Pi is IRRATIONAL: simplest proof on toughest test
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In the last video of 2017 I showed you Lambert’s long but easy-to-motivate 1761 proof that pi is irrational. For today’s video Marty and I have tried to streamline an ingenious proof due to the famous French mathematician Charles Hermite into the hopefully simplest and shortest completely self-contained proof of the irrationality of pi. There are a few other versions of this proof floating around and we’ve incorporated the best ideas from these versions into what I’ll show you today; I’ll list some of these other versions below. I also talk about the problem of pi + e and pi x e being irrational at the end of the video, really nice stuff.
Articles to check out:
Ivan Niven’s simple proof (we essentially use his sequence of integers) projecteuclid.org/euclid.bams...
A really nice article by Li Zhou arxiv.org/abs/0911.1929. Among many other things this article, has an introductory section on the origins of the strange integral at the core of the proof that I am presenting in this video.
Another article by Li Zhou arxiv.org/abs/0911.1933. (the 6 line proof of Theorem 2 is essentially what I am presenting in this video).
All these articles are great, but I think the one article that deserves most credit for having brought Hermite’s beautiful proof to the attention of the wider mathematical community is this article by Jan Stefvens: Zur Irrationalität von pi, Mitt. Math. Ges. Hamburg 18 (1999), 151-158. This one also has a very nice account of Lambert’s and Niven’s proofs.
In the video I mention that another version of this proof made an appearance in the toughest Australian maths exam in 2003; here is the link to this exam
educationstandards.nsw.edu.au/...
As usual thanks to Marty and Danil for all their help with this video.
Enjoy!
"333 Only half evil" that's a great shirt, but I prefer: "25.807... The root of all evil"
Gotto get myself a t-shirt like that too :)
I guess that means 25.807=money
It's _the love_ of money that is the root of all evil.
@@burrbonus English proverbs always seem to get shortened so their original meaning gets lost. And the original meaning is usually the exact opposite of what people think. You said one of them, and if anyone is interested you can look up "Jack of all trades, master of none" and "Blood is thicker than water".
What does that mean?
Host: "Everything so far has been pretty easy..." Me: "IT HAS?!"
Hey, fun fact, I got the highest mark on this exam in NSW. I enjoyed this question a lot! It seemed funny / expected since the previous year's exam asked you to prove that e was irrational.
wow, i hope to be able to achieve close to your maths accomplishments in the following year. i need your blessings for my ext 2 exam 🙌🏼🙌🏼
Well done my man. What's the name of the exam? Id like the see the other questions
Impressively impressive.
@@richardjehoe8841 You can find a link to the exam at the bottom of the video's description.
Congratulations 👏🏻!
Proof that x^n/n! converges to zero (cheat version): The series e^x=sum(x^n/n!, n from 0 to infinity) converges so the sequence x^n/n! converges to zero. QED
no no no how can you prove that the Taylor series for e^x always converges the Taylor series for ln(x) at a=1 only has a radius of convergence of 1 it diverges everywhere apart from (0,2] your logic is actually circular
@@DendrocnideMoroides My comment was meant to be circular, hence the "cheat version"
@@Cubinator73 oh I did not understand what (cheat version) means. I thought that you did not figure it out by yourself and looked it up
I dont think it is circular, I think the series formula of exp can be derived from the definition of e with a limit. That is: exp(x)=lim (1+x/n)^n with n->oo. It still feels like overcomplicating it though
😂😂😁😁🤣🤣🤣
Would really appreciate an 'algebra autopilot' for my exams >
Then you'd have to practice. A lot. The reason why it's so automatic for so many people is because they've been solving it for years.
930 8323 I wasn't saying arithmetics was challenging. My point is that some derivations take pages while the maths behind them is trivial.
Oh, so that's what you mean. My apologies.
Isn't this a TI-89?
Practice a lot and you’ll learn how to do that.
I’m so glad that I have discovered your wonderful channel. As an economist who uses quite a bit of advanced mathematics, I not only find your videos edifying but exceptionally entertaining. Sometimes, I literally make some popcorn, and sit down and view them. And I’m jealous of your shirts. :)
That's great :)
Please tell me what maths you use in economics. I am really curious!
I love math competitions, but when it has time constraints as small as 3 hours to do 8 problems it makes me vexed. For only one of the problems you'd have to do the thinking that took Mathologer 19 minutes to explain and then acutally write all of details that he quickly skipped and not make one mistake while doing it. All of it in under a half hour. Math in the real world is only constrained by time in the way that you only have limited time being alive.
You just get marked down when you make mistakes. It's not like they fail you if you make a mistake.
Yeah, the main point is that if you want to win this kind of competition you don't train by improving your understanding of math, instead you have to train the speed with which you do computational problems and the accuracy of these computations. I don't really have a problem with doing computations by hand per se, but with the fact that this kind of math competitions don't test your mathematical talent, but the speed of computations which is far less important. Mathologer in the video under which I am writing this comment for didn't focus on how to do these computations, because if he did this video would be twice as long and uninteresting. He just quickly shows the computations for a few seconds here and there while focusing on the meritum.
This is actually not a math competition, this is a regular school exam :)
Not a lot of difference. One should have harder questions, but both should have enough time to ponder about the questions and try multiple approaches. International Mathematical Olympiad has a really good format with 6 problems and 9 hours divided in half for two days.
I personally don't like exams as a way to test students in maths, just often there is no real viable alternative :)
I know I'm a bit late but here's my take on lim as n-->inf of a^n/n! And by "my" I mean "my memory of some guy on StackExchange's". f(n)=a^n/n! f(n+1)=a^(n+1)/(n+1)! f(n+1)/f(n) = ( a^(n+1) *n!) / ( (n+1)! * a^n ) = ( a^n*a *n!) / ( (n+1)*n! * a^n ) = a/(n+1) lim n-->inf a/(n+1) =0, therefore f(n) is decreasing in absolute value and it is decreasing until 0. Not sure if that last sentence has any logic holes in it but it looks good enough to me.
pi=e=3
Brilliant
:)
engineers are the best
e^π-π=20 5(e^π+π^e)=228
Sergio Korochinsky (e^π)-π?
Another magnificent piece of work from my favourite math mystic. You transform mathematics into performance art! I particularly like the argument that generalizes to "At least one of (pi+a) and (pi*a) is irrational, where a is any real number." I look forward to verifying the steps of Hermite's proof on my own!
If possible, i would like to watch a video where you read or analize a math paper. I think it can be interesting because if we had some more tools to understand those things, our self learning will increment
I think that's a *great* suggestion! It's perhaps not really Mathologer style, but maybe he can start a Herr Professor Mathologer channel.
Great video! Although this test question can't compete with question 6 of the IMO that year in terms of difficulty. The problem asks you to prove that for all primes p, there exists a prime q such that for all integers n, n^p-p is not divisible by q. This problem also has a nice solution.
How patiently and diligently you explain that counts.
Why x^n/n! tends towards zero: When you add one to n, the numerator increases by a factor of x, but the denominator increases by a factor of whatever n now is (by the definitions of exponentiation and the factorial). Thus, adding ons to n is the same thing as multiplying the whole fraction by x/n. As n gets arbitrarily large, x/n approaches zero. This means the numbers in the sequence will get smaller everh time once n>x and approach zero.
The proof was quite beautiful and simple, but I agree that the motivation is very hard to decipher. I would love to see how Hermite constructed that sequence
Thank you for your excellent work! Your videos are pure gold :)
Beautiful explanation! Congratulations.
7:22 Roughly speaking; (x^n)/n! can be expanded out to the form: (x*x*x*…*x*x)/(1*2*3*…(n-1)*n), where the number of x’s in the numerator is equal to the number of integers being multiplied in the denominator; which, of course, is equal to n. Each integer in the denominator is exactly 1 bigger, than the last (so, their size is steadily increasing), while the x’s stay constant. This means that, after the xth coefficient, each additional number in the denominator is larger, than the corresponding x in the numerator, by a steadily increasing amount. We can, then, rearrange the ratio into this product of ratios: (x/1)*(x/2)*(x/3)*…*(x/(n-1))*(x/n). As n tends to infinity, the ”last” ratio x/n approaches 0, which can be considered the limit of x/n, as n -> ∞. Any finite number, however arbitrarily large, multiplied by 0, equals 0. Therefore; for any finite value of x, the product of the ratios (and therefore, the original ratio: (x^n)/n!) approaches (x/1)*(x/2)*(x/3)*…*(x/(n-1))*0 = 0, as n -> ∞. Even shy of infinity; as n gets larger, the ratios after x/x get increasingly smaller; and thus, the whole product (which really is our ratio (x^n)/n!, in disguise) gets tiny, and approaches 0. Therefore: (x^n)/n! -> 0, as n -> ∞. 🙂
The way he uses colour to represent the meaning or character of something is ingenious! I refer to such as the formula and graph at 5:15
I recently understood how big of a nerd I am when I started liking these videos
You asked for interesting questions from exams. This was posed to me as an extra credit question X + y + z = 1 X^2 + y ^2 + z^2 = 2 X^3 + y^3 + z^3 = 3 Either find x, y z or show it can't be done
One way: minimize distance from x³+y³+z³=1 to origin and see that it is ∛3, which is greater than √2. Therefore, it cannot intersect sphere x²+y²+z²=2. Is there a better way?
After some algebra, we find: x+y+z=1 xy+yz+zx=-1/2 xyz=1/6 Therefore, now we can build a cubic equation with x,y,z as roots. x³ - x² - x/2 -1/6 = 0 (eqn 1) The roots of this equation will be the solution. After solving, the roots are (approximately), 1.4308 -0.21542-0.26471i -0.21542+0.26471i And they are the values of x,y,z. Or alternatively, if they have asked only about real solutions, you can differentiate eqn 1 and find the critical numbers (points where the slope is 0). In this case, we find 2 critical numbers (one local Maxima and another local minima) . We can differentiate again, to find which one is the Maxima. Because the coefficient of x³ in eqn 1 is positive, in the long term, it rises on the right and falls on the left. If we substitute the value of x we found for the Maxima, we find that the value of the function is negative at that point. Therefore, there is only one real root for the equation as the graph cuts the x axis only once due to the Maxima being below the x axis. So there are no 3 real numbers which satisfy these equations.
@Peter Flom (your question continued) x^4 + y^4 + z^4 = 4 And then find x^5 + y^5 + z^5 = ?? Actually this question is one of the recent videos of Mind Your Decisions KZhead channel by Presh Talwalkar.
Very cool proof! It's the first time I see it. Thanks for sharing
Great video! Love your channel
The proof at the end: that generalises to any pair of irrational numbers where at least one is not algebraic, doesn't it?
For the puzzle at around 7 minutes in, as you said, express it as (x^n)/(n!) . This means we have n x's on the top, and n terms on the bottom, so we can expand to look something like (x/n)(x/(n-1))(x/(n-2))....(x/2)(x/1) . If we take the limit as n approaches infinity, all of the fractions with smaller denominators are still arbitrary fractions, but the one on the left, (x/n), approaches zero. When you multiply this fraction with the other arbitrary fractions, that product also tends to zero. I know that the other fractions close to (x/n), such as (x/(n-1)) also approach to zero as n approaches infinity, however I'm not sure how important it would be to state that. The idea is that as n increases, the other fractions are multiplied by smaller and smaller fractions, so it must approach zero. Sorry if my answer is unclear, not sure of the exact wording I should use.
Thank you. Greatly enjoyed this video.
What brought you to Australia? (Don't say "a plane." (If it was, I bet it was a complex plane.))
I love you mathologger!!!!!
I'd really like all your t-shirts. You should have a store.
oh man, im way out of my depth here
the reason x^n/n tends to zero is as follows; x^n can only have factors of x. When n gets larger than x, it will constantly grow faster. You’ll have until x=n or until it’s x rounded down. as one fraction that will be a factor of all numbers that apply to x^n/n. Call that number z. So, b steps after x rounded down plus one looks like this; z•(Floor(x)^(b-n))/[(Floor(x)+1)(Floor(x)+2)...(Floor(x)+b)] The bottom is larger terms than the top each time. As a result, whenever we add a new term after floor(x), it gets smaller, and thus after that point we have a constant times a decreasing number, ergo it tends to zero.
Ok I know you linked the articles in the description but it would have been super great if you gave a tiny explanation of where the hell those I0, I1 and In terms came from. I'm totally lost on how they have anything at all to do with pi, especially since the integral of that sin component from 0 to pi equals 2...
Please record a video explaining Set Theory! I would love to see how you explain it intuitively!
I think everybody should watch this video. Thank you for your effort.
Thank you Mathologer for all your videos! They are always amazing and very very interesting. Just a clarification: towards the end of the video, when you talk about the irrationality of e+pi and/or e*pi, you say (18:47) “we don’t know which” and that seems to assume that one of them is irrational while the other one is rational. But as you said just a few seconds before, we know that at least one of them is irrational, which means that either one is rational and the other one is irrational or they are both irrational. I’m afraid that this last scenario has not been really understood by everyone.
About ideas for other videos; What about Cook's theorem on NP-Completeness? It's the first and only problem directly proved to be NP-Complete and question to my junior year Algorithms exam in university. I still (5 years later) remember the details because it was very nice.
The proof of the corollary that either pi*e or pi+e is irrational did not appear to use any properties of e, so actually shows that either pi*x or pi+x is irrational, for any x. And the only property of pi that it used is that pi is not the solution of a quadratic polynomial with integer coefficients, which is true of any transcendental number. So this really establishes that for any transcendental number t, and any number x, with t*x or t+x must be irrational.
Outstanding video!
In the last video of 2017 I showed you Lambert’s long but easy-to-motivate 1761 proof that pi is irrational. For today’s video Marty and I have tried to streamline an ingenious proof due to the famous French mathematician Charles Hermite into the hopefully simplest and shortest completely self-contained proof of the irrationality of pi. There are a few other versions of this proof floating around (see the description) and we’ve incorporated the best ideas from these versions into what I'll show you today, really nice stuff. Enjoy! As usual if you'd like to help out with Mathologer please consider contributing subtitles in languages other than English and Russian :)
I have a very cool and challenging question that I faced (as practice fortunately) when I was preparing for the exam for admission to the University. I really would like to share with you but I don't have your e-mail!
burkard.polster@monash.edu
In the last couple of videos I've definitely been pushing it :) Still trying to figure out exactly how much I can get away with. For example, a proof that e is transcendental is very much on my wish list ...
Mathologer Please shoot me!
would be very nice:) please do it:) (i mean the proof of e is transcendental)
You can show that the expression tends to 0 when "n" tends to infinity by showing that the exponential function does not grow as fast as the factorial function, which means that as "n" grows larger and larger the top side will get smaller and smaller compared to the lower side, thus the fraction will always get close to 0 as long as you keep making "n" bigger.
Be careful with these handwaving notions like "does not grow as fast as", they fooled many.
proof of thing at 7:00. first we do what you suggest and instead, try to prove x^n/n!. We write it as the product x/n * x/(n-1) x/(n-2) ... x/1. then we consider the situation of n=x^2, and split it up into 2 parts: all terms > x/x (aka 1) and all terms < x/x. We now reorganize our terms so that start multiplying the largest term with the smallest term the second largest with the second smallest and so on (1,1/(2x-2),1/(3x-3) etc) as you can probably see for no values of x will this every be < 1. Now with this we have left off a bunch of terms but that doesn't really matter since those terms are all < 1. so now that we know that x^n/n! for n=x^2 is < 1 we can finally prove the lim of x^n/n! as n->infinity fairly easily since x^n/n! for n=a will always be less than x^n/n! n=a+1 (we get this from the product x/n * x/(n-1) x/(n-2) ... x/1, and how if you increment n you get the same product * x/(n+1)) so the limit has to be some constant =
Years ago on a math contest [one that I won (: ] there was an interesting question that involved a power-tower. A power-tower is something like X to the power of X to the power of X to the power of X, etc. carried out infinitely. In the question on the test, X was set equal to the square-root of two; and we were asked to evaluate the infinite power-tower. It must be noted that there is an ambiguity in how I just defined the power-tower (something that the examiner might have missed). We can think of the power tower as a Sequence of infinitely many terms (Ni), defined as a recursion relation. Definition 1: N(0) = Sqrt(2) N(1) = N(0) ^ Sqrt(2) N(i) = N(i-1) ^ Sqrt(2) In this case, the power-tower explodes and quickly increases without bound. Definition 2: (more interesting) N(0) = Sqrt (2) N(1) = Sqrt (2) ^ N(0) N(i) = Sqrt (2) ^ N(i-1) In this case, the power-tower (p-t) approaches a limit. Can you see what it is? Set the power-tower = M. Then the p-t can be written as Sqrt(2) ^ M = M since any infinite subset of the p-t can replace the entire p-t. Simple huh. Obviously M = 2 solves this equation, so the p-t = 2. But wait M = 4 also solves the equation. Not hard to show that this solution can’t be correct. But why does a second solution show up at all. (see below?) Even more mysterious (to me) is why the extension to other power-towers doesn’t work. In particular, if X is equal to 3 ^ (1/3) should also be easy since CubeRt(3) ^ M = M can be solved by M=3. But the power tower actually converges on a different value. A couple hundred terms shows it converging to approx 2.47805268 So, what gives? Also X= 4^(1/4) now gives both 2 and 4 as solutions (not surprising since 4^(1/4) = sqrt(2)… so that’s where the second solution above (4) comes from. What about X = 5 ^ (1/5). Is the p-t = 5? Absolutely NOT… but why not??? Again, some numerically found values says it converges to approx. 1.76492. What is that number??? Any general conclusions???
I'll bite. (n^(1/n))^x = x we can rewrite to n^(x/n) = x. As you point out, it has a solution at x=n. Note y=n^(x/n) is just an exponential so can intersect y=x at at most two points. One point of intersection is at (n,n). There may be another point of intersection as well. In the case of n=3, it's at (2.47..., 2.47...) and in the case of n=5, it's at x=(1.76..., 1.76...) . For n > e, I compute that dy/dx>1, meaning the second solution definitely exists and is smaller than n. Proving the tower converges to the smaller number for all n>=3 (if that's even true) would take someone either smarter or more motivated than I.
Integration by parts is actually carrying derivative from one multiplier to the other multiplier. It's reverse operation of derivative of multiplication of two items.
You are absolutely right. In other versions of this proof integration by parts is avoided this way. See, for example, Niven's short proof (link in the description :)
Her's my super simple proof of why x^n/n! goes to zero when n goes to infinity for any x. So, we can simply write x^n/n! = PI(x/i) for i = 0 to n for x > 0 and x^n/n! = PI((-1)^n * abs(x)/i)) for x < 0. In both cases we have a product of x/i 's . We know that each of these individual terms are finite because it's just x divided by an integer, so no worries of weird indeterminate infinity times zero things. Then in both x < 0 and x > 0 cases we know that one of the terms in the product will be x/n. and simply lim as n goes to inifity of x/n = 0. Thus the whole product would equal zero. And of course x = 0 is a trivial case.
Since v, π are fixed, positive integers, π(vπ)² is a positive constant. Let x be this constant, the required expression is now x^n/n!, which was also given. For any value of n, the nominator is always x*x*x...*x, where there are exactly n terms of x; (A short proof of what is about to be utilised follows. It should, however, be common knowledge, so you can skip it if you want) (There is exactly 1 value counting from 1 to 1, i.e. 1; there are exactly 2 values counting from 1 to 2, i.e. 1, 2; there are exactly 3 values counting from 1 to 3, i.e. 1, 2, 3; this is applicable to any counting from 1 to k, and from 1 to k+1; it follows that, by mathematical induction, there are exactly n values counting from 1 to n.) the denominator is always n*(n-1)*(n-2)*....*1, where there are exactly n terms of decreasing integers. Thus, the expression can be rewritten as the following, x/n * x/(n-1) * x/(n-2) * x/(n-3) * ... * x/1, where exactly n terms are being multiplied together. As n → ∞, it follows that, as x is a constant, the first term → 0; Thus, it is concluded that the expression π(vπ)²/n! can be rewritten as 0 * ... * x = 0 as n → ∞. QED
I've had a look at the maths extension 2 papers. Yes they are very challenging I must say but the STEP math exam papers are as much daunting as them if not more. You should certainly check them out.
I've seen a few of these and I definitely agree that the STEP exams are very challenging. If you know of any specific questions on theses exams that you think would interest me please let me know :)
1:50 is my reaction to 1:02
Hello, I have a question. This method works for pi but why won’t this work out for an actual rational number like 4?
Pi is the upper limit of the integral, so if you change it for another number, then the series that approaches 0 with pi, will not approach it if you use a rational number
Integral of sin to a rational number is irrational/transcendental.
7:15 After a certain point along the road to infinity, exponential expansion is dwarfed severely by factorial expansion. Sure, for the first few terms (x^2 / 2!) or (x^3 / 3!) or (x^5 / 5!) is still indicative of a possible trend towards infinity, but while exponential growth has a steeper gradient in the beginning (increasing by a factor of x), it is eventually overtaken by factorial expansion which grows logarithmically (by Euler's number). Did I uh, did I get that right?
perfect T-shirt. (also great maths as always!)
For 7:24 puzzle, I tried using the squeeze theorem. If x = 1, then the limit as n approaches infinity of the given expression is zero (I don't know if I have to show this, but I figure this is trivially true). This equation is written as lim_{n -> ∞) π/n! = 0 If x = 0, then luckily, the limiting expression is zero. Important for this is that lim_{n -> ∞) 0/n! * π = 0 Therefore, for 0 ≤ x ≤ 1, by the squeeze theorem, the limit as n approaches infinity of the given expression must be zero. The problem I have is for all x ≥ 1. If anyone can give me a hint for that, it would be great. If x ≤ 0, because x^(2n) is an even function, any negative x will be the same values as the positive, so we can precisely say that the limiting expression is equal to zero for any -1 ≤ x ≤ 0. I may need another hint here as well. I don't know if this is exactly rigorous since the most of calculus I know is from Calc 1.
Hmmm... the power-tower of "i" is not (1 + i)/sqrt(2) as I wrote in my previous comment. Rather the value can be calculated from the requirement that i^Z = Z, where Z is the complex number "a + bi" with "a" and "b" real. This leads to the 2 simultaneous equations: a = sqrt[(e^(-pi*b) - b^2], and a^2 + b^2 = e-(pi*b). This leads to b = sin{pi/2 * sqrt[(e^(-pi*b) - b^2} which I solved numerically to get b = Im(Z) = 0.360592471872082.... and a = Re(Z) = 0.43828293672833....
At 15:25 it is stated that this is the first time we use the assumption of pi being rational, but I think we make this assumption from the beginning since v is in the expression for I_n
Before that point u is not necessarily assumed to be an integer
Nice video!
Here is how to quickly proof x^n/n! tends to zero (requires Stirling formula). n! ~ (n/e)^n * sqrt(2πn) so x^n/n! ~ (x*e/n)^n * 1/sqrt(2πn) Obviously 1/sqrt(2πn) tends to zero. Because n tends to positive infinity, we can ensure that a value of n bigger than whatever we want will exist (for example x*e). Let n0 > x*e. So if we choose a n > n0, 0 < x*e/n < 1. So (x*e/n)^n tends to 0. The two terms tends to 0 so we can conclude that x^n/n! tends to 0.
You can instead estimate n! from above as n^n, and from below as (n/2)^(n/2).
Thank you for your reply for throwing light on me.
Simplest???????
Yes, does not get any simpler than that. If I were to tell a mathematician about this proof, I'd cover the whole thing on half a page. I actually link to a paper in which this is done. Quite a few things that I spend time on nutting out details in this video are absolutely trivial as far as anybody in the know is concerned :)
I was just kidding. I guess there should not be another simpler one: but this one is still really difficult for laymen in mathematics as me. Thanks for answering.
Extension 2 Maths is no joke. If it makes you feel any better, this question was made for 17/18 year-olds lol.
Well, there could be another simpler proof. Maybe we just have not found it yet. I think everybody interested in this sort of problem would really like to see a proof that's a bit more geometric/intuitive, like some of the proofs for the fact that root 2 is irrational :)
Thanks again for Answering. :)
math is more about hardwork, creativity and passion. we have to be mad about things. this is true for any subject like physics.
Just an idea, can't we assume immediately, that I_n is increasing due to the behaviour of the function on (0,pi) therefore concluding the proof with a contradiction that I_n has two different limits as n goes to infinity? Do we actually have to show the reccurence relation? The reccurence relation holds term with (4n-2)*I_(n-1) which then as n goes to infinity is infinity regardless of I_n or v. The minus term: u^2v^2I_(n-2) is fixed by u,v.
Not entirely sure I understand everything you are saying here. However, if you can also prove that I_n is increasing that would definitely also establish a contradiction :)
Hi, great video! Well explained
kremmanie he states earlier that the proof in the video can be altered to show that pi isnt the root of any quadratic polynomial
@@thiantromp6607 right, now it makes sense. Thanks!
I got lost as soon as the first integer sequence was brought up after making the assumption that pi is a ratio of integers. Still a fun video to watch (I always like the animations :D ), but I'll have to try again some other time.
I'd say keep coming back to this video. Really getting this proof is well worth it :)
How to prove that if I_0 = 2, I_1 = 4v^2, then I_n = integral
It's really the other way round. Define I_n = integral and then convince yourself that you get I_0 = 2, I_1 = 4v^2 as I do later in the video :)
I also don't get the jump from (assuming) pi is a ratio of integers to the sequence I0,I1,I2... approaching zero. How does this sequence of integers relate to the first assumption? Is it supposed to say that if pi is rational then the there must existing a sequence of integers that approach zero? How is this different than saying if 1/2 is rational there must be a sequence of integers that approaches zero?
Mitch Dzugan Yeah, I was admittedly half-asleep when I tried watching it earlier today. I know I missed some crucial detail...
Regarding the question at 16:20, are we supposed to get a "continued fraction", or a series of related recursively defined fractions? Because, to be honest, I feel a bit stuck with the former task. (On the other hand, almost out of the blue, the coefficients in this recurrence are very similar to Lambert's partial numerators and denominators for tan, especially after a few equivalence transformations of the latter.)
Here is a hint. The golden ratio is the positive solution of x^2-x-1=0. Rewrite this as x=1+1/x and then repeatedly substitute this equation in itself: x=1+1/(1+1/x), x=1+1/(1+1/(1+1/x)), ... If you keep going this gives you the simple continued fraction expansion of the golden ratio 1+1/(1+1/(1+... :)
Well, thanks, but I am still stumped. The q = ... 1/q part is apparent indeed, but the (2n - 1) term is the issue. The coefficients of the continued fraction, as in Lambert's case, are assigned per level of the fraction. (May be a definition difference?) Here, the (2n - 1) term depends on the index of the recurrence. So instead of q_i = a_1 + b_2 / (a_2 ... + b_i / ( a_i ..., you have q_i = a_i + b_i / q_{i-1}. If the coefficients are constants, as per Fibonacci, the forms should be interchangeable. Please, remark if you find that my definitions/expectations are bugged. Otherwise, let us not spoil the exercise for the rest of the viewers. Thanks.
Don't expect to get all the way in terms of the continued fraction in Lambert's proof. I was really just after you translating the recurrence relation into some/any infinite continued fraction to indicate a loose connection between these two proofs that at first glance look so different. To get to the real connection takes quite a bit more work :) Oh, and solving for q_{i-1} instead of for q_i helps, to make the continued fraction develop in the right direction.
I understand. Indeed, it is more like 1/q_n = a_n + b * q_{n-1}, or q_n = 1 / (a_n + b * q_{n-1}). I will avoid further disclosure of the coefficients, but the point is - it makes sense. Thanks. Also, the analogy of the coefficients with Lambert is indeed curious to a laymen (like myself), because the proof at first sight appears to use completely original construction (with the "rabbit out of a hat" integral originally involved). But I believe you, in that it is not trivial to explore. Finally, one can (probably naively) expect that there is a general principle connecting Lambert and Hermit style of proofs. Anyway. Thanks for the clarification. Best wishes.
All under control then :)
7:23 let m be some natural number such that m>x (this is guaranteed to exist since x is finite). Then factor out x^m/m! (which is clearly finite as well) and we're left with x^(n-m)/(m+1)(m+2)(m+3)... etc. This expression is definitely less than the same expression with all denominator factors =m. x^(n-m)/(m+1)(m+2)(m+3)...x, x/m
Thanks! But for future reference, the colors you used @9:50 are confusing for at least colorblind folks like me. I pretty much only see two colors there. What I guess you called the orange and the green look very similar, and what you call the pink and the blue look very similar
Wow, what do you see? Are they both green, or both orange? Are they both pink, or both blue? They actually look starkly different. You can't see red maybe.
I think the illustration of an Eagle 🦅 , used by insurance companies, financial businesses, and gun manufacturers, was based upon Hermite’s hair.
I just did that question and it took me almost 3 hours to do it alone. An entire test full of questions like that for 3 hours? Wow
What about a more technical write-up of the Question A6 on the 1992 Putnam Test, 3blue1brown did a video about, as an idea for a future video?
Can someone explain 3:15 where this sequence of numbers is coming from?
I warned you ... black magic. If you are interested in some more details as to where these integrals come from have a look at the articles that I linked to in the description :)
Extremely simple version of proof that ((v*pi)^2n)pi/(n!) approaches 0: since v, pi, and 2 are all constants, you can replace the entire expression in the numerator with just c^n, for some constant c. (Mathologer used 'x' in the video) The expression simplifies to (x^n)/n!, and if we can show that the expression in the denomiator grows faster than the expression in the numerator, the fraction will get closer and closer to 0, hence the term APPROACHING 0. Here's the key idea: Both the top and bottom rely on a variable n, and the top is a quadratic expression while the bottom is a factorial. As factorial functions grow faster than quadratic ones, it is easy to see why the function ((v*pi)^2n)pi/(n!) approaches 0.
As far as "hard proofs at school" go, the Gaussian primes proof for Fermat's Christmas theorem was a big mathematical stepping stone for me. It was the first proof I encountered where you work in a different ring to say something about the integers. (I'm fully aware that its a little advanced for a video like this)
Nice one of course, but did they actually showed you this one in school ? :)
Does second year of undergrad not count as school? :P
Australian test Charles hermite died in 1901 Australia became a federation in 1901 COINCIDENCE? I THINK NOT.
I can't remember highschool math being even close to this level. Never had any proofs there 😅
You never had a geometry class?
@@MusicalInquisit without proofs
@@hansisbrucker813 Ourch, that sucks
I feel you. We had proofs, but they were kind of handwaved, spoon-fed and never explained with any sort of reasoning. Just "Here, therefore this, so that...q.e.d." Recently I stumbled upon an actually consistent proof that sqrt(2) is irrational, and I found that to be beautiful. I wish we had discussed this consistently like that in high school.
@@monkeybusiness673 how about when they say it is a postulate and then you have like no idea why it is a postulate? Also, what proof is that? The proof by contradiction?
Hello, for the very latest demonstration, it is true for any couple of irrational numbers. especially for PI and 1/PI, in this case we have the choice between Pi + 1/PI or Pi*(1/PI). If we take twice the same irrational R we end up with the conclusion that R² is irrational or 2R is irrational. Let's define SQ as the following: for (a, b) E N², SQ(a,b) = (sqrt(a)/ sqrt(b)) Only a subset of SQ can verify R² is rational but 2R is irrational. But if 2R is rational then R² is rational. then all irrational numbers not icluded in SQ (transcendantal numbers) are verifying R² is rational and 2R is irrational...... am I right ?????
The Pi + 1/PI or Pi*(1/PI) is observation is great. Then, what I say in the video is true for any number N and pi, so we can be sure that pi + N or pi x N is irrational. Twice an irrational number is always an irrational number and for an irrational number squared to be rational it has to be of the form root (a/b) with a, b integers :)
I'm your huge big big fan sir.. Sir aapki depth hai maths me...Very sound knowledge in Maths. Great..Love from #MathsPathshala
What is the expected score of an average student on this exam? I've taken exams where the professor more or less expects us to only complete one or two problems. This implies that we should just choose the problems that we think we can do within the exam time. Honestly, given the instructions on the question's problem statement, I think I could have done most of it, but I not 8 of these in 3 hrs.
In physics, we have two journals (AJP, EJP) where interesting explanations like yours - but about physics - can be published for the benefit of students and teachers alike. Is there a maths equivalent of our two physics journals?
There are a couple: The College mathematics journal, the Mathematical intelligencer, the mathematics magazine, the (British) Mathematical gazette, the American mathematical monthly, ... :)
Mathologer Thank you for your reply. And so prompt too! Even more impressive than your very beautiful and fascinating videos.
Well, obviously, as far as x is a finite number, x^n will grow slower than n! at n -> inf since no matter how big x is, at some point some sub product of n! will start going faster than x^n since x is finite (we assume that x is natural but that doesn't really matter since, if it works for, say, x = 12 then it will work for x = 11.2989809580234785034976 too): (if n is >> than x) n! = 1*2*...*(x-1)*x*(x+1)*...*(n-1)*n with n terms vs. x^n = x*x*...*x n times As we can see from term (x+1) of the first expression every single next term is bigger than the corresponding term of the second expression. While the sub product to the left of (x-1) is smaller than that of the second expression, this sub-product is finite, so it doesn't contribute to the final result, n! grows infinitely faster, at some point takes over x^n and keeps going. Therefore, x^n / n! has a limit of 0 at +inf
I habe a question. Why at 15:40 you just can replace I_3, I_4 and I_5 with I_n-2, I_n-1 and I_n? We only showed it for 3, 4 and 5 and not for all numbers. Or did I miss a part of the proof? *thinking*
The same argument can easily be generalized to any n
if every thing on the right side of the equation at 12:24, evaluated at n=3, is an integer, why must one integrate, etc., to show the general case?
Hi! I'm sure you have been asked this a million times, but what software do you use to animate your equations? Thanks.
To animate the equations I mostly just use Apple Keynote and a bit of Adobe Premier as part of the overall gluing together of the video of me and the slideshow. It's an extremely time consuming process, but it gets me the results I want :)
Mathologer Thanks man! Wow I really thought you had a software doing it automatically. Keep up the good work.
From the two videos on Pi's irrationality, we can conclude that the true reason Pi is irrational is because you can't build an infinite decreasing sequence of positive integers. In other words, because starting from a finite number you can always count down to zero in a finite number of steps. Pretty obvious, right? And that's what I'm going to answer from now on when asked about Pi's irrationality. :)
x^n/n! tends to zero for every x because after some n that's greater then x numerator is increasing only multiplication by x, and denomenator increases by multiplicating larger number, and larger with every next n, so it increases faster, and if we go up to infinity, there's no matter how big x was at the beginning, because numerator is larger then denominator only for some finite starting amount of steps, and lower in following infinite cases. (Wow, my English is SO bad).
Илья Лагуткин that’s an empirical proof,it’s not rigorous,try using the stoltz cesaro theorem for limits of sequences in the infinity/infinity or 0/0 situations
Matisan Andrei I understand what you're talking about, my proof is intuitive and simple, not truly rigorous in mathematical sense, but I think it's "rigorous" enough if you can't disprove it easily or find some fallacy in its logic.
I fail to see why this proof is not rigourous.
Илья Лагуткин in that sense yeah
densch123 try stoltz cesaro is way shorter
How about this one? Take a sphere of unit radius, then remove two cylinders of radius 1/2 that are tangent to each other along the z-axis (and so also tangent to the sphere on the boundary). What is the volume of the remaining sphere-with-holes? The final answer is surprisingly simple.
did you already show that pi is transcendent ? I've never attended the algebra lection and would really like to see this one.
Why we can multiply a and b but can't add a & b? Please explain its physical meaning
Proof that a^m/m! tends to 0 as m tends to infinity: Each of a^m and m! is a product of m factors. Arrange these into m fractions of size a/i, and these go to 0 as i goes to infinity.
This is actually used in the proof that e is transcendental as well. Here it's used in the form, lim (e^n n^p (n!)^p / (p-1)!) as the primes p tend to infinity. The constant (e^n n n!) is easily put out of the way since lim KX = K lim X, where K is constant and X is some function of the independent variable of the limit. In particular, if lim X = 0, then likewise lim KX = 0.
i is always equal to root(-1)
The ratio of the consecutive terms of the sequence u(n)=x^n/n! I.E u(n+1)/u(n)=x/(n+1) tends to 0 as n tends to ♾️ , causing the limit of the sequence u(n) to be 0.
If there really is a smallest possible distance (i.e Planck length) in the universe, does that mean that every circle (not matter how big) can be constructed with a Pi with a finite number of decimal places because at some point adding another decimal point wouldn't make the approximation of the circle any better than the circle is in reality?
x^n/n! -> 0 for any fixed positive x: Fix any N>2x and consider any n>N. Then x^n/n! > x^N/N! * x^(n-N)/N^(n-N) because n! = N! * (N+1)(N+2)...(n-1)n > N! * N^(n-N). As x/N < 1/2, this upper bound tends to zero.
I'm trying to understand the thought processes from 3:06 to 3:59. "Based on this assumption [of u/v = pi] we conjure up an infinite series of numbers" -> How? In what way do you take u and v and "conjure up an infinite series of positive numbers that gets closer and closer to zero"? Is it just u = a, v tends to infinity? So the ratio, a/v tends to zero? "But then, ... it turns out that the numbers in the sequence are actually all integers" -> They are? So then we can't be talking about a/v as the series, cuz that can get arbitrarily close to zero. So what are we talking about? And then finally, we get this epic logical craziness, "but since nothing false can following from something true, this means our assumption has to be false to start with" ... HUH!? So, if A is not B (and this is true), and B is not C (and this is also true) then C cannot be A, which is nothing but a logical fallacy. It's very easy to build false statements from true statements. My rabbit is white. My car is white. My car is a rabbit. Huh? I hope this didn't come off too condescending, but this whole bit seemed to make exactly zero sense! Thank you, EDIT: Oh, you start breaking this down in the very next sentence. I thought that was supposed to be one idea, and the main video was talking about something related. Anyways...
Thank you sir.Any division which gives non terminating and non recurring decimal is irrational number.Here no even number odd number relation.I understood.
17:20 - You mention that the proof can be altered / enhanced to show that pi also cannot solve a quadratic polynomial. Can you please give a hint / reference how this enhancement works?
Mathologer are you in Australia? You sound German. Kind regards and thanks for the wonderfully-insightful videos.
He was born in Germany, teaches in Australia. Lucky Australian students.
Please, do a video on the new 'pictorial math language'. Cheers!
Yes, that would be an interesting project :)
x^n/n! = (x/n)(x^n-1)/(n-1)! x/n is approximately equivalent to zero as n approaches infinite. The equation can then be even further 'divided' for an infinite amount of times: x^n/n! = (x/n)(x/n-1)(x/n-2)(x/n-3).....(x^1)/(1!) and we can see, although the last terms are not near zero, the first several terms are approximately zero (any value over infinite), more so that they are being multiplied with each other.
Really nice video, thank you for your hard work in making it.
Simple proof of x^n/n! tends to 0 when n tends to infinity: Let p > 2x an integer. Let u(n) = x^n/n!. For all positive n, 0 < u(p+n+1) = x/(p+n+1)*u(p+n) < 1/2*u(p+n), with recurrence, 0 < u(p+n) < 1/2^n*u(p). Given that 1/2^n tends to 0 when n tends to infinity, x^n/n! tends to 0 when n tends to infinity too.
(x^n)/n! proof that this goes to zero as n goes to infinity; define the succession: a(n) = x^n/(n)! if x= 0, trivial; if x>0; every member of the succession a(n) is positive; so the succession is bounded (a(n) > 0 for every n); for n sufficiently large (n > x, and this n is always possible to find as the set ℕ has no sup) what happens is that a(n+1) = x^(n+1)/(n+1)! = (x/(n+1))*(x^n/n!) < (x^n/n!) = a(n) so the succession has a limit as n goes to infinity, as it is decreasing function bounded inferiorly; also we know that lim(n->+∞) (a(n)) = inf{a(n), n€ℕ, n>x} ≥ 0; (can't be less than the bound 0, as if it were negative we would have a succession of positive numbers having negative values after a while); then let's call this limit A; A = lim(n->+∞) (a(n)) = = lim(n->+∞) (a(n+1)) = = lim(n->+∞) ( (x/(n+1))*(x^n/n!) ) = lim(n->+∞) (x/(n+1)) * lim(n->+∞) *(x^n/n!) )= lim(n->+∞) (x/(n+1))* lim(n->+∞) a(n)) = 0*A; as we know x/(n+1) tends to zero for n large; so A = 0*A = 0 and there we go, limit as n -> +∞ of xⁿ/n! is zero if x
Why does that look so complicated? Let |x|−1 < i < n, then xⁿ/n! ≤ (xⁱ/i!) (x/(i+1))ⁿ⁻ⁱ → 0 as n → ∞.
Nice... I agree this is a simple proof. Not a natural proof whose "why" can be explained easily; but it works. Also, i feel that at 16:01 you should have put a warning that what you did was to only verify the proposed recursive relation for n=5; some people may mistakenly believe that what you did was sufficient proof for the general case.
Well the example I work through has to serve three purposes: 1) a mini intro to/reminder of integration by parts, 2) it has to be "general enough" so that anybody who is interested in writing out the the calculation for general n can just copy what I did, 3) keep the overall argument as uncluttered and at the same time as complete as possible. Also, if this proof would appear in a maths paper it would just feature the recurrence relation and would usually not contain any of the details of the integration by parts bit. Trivial calculations are usually not part of write-ups. In fact, I'd say written up for a maths journal this proof would take up between 5 lines and half a page. Check out some the articles I link to in the description to see what counts as a proof in this respect :)
Fair enough. In a paper, the proof would be left to the reader. I was thinking more along the lines of an exam or test question.
That was the simplest PROOF? Wow! I wonder how the other proofs looks like.