Fermat's Christmas theorem: Visualising the hidden circle in pi/4 = 1-1/3+1/5-1/7+...
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- Опубликовано: 12 сен 2024
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Leibniz's formula pi/4 = 1-1/3+1/5-1/7+... is one of the most iconic pi formulas. It is also one of the most surprising when you first encounter it. Why? Well, usually when we see pi we expect a circle close-by. And there is definitely no circle in sight anywhere here, just the odd numbers combining in a magical way into pi. However, if you look hard enough you can discover a huge circle at the core of this formula.
Here is a link to the relevant chapter in Hilbert and Cohn-Vossen's book Geometry and the Imagination (Google books). I am pretty sure that the idea and proof for the circle proof of the Leibniz formula that I mathologerise in this video first appeared in this book and is due to the authors: books.google.c...
Here is a link to a video in which 3blue1brown about the same hidden circle in Leibniz formula:
• Pi hiding in prime reg...
And another video by him about a hidden circle in the solution to the Basel problem:
• Why is pi here? And w...
There is also a neat generalisation to what we talked about in this video to the solution of the Basel problem - in terms of the lattice points in a 4-dimensional sphere and the 4-square counterpart of the 4(good-bad) theorem. If you are interested in some details have a look at the last proof in this write-up by Robin Chapman: empslocal.ex.a...
Links to two Numberphile videos about the one-sentence proof by Don Zagier featuring Matthias Kreck: • The Prime Problem with... (intro), • The One Sentence Proof... (the math)
Link to the original Jodocus Hondius engraving of Jodocus Hondius that Google tries to pass of as a portrait of the mathematician Albert Girard
www.swaen.com/...
Thank you very much to Marty for all his help with polishing the script of the video and Karl for his idea for the 2019 Easter egg.
Today's t-shirt: google "spreadshirt pi+tree+christmas+math"
Enjoy :)
Mathologer recently hit 500K subscribers and I would like to thank you all for your interest and your support over the years.
Since I started the channel four years ago, it has pretty much been a one-man labour of love. However, maybe now is a good time to take Mathologer to the next level and hire someone to assist with editing the videos, preparing subtitles, etc. In preparation for this, I recently monetised the videos by switching on the least annoying ads on RUclips. I also just launched a Patreon page a couple of minutes ago:
www.patreon.com/mathologer
If you enjoy these videos and you can afford it, please consider taking out one of the Patreon memberships or making a one-time donation via PayPal:
paypal.me/mathologer :)
My plan is to also use this Patreon page as a platform to share more Mathologer materials with you, get into live chats, etc. Let’s see what’s possible and what makes sense here. Merry Christmas :)
Knowing that Mathloger might expand was my best recieved Christmas gift so far
I'd love to see this channel growing big. All the immesureasurable effort put into your videos is just awesome. Merry Christmas :)
This channel reminds me why I love maths. The most beautiful and perfect stuff in the universe and beyond!
It will become...1M.....very very.....soon
Hey
I found 2019 in your π-shirt after the 244th digit after the decimal
#yay
Hey, Mathologer is back after a short interruption of service. Nice to see you again!
Merry Christmas!
I think the easiest way to show pi = lim N(r) / r^2 is to say that the squares centered in grid points = N(r) / r^2 >= pi * (r - 1)^2 / r^2 and since both leftmost and rightmost side of this inequality approach pi as r->inf, we get what we wanted
Squeeze theorem... Gotta love the name
Well I call it sandwich theorem :D
Policemen Theorem!
Three sequences theorem...? I was thaught boring names.
@@peterw9006 "Théorème des Gendarmes"
Let's all just forget about yesterday shall we? Great to see the videos are back and I really hope they work their situation out (IN PRIVATE)!
At dawn somewhere deep in the woods. My personal choice of weapons is swords :)
@@Mathologer I would go for a shotgun... better safe than sorry
@@Mathologer just enable 2 step verification in your google account, no 'swords' needed ;)
@@Mathologer That was ugly.
Way too much drama as far as I am concerned. I llike my math to avoid it.
Thanks for recognizing Indian Mathematician Madhava for his work on Calculus and infinite series relating to Pi. More than the credit, I am interested in bringing in awareness among talented Indian youth towards Maths which is currently more influenced by cash-generating Wall-street jobs and related courses.
Madhava is an Indian (Keralite) mathematician. Actually I am watching this video miles away from his native place (Irinjalakkoda, sangamagrama as you said). Nice work sir👏
@@sachinnandakumar1008 ya me too
Love to India, from Finland 🇫🇮❤🇮🇳! 😌
Odd factors of 2020: 1,5,101,505, all good. This tells us there are 16 pairs.
2020 = 16^2 + 42^2 = 38^2 + 24^2. The two permutations and four sign choices yield all cases.
The problem is that you can only choose two signs at a time. There are four cases involving 16, 42 and their negatives and also four cases with 38 and 24. There have to be two more pairs of positive integers whose squares add up to 2020.
@@cheshire1 : There are eight for each. We can also swap the order, which is what I intended by permutations. Thus the other two are 42, 16 and 24, 38.
The smallest number with this pattern of dots is 5*13=65, which one can quickly verify has precisely two pairs in the first half-quadrant: (8,1) and (7,4).
@@Utesfan100 no there is only 4 ways for each and eight ways combined both because commutative property says that a+b = b+a and here this is what you are talking about changing orders . But in that 4 way every term has different sign so they are different but a+b is same as b+a
@@Abhishekkumar-gu5gi We're searching points coordinates so (16,42) and (42,16) are two different points thus here permutations must be taken into account. This wasn't obvious in the video because most examples were small and a was equal to b but rewind and look at 7:04 for 5 = 1²+2², you have only one *pair* of positive numbers whose square add to 5, but you have eight *couples* since (1,2) and (2,1) are different.
Hooray I was sure that the videos would need to be reuploaded from scratch. Very glad to see the old comments were not lost.
I think they were just private.
The disaster has been averted. Glad you’re back!
hope you sue the fuck out of sob... he commit several offenses and probably even criminal
Yupp
Instablaster...
We dont celebrate Christmas in my country but I celebrate any Mathologer video anytime
A big thank you for acknowledging work by an Indian mathematician! First video doing so. You are a gifted teacher.
The good, the bad and the even
What an odd way to put it.
Dear Mr Burkard Polster,
I sincerely appreciate the fact that you give credit to the right person for the discovery of a mathematical identity/formula (in this case, the Indian mathematician, Mādhava). Earlier also, I learned through you that the fact that e^ix = cos x + i×sin x was known to Roger Cotes before Euler.
Cheers and merry Christmas!
Surprised you didn't call good and bad numbers nice and naughty numbers respectively.
:)
Now I can hear him laughing after calling them nice and naughty
Number of presents left under tree = number of good children - number of bad children
Or impish and admirable
@@gcewing Number of children - Number of bad children maybe ;)
My mnemonic for this infinite sum now will have to be that it's related to the number of integer points in one quarter of an infinitely large circle. Pi being the area of the whole unit circle. With the Pythagorean theorem, that connects it to the sums of squares. Beautiful proof. I just rewatched this one. You presented it simply enough for a high school math guy like me to understand.
I am really happy to see your videos back. We don't know the value of what we have untill we lose it.
Great to see you back!
My proof that bad odd numbers can't be written as a sum of integer squares:
Let our bad odd number be 4n+3, and let's say it's equal to j²+k², for some integers j and k.
However, since j² and k² add to an odd number, one of them must be even and the other must be odd. Let's choose j (arbitrarily) to be odd and k to be even. Hence,
j=2a+1 and k=2b (for some integers a and b).
So 4n+3=(2a+1)²+(2b)².
Expanding and rearranging, 4(n-a²-b²-a)=-2
So 2(n-a²-b²-a)=-1.
However, this is saying that 2 times an integer is -1, which is something only a drunk man would agree with.
Hence, 4n+3 can't be a sum of 2 integer squares.
QED.
You can also say that because all squares are congruent to 0 or 1 modulo 4, the sum of two squares must be congruent to 0, 1, or 2 modulo 4.
To see that all squares are congruent to 0 or 1 modulo 4, notice that 0^2 = 0, 1^2 = 1, 2^2 = 0, and 3^2 = 1 (mod 4). These are called the quadratic residues modulo 4.
@@aypfvn Isn't that the same as what was said in the video?
The Religious Atheists Whoops.. I was reading comments half way through the video and didn't make it to the end yet. Sorry.
@@aypfvn No probs
Since 2020/4 = 505 which has prime factors 5 and 101, we just have the power set of {5,101} which is {1,5,101,505}. Since both 5 and 101 are congruent to 1 mod 4, they are both good and so are all their products. You definitely gave us an easy one. So 4 good - 0 bad gives us 16 ways of writing 2020 as the product of squares.
To find them, we'll first use the usual trick of dividing out the 4's. We'll find the ways to write 505 as squares and multiply each component by the square root of 4. True confession time: I'm adapting this from work by Dario Alejandro Alpern, whose fsquares program I ported to Gnu GMP.
We can find the ways to write 505 by finding the way to write its factors and using the fact that
(a^2+b^2) (A^2+B^2) = (aA+bB)^2 + (aB-bA)^2
We'll find the solutions in positive integers, and then convert each such solution (a,b) into 4 solutions, {(a,b)(-a,b)(a,-b)(-a,-b)}.
We know that every prime congruent to 1 mod 4 is the sum of two squares. For 5 this is easy: 5 = 1^1 + 2^2 and that's all. For 101 it's not hard either: 101=1^2+10^2, and this confirms that you are pitching us your softest softball. A quick check vs {49, 64, 81} confirms that this is the only way to write 101. Again, this is just the positive/positive solutions.
So we have:
(1^2+2^2)(1^2+10^2) = (1+20)^2 + (10-2)^2, which gives us:
505 = 21^2 + 8^2
2020 = 2*21^2 + 2*8^2 = 4*441 + 4*64 = 1764 + 256
and consequently 3 other solutions,
2020 = (-42)^2 + (16)^2 = (42)^2+(-16)^2 = (-42)^2+(-16)^2.
We also have:
(1^2+(-2)^2)(1^2+10^2) = (1-20)^2 + (10+2)^2 = (-19)^2 + 12^2 = 361 + 144
Thus we find 8 ways of writing 2020:
2020 = 42^2+16^2 = -42^2+16^2 = 42^2+-16^2 = -42^2+-16^2 = 19^2+12^2 = -19^2+12^2 = 19^2+-12^2 = -19^2+-12^2
According to the formula, there must be 8 other solutions out there, but I'm not seeing the permutation of these equations that gives them.
Ah, I think I found the other 8. Since in the solutions for 18, we're counting both 3^2+-3^2 and -3^2+3^2 as separate solutions, we seem to be allowing a^2+b^2 and b^2+a^2 as distinct solutions.
Thus the other 8 ways to write 2020 are the flips of the ones I gave, starting with 2020 = 16^2+42^2 etc.
@@Tehom1 The other possibility is that since 24^2 + 38^2 = 2020 too, this gives the other 8 with the permutations of signs that you illustrated above? Though of course they don't come from 505...
edit: corrected 16 to 24
@@dlevi67 Excellent. I was too focused on 505.
Tehom Yes, that's right. It's because we're looking for the points on an integer lattice. So (16,42) and (42,16) are different. Took me a while to realise that too.
@@dlevi67 16^2 + 38^2 = 1700
Best thing about you, you always tell the real mathematician of the theorem. 😀👍🏻
BTW merry Christmas🥳🥳
Hi
@@mariomario-ih6mn Hi
I couldn't wait for the video proof of the Christmas thoerem and found something AMAZING on mathoverflow that I wanted to give you a heads up about in case you had not seen it before. I think the famous one sentence proof can be Mathologerized using the incredible geometrical interpretation of the involution in the answer by Moritz Firsching to the MathOverflow question titled "Zagier's one-sentence proof of a theorem of Fermat". I have read many books/papers on polynomial automorphisms and never saw anything like that before in my life. After going through a bunch of proofs I think the one-sentence version is the best hope to do at the level of your videos using that geometry trick to explain where the involution comes from. There is also a wonderful numberphile video on this topic with an outline of the proof.
Very nice. Will ponder this a bit, may well be worth doing a followup video on this insight :)
I found TheOneThreeSeven!
God save internet! I can not even imagine a life no more without having access to these wonderful, interesting, and enlightening videos like the ones you present every now and then!
Herzlichen Dank, fröhliche Weihnachten, und bitte weiter so.
Glad the issue has been resolved. This is one of the best channels on the internet. I do note the updated 'About' page, removing the GG reference...
Yes, and I guess you also know why. If you want to see what the page looked like in the past, just look it up on the wayback internet archive.
OMG! You are a fabulously knowledgeable and effective teacher. I am deeply humbled by your excellence. Thanks so much!
(Dr. Mike Ecker is a rational skeptic and a Ph.D. mathematician - CUNY 1978, Ph.D. Summa Cum Laude - who researches and writes prolifically. He is also a retired PSU mathematics professor and former computer journalist.)
Welcome back! So happy to see everything back to normal!
One video is still missing.
@@mwalton9526 Which one is it?
At worst, you might find a copy on Bilibili
Mathematician looks at 3:10 - "hey, neat, the area proof of pi!"
Engineer looks at 3:10 - "hey, look, nuclear reactor tiles!" ;)
Thanks alot for coming back!!
This video reminded me of Grant Sandersen's (3Blue1Brown) video entitled: "Pi hiding in prime regularities"
However I find this video more straightforward.
Thanks for the present and Merry Christmas!
I think this one seems more straightforward because it proves less. 3b1b's video only left the "all primes congruent to 3 mod 4 are gaussian primes and all primes congruent to 1 mod 4 can be factored into a pair of conjugate gaussian primes" to be proven. I suspect that the proofs of this video's lemmas Mathologer is trying to adapt are of similar complexity, but they're a bit more integral to the overall proof using this path.
Given how jumpy people get around the complex numbers, I suspect it will take a while before we see them.
@@timh.6872 3b1b's video also doesn't prove that prime factorizations under the gaussian integers are unique.
mwalton probably because they aren’t unique... (1-2i)(1+2i)=5 and (-1-2i)(-1+2i)=5
@@captainsnake8515 unique aside from multiplication by i,-i,-1.
(1-2i)*(-1)=-1+2i
(1+2i)*(-1)=-1-2i
mwalton (1-2i)(1+2i)=(2-i)(2+i)=5 I don’t even know why I’m arguing this point, it’s the worlds dumbest argument ever. I knew what you meant in your first message. I’m just being dumb lol.
First the Mathvengers: Eulergame, then Numberphile hit pi million subs, and now a new Mathologer video?! Is it Christmas? Oh wait, it is.
Been following your videos for 4 years and I have never been disappointed. Great stuff mate 👍!
You are one of the Great knowledge provider....
Thanks from India...
I really appreciated how you saved the explanation for "4 good minus bad" theorem until the end of the video. It cut off almost 9 minutes from the main proof and made it much more manageable. Good job
As a colorblind, can you not color like you did with good and bad numbers?
It may be very clear to people without vision problems, but green/red is the most common colorblind problem, using blue/red, blue/green, yellow/red is not a problem, even a strong red with green is generally okay.
Great video again, nice job!
YAY! The videos are back!
So the vids are back
Yeaaaaah!
Thank god, my favourite videos are back! Thank you, mr. True Mathologer.
The fact that such a channel can hit 500K+ subs, and maths videos with millions of views, gives me hope in human kind.... happy to +1 on Patreon
Great :)
I think an easy proof of equality in the limit uses Pick's theorem. Since we know that we can approximate the area of the circle with regular inscribed and circumscribed polygons, bound the area of the circle by those. Then, apply pick's theorem to both of those, and the error term grows like the number of lattice points on the boundary, which since these are line segments is bounded above by the arc length (since every line of length 1 can hit at most, say, 2 lattice points), which is bounded by something proportional to the circumference. So |pi * R^2 - N(R)| = O(R), which gives the desired result easily.
I'm glad that you are back :)
Thank you for all the lovely videos you made this year! I wish to you and Marty all the best in the next year, and I wish myself many more of your great videos... Merry Christmas!
Glad to see things are back to normal.
If you ever need anything from me (not sure what that would be, but I thought I would offer), just let me know!
Hmm, would you be interested being part of a proofreading team? If there wasn’t the self-imposed pressure to deliver a video at least once every four weeks, I would keep fiddling with these tricky topics that I am now specialising in much longer. Anyway, those four-week deadlines have the effect that invariably I am running out of time and that the shooting and editing happens last minute and little mistakes can slip in at this stage. Not really a problem but I am thinking of making a preview available to four or five math savvy friends who can have a quick look over the video to see whether they can spot anything. This preview could arrive at some untimely hour and nobody in the group should feel obliged to do anything if the short response time doesn't suit them. As long as I get feedback from one or two in the group each time that should be plenty. If you are interested please just send me an e-mail at pi.e@aol.com? No problem at all this is not for you :) At any rate thanks again for all your help with patiently answering questions over the years.
@@theongrejoy5645 I don't* :)
I am very happy that Your channel continues to work. Your wonderful deeds, o Lord.🙊🙃
Mathematics is like an intellectual equivalent of Willy Wonka's Everlasting Gobstopper -- it is harder than anything else on Earth, it tastes wonderful and no matter how long you work at consuming it, it never diminishes.
Not bad Ray, but as a theory it sucks.
@@petergregory7199 It's a theory you only get, if you suck at it ... kind'a like a paradox.
That bell between chapters ..... it gave me goosebumps !
Merry Christmas to Mathologer and to all!
I wonder who the hell can dislike such videos.
Love your work though!
Your videos are an inspiration to me sir🙏🙏🙏🙏
A) 18 = 3² + 3² and parity copies
B) 16 ways to right 2020 as sum of squares (1,5,101,505 are all good)
C) 2020 = 24² + 38² = 42² + 16² and parity and swapping copies
D) floor(r²/k) = r²/k - frac(r²/k). Dividing by r^2, we get 1/k + error which grows smaller as r becomes larger because frac is limited to between 0 and 1
Welcome back!
I came across Zagier's one-sentence proof a while ago. I couldn't understand it until I read a longer explanation in "Proofs from the Book". It's really neat!
Greeting from China!
So nice to see you back~
Kyo
Waiting for some hardcore video on Abel-Ruffini theorem :D
Phew! I hope the problem has been solved for good. I was so saddened yesterday.
Brilliant! Thank you! Love your approach to explaining things! ( _Merry Christmas! ... let's make 2020 the best year so far!_ )
This comment did not age well
@@braydentaylor4639 Nor 2021, but let's be as optimistic as we can be with 2022... it's a decision, a choice... make the best out of whatever happens.
@@gheffz Bro, 2021 was an okay year, eons better than 2020
Brilliant and Fantastic steps those few from 15:10.. looked like some magic done
Merry Christmas and happy new year! Congratulations on the success of your channel. It's well deserved.
I love following along and getting that giddy feeling when I can tell where I proof is going. Merry Christmas! Love the Channel since day 1!
That's great. I wish RUclips would give us a list of all subscribers in the order they subscribed. Would be nice to know :)
19:00 Zagier's "One Sentence Proof".
In case you want to investigate a bit I'll leave here a rough explanation of what an involution is.
Imagine a process that pairs elements so if you have element A it returns element A' and vice versa: for element A' you get A. One example of this would be when you get a direction, say North (in Projective Geometry we called it a point at infinity). If we had a parabola tangent to the Y axis, drawing lines parallel to the North would give us either 2 crossing points, 1 unique double-crossing point (the tangent to the parabola through the Y axis) or 0 points. These crossing points form an involution since each relate to another and vice versa through some crossing and all involutions behave in a way that you always get a double point that is related to itself, which we can call a fixed point.
This fixed point is what assures that we have an odd number of answers, that we can then interchange y and z in the involution and then get 4y^2 = (2y)^2 which summed to x^2 yields p.
In the proof, the involution is crafted very delicately so it proves Fermat's Christmas Theorem, but I've not spent the time to look into how it exactly works.
I hope this helps understand that part of the video.
Just to clarify/correct this a bit...
1) An involution is just a name for a function that is its own inverse i.e. f(x) = y => f(y) = x, or alternatively f(f(x)) = x (definition on wikipedia). Note that y can equal x in which case this would be a/the fixed point but...
2) "...all involutions behave in a way that you always get a double point that is related to itself,..." is blatantly wrong. This only holds if a) the involution is on a finite set; and b) the set has an odd number of elements. Both conditions are necessary to guarantee a fixed point.
There are many simple counterexamples, for instance S = {0,1,2,3} with f: S -> S, f(x) = x + 2 (mod 4). Writing it out, f(0) = 2, f(2) = 0, f(1) = 3, f(3) = 1 so it is clearly an involution but it has no fixed point.
Also, there is no guarantee that you have a single fixed point (just from knowing that its an involution). In the one sentence proof, it is guaranteed that the involution (the more complicated one; if it's not clear the proof involves defining 2 involutions) has a single fixed point but that is not the case in general.
In Zaiger's proof, he defines 2 involutions, one trivial, and one which has that complicated, 3 line definition. He then "shows" that 1) the complicated one always has a single fixed point and 2) that the 2 involutions have the same parity and thus a solution exists to p = x^2 + y^2.
1) is the tricky part of the proof and is why the definition is as complicated as it is but 2) is quite simple. For any 2 involutions on the same set, the number of fixed points must have the same parity (i.e. either they both have an odd number of fixed points or they both have an even number of fixed points; but they don't necessarily have to have the same number of fixed points, only the parity is guaranteed; you can probably understand why by thinking about this for a bit). Hence by showing that the first definition has a single fixed point (thus it has odd parity) the other one must as well and hence, since 0 is even, there is at least 1 fixed point in the second involution and thus at least 1 solution to p = x^2 + y^2.
The first thing I did this Christmas was watch this whole video! Fantastic as usual thank you!
This is the best possible christmas gift 😁😁 Thank you MATHOLOGER
Merry Christmas ❤️❤️
Perfect... a mathematician that speaks with a small german english accent, to me classroom example of a professor! Grusse von die Niederlande!
I watch your videos because currently I am learning German, and I love math and would pursuit a career of Engineering in Germany.
Currently also live in Melbourne.
Merry Christmas to the Mathologer, and to my son, a mathematics major in his first year, who will receive a copy of "A Dingo Ate My Math Book" tomorrow morning from Santa. I'm hoping he enjoys it as much as I have. Thanks, and Happy Holidays to all your viewers too!
Marty and I just had a chat about a possible part 2 of the Dingo book. Still got lots of material left over from those articles in the AGE for at least three more books :)
Glad to see you're back!
BEST MATH CHANNEL! MERRY XMAS TO THE MATHOLOGER AND HIS FAMILY!
One of the best from mathologer. This proof really blowed up my mind. Amazinnnnggg
Great video, Love it. I actually found out about those theormes recently when I was trying to solve an olypmpiad problem about sums of two squares. It was pretty difficult problem and I'm really proud I could solve it.
Merry Christmas 🎄🎄🎄
Excellent video! Very nice!
I've encountered the result about sums of squares for _primes_ before, because that gives us a test for which integer primes remain primes in the Gaussian integers. It's nice to see that there is a much more general version of the result covering _all_ nonzero integers and telling _how many_ ways there are to write as sums of primes.
Also, that's a beautiful way to get that infinite series formula :)
Merry Christmas!
as r -> Inf, N(r) -> Pi*r^2; Therefore lim r->Inf (N(r)/r^2) = lim r->Inf (Pi*r^2/r^2) = Pi
Although not sure if this is a circular argument. It depends on whether we're allowed to assume that the area of the approximation approaches a circle towards infinity.
Yes, does not quite work as a proof I am afraid :) I'd say have a look at some of the solutions submitted below.
I enjoyed this holiday gift, from the very best maths channel ever!
I had to try this out and I made myself a spreadsheet table.
I made it to begin with 1 on line 1 and each successive line would add 2 to the previous line's value so every line starts with an odd number as a seed.
Every line calculates the reciprocal of its seed.
Every even line subtracts the seed's reciprocal from the previous line's result.
Every odd line adds the seed's reciprocal to the previous line's result.
Well, now I found early on that I may have to take a shourtcut.
I made every line to also calculate the average of its result with the previous.
All results were obviously multiplied by 4 to get approximations of Pi.
I iterated the list up to n=20001.
Then I set myself some boundaries. I seeked the list for as close as possible approximations of Pi at different decimal lengths, and looked for the first number in every list (result of last subtraction, result of last addition, and the averages) that exceeds but rounds up to equal the approximation of Pi at said decimal precision.
3,1 (1 decimal) - line 5 (avg, n=9) and line 26 (sub, n=51)
3,14 (2 decimals) - line 14 (avg, n=27) and line 622 (sub, n=1255)
3,142 (3 decimals) - line 24 (avg, n=47) and line 1103 (add, n=2205)
3,1416 (4 decimals) - line 94 (avg, n=187)
3,14159 (5 decimals) - line 435 (avg, n=865) and line 464 (avg, n=927)
3,141593 - none reach this high
3,1415927 (7 decimals) - line 2280 (avg, n=4559)
3.14159265 (8 decimals) - getting close but seems to take thousands of more iterations
So it seems that averaging two consecutive iterations gives closer approximations at earlier stages.
Awesome video once again. I love the way you visualise everything to make the math understandable.
Really love your channel. The math is great, amazing. But the mathologer is even better!Congratulations on hitting 500 000 subs. Merry Christmas and Happy New year to you. :)
Fröhliche Weihnachten :)
Und einen guten Rutsch in neues Jahr!
Fröhliche Weihnachten!
Thanks again!!! It was a revelation for me again!
I was wondering for a while why pi/4 can be expressible with this inf. series.
And thanks for the recommendation of the book of Hilbert and Vossen.
So far the best Christmas present! Thanks "Manta" (Math Santa) and Merry Xmas!
We can also go by an another method I = 1-1/3+1/5-1/7....
Because we have integers in the denominator we can think they came after integration . Now why that Int x²dx = x³/3 +c. so considering a integral I = Int from 0→1 (1-x²+x⁴-x⁶+x⁸...∞)dx
We can directly integrate each term
I = x - x³/3 + x⁵/5 - x⁷/7 + x⁹/9.... and now substituting the limits of integration from 0 to 1
I = [1 - 1³/3 + 1⁵/5 - 1⁷/7 + 1⁹/9...∞] -[ 0 - 0³/3+0⁵/5-0⁷/7...∞]
I= 1-1/3+1/5-1/7+1/9...
But by another way we can say the integrand (1-x²+x⁴-x⁶...∞) is an infinite sum of a Geometric Progression with first term 1 and common ratio -x² and it will converge for x²
Glad to have you back Mathologer!
Just found this video. Amazing way to visualize the proof. Thanks for the hard work!
I think I am able to parse the pi in its series form, pi is not a circle or parts of circle, but is a ratio between a circle and triangle, value that represent the enclitic-ness of the determined. A series is a type of succession, or type of journey. A series' sum is what you get at the end of the journey. It also the product of movement that refers to an origin that it can return to and revolves around an origin that it cannot (or don't want to be) in order to remain consistent in at least one known dimension.
Quadratic reciprocity?! My old professor, way back, wrote a book on number theory, in which he and his coauthor proved the law of quadratic reciprocity several different ways. Then he included a question at the end of the chapter asking the reader to find their own proof! At the time he was teaching analysis but he often slipped in number theory themes in his lectures. I never picked up the book because I wasn't interested in number theory at the time and frankly I was not any good, but I always thought he was special, a genius even, who loved his subject. You remind me of him. Happy New Year!
I think the proof of "the limit of approximation is pi" is as follows,
Consider we have a circle of radius of 1, then N(1)/1^2=4. Algorithm in the video says increase radius of circle, but instead, we can just decrease distance between two points and count how many squares are in the circle. How? let's say we want to increase radius to 10, then decrease the distance between two points to 1/10. Then, we will have exactly the same number of squares if we have just increase radius to 10. Now, what N(10)/10^2? N(10) represents is how many squares are in the circle. But we lowered each squares' area by factor of 1/100. Then sum of these squares' area is N(10)*1/100=N(10)/10^2. If we keep doing this process we will see that what we actually do is Riemann integration. And by definition, area of circle of radius 1 is pi.
Thank god they are back :)
2019 hidden in your t shirt ...thanks for this Christmas Present ...Merry Christmas Mathologer and all
The odd factors of 2020 are 5 and 101, both are "good" factors so there is 8 different ways to write 2020 as a sum of two integer squares:
16^2 + 42^2 =
16^2 + (-42)^2 =
(-16)^2 + 42^2 =
(-16)^2 + (-42)^2 =
24^2 + 38^2 =
24^2 + (-38)^2 =
(-24)^2 + 38^2 =
(-24)^2 +(-38)^2 =
2020
Hey
I found 2019 in your π-shirt after the 244th digit after the decimal
_I am a man of culture_
Found it myself and had to scroll down VERY far to this spoiler ;-)
2^2 + i^2 = 3 which means that we can do the "bad" numbers using sums of complex squares since that's (2 + 0i)^2 + (0+i)^2
Merry Christmas. Your channel is a treasure.
For the limit of floor(r^2/3)/r^2 ;
We have the identity :
x-1
That's it :)
9:25 challange accepted. the four ways to write 18 as a sum of 2 squares are:
3^2 + 3^2
(-3)^2 +(-3)^2
3^2 +(-3)^2
(-3)^2 + 3^2
And the next number of ways to write 2020 as a sum of 2 different squares is 16, they are
16^2 + 42^2
42^2+16^2
(-42)^2+16^2
16^2+(-42)^2
(-16)^2+(42)^2
42^2+(-16)^2
(-42)^2+(-16)^2
(-16)^2+(-42)^2
And the same combinations are followed with 38 and 24.
The 4 ways to write 18 as the sum of 2 squares are the 4 ways of (+/-3)^2 + (+/-3)^2
For 2020, there are 8 ways (odd factors are 5 and 101, both are 'good'). The 8 ways are composed of the squares 16^2 and 42^2, with all combinations of signs and both orderings (ie 16^2 + 42^2 is a different solution from 42^2 + 16^2).
Half right. You missed some odd factors: 1 & 505. Thus there are also 8 solutions for (24,38).
just spend about 2 more seconds thinking about it and you should see that this predicts that pi/4 is the true fundamental circle constant.
it was multiplied by 4 artificially all along, so dividing it by 4 at the end is actually just correcting that initial mistake. this is further corroborated when you notice that pi/4 is the ratio of the area of any ellipse to the area of the enclosing rectangle as well as the ratio of the circumference of any ellipse to the perimeter of the enclosing rectangle (numbers don't normally correlate both length and area this way, and per usual pi itself doesn't, which should instantly raise some eyebrows that pi/4 is somehow more special than pi). there's also the fact that the prefix 'co-' on trig functions means 'reflected across pi/4': cosine = sine reflected across pi/4; cotangent = tangent reflected across pi/4; cosecant = secant reflected across pi/4. additionally, the historic candidates for fundamental circle constant, pi/2, pi, and tau are all just power of 2 multiples of pi/4 which simply correspond to the various symmetries of a circle inscribed within a square, with pi/4 representing the smallest division found thusly (whole figure = tau, fold in half either edge to edge, or corner to corner = pi, fold in half again, keeping circle center as a corner of the figure = pi/2, fold in half one last time, keeping circle center as a corner of the figure = pi/4, no further iterations are possible).
beyond this there are quite obvious paradoxes implied if pi were the fundamental circle constant, such as an infinite regress paradox if we attempt to generalize the form of a=pi*r^2 to finding the area of all 2-dimensional shapes. played out, the problem here comes when you try to find the area of a square this way, because that area should be 4*r^2, where r is half one of the side lengths, but solving this requires finding the area of a square 1/4 the size of the original square in question, and if we can find its area surely we could also find the original square's area the same way, so we must now apply 4*s^2 to find the area of the square with side length r, where s is half of r, ad infinitvm. clearly this is ridiculous and we don't find the areas of squares this way, but in the same vein we shouldn't find the areas of circles this way either, instead we should use the proper general form a = pi/4 * d^2 (a = pi/4 * x * y for any ellipse).
this is proper for many reasons, but probably the most intuitive is that it considers the area of the circle as a fraction of the area of the circumscribed square, and thus the corresponding area equation for a square is simply a = 1 * d^2 (a = 1 * x * y for any rectangle). this corresponds term for term to the area equation for a triangle, a = 1/2 * x * y. and thus we get constants corresponding to various shapes, which all must lie between 0 and 1, where 0 means 'none' and 1 means 'all'.
this shape constant occurs in generic exponentiation as well. it is the 1 in 3^0 = 1, and the 1 in 3^-4 = 1 / 81, and it is an invisible, forgotten feature of 3^4, and from this we can get a clear sense of the semantics of the elements of an exponential expression: start with 1 and multiply it by the base exponent times. thus:
- 3^4 = 1 *3) *3) *3) *3)
- 3^0 = 1
- 3^-4 = 1 /3) /3) /3) /3)
note that the negative symbol in math always just means 'opposite', and the opposite of multiplying is dividing and thus negative exponents result in the initial 1 being repeatedly divided by the base. we have also established a definition that 0^0 must be 1, and this is a definition couched both in philosophy more sound than classical philosophy of maths and in demonstration from experiment with the consequences of that new philosophy.
I love your content and am proud to support your Patreon at the π level!
That's great, thank you very much :)
I like it when I see more than one way to prove something. I have heard, for example, of 5 proofs of the infinitude of primes. The usual way to prove this is to divide 1 by 1+x^2, integrate the result by x, and substitute first 0 and then 1 for x, and pi/4=1-1/3+... appears. Good and bad numbers? Yellow numbers are definitely bad, but some good numbers are deceiving. 17 is OK. It's definitely a good number. But 21 is a wolf in sheep's clothing. It may look good, but it is the product of two bad numbers, 3 and 7. It's a double negative. Two wrongs don't make a right, and so 21 is not the sum of two squares,.
Return of the King
Bitte mach mehr von deinen Mastervideos. Ich studiere Mathe im ersten Semester und liebe die langen Videos die richtig tief gehen, je tiefer desto besser ;)
Thank you for your videos. Happy New Year.
$1 pledge level? No brainer.
I made a comment on a recent video (maybe the most recent) about wanting to see more about how theorems have been used to solve other theorems. I'm not sure if you meant to do that here, but I really like learning about putting together these "looks suspiciously related" theorems to build another result. Thank you very much!
A great pleasure during Christmas for me. Thank you!
Thank you to acknowledge Madhavas contribution- He is from my place
Reminds me of "Pi hiding in Prime numbers" from 3Blue1brown. That was the video that inspired me to do maths. Lovely job, Mathologer
5:10 Idea for the proof of equality in limit: The error from N(r) vs actual area is at most 1 at each grid point on the edge. There is at most 4r grid points on the edge in total. so we have (N(r) - 4r)/r^2