The Basel Problem Part 2: Euler's Proof and the Riemann Hypothesis

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In this video, I present Euler's proof that the solution to the Basel problem is pi^2/6. I discuss a surprising connection Euler discovered between a generalization of the Basel problem and the Bernoulli numbers, as well as his invention of the zeta function. I explain Euler's discovery of the connection between the zeta function and the prime numbers, and I discuss how Riemann's continuation of Euler's work led him to state the Riemann hypothesis, one of the most important conjectures in the entire history of mathematics.
If you would like to support the production of our content, we have a Patreon! Sign up at / zetamath
Sections of this video:
00:00 Intro
01:24 Euler's Basel proof
23:20 The zeta function and the Bernoulli numbers
32:01 Zeta and the primes
48:15 The Riemann hypothesis
--
Further viewing from 3Blue1Brown:
Why is pi here? And why is it squared? A geometric answer to the Basel Problem • Why is pi here? And w...
Taylor series • Taylor series | Chapte...
--
Further viewing from Mathologer:
Euler's real identity NOT e to the i pi = -1 https:/ • Euler's real identity ...
Euler's Pi Prime Product and Riemann's Zeta Function • Euler’s Pi Prime Produ...
Ramanujan: Making sense of 1+2+3+... = -1/12 and Co. • Ramanujan: Making sens...
--
Thanks go to Keith Welker for our theme music. www.lunarchariot.com
Some of the animations in this video were created with Manim Community. More information can be found at manim.community

Опубликовано:

16 июн 2024

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Комментарии : 136

@damland1357 2 года назад

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.

@zetamath 2 года назад

I wonder if he was more excited by figuring out the value was pi^2/6 or figuring out how to prove it!

@michaeltrungold86 3 года назад

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!

@JamesWylde 2 года назад

He just needs to drop the mid roll ads, if he did I would probably support it.

@ingenuity23 Год назад

@@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

@lorendisney5068 Год назад

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.

@AkamiChannel 10 месяцев назад

@@JamesWyldesuck it up and pay for yt premium

@nibn4r 9 месяцев назад

I agree it’s excellent

@carlosayam 2 года назад

You made this as clear as possible while still keeping a healthy dose of rigour and detail. Awesome work!

@cblpu5575 Год назад

Seeing what has happened to youtube math has been very incredible. I hope all of these channels get the recognition they deserve

@omerelhagahmed551 2 года назад

This should be one of the most famous math channels soon, because definitely it's one of best

@maxfred1696 2 года назад

This is amazing! You have a rigoros and fun way of describing :) We need more in depth math like you on RU-vid

@zetamath 2 года назад

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!

@davictor24 Год назад

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.

@milenamarquez Год назад

If not for you, I would have lost my confidence in math. Thank you so much!!

@studentofspacetime 3 месяца назад

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.

@TheOneSevenNine 2 года назад

this video had more narrative tension than most movies i've seen. feel like i should cheer.

@ANTOINETTE-nk1tm 9 дней назад

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.

@jedb872 2 года назад

This is a superb explanation. My understanding increased tremendously. Thank you, imagine it takes tremendous work to put this together.

@paulbooker 2 года назад

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.

@xyzct 2 года назад

Everything about your presentation style is AWESOME! That was so enjoyable, and _crystal clear._

@SmartHobbies 2 года назад

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.

@richarddizaji7848 Год назад

Absolutely amazing how it all comes together in each one of these videos.

@pythagorasaurusrex9853 2 года назад

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!

@georgeorourke7156 2 года назад

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!

@angeluomo 2 года назад

Excellent video. Provides illuminating depth on the relationship between Euler's and Riemann's work. I am surprised this does not have more views.

@bored_abi 2 года назад

this is byfar one of tte most amazing proof walkthroughs I've ever seen

@iansragingbileduct 2 месяца назад

Your long form deep-dive videos are gorgeous. Thanks!

@TheFarmanimalfriend 2 года назад

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.

@NicolasMiari 9 месяцев назад

This is the most amazing math video I've seen in a while. And I've seen quite a bit!

@NotBroihon Год назад

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 RU-vid. Love it. Now I can only hope that in the final part you either prove or disprove the Riemann Hypothesis ;)

@Number_Cruncher 2 года назад

Thank you for your efforts to introduce the zeta function and its relation to the distribution of prime numbers.

@rajendralekhwar4131 2 года назад

Just excellent Thank you for creating this channel , at the first place..!!

@keinKlarname 2 года назад

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 RU-vid. Hopefully you can do more math presentation of such form, it would be highly appreciated. Thanks a lot, zeta!

@abhijeetsarker5285 2 года назад

This video was so beautiful.....i got emotional by watching it.Very well done zeta math keep it up!!!!

@xulq 2 года назад

this channel is a hidden gem

@jayvaghela9888 Год назад

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 👏

@jmathg Месяц назад

That moment at 19:17...jaw-dropping! Such a good lesson in persistance - it's incredible that Euler came up with this!

@ydw3284 Год назад

i do not why such a fantastic channel only got fans less than 10k, i hope more people can find this...

@GlenMacDonald Год назад

This video was so well-done that it motivated me to become a Patreon supporter. This channel deserves far more support!!

@zarkalonbenha Год назад

I agree. Can't wait for another video

@user-yl7wn2fz1t 2 года назад

Outstanding explanation.

@seanoneill2098 Год назад

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

@pointfive7101 11 месяцев назад

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.

@wenzhang365 2 года назад

Simply great! Thank you.

@lorenzodavidsartormaurino413 2 года назад

I am getting goosebumps. This is my favorite channel ever. I love you

@zetamath 2 года назад

That makes me very happy to hear!

@wallstreetoneil 3 года назад

Thank you so much for this - it was amazing

@lorendisney5068 5 месяцев назад

Great video! Even better the second time.

@joeeeee8738 Год назад

One of the greatest videos ever !

@jamesknapp64 Год назад

Animations were superb well done. Though I did know most of this already the animations made it very enjoyable.

@jessstuart7495 Год назад

Awesome video. Thank you!

@minipashki Год назад

Great content! Just great!

@malicksoumare370 Год назад

Very beautiful work here

@wallstreetoneil 3 года назад

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.

@zetamath 3 года назад

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!

@nahoj.2569 3 месяца назад

I stayed up until 1am watching your damn videos. good job.

@mMaximus56789 2 года назад

Love your content

@ki-ka Год назад

Amazing. Thank you so much.

@arjunbhandari3693 3 года назад

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....

@RSLT Год назад

Very Interesting Great Job

@charlesdarwin1040 8 месяцев назад

great video!

@michaeledgley7570 2 года назад

really interesting. keep going.

@darkseid856 2 года назад

Eagerly waiting for your next video . Please upload fast :( .

@francoisamman2620 3 года назад

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?

@zetamath 3 года назад

Thank you for the positive feedback! Your question will be the topic of a future video, so stay tuned.

@mpalin11 Год назад

This is some seriously cool stuff.

@jennyone8829 Год назад

Thank you 🎈

@studentofspacetime 3 месяца назад

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.

@Zeitgeist9000 3 месяца назад

Thanks!

@MrRibeirobr 11 месяцев назад

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.

@miloszforman6270 17 дней назад

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.

@nitindhiman4600 2 года назад

Math is so beautiful

@prakharrakhya7964 3 года назад

when is the next video coming out loved this one

@zetamath 3 года назад

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!

@pawebielinski4903 Год назад

Amazing! Will we get more on Bernoulli numbers?

@zetamath Год назад

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!

@inkognito8400 13 дней назад

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.

@elihowitt4107 Год назад

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.

@zogzog1063 Год назад

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.

@jaafars.mahdawi6911 Год назад

True MatheMagic!

@maximiliansans8257 Год назад

all those 3b1b references, i love it

@bowtangey6830 Год назад

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!

@phenixorbitall3917 5 месяцев назад

Nice theme music :)

@ieatbananaswiththepeel4782 2 года назад

46:25 when I put that sum in a calculator, it says it that’s log(p)/1-p^s, not p^-s. Was that a mistake on your part, or am I just really, really stupid?

@Jaylooker 4 месяца назад

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.

@miloszforman6270 2 месяца назад

_"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).

@msamadzad 2 года назад

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!

@ericbischoff9444 2 года назад

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.

@zetamath 2 года назад

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.

@ericbischoff9444 2 года назад

@@zetamath aaah it means that for example 1/ 7² = 1 / 49 is in the sum too if x is 100? thanks for the explanation!

@ozaman-buzaman9300 2 года назад

unbelievable

@burrbonus 2 года назад

32:35 -- sum of reciprocals of primes

@bowtangey6830 Год назад

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?

@miloszforman6270 19 дней назад

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.

@davidmwakima3027 3 месяца назад

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...

@miloszforman6270 2 месяца назад

What exactly is the question? Do you want to use the Euler-Lagrange method on ∑1/n^4?

@hh46465 2 года назад

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

@NevinBR 2 года назад

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).

@dougrife8827 Год назад

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.

@zetamath Год назад

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.

@dougrife8827 Год назад

@@zetamath Thanks! Would never have guessed the answer. Then, every other partial sum is real to infinity? That's amazing.

@gurk_the_magnificent9008 2 года назад

“Euler didn’t prove this either” If it’s good enough for Euler it’s good enough for me

@wawangsf Год назад

Did anyone workout the proof for Σ 1/n^4? It took much longer than I anticipated

@davidmwakima3027 3 месяца назад

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?

@shawnouellette1953 7 месяцев назад

Particles possible paths through zero?

@miloszforman6270 6 месяцев назад

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"?

@zetamath 4 месяца назад

This was me trying to avoid discussion the trivial zeroes and the non-trivial zeroes (discussed later).

@Waferdicing Год назад

❤

@physira7551 Год назад

Is it just me? Or this is actually like seeing nature unraveling it's most intimate mysteries!

@gebruikerarjan Год назад

Wow great, wouldn't it be great to see a vid about why there is no algabraic formula for x^5+...

@zetamath Год назад

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.

@MrFran007 7 месяцев назад

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 😅

@miloszforman6270 6 месяцев назад

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).

@thisisnotmyrealname628 Год назад

Wooow

@yagor144 11 месяцев назад

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}}.

@marciliocarneiro Год назад

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)

@zetamath Год назад

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).

@marciliocarneiro Год назад

@@zetamath ok

@yoavshati Год назад

4:23 That's not true... If q(x)=x^2+1, for example, p still has only 3 real roots

@NotBroihon Год назад

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.

@PeterParker-gt3xl 2 дня назад

Euler/Euler/Euler...

@riotcelestian4587 Год назад

53:50 how can you put zeta zeroes in real plane zeta zeroes are imaginary number.

@jean-francoisbrunet2031 Год назад

I am not a mathematician and I am just trying to understand a few basic things. However, I am sensitive to language and logic and I stopped listening to this video at 1:30: "Imagine we are Euler and our goal is to prove that this sum has the value of pi to the square over six". But the only thing I can imagine, is that Euler's goal was NOT to prove that the value of this sum was pi to the square over six (a value that somehow he would have figured out in advance, before proving it), but to DISCOVER the value of this sum (and he found that it was pi to the square over six). Right?

@miloszforman6270 19 дней назад

He first discovered that the value was apparently π²/6, and then he tried to prove it, and he succeeded. What kind of nonsense are you making out of this?

@jean-francoisbrunet2031 16 дней назад

@@miloszforman6270 What is the difference between discovering the value of something and proving that it has this value, according to you? (who are so much better than me at making sense).

@miloszforman6270 16 дней назад

@@jean-francoisbrunet2031 The difference is that he first _presumed_ that this value was really π²/6, because he _discovered_ that it was very close to π²/6 (he calculated it for 16 or 17 decimal places). This _discovery_ made it quite likely that he would succeed in _proving_ this assumption.

@Isaac-Mor 3 года назад

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.

@necrosudomi420thecuratorof4 Год назад

who want to peer reviews my answer to the longstanding hypothesis

@marcellomarianetti1770 Год назад

Just a small detail: you cannot prove that lim x-> npi of sinx/(x-npi) is (-1)^n with L'Hopital's theorem because taking the derivative of sinx is possible only AFTER knowing the limit of sinx/x geometrically (which is 1), but it's easy noticing that if t = x - npi, then x = t + npi and sin(x) = sin(t + npi) = (-1)^n sin(t), and if x approaches npi, then t approaches 0. The limit then becomes: limit as t -> 0 of (-1)^n sin(t)/t which is (-1)^n

@miloszforman6270 6 месяцев назад

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