer simplification of the Riemann Hypothesis idea on the internet
I already know all of this but I just love to watch the Riemann Hypothesis every time I see a new video about it
btw at the beginning of the video the Basel Problem explanation was a bit overkill which made it less good for me
you should also check m.y. v.i.d.e.o.s.
❤