Euler’s Pi Prime Product and Riemann’s Zeta Function

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  • čas přidán 7. 09. 2017
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    What has pi to do with the prime numbers, how can you calculate pi from the licence plate numbers you encounter on your way to work, and what does all this have to do with Riemann's zeta function and the most important unsolved problem in math? Well, Euler knew most of the answers, long before Riemann was born.
    I got this week's pi t-shirt from here: shirt.woot.com/offers/beautif...
    As usual thank you very much to Marty and Danil for their feedback on an earlier version of this video and Michael (Franklin) for his help with recording this video..
    Here are a few interesting references to check out if you can handle more maths: J.E. Nymann, On the probability that k positive integers are relatively prime, Journal of number theory 4, 469--473 (1972) www.sciencedirect.com/science/... (contains a link to a pdf file of the article).
    Enjoy :)

Komentáře • 659

  • @Kabitu1
    @Kabitu1 Před 6 lety +593

    "Euler pushed that infinite sum to the limit.."
    You deserve a beating for that joke.

    • @koenth2359
      @koenth2359 Před 6 lety +3

      Even better, he said 'to the absolute limit' to make it more complex. Is that good enough for a total beat-up?

    • @christopherellis2663
      @christopherellis2663 Před 5 lety +3

      From the Inner Mind, to the Outer Limits!

    • @tejnadkarni131
      @tejnadkarni131 Před 5 lety +6

      That was a Parker Square of a joke

    • @davutsauze8319
      @davutsauze8319 Před 3 lety +2

      @@tejnadkarni131 Ah, I feel like you're a traitor, even though I do like Numberphile too

  • @jameshoffman552
    @jameshoffman552 Před 3 lety +90

    This channel and 3Blue1Brown have fantastic visuals that are very helpful. Both also have excellent, clear presentation. No other mathematics channel I've found come close to your skill in communication, which obviously involves a great deal of work to create all the graphis.

    • @xarmanhsh2981
      @xarmanhsh2981 Před 9 měsíci +1

      Best channels for insomnia so relaxing

    • @bigmouthfisheyes
      @bigmouthfisheyes Před 8 měsíci

      The expressions 6n +/- 1 produce all prime numbers greater than three, and many more composite numbers. If we knew exactly where the composite numbers would appear in these sequences, we could infer the location of all of the prime numbers. Am I understanding this correctly? Of what use would this be to anyone?

  • @naomioverton7893
    @naomioverton7893 Před 4 lety +42

    My mind is so blown that I might cry a little later at the indescribable beauty of what I just learned. Thank you for existing Mr. Mathologer.

    • @aaronleperspicace1704
      @aaronleperspicace1704 Před 4 lety +6

      "Take the differences between 1 and the square of reciprocal of every prime number. Multiply them out to each other and take the reciprocal of this product. Multiply this reciprocal by 6 and square root the result. The value thus obtained is the ratio of the distance all around a circle to the distance across it".

    • @hybmnzz2658
      @hybmnzz2658 Před 3 lety +3

      @@aaronleperspicace1704 something interesting about this is how mystical pi seems when it is classically defined in terms of circles. No wonder hardcore analysts think of pi as defined based on infinite series and do not appeal to geometry.

  • @maurolocatelli3321
    @maurolocatelli3321 Před 6 lety +27

    14:00 with the primes < 1.000.000 you get 3.14159254713, 6 decimals correct!
    (Used Python...)

  • @giladzxc17
    @giladzxc17 Před 6 lety +31

    I love how this shows yet another intuitive explanation on 1is neither a prime nor a non-prime.

    • @viharsarok
      @viharsarok Před 2 lety +1

      By all means it's a non-prime.

    • @PC_Simo
      @PC_Simo Před rokem +1

      @@viharsarok So, it has other factors, beside itself and 1 (which is itself)? I’d like to know more about these mysterious factors. 🤔

  • @PrincessEev
    @PrincessEev Před 6 lety +152

    This might be my favorite video of yours yet. So much amazing information - especially how the prime product connects to the zeta function. That has always confused me - and I was probably too lazy to look it up - but that it's so simple is just ..beautiful in a weird way. XD

    • @davutsauze8319
      @davutsauze8319 Před 3 lety +1

      "especially how the prime product connects to the zeta function. That has always confused me - and I was probably too lazy to look it up - but that it's so simple is just ..beautiful in a weird way." Ah, me too, I'm glad that I understand it now

    • @reggyreptinall9598
      @reggyreptinall9598 Před 2 lety +1

      I solved the Riemann Zeta function! I posted the solution, check it out!

    • @reggyreptinall9598
      @reggyreptinall9598 Před 2 lety +1

      I solved and proved the Riemann Zeta function. I just posted the solution check it out!

  • @rudilapa6569
    @rudilapa6569 Před 6 lety +7

    Professor, these are perfectly paced. They don't go so fast that I lose the chain, but never so slow that it becomes unexciting. It's high level, so there's room for mulling over what's needed to ground the proof. There's always hints of ways to take it further. And the use of the animated background blackboard to do the steps while you explain the ideas keeps it flowing.
    If you offered this type of freewheeling 'class' at your university it would be packed.
    And PI is related to the primes?! This is at least as amazing as e^(i*π).

    • @Mathologer
      @Mathologer  Před 6 lety +1

      Glad these videos work for you exactly as planned, and thank you very much for saying so :)

  • @MrQwefty
    @MrQwefty Před 6 lety +6

    Mate, your pressentation is STELLAR! Seriously, the transitions, coloring, everything! Great job :D

  • @abdullahalmosalami2373
    @abdullahalmosalami2373 Před 4 lety +19

    12:00 Woah!! That was awesome!!! I just finished taking my Random Processes course as an Electrical Engineer last semester, and I'm so glad I watched this video after having taken that, cuz that was brilliant!

    • @pro4skill
      @pro4skill Před 3 lety +1

      Please tell me how it was relevant! I'm doing math and physics research

  • @gerardomoscatelli8584
    @gerardomoscatelli8584 Před 4 lety +26

    Hello there, great video. I used a fast function I found using a boolean sieve to list primes in Python and listed primes up to n=1,000,000,000. There are 50,847,534 primes and the last one is 999,999,937. Using this list estimated pi with Euler product at 3.1415926536126695 accurate up to the 9th decimal. Can't find more than up to n=1,000,000,000 prime, memory error. It would be interesting to graph how the accuracy of pi evolves with the number of primes.

  • @KazimirQ7G
    @KazimirQ7G Před 6 lety +11

    I liked this format... with you floating around the equations. Much better than you presenting into a board, or scaring me with sudden sound effects. Keep it up!

  • @darklink1113
    @darklink1113 Před 4 lety +5

    Your graphics make these lectures.
    I foresee a time when all math is taught like this.
    I can't follow half of it, but I can at least follow.
    Thanks for these

  • @calma5869
    @calma5869 Před 4 lety +4

    I was deeply astonished by Euler's geniusity!! Thank you for introducing his research to me

  • @richardschreier3866
    @richardschreier3866 Před 6 lety +6

    Thanks for another fun video! It is great that you still make the time to do this work even though the crush of the school term is once again upon you. Looking forward to more fun with zeta in the next instalment!

    • @Mathologer
      @Mathologer  Před 6 lety +3

      If only I had a bit more time. It would be great to be able to make more of these videos :)

  • @MrWkuling
    @MrWkuling Před 6 lety +4

    Thank you... beautiful video, your explanations demonstrate deep understanding and passion at the same time!

  • @christophersewell6611
    @christophersewell6611 Před 6 lety +5

    Awesome video! This channel is probably my favourite math subscription :)

  • @enzogiannotta
    @enzogiannotta Před 6 lety +3

    You are my favourite math channel. I learn a lot from your videos!!!

  • @Meine_Rede
    @Meine_Rede Před rokem +4

    Hello Mathologer,
    thanks you so much for your video, which is the most impressive one I have seen in regards to the relation between Euler's Pi Prime Product and the Riemann Function.
    Just one small remark: I think on 6:21 it should say 1/27^z + ... and not 1/28^z + ... as stated in the video, simply because 3*9 = 27.

  • @Mathologer
    @Mathologer  Před 6 lety +57

    Greetings from beautiful New Zealand. If you happen to be in Auckland today as part of the Maths craft festival I'll be giving a talk about the best ways to lace your shoes at the Auckland Museum at 5.15 p.m. Come and say hello :) www.canterbury.ac.nz/news/2017/maths-craft-festival-2017-hits-auckland.html
    As usual if you'd like to support the channel please consider contributing subtitles in your native language.

    • @JTX8000
      @JTX8000 Před 6 lety +2

      Welcome to Nz, hope you have an enjoyable stay

    • @JRush374
      @JRush374 Před 6 lety +2

      Please do some videos on fractional calculus!

    • @Mathologer
      @Mathologer  Před 6 lety

      +Josh I'll put this topic on my list of things to ponder :)

    • @michaels4340
      @michaels4340 Před 6 lety +1

      Mathologer I'm not sure whether your proof (with conclusion at 5:16) is complete. Is that really a contradiction? Even if one function is bounded above by another, couldn't they still converge to the same value, as long as the difference between the sums converges to 0?

    • @AnarchoAmericium
      @AnarchoAmericium Před 6 lety

      You should consider doing a video taking infinite products to infinite fractions (which is again due to Euler), especially condiering you already have videos on infinite fractions you can reference.

  • @SimonFrank369
    @SimonFrank369 Před rokem +1

    IMHO your math videos are some of the greatest to be found on CZcams!! Thx a lot for your work, Mathologer!
    And how come you speak German that perfectly!??

  • @yash96819
    @yash96819 Před 4 lety +1

    Great video, amazing explanation of how Euler arrived at that form involving just primes.

  • @FaissalChahrour
    @FaissalChahrour Před 6 lety +1

    many thanks for this video. it's the first tim i find the Euler's formula so beautiful and could really understand them

  • @koenth2359
    @koenth2359 Před 6 lety +1

    I think 32 years ago I was playing about with numbers and I discovered this relation between the prime product and the integer power sum (for real s>1) from a sort of sieve argument. I did not know this was a known result, nor had I ever heard of the Euler prime product or the Riemann Zeta function, but thought it was an interesting result. It was in the days before you could just google things. My father persuaded me to hand over my scriblings, so he could show it to a friended mathematician. Of course I got it back with the flaws in my 'proof' pointed out, without any hint that the result was known. My intuition was right, but I had no idea what a proof required.

  • @eliesimsch3491
    @eliesimsch3491 Před 6 lety +1

    Hi Mr. Mathologer(for lack of a better name), while I was doing my math homework today, I remembered a problem my math teacher gave me and my friend in 5th grade after we'd finished the work a little early. The problem was as follows: the king's daughter is to be married, and to choose the prince, the king has decided to seat the knights around a very big round table. Then he will start at the first knight, and say "you live." To the second knight, he says "you die." The last surviving knight is the one who wins. The challenge was to make a formula so that you could find which seat you wanted to sit at, to live(given between 1 and infinite knights). I made a little progress, finding that (obviously) even seats are bad, and that the good seat starts at 1, then counts up odd numbers to 2 higher than before, then resets to zero, but I only had a little time and I'm not good at math. In retrospect, I don't think I've heard of any similar problems, which is why I still find it curious.

    • @Mathologer
      @Mathologer  Před 6 lety

      +Elie Simsch This problem is called the Josephus problem :)

  • @deathlord109
    @deathlord109 Před 6 lety +3

    I really appreciate your effort in doing these videos :)

  • @rustyshackleford1964
    @rustyshackleford1964 Před 3 lety +1

    Probably my favorite video! Thank you mathologer!

  • @xdingo93x
    @xdingo93x Před 6 lety +2

    Fiiiinally, now I understand how all the prime numbers got in there! Thanks for this video :)

  • @joelhaggis5054
    @joelhaggis5054 Před 6 lety +1

    I love the Euler product formula. Glad you did a video on it.

  • @davidherrera8432
    @davidherrera8432 Před 6 lety +1

    Beautiful, I'm seeing stars and little numbers arround my head, this is the stuff that make me love math!

  • @yairalkon4944
    @yairalkon4944 Před 6 lety +1

    An absolutely great video. Keep up the amazing job!

  • @valor36az
    @valor36az Před 2 lety +2

    Amazing explanation I have been trying to understand this forever

  • @-fitzy-3335
    @-fitzy-3335 Před 6 lety +12

    Really great video man, keep it up :)

  • @balern4
    @balern4 Před 4 lety

    I have watched many of your videos and I think this was the best one.

  • @jacksonstarky8288
    @jacksonstarky8288 Před rokem +1

    The number of views of this video compared with, say, the number of views of your Harmonic Series/Gamma video suggests to me that this video is criminally underviewed... which might be why it seems like we're still waiting for that Riemann follow-up video you talked about at the end. I would love to see something like what you hint at, explaining the connection between the Riemann zeta function and the Euler gamma constant.

  • @mrsebakuna
    @mrsebakuna Před 6 lety +1

    beautiful story like always, love your presentation. strong logic and very valuable historical facts.

  • @erikcools891
    @erikcools891 Před 6 lety +2

    brilliant video, as always. Keep up the good work.

  • @pyrrho314
    @pyrrho314 Před 6 lety +3

    your channel and videos are awesome, thanks!

  • @javierrivera7685
    @javierrivera7685 Před 6 lety +2

    Awesome video, very instructive and clear. I specially liked the probability connection of the zeta function

  • @alexandersanchez9138
    @alexandersanchez9138 Před 6 lety +8

    10:20 The killer instinct. Mathologer isn't afraid to strike at the heart.

  • @1DR31N
    @1DR31N Před 4 lety +1

    Nice and neat explanation. I would have wished to have a Math teacher like you when I was at school.

  • @billbill1235
    @billbill1235 Před 2 lety +2

    You should do a full video on the reimann heipothesis

  • @abdeljalilpr2033
    @abdeljalilpr2033 Před 6 lety +1

    Salamo aalaikom..you are posing a very important maths issues ..i want to say that resect you and you have to continue ..respect from morocco

  • @forestpepper3621
    @forestpepper3621 Před 5 lety +1

    Comment on 10:45 in video. The concept of a "random integer" is tricky. When you say, "choose one of these items randomly", if there are N items, then usually each item is chosen with probability 1/N. However, the word "randomly" assumes that you have defined a probability function on the items. 1/N is the the "uniform" probability function, but a different probability function would make it more likely to select certain items than others. *PROBLEM*: The integers can not be assigned a uniform probability function, with all integers equally likely, because there are infinitely many. The sum of the probabilities assigned to all items must equal 1, and if all integers had the same probability, these probabilities would sum to infinity. So, in choosing an integer "randomly", you must specify which probability function has been defined on the integers. There is an unlimited variety of probability functions that may be defined on the integers.

  • @sylviaelse5086
    @sylviaelse5086 Před 2 lety +1

    When z = 1, the manipulations required to get the prime product result require multiplying infinity by integers, and then subtracting one infinity from another, and finally dividing one by zero, just for good measure. The conclusion at around 9:27 that there are infinitely many prime numbers is certainly correct, because we know that to be true, but I'm not comfortable that reasoning there is valid.

  • @Mar184
    @Mar184 Před 6 lety +3

    8:02 wow, that pronunciation was perfect - well done!

  • @saahilmehta
    @saahilmehta Před 3 lety +2

    I remember our math teacher gave this to us in a probability question(Find the probability that 2 randomly chosen natural numbers are coprimes). The answer is 6 by pi square

  • @maciejkozowski6063
    @maciejkozowski6063 Před 6 lety

    one of the best youtube chanel about math

  • @PhilipBlignaut
    @PhilipBlignaut Před 6 lety +2

    Just love your videos!

  • @zubmit700
    @zubmit700 Před 6 lety +3

    Best math channel!

  • @mattdarcy6891
    @mattdarcy6891 Před 6 lety +2

    unreal video keep it up man i love this

  • @obsoquasi
    @obsoquasi Před 6 lety +1

    very nice presentation. This is how math should be taught! Herzlichen Dank!

  • @zixuan1630
    @zixuan1630 Před 5 lety +23

    I really hoped that you could prove the Riemann Hypothesis.....

  • @jennyone8829
    @jennyone8829 Před rokem

    Thank you for existing 🎈

  • @rhc-weinkontore.k.7118
    @rhc-weinkontore.k.7118 Před 2 lety +1

    Toll erklärt und sehr instruktiv. Ich danke dafür!

  • @silasrodrigues1446
    @silasrodrigues1446 Před 5 lety

    For this, you've just got another subscriber!

  • @nowhereman8564
    @nowhereman8564 Před 3 lety +2

    Riemann what a genius was involved in general relativity and prime numbers

  • @Hythloday71
    @Hythloday71 Před 6 lety

    AWESOME insights ! Expanded my knowledge / understanding INFINITELY ?

  • @AltonMoore
    @AltonMoore Před 4 lety +1

    This is a really good channel.

  • @martinullrich655
    @martinullrich655 Před 7 měsíci

    Thanks a lot for this wonderful lesson!

  • @tutordave
    @tutordave Před 5 lety +1

    Very informative. Thanks!

  • @nezihokur
    @nezihokur Před 6 lety

    I watch your videos with a lot of wow!s. Great fun!

    • @Mathologer
      @Mathologer  Před 6 lety

      Great, looks like "mission accomplished" as far as you are concerned :)

  • @satyanarayanmohanty3415
    @satyanarayanmohanty3415 Před 4 lety +1

    You are amazing sir.

  • @jweezy101491
    @jweezy101491 Před 6 lety +1

    This was a really great video.

  • @Achill101
    @Achill101 Před 3 lety +2

    I've been writing questions in my comments so far, but I want to say also Thank you, Thank, Thank you, for your videos. It's a delight to watch them, especially this one. I hadn't heard it before, because I'm interested in geometry and calculus, not in primes, but I must admit this is a beauty.
    Now, to my question - sorry I can't let it be ;-) I would be grateful for an answer by everyone, of course, not only by Burkhard.
    @12:13 - it was said: "at least two of the ingredients of this proof need a little bit more justification and, well, can you tell which?" I cannot tell it now. Could anybody tell me, please, which two ingredients that could be?

    • @Achill101
      @Achill101 Před 3 lety

      I like this video a lot.
      At @9:46 the video seems a bit unclear of what is required to show for a second time that there are infinitely many primes, using equation:
      pi^2 / 6 = 1 / ( (1 - 1 / 2^2) * (1 - 1 / 2^3) * (1 - 1 / 5^2) * (1 - 1 / 7^2) * ... * (1 - 1 / prime(max)^2) ) with an assumed maximum prime number, prime(max).
      On the right side is a rational number, and it would follow that pi^2 is rational. But that pi^2 is irrational is a stronger statement than pi being irrational, like the square root of 2 is irrational, but its square, 2, is rational. Therefore, it's not enough to show that pi is irrational, which is difficult enough, but we would have to show that pi^2 is irrational for showing that there are infinitely many prime numbers. I personally find the proof using zeta(1) more beautiful.

  • @jimjackson4256
    @jimjackson4256 Před 3 lety +1

    It is about time I subscribe .

  •  Před 6 lety +1

    This is so satisfying!

  • @holyshit922
    @holyshit922 Před 5 měsíci

    13:16 To approximate pi i used Euler's formula for acceleration of series convergence to the arctan(1)
    After acceleration each iteration sets one bit of approximation
    (10 iterations gives three correct digits)

  • @matthaiosghanim562
    @matthaiosghanim562 Před 4 lety +2

    I've seen no videos regarding the odd powers of this function. Can you do a video on it?

  • @AngryArmadillo
    @AngryArmadillo Před 6 lety +2

    Fantastic video!

  • @ronrubin5601
    @ronrubin5601 Před 4 lety +1

    Hi - your video lecture is truly amazing. I am thinking about adding a video component to some physics courses I teach at Boston College in the US. Could you advise as to what software you use to create these amazing lectures. Thanks a ton if you can!

  • @raopsepol
    @raopsepol Před 2 lety +4

    After watching this video i'm an inch closer to winning a million dollars. :D

  • @xyz.ijk.
    @xyz.ijk. Před 4 lety

    This is even better the second time around.

  • @Mateusz-Maciejewski
    @Mateusz-Maciejewski Před 4 lety +2

    Great lecture. The only problem is with the probability of two random numbers to be relatively prime. You didn't define the probability space, namely the probability p_k of choosing a natural number k. It is an interesting question on its own, can p_k be chosen in such a way that
    P({k, 2k, 3k,...}) = 1/k, as is stated in the video.

  • @tvalladaress
    @tvalladaress Před 5 lety +1

    Wonderful work. Congratulations! By the way, what software do you use to make this animations? I'd like to use it to prepare my lectures. Sorry for my wrighting. English is not my mother language.

  • @SmileyMPV
    @SmileyMPV Před 6 lety +1

    Since you're a mathematician and you like Rubik's cubes, have you ever looked into the fewest moves challenge? You have one hour to come up with a solution to a scramble, and the goal is to use as few moves as you can. I recently started to look into this, and there are some wonderful mathematics that have helped people achieve averages of less than 30 moves! There is some really interesting group theory involved. There is a technique called insertions, which uses commutators and conjugates. But the most interesting thing I have yet found is NISS, which is also based on group theory.

    • @Mathologer
      @Mathologer  Před 6 lety

      I looked at this a while ago, but from your comment it would seem that there have been some new developments. Do you know of some good references? I should really do another Rubik's cubed video sometime soon .... :)

    • @SmileyMPV
      @SmileyMPV Před 6 lety

      Well I learned most things from this video series on youtube:
      czcams.com/play/PL0yL0AZiHw10Kx5um2l4MdOAICtRqJwac.html
      But this video series only explains how to apply some techniques. I came up with explanations for the techniques myself and you probably could too, but I later found this pdf:
      fmcsolves.cubing.net/fmc_tutorial_ENG.pdf
      This pdf goes more in-depth and has some explanations for the techniques it discusses as well. It also describes even some techniques not discussed in the video series.
      I do think the video series is still very good and will probably help more with visualizing than the pdf, even if the pdf has pictures and you can follow what the pdf is talking about by applying examples to one of your own Rubik's cubes. But I also recommend reading the pdf, even if it is pretty long, because this stuff is pretty interesting.

  • @weidox
    @weidox Před rokem

    Great. Only that I'm unable to find the next video which was advertised at the end.

  • @travelion5359
    @travelion5359 Před 6 lety

    title seen - and already hyped

  • @aaronleperspicace1704
    @aaronleperspicace1704 Před 4 lety

    "Take the differences between 1 and the square of reciprocal of every prime number. Multiply them out to each other and take the reciprocal of this product. Multiply this reciprocal by 6 and square root the result. The value thus obtained is the ratio of the distance around a circle to the distance across it".

  • @aaronr.9644
    @aaronr.9644 Před 6 lety +2

    Awesome video!

  • @adamaenosh6728
    @adamaenosh6728 Před 3 lety +2

    I would really like to know what the Riemann hypothesis would show about the way the primes are distributed

  • @pronounjow
    @pronounjow Před 6 lety +2

    Looking forward to the next video! I think I'm finally starting to understand the Riemann Zeta Function!

    • @dlevi67
      @dlevi67 Před 6 lety

      Wait until it becomes complex... (or rather, its argument becomes a complex number)

  • @hamidkh5488
    @hamidkh5488 Před 3 lety +1

    Thank you very much .

  • @sarojsi890
    @sarojsi890 Před 5 lety +1

    mind blowing excellent sir keep it up

  • @SamuelAndradeGTutos
    @SamuelAndradeGTutos Před 6 lety

    Something quite intersting. If we try to count the prob. of taking any random number and it is a prime number is 1/zeta(1)=0. It means that we have infinity much more composite numbers than prime numbers.

  • @monkerud2108
    @monkerud2108 Před 2 lety +1

    I rewatched this and came up with an elementary proof the sum is infinite. Just take 1/2,1/4,..,1/2^n out by collecting all of these terms in a bag or pouch. Do the same for 1/3,1/6,1/12 ect, then 1/5,1/10,1/20,.. all the different geometric series you can grab basically. Which gives you one for all odd numbers because all the series are geometric and infinite and in all instances its the series 1/2+1/4+...+1/2^n=2 times a constant A for each, the 1/5th series its A=(2/5) which gives
    (2/5)*(1/2+1/4+1/8+...+1/2^n) =
    1/5+1/10+1/20+...+1/(2/5)2^n = (2/5)*2
    So we get these coefficients times 2 for each odd reciprocal and two.
    2*(1+ 2/3 + 2/5 + 2/7 +...+2/2n-1)=
    4*(1/2+1/3+1/5+...+1/->)+2=
    4*(1/1+1/3+1/5+1/7+...+1/2n-1)=
    (1+1/2+1/3+...+1/n)
    We can separate it out into two series
    A=(1/2+1/4+1/6+...+1/2n) and
    B=(1/1+1/3+1/5+...+1/2n-1)
    We get
    4*B=A+B->A=3B
    As you said term by terms B

  • @trentmarcus
    @trentmarcus Před 6 lety +2

    Using the first 168 prime numbers, the approximation for pi is 3.14139318478793 (done using Microsoft Excel).

  • @vadimkhudiakov526
    @vadimkhudiakov526 Před 4 lety

    5:30, if there is strict inequality between 2 sequences a_n < b_n, it implies only non-strict inequality between limits lim a_n

    • @sergey1519
      @sergey1519 Před 4 lety

      It doesnt?
      Assume your statement is true, then:
      A = 1/1 + 1/3 + 1/5 +...
      B = 1/2 + 1/4 + 1/6 +...
      Let C= 1/3 + 1/5+...
      D = 1/4 + 1/6 +...
      By your statement C>=D.
      A -1/1 = C
      B -1/2= D
      A - 1/1 >= B - 1/2
      A >= B + 1/2
      A > B

  • @BloonMonkeyStudios
    @BloonMonkeyStudios Před 6 lety +13

    I had a math team practice about a week ago with the derivation of one of the (easier) identities being involved in the answer..
    the question was:
    Consider a infinite number of concentric squares, with side length 1, 1/2, 1/3, and so on, with the nth square having side length 1/n. Starting with the outermost square, the regions between the square are colored alternating blue then red. Given a dart thrown at the square lands inside the square, what is the probability the dart strikes a red region?

  • @samuraimath1864
    @samuraimath1864 Před 6 lety +2

    9:37 Even Euler "saw" that one coming :D

  • @user-cb6rp1ed8k
    @user-cb6rp1ed8k Před 6 lety +73

    6:22 it's supposed to be 1/27^z not 28

  • @SquirrelASMR
    @SquirrelASMR Před 2 lety +2

    That was really cool

  • @VSP4591
    @VSP4591 Před 3 lety +2

    Very interesting. Euler was a genius.

  • @diegomontalvo9173
    @diegomontalvo9173 Před 6 lety +1

    Great video!

  • @billybolton6555
    @billybolton6555 Před 4 lety

    i would say that zeta(1) is infiniti + Euler-Mascheroni constant... as if you minus 1/(x-1) it approaches the mascheroni constant

  • @pierrecarrette4976
    @pierrecarrette4976 Před 3 měsíci

    Illuminating video. Have you ever shooted the video on the alternate inverse squares series? Any link? Thanks.

  • @LukePalmer
    @LukePalmer Před 5 lety +4

    Did you ever make the video you said you were going to make at the end? I am very interested in this

    • @wallstreetoneil
      @wallstreetoneil Před 3 lety

      czcams.com/video/YuIIjLr6vUA/video.html (he made this - it addresses some of it)

  • @yousefalyousef59
    @yousefalyousef59 Před 4 lety

    Let the equation below accept a single solution(n) specify both(a,b) in terms of (n)
    f(x)=X^2-(a+b+1)X+(ab)=0
    since f(x)=0 is equivalent to
    x= (x-a)(x-b)
    I think in this space there are zeros of the zeta function .

  • @josephyoung6749
    @josephyoung6749 Před rokem +1

    one of your easier to follow videos for me (just a simple layperson)...

  • @oneofmant
    @oneofmant Před 6 lety

    amazing video!

  • @manuelrodriguez4952
    @manuelrodriguez4952 Před 9 měsíci +1

    One error, minute 6:24 the 28 should be a 27. Nice video, thanks for the info.