If you start with 1, the sequence of partial fractions is constantly 1, thus converges to 1. If you start with 2, same thing just with 2. If you just look at the infinite fraction, you get a quadratic equation by calculating its value. This equation has two solutions: 1 and 2. So the infinite fractions are basically the same, but their values depend on the value you start with.
Define a series: a[1]=2, a[n+1]=2/(3-2/a[n]) for n=1,2,.. For example, three following elements are: a[2]=2/(3-2/a[1]) --> a[2]=2/(3-2/2)=1, a[3]=2/(3-2/a[2]) --> a[3]=2/(3-2/1)=2 a[4]=2/(3-2/a[3]) --> a[4]=2/(3-2/2)=1 It's easy to prove using induction that a[n]=2 for odd n and a[n]=1 for even n. Element a[n] in this series can represent the n-th step in the finite construction of the infinite fraction. The series does not converge and the question "what it converges to" has no meaning.
I consider this as a sequence: start from a_0, cook up a_1 as a_1= 2/(3-a_0), and in general, a_{n+1} = 2/(3-a_n). This sequence has the general term a_{n} = ( 2^{n+1} - 2 - (2^n - 2) a_0 ) / ( 2^{n+1} - 1 - (2^n - 1) a_0 ) = ( 2^n (2 - a_0) + 2 a_0 - 2 ) / ( 2^n (2 - a_0) + a_0 - 1 ). So, when a_0 != 2, the sequence converges to 1, and when a_0 = 2, the sequence converges to 2. The 'usual' way to chop off a continuous fraction is to take a_0 = 0, and consider the continuous fraction as the limit of the above-mentioned sequence. So the 'correct' value of the continuous fraction is lim_{n->\infty} a_n (with a_0 = 0), which is 1. Sorry if it appears twice, I cannot see my initial post.
I think property holds only for + operation not for negative operation and may be there is quadratic equation associated with for two different values.....
The equalities shown at the beginning (i.e. "1=..." and "2=...") were correct until the fractions were made infinite, at which point the value of the expression itself changed. 1 does not equal that infinite fraction, and neither does 2. Therefore it does not follow that 1=2.
A comment passing by. I think the 1=2 is about importance of notation, like the 1 / 0.414213... is at the bottom of the partial fraction at 7:39. Obviously for irrational numbers, the decimal place continues forever, so when you write them in the partial draction form, I think it's "correct" to write 1 / 0.few_decemal_points_of_the_irrational_number to make it less CONFUSING. At 7:39, sqrt(2) is represented by partial fraction with numbers 1, 2, 2, 2, 2. And I could just make up some irrational number now whose partial fraction representation starts with 1, 2, 2, 2, 2, which equal to first five of sqrt(2) then it can be some random sequence of infinitely many numbers. Both the sqrt(2) and the random irrational number I made share same first five "coefficients" so when they are abbreviated to partial fraction with five coefficients and dot dot dot, they both have the notation but they obviously are different numbers. So to reduce confusion, you'd write the 1 / 0.decimal place for the irrational numbet. Just like this, 1 = 2 / (3 - 1 / (3 - 1 / ...)) and 2 = 2 / (3 - 1 / (3 - 1 / ...)) should be notated as 1 = 2 / (3 - 1 / (3 - 1 / (... / 3 - 1) )) and 2 = 2 / (3 - 1 / (3 - 1 / (... / 3 - 2) )) So according to my theory on the paradox is that the notations for 1 and 2 is that the notation missed on important terms that made two different numbers look the same, just like how 10 and 1000 look the same without "0"s.
So, basically it's like saying 10 = 1 * 10 and if you replace 10 with 1 * 10 few times, it's 10 = 1 * 1 * 1 * 1 * 10. And if you do it infinitely, it's 10 = 1 * 1 * 1 * 1 * .... And for 100 = 1 * 100, 100 = 1 * 1 * 1 * 1 * 100 becomes 1 * 1 * 1 * 1 * ... Oh, wait, both 10 and 100 have same 1 * 1 * 1 * ... part, so they must be equal??? What I'm saying is that, if you want to do that, you should 10 = 1 * 1 * 1 * ... * 10 and 100 = 1 * 1 * 1 * ... * 100.
You showed how there is a clear pattern in the integer parts of the infinite fraction representation of e. Is there such a pattern in pi as well, or is it as random as the decimals are?
Pi 's simple continued fraction is "random". However, there are other continued fraction representation of pi that exhibit nice patterns, e.g. functions.wolfram.com/Constants/Pi/10/0002/MainEq1.gif
A quadratic approach... Let, x=2/3-2/3-2/3-..... Therefore it can be written as X=2/3-x 3x-x^2=2 Solving out, We get, (X-1)(x-2)=0 X=1 or x=2.. Inference: That continued fraction will be equal 1 OR 2...if we solve out a quadratic problem we get two roots but that doesn't mean we compare both roots..and come to a contradiction..the same is with the case of this continued fraction.... Please comment about my proof..and if there is something wrong.Please be kind enough to point out the mistake.. Thank you
Is the mistake because the infinite fraction representing 2 cannot actually be written in the way you did, because there is a final step that is not taken into account with the . . . ? In other words, the . . . represents you performing a function that returns itself an infinite number of times in the middle of an existing expression, rather than actually representing the entire expression? Like how if you started with 1 = 1* (1) and expanded that to 1 = 1* (1* (1* ( . . . and also had 2 = 1* (2) and expanded that to 2 = 1* (1* (1* (1* . . . the two expansions would be written the same way, but wouldn't be equal, since the latter is correctly written as 2 = 1* (1* (1* . . . (2).
The sequence of partial fractions does not converge to any specific number. This is because 1 and 2 are rational, thus their fractional representations should be finite.
+Birric Forcella I wouldn't say something that is post-modernistic is reasonable. More like it's not reasonable but who cares it doesn't matter, if I feel it's right then it is right in some form that no one would even bother
My answer to the riddle: Is it = 1 or = 2? In that continued fraction, as you write it, you alternate between ending with a numerator and ending with a denominator. If you stop to evaluate it, ending always with a numerator, that sequence converges to 2 - in fact, it IS just 2, every time! But if you always evaluate it ending with a denominator, the sequence converges to 1. Of course, if you stop at every numerator and every denominator, the sequence fails to converge. So the meta-answer is, you have to define how you mean to interpret the CF; the usual way being the 2nd - always stop at a denominator. That definition makes the answer, 1.
Thanks! On further contemplation, it seems that you will run into this conundrum when and only when you allow numerators to exceed denominators, in a non-vanishing fraction of the terms. Or some condition close to that. Which may be one reason that the 'all numerators must = 1' case is of special interest.
Basically, no matter how far you carry the calculation, whether you stop at a 2 or a 3 radically affects the end value. This is basically saying that the fraction doesn’t converge, which is what Mathologer said is part of recognizing that the fraction is not meaningful, since the truly infinite version does not stop at any point.
Hey @@ffggddss , great answer above. Although I think the theory of numerators exceeding denominator might not be right. Consider (1/2)/((3/2)-((1/2)/(3/2)-((1/2)/(3/2)... ( This was the best font I could come up with in RUclips comments). This above sequence converges to either 1 or 1/2 depending on your theory
That can be regarded as the most rational irrational number, since it can be approximated better and better by rationals the deeper you go. It equals I0(2)/ I1(2) = 1.4331274..., where the In are modified Bessel functions of the first kind.
@@landsgevaer Hi, I’m not good at math and I know I’m a month late, but would you mind explaining it in a way where I wouldn’t need to know Bessel function is? Or is there a way of expressing this number as a radical?
Let the infinite fraction from the first problem equal x. Now replace the "infinite part" with x to get x=2/(3-x) Rearrange ==>. 3x-x^2=2 x^2-3x+2=0 (x-2)(x-1)=0 Therefore there are two separate solutions x=1 and x=2 but since they are separate, 1 does not equal 2! Maths is not broken! Rejoice!
+Eden Lumbroso no, because a polinomial of nth degree has n solutions(zeros). Damien ended up with a polinomial of the 2nd degree with two solutions x1 = 1, x2 = 2.
+My Name is Chef still doesn't answer the question how come 1 and 2 both equal the same thing (the fraction. not x). a guy wrote 2 is not a valid answer because the fraction does not get closer and closer to it like it does with 1.
+Eden Lumbroso it does, substitute x=1 and x=2 into the original equation and they both work but not at the same time, it is either one or the other. This is due to the nature of quadratics and polynomials in that there will be n roots where n is the highest order of the polynomial (n=2 for quadratics) but those roots are not equal to each other and with the rearrangement I did, I proved the the infinite fraction is just a quadratic in disguise and will follow the same rules as polynomials.
It does; it shows that for any single-repeating-digit continued fraction there are two valid solutions. In general for q = 1/(a+(1/a+(1/a + ...))) we see that: 1/q - a = q so 1 - qa = q^2 q^2 + aq -1 = 0 Without factorising, we see that it has two solutions because it's order two.
Step 1: Turn into x = 2/(3-x) Step 2: Solve for x to get x = 1 or x = 2 Step 3: Cry because this simple method that you taught me doesn't work and I suck at maths
Both 1 and 2 work for that equation (and that infinite fraction above), but that doesn't mean they are equivalent. x=1 OR x=2. Not 1=x=2 => 1=2. I believe that is the problem answer.
That's an interesting observation. I should add that if you've got three consecutive best approximations they are always related like this a/b, c/d, (a+kc)/(b+kd) for some k. Try it :)
That’s because the last number in the finite fraction is a 1. For example, phi has a continued fraction of all 1s, and its finite approximations are ratios of consecutive Fibonacci numbers, which always have that property. In fact, this works for any pair of approximations, except you have to multiply top and bottom of one of them by the last number of the fraction before adding things up; it’s just more visible when that last number is 1.
Me too, and I got REALLY disappointed. First, I basically knew everything he said, and second, the explanation of why phi is the most irrational number was just wrong. I explained it in this comment: ruclips.net/video/CaasbfdJdJg/видео.html&lc=Ugxik2-iu5Y5EYQQj5R4AaABAg
A couple of interesting answers to the puzzle already, but I'll refrain from making any comments myself until tomorrow (here in Australia). Have fun until then :) Also, thank you to Zacháry Dorris for contributing English subtitles to the last video (Riemann's paradox), Rodrigo Naranjo for contributing Spanish subtitles and Étienne Leb for his French subtitles! Added the next day (in Australia). Quite a few answers that are spot on. Before I give my own answer to the puzzle just two hints: 1) The continued fractions in the puzzle are a bit different from the simple continued fractions that I discuss in the rest of the video. 2) What I say in the interlude is crucial for getting to the bottom of what is going on here.
I think that the simplest answer is that a rational number can't be expressed as an infinite fraction, only irrational numbers can be expressed that way, so I think you did say the answer in the video, you tried to got us! 😆
I don't think so. If you use the method explained in the video in order to get a continued fraction of either 1 or 2 the continued fraction will be finite. I don't think it means you cannot get an infinite fraction of a rational number.
+yotam vaknin But he himself says that for a rational number, there's no infinite fraction, it can be expressed as a fraction, but not an infinite one.
+Ⲇⲁⲛⲓ Φi No he said that using the methode to create a fraction will end, not that any given fraction representing that number has to end (aus far as I understood)
I believe the answer to the puzzle is in the interlude. You can only place the equals sign if the partial sums converge to the answer. But in the expansion for 2 they don't, so you cannot say 2 = (the written infinite fraction) = 1, because it only converges to 1.
Both identities and infinite fractions are, in fact, correct. The nuance here is at 1:23 when he states the right sides are both identical; they are not, because in the case of 1 we have been replacing a 1 the entire time, in the case of 2, a 2. The "last digit" of this series is, in fact, quite important, because it is the number we have been substituting for this entire time. The "dot dot dot"s are not the same at all, as one has a value 1, the other, 2.
You have to remember that in one series, 3-2=1 and in the other 3-1=2. They both are technically correct in and of themselves as they go on towards infinity, but the part that is being replaced is different even though they appear the same at a higher level. Also, keeping the above in mind, you can end either series at any point and so they are not an irrational infinite series, more of a faked one.
The author just uses dot dot dot to replace the series, but the dot dot dot in each series are not the same. You explained it in words much better of course.
The partial fraction does converge to both 1 and 2 but it depends on the way one decides to evaluate the partial fraction. If we evaluate it by just considering the fraction before the negative sign then it will converge to 1, but if we were to evaluate it by removing everything below the denominator then the fraction will always be 2!
+Not Broihon I think that if you use the partial fractions for the "=1" part you have to end with 1 and for the other one you have to end with 2 I will give try when I arrive at home
2/3 = 2/3 2/(3-2) = 2 2/(3-2/3) = 6/7 2/(3-2/(3-2))= 2 etc Depending on how you take the partial sums, it'll either converge to 1 or always be 2. EDIT, I believe the way to take the partial sums that lead to 2 are false.
There are no partial sums but fractions ;). He also mentions that you chop the seq. before the further addition for partial fractions, which only makes sense as you can´t add the -2 without aknowledging that it functions as a numerator thus cant be added without adding the denominator, as mult. has precedence over add.
I think I've got it. Mathologer, I'd love to know if I am right or not. The interlude is really important because you explain what the "equals" means here. The infinite series has to converge to the number in question for us to label it as equals. If we start taking the partial sums by cutting off at the minus signs, we first get 2/3 (0.6667). The next term is 2/(3-2/3) which is 6/7 (0.8571) Third term is 2/(3-2/(3-2/3)), which is 14/15 (0.9333) The fourth term is really nasty to write out, but it gives us 30/31 (0.9677). The pattern emerges to show that each term can be determined by its preceding term. The numerator of any term n will be double the denominator of term n-1, and the denominator of term n will be 1 greater than that. While we see this series does converge to 1, we can prove it using calculus. This series is just a faster converging series of the form n/(n+1). The nth term test gives us infinity/infinity, but L'hopital's rule says we can take the derivative of the top and bottom, perform the nth term test again, and find out what the series converges to. The derivative of n is 1, and the derivative of n+1 is 1. The limit of 1/1 as n approaches infinity is 1, so the series n/(n+1) also converges to 1. We can use this information to also conclude that the infinite fraction also converges to 1. Thanks for leaving us this problem to work out. Great video!
I think that the last value in "..." should be 1 to equal 1 and 2 to equal 2. To illustrate this, let's consider the following recursive functions: f(0) = 1 f(k) = 2 / (3 - f(k-1)) g(0) = 2 g(k) = 2 / (3 - g(k-1)) For any n>=0, we have the following identities: f(n) = 1 g(n) = 2
I think that the problem occurs when you transform the fractions into infinite fractions. You took a really important part away. The fraction with 1 always had a 1 to be substituted again. And the fraction of 2 always had a 2 to be substituted. So the fraction of 1 has a 1 in the end (that is hidden by the elipses) and the fraction of 2 has a 2 in the end (which is also hidden by the ellipses). And with that in mind we can say that they aren't really equal.
Well, right. The issue is that the "1" version needs a 1 at the end to maintain the identity. Filling it in with the ellipsis actually breaks the identity by changing it to a sequence of partial fractions which converges to 2, not 1.
not sure if i'd agree that phi is "more irrational" than pi, although I guess it depends on the context of what you're asking. phi after all is algebraic and pi is transcendental so i would have thought that pi would have been "more irrational" in that sense.
imagine if you will two line graphs with a gradient of the natural number you want to test y = phi*x which can be compared to another, your example includes pi so y = pi*x the closer either line graph is to the cardinal points, the "more" rational it is. This happens when their convergent fractions come very close to approximating their value. ie in each convergent fractions second iteration ( pi ~ 22/7 as opposed to phi ~ 3/2 )
Absolutely correct. More formally, the Liouville-Roth irrationality measure of the golden ratio is 2, while that of any Liouville number would be infinite.
The problem is that the notation of the first example with three dots isn't even rigorously defined at all. We should do: Lets define f(n)=2/(3-n). {a(n)} and {b(n)} are 2 recurrent sequences defined by a(n+1)=h(a(n)) and b(n+1)=h(b(n)), but a(0)=1 and b(0)=2, which means the 2 sequences aren't the same one. That implies that their limits aren't necessarily the same. An expression of the form in the video doesn't give us an clue of the original term/seed of the sequence in question and isn't well defined because of that. The notation works for the rest of the video because the sum (and so the limit of the sequence) is actually done in the other way: from left to right. So, it actually has a very well defined first term.
Thanks for the interesting video! I've also been watching videos about "distributed mode loudspeakers" which are essentially a flat panel (of wood, foam, etc.) with a small speaker driver attached to the back. One of the recurring problems with DML construction is resonance - depending on the dimensions of the panel and the position of the driver certain wavelengths will cause the panel to resonate. For example, with a square 1m x 1m panel, and the driver placed in the middle, there will be resonance at frequencies that correspond to a wavelength of 50cm, as the wave will reflect off the edges of the panel and constructively interfere with itself. But it's worse than that, as as well as all these simple reflections along a single axis you also get more complex 2D resonances - all the different patterns you that Chladni plates show are visualizations of all the different resonances that we want to minimize. A rule of thumb I've heard from DML designers is to use a panel whose side lengths are not a simple ratio of each other, and to place the driver roughly ⅖ from two adjacent edges and ⅗ from the other two edges. I wonder if the golden ratio has applications here? Specifically, I hypothesize that the amplitude of resonance peaks may be minimised by a panel whose side lengths have the golden ratio to each other, and the driver is positioned according to the golden ratio - I think (1/φ) from two adjacent edges and (1 - 1/φ) from the others? It should be possible to simulate this but I can't find any existing Chladni plate simulators that allow for non-square plates, and I don't think they model the position of a 'driver' either.
The identity at the start of the video only holds for finite expansions of 1 and 2. If we do not end it with a 1 or 2, then we are essentially breaking the equal sign. On top of that, if you compute the sequence of partial fractions for these two sequences, you will find it converging to 1. For example: try 2/(3 - 2/(3 - 2/(3 -2/(3 - 2/3)))).
In 1=2 riddle, it depends wether we end the continued fraction with 2 or 1. If continued fraction ends with 2 in its tail, it collapses to 2 while if continued fraction ends with 1 in its tail, it collapses. Since, the 3 dots doesn't tells us anything about term on tail, the continued fraction doesn't have a particular value. This is similar to infinite series 1-1+1-1+1... whose sum is either 1 or 0 depending on the last term and since this doesn't conclude that 0=1, our ICF paradox doesn't conclude 1=2.
@@1paper1pen63 (1 + x)^-1⠀⠀when x = 1 (which of course wouldn't be a "valid" expansion but it does work out to 0.5 if you just evaluate (1 + 1)^-1, but the expansion becomes 1-1+1-1+…)
as other people have noticed, x=2/(3-x) has 2 solutions: x=1,2. however, for the iterative process x_{n+1}=2/(3-x_n) to converge to either 2 or 1, we need to look at the derivative of the iteration. the derivative of 2/(3-x) is 2/(3-x)^2, which is 0.51 for x=2. That is why we approach 1 with the continued fractions (derivative less than 1). if we play with the equation a bit and rewrite it as x=3-2/x, then the iterations (not similar to continued fractions so much anymore) would actually converge to 2: 3, 7/3, 15/7, 31/15,... this is because the derivative of 3-2/x is 2/x^2, which is 2(>1) for x=1 and 0.5(
11:40 I used to think you could 1-up φ, by choosing 0’s in all the denominators (because there’s still the infinite ”+(1/[stuff])” -tail, also, in each denominator; it’s not division by 0). You *_CAN_* do that. However; it turns out to be anything but irrational: 0+(1/(0+(1/(0+(1/(0+(1/(0+…)))))))) = 1/(1/(1/(1/(1/(1/(1/…)))))); which, actually, collapses to just 1 (because it’s just: ”1 over 1 over 1 …”, over and over again, it stays, at 1, through all the infinitely many truncations, we get out of it. So, trying to get too greedy, results in the complete opposite of an irrational number: An Integer; and a basic one, at that: *1.* 😅
My take on the puzzle: If you consider the function f(x) = 2/(3-x), it has two fixed points where f(x) = x: x=1 and x=2. However, x=1 is an attracting fixed point of the function, and x=2 is a repelling fixed point. That is, if you take some value close to x=1 and then iterate the function, calculating f(x), f(f(x)), f(f(f(x))) and so on, the values will get closer to 1; but if you start with some value close to 2, the values will run away from 2. This is because the function's slope is shallower than 1 at x=1, but greater than 1 at x=2. I am thinking this is a thing to be wary of when working with continued fractions.
...and the Mathologer's "interlude" about partial fractions about halfway through really gives the key. An infinite fraction expansion only really makes sense if the sequence of partial fractions converges.
Similar to what 3Blue1Brown said in his “Things They Won’t Teach You in Calculus” video (ruclips.net/video/CfW845LNObM/видео.html ) about why the proper solution for 1+1/1+1/1+1/1+1/.... is phi, rather than its negative counterpart, even if that also satisfies x=1+1/x.
For the 1=2 thing, you can try to solve for what 2/(3-2/(3-... equals by setting it equal to x. So you get x=2/(3-x), or x(3-x)=2, and you get the quadratic x^2-3x+2=0, and when you solve for x you get x=1 or x=2, so both answers are correct. They differ depending on what you interpret the dot dot dot as
I think the problem is that 1 is the loneliest number, and since 2 can OBVIOUSLY be as bad as 1 (because its the loneliest number since number 1), there is no way to say they are completely equal. Not to mention the that fact that 1 + 2 = buckle my shoe, and shoes don't have buckles anymore, so it's clear that something weird is going on. Either way, it's been 1 week since you looked at me, and even though I TOLD YOU I had 2 tickets to paradise, you INSIST that I start back at 1, just to make your dreams come true.
lol I thought it was because they were both rational numbers that couldn't be applied to an infinite fraction, but after reading your post I realize that you were right all along. It just makes so much sense!
Well, you are right, the heart will go on. I would say it's a shot to the heart, and that you're just a bit too late, but I would just be left with an akey-breaky heart, which is never a good thing. But if you could just look into your heart shaped box and see that a total eclipse of the heart is bound to happen when you check in at the heartbreak hotel, then you would see that you're terrin up my heart when I'm with you, and that sgt pepper and his lonely band just want to have fun.
You can't ... it since there is - not a +. This means you don't get smaller more meaningless numbers out. Only number you want to know is the last number 1 or 2.
Well 2/3-2/3-2... continued fraction. Let it be x, so x= 2/3-x or x(3-x)=2 or x²-3x+2=0 or (x-1)(x-2)=0 Which means this quadratic equation has 2 solutions 1 and 2. That explains that 1≠2 rather x=1,2. Or we can say that our continued fraction has 2 solutions 1 and 2.
Using the simple form: 1's in the numerator and addition of positives in the denominator --ur assertion is true. That is, a simple continued fraction terminates if and only if it is an expansion of a rational number. However the form here is not simple. And so has nothing to do with that fact.
That seems legit to me. Also, choosing 2 over (3-1) seems somewhat cheaty to me. Like, you could possibly do this with any number and get an identity. Like in the video, 3 = 3 over (4-3) having the 3 in the denominator again, thus also the infinite fraction.
Okay, I'm gonna try and answer this now. Here's the problem - the way we represent continued fractions and the numbers used obscure the problem a bit. The thing is, infinite fractions don't automatically have any sort of meaning, we need to define what they mean. Otherwise we can just throw whatever numbers we want in and have no idea whether it actually represents a real number or not. That's why we use convergence as our definition - in the fraction given the sequence we get is 2/3, 2/(3-(2/3)), 2/(3-(2/(3-(2/3)))) and so on, the nth term of the sequence evaluating to (2^(n+1)-2)/(2^(n+1)-1), which tends to 1 as n tends to ∞. Notice, however, that we took only the integer part of each denominator in our partial terms - why? It's purely definition. We take the integer parts as they're what defines a continued fraction, at least in the case that the numerator is only allowed to be 1, so we continue with this definition when extending to more general fractions. Furthermore, the (cont)
fractional part should be overwhelmed by the integer part - if instead of just including the integer part, we include the numerator of the fractional part too, the terms we get aren't really representative of the actual continued fraction. In an infinite continued fraction with numerators set to 1, this shouldn't really matter, but when we're working with fractions which don't have this regular behaviour and you start treating what should be a fraction as an integer when looking at convergence? That doesn't really make sense. However, if we take this approach when looking at the given fraction we end up with the sequence 2/(3-1), 2/(3-(2/(3-1))), 2/(3-(2/(3-(2/(3-1))))) and so on, which evaluates to an sequence of 2's. This is pretty obvious as in this case the numerator of the fraction in question actually is equal to the fraction as a whole! (That is, 2=2/(3-1)). I think that for this reason the definition of what counts as a continued fraction is somewhat obscured, but it (cont)
+Hwd405 could explain why two different values emerge. It's all about definition and which way we approach a continued fraction, but usually we take the former approach and the result is convergence to 1. I'm not sure either result is "more correct" than the other though, but it's important to choose a mode of convergence and stick with it so that we don't end up with concepts that aren't well defined.
I see what you're saying, of course you can define things however you like, although you don't have to add any definitions. Sometimes it's alright for things to equal two different values (example bellow) without it breaking maths. Both values of the continued fraction can be found without looking at convergence of partial sums (but I go into that more in my main reply to the video) y = f(x) = √x for x = 4 y = 2 or -2 f(2) = 2 f(2) = -2 f(2) = f(2) -2 = 2 Hopefully thats an illustration that the problem here isn't that the continued fraction doesn't equal 1 and it isn't that the continued fraction doesn't equal 2. The problem is in the fallacy that because it can equal 1 and 2, means that 1 equals 2.
+oliver maclean that's not a valid example, a function can only have one output given any input. f(2) must take the value of 2 OR -2, not both. Indeed, we usually define the square root symbol to mean the principal root (the positive one). Furthermore, the repeated fraction notation isn't a function - it's notation. If it's consistent well defined notation it shouldn't be equal to two different numbers, otherwise that breaks the transitivity of the equals relation.
the answer in my opinion is that the fraction should always equal numbers between 0 and 1 and to get other number you add whole numbers like you took the whole part of pi and the golden ratio and you always take the entire fraction out so it would be x= x=2/3 x=2/(3-2/3) x=2(3-2/(3-2/3)) so, x just gets closer to 1 until it becomes 1 so x=1 i am still in school and i my "natural" laguage is not english so i don't understand everything and am probably wrong
The answer is this: The result of the infinite operation depends on the definition of "partial continued fraction." Such a fraction as written means nothing more than "the sequence of partial fractions of this form (form is to be understood after a certain definition that's not been made precise here) converges to such and such a number." The devil is in the precise definition of how to extract the partial fractions from the general formula with "...". Of course, "..." means "and so on" but this is not a precise mathematical phrase. The fact that we often write "..." assumes that we KNOW what we mean by this. Here, it's not clear what it means. It could mean a sequence 2/3, 2/(3-2/3), 2/(3-2/(3-2/3))... or it could mean 2, 2/(3-2), 2/(3-(2/(3-2))... The first sequence will converge to 1, the second converges to 2 (is in fact 2 all the time). The answer is therefore a question of definition of the partial operations. It can (most likely) be proved that the first sequence is always < 1 and strictly monotonic. I say "most likely" because I have not (yet at least) made any proof :)
@Mathologer: Very neat! Continued fractions are also an interest of mine, and here's another twist on "most irrational." What you've shown us is the traditional way to do CF's with unit numerators. But notice that when a "1" occurs in the string of denominators, this is a signal that the CF up to (just before) that point, is pretty crummy. (This can be quantified by defining a figure-of-merit (FOM), that compares the fractional error |x - (n/d)| / x with the "size" of both num. and denom. - actually, with the reciprocal of the product of the two, 1/[nd] ). The rationale here is that the product, nd, is the "outlay" you make to get the fractional precision you obtain. FOM = (1/[nd]) / (|x - (n/d)| / x) = x / (n |dx - n|) Those "crummy" approximations can be eliminated by allowing negative denominators, d, but restricting them to |d| ≥ 2, with the additional proviso that when d = ±2, the next d must have the same sign. On a calculator, e.g., these denominators are found iteratively: • subtract the *nearest* integer to the current result [that integer is the current denominator] • then invert that difference to get the next result This technique gives, for π, {3; 7, 16, -294, 3, -4, 5, ...} rather than {3; 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, ...} - these are given to the same rational value: 5419351/1725033; integer + 6 fraction coefficients vs integer + 11 fraction coefficients, for the same precision. For e (allowing a little exception to the rules for the first couple terms for the sake of mathematical beauty), it gives: {1; 1, -2, -3, 2, 5, -2, -7, 2, 9, -2, -11, ...} rather than {2; 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, 1, 1, 10, 1, ...} - these, too, are shown out to the same fraction: 566827/208524 - The sequence without the exception is {3, -4, 2, 5, -2, -7, 2, 9, -2, -11, ...} integer + 9 fraction coefficients vs integer + 15 fraction coefficients, for the same precision. For φ, it gives: {2, -3, 3, ...}, with [-3, 3] being the infinitely repeating part. This makes φ seem not so very irrational. But for √2, it gives (as does the traditional method): {1; 2, 2, 2, 2, ...}, and now √2 takes over the title of "most irrational," because under the new rules, this is the "crummiest" possible convergent sequence. This can be verified using the FOM, which actually just gives a value very close to the (abs value of the) next result in the iterative procedure described above ("On a calculator, ..."). Any thoughts? Addendum: Under the "new rules," the FOM for φ is φ+1 = φ² = 2.618... And for √2, the FOM is 1 + √2 = 2.414... So the contest for "most irrational" is actually pretty close! Also worth noting, is that the FOM for a general irrational number, bounces all over the place as you proceed down the CF. Heck, just look at π, e.g.!
You do such great work here!Have you ever considered doing something about Turing's halting problem and the associated Busy Beaver numbers?Just a thought- I eagerly await you next video, whatever the topic may be.Thanks again!
You should have done rationalising the denominator for The sqrt2 Continued fraction because it’s not clear why the decimals are the same (sqrt2-1)(1+sqrt2)=1
*Mathologer*, thank you so much for that introduction into *infinite fractions*. My answer to your puzzle is that *infinite fractions* need a _mathematically strict_ definition (just as *infinite sums* do). In the sense of definition you gave at 6:50 the fraction being considered converges to 1 and to 1 only. Again as one did with *infinite sums* he may define it as a limit of _sequence of numbers_ -- *partial fractions* (compare to *partial sums*). When you try to mess around with such objects only relying on , you may face (like with conditionally convergent series). However one may come up with another definition. When you calculate _N_ th partial fraction you may replace _diagonal dots_ with some particular number _d_ instead of zero. If you choose _d_ = 2 it converges to 2. For all other values of _d_ but 3, 7/3, ... when division by zero takes place (see comments below) it converges to 1. One may also notice that if such fraction has a value it _should be_ the root of x(3-x)=2. However such trick only works for recursive fraction expansions (e.g. not for pi). P.S. I'm looking forward to watching your video on Phyllotaxis phenomenon since I'm only familiar with Physicists' explanation of it. I would like to share it with you. Here are the lecture slides www.mit.edu/~levitov/FibonacciPhyllotaxis.pdf and here is the experimental proof arxiv.org/pdf/1002.0622.pdf.
Great spot-on comment and thank you very much for the links. There was also a great feature article on the American math society site on all all this. Just remembered that I want one of those magnetic cacti !
Small clarification: If you choose d = 2 it converges to 2. If you choose d = (2^(n+1)-1)/(2^n-1), that is { 3; 7/3; 15/7; 31/15 ... } it is division by 0. (If you define 2/0 = infinity and 2/infinity = 0, it still converges to 1) For all other d it converges to 1.
+David Saykin Are you sure it diverges for d > 2? What about 1000? The resulting term of the first iteration is very close to 0, which leads to a term very close to 2/3, and for d < 2 it converges to 1, right? Also try d = 3 or d = 7/3 or d = 15 / 7 etc. :-) My favorite response to this video is the one from some guy who compared this proof to the following one: 1 = 0 + 1 substitue 1 by 0 + 1: 1 = 0 + 0 + 1 substitute endlessly: 1 = 0 + 0 + 0 +..... = 0
+Kamil Musiał thank you. I wasn't careful enough. +earthbind83 I made a mistake. You're right. Your example shows how trick was done pretty well, I myself also find it similar to infinite sums.
The problem with the identity is that the second part of the identity is false. 2 is not equal to 2/(3-2/(3-... If we look at the partial fractures, 2/3, 2/(3+2/3), 2/(3+2/(3+2/3), ... , we see them converge to 1. The whole fracture only becomes 2 if you put a 2 at the end. In the infinite fracture, there is no 2 at the end, because there is no end, so the infinite fracture converges to 1. So the second identity claimed is false.
Hi, Mathologer. Why do you say: "Good answer"? Yes, it is a very good answer, but I think that you say this because this answer, as thoughtful as it is, does not reveal the real answer. And you want the secret, the real answer, kept hidden and the discussion to keep going. So you say "Good answer" because this answer hints at the answer and does not really reveal it. And you want that. And I do not like that you want that (if, of course, I am correct). I guess that others that talked about attractors, basins, attracting and repelling fixed points, recursive calculations, nonlinear recursion x(n+1) = 2/(3-x(n)), f(f(f(x))), etc. really have the correct answer. The correct answer as to why Mathematicians choose to equate this to 1, instead of: a) 2 b) either 1 or 2 c) not convergent, because it alternates between 1 and 2 d) something else, like the average of 1 and 2. So, am I correct or have I misjudged you?
Can’t the square root of 5 be approximated by finding two perfect squares whose ratio is approximately 5 (e.g. 81 and 16)? Based on that, isn’t 9/4 a fairly good approximation of sqrt(5), and thus 13/8 a good approximation of phi?
You can do that trick with any √n, where n > 0 is a non-square. It's a Diophantine problem (solving algebraic equations in integers) called Pell's Equation: b² = na² + 1 Every +ve non-square n has a "primitive" solution (a₁, b₁), from which an infinite number of larger solutions can be generated by the recursion a₊ = b₁a + a₁b b₊ = na₁a + b₁b For n = 5, (a₁, b₁) = (4, 9); the one you quote. The next couple of solutions are (72, 161), (1292, 2889). Of some interest, you can get more "best fractions" for n=5 because it's one of the n's that also has a solution to the "Associated Pell's Equation": b² = na² - 1 namely, (1, 2), which can be used to generate more "best fractions," every other one being in the above list, so that they alternate between satisfying P.E. and the A.P.E. : (1, 2), (4, 9), (17, 38), (72, 161), (305, 682), (1292, 2889), ... For general n, the primitive solution, (a₁, b₁) jumps around erratically. Some are easy to find by trial-and-error; some are next to impossible that way. n = 13 is somewhat challenging; n = 61 is very hard. Luckily, there's a procedure involving - wait for it - Continued Fractions! - that will always yield the first solution. Fred
It makes more sense to me to call phi the most *rational* irrational number. Think e and Liouville's constant. These are quintessential transcendental numbers, and the reason we can prove they are transcendental is that they are well approximated by rationals. Does this make them similar or dissimilar to rational numbers? Well, are rationals well approximated by rationals? If we can use the number itself as an "approximation" then yes, but if we can't use the number itself then rationals are the worst to approximate by (other) rationals. So if we don't count numbers as approximations of themselves we get rationals on one end of the spectrum and e and Liouville's constant on the other end, as it should be. From this point of view phi is the most rational irrational number.
IT IS ACTUALLY NOT THE SAME BECAUSE "THE 3 DOTS" IN THE "FRACTION OF 2" EQUALS 2 AND "THE 3 DOTS" IN THE "FRACTION OF 1" EQUALS 1. THOSE ARE NOT THE SAME 3 DOTS AND THEY DON'T EVER CHANGE. THEY DON'T CONVERGE TOWARDS 0.
Well, the problem is that you can't really keep the equal sign when you do that with two because the sum doesn't converge to 2, but to 1 [2/3;6/7;14/15;30/31;...]. If you call the infinite fraction x, you can get this equation because of the pattern: x=2/(3-x) and the equation roots are 1 and 2. But, even though 2 solves the equation, it breaks the condition of the partial fractions converging to the number, so it have to be discarded. The necessity of discarding one of the possible solutions becomes more obvious if you try to solve the square root of two equation, because you'll get sqrt(2) and -sqrt(2) and the negative answer is clearly false because you are summing only positive numbers.
Assume that the infinite fraction converges. Then there exists a limit a. Then a = 2/(3-a). Solve for a to get a1=1,a2=2. This contradicts that the limit is unique and therefore the infinite fraction diverges.
It would be interesting to see continued partial fractions of complex numbers. Awesome video btw, too bad our brain is used to simple summation, and not fractional one. I have read about a guy who could do them as easily as we normal people do ordinary summation. Incredible!
Infinity is not a number, it is not included in the set of real numbers or any number set. It is the limit of the real numbers set but not part of it. That is why we have to use the lim operation whenever we have something that "continues to infinity". Not every infinite limit converges. If you are given that infinite fraction without being told how it is produced, then you cannot find the limit because it diverges. Sorry if that doesn't make much sens e, English is not my first language
Hey Mathologer ! Nice video, as usual. I was wondering if there's a "true definition" (conventional definition, rather) of "more irrational" ? If not, don't you think the definition we invent should be such that phi is less irrational than pi, since the latter is a transcendental number, whereas phi is one of the "most algebraic" numbers there are, so in this regard, phi is "more rational", than pi ?
Pi is both trascendental and irrational; if it's trascendental it needs to be irrational, but it doesn't mean the trascendentality makes it more irrational (trascendentality only speaks of our inability to express that number algebraically, this of course means we can't express it as a finite fraction, but it says nothing about how badly we can approximate it). If we extrapolate the definition of irrationality to be different depending on which irrational number you chose, you'd have the definition seen in the video: an irrational number m is more irrational than n if it takes more digits in the denominator to approximate the same number of digits in m as in n. This makes Φ the most irrational number (over any other irrational *and* trascendental number).
Also, here's an excerpt from the description: "If you are reasonably clued up mathwise have a look at the following VERY nice textbook chapter on infinite fractions by Professor Paul Loya from Binghampton University: [LINK] In particular, check out section 7.5.1. The mystery of π and good and best approximations. *I use the definition of "best rational approximation" given there.*"
The answer to the puzzle lies in the jump from finite to infinite. You could replace that fraction 10^10^10^10^10 times (just to mention a staggering number) and it would still equal 1 or 2, depending on the innermost digit. But as soon as you jump to an infinite sequence, it becomes an irrational number, which cannot be either 1 or 2. This reminds me of geometrical paradoxes. Take a square of side x and draw a stepped diagonal through it, like a staircase from one corner to its opposite. No matter how many steps-and how small-you draw, its length will always be 2x. Now jump to infinite steps, each infinitely small, and you get the usual diagonal, whose length is, unsurprisingly, an irrational number: x√2
Because that's what irrational (computable) numbers are: they are "placeholders" for what you would get it you could run a given algorithm forever. π, e, √2 are all such objects. In this sense irrational numbers don't really exist in the real world, because you can't run computations forever. They also have intrinsic fallacies. For example you can't always compare two algorithms for equality or ordering (it's undecidable in the general case) so it follows that equality and ordering aren't always defined for irrational numbers as well. But wait, there's more. According to current orthodox math, 100% of "real" numbers are not only irrational, they aren't even computable. So you can't even give them a name or identify them, except in a few rare cases (such as Ω, defined over the halting problem, which is uncomputable.) Conclusion: don't believe what math books tell you about numbers and only use fractions!
Mathologer Yes, the conclusion was a bit of a hyperbole. But I'm not the only one to thing the current view of real numbers is nonsense. See for example Prof. N. J. Wildberger here: njwildberger.com/2016/01/01/uncomputable-decimals-and-measure-theory-is-it-nonsense/ «This is modern pure mathematics going beyond parody. Future generations are going to shake their heads in disbelief that we happily swallowed this kind of thing without even a trace of resistance, or at least disbelief.»
Very smart guy and definitely a real mathematician. However, I think it is important to stress that he is really only speaking for himself when he criticises the foundations of mathematics and in particular the real numbers. He is a professor of mathematics and so am I and neither I nor any other professional mathematician I know agrees with him in this respect. Having said that, I find a lot of what he has to say very interesting and thought provoking. As well, his videos are a great resource for learning a lot of great mathematics. So, despite of me disagreeing with him in many way I consider him to be one of the "good guys" :)
Fair enough. In any case I was wrong: the "infinite" fraction 2/(3 - 2/(3 - ...)) is nothing but a fancy way to write the equation x = 2/(3 - x) which has exactly two solutions, 1 and 2. Nothing irrational here.
Evan490BC but if you noticed anything past "numbers", you'd see that you can make any version of this "problem" with any numbers you want, like I did where I changed it from 1 and 2 to 3 and 4, making it ?=12/(7-?)
I am unaware if you are still looking at the comments, but with the 1=2 madness in the beginning, we can use the partial sums like you say in the interlude. If we do this, we get 2/3, 6/7, 14/15, 30/31, 62/63, so on so forth. This seems to converge to 1, so I believe the final answer is 1
it's the argument wether a converging infinite Series/Sequence etc is really equal to the Number it converges to, basically if you can really treat that infinite series as a number itself.
+Mathologer sir, i would also like to know the prime number theorem of ramanujan of what proof he did... also the partition theorem .. please post a video on these topic sir... eagerly waiting..
thank you sir... there was no direct proof of his work in the internet.. so, number theory lovers are finding it hard to know about this great man.. please do it sir.. thank you
Burkard Polster. I studied Philosophy and Math at University in the 70's and look at at least one math video every day. I just recently discovered your channel and have subscribed and started to put thumbs up on your shows (although I suspect this is the reverse of what the ancient Romans meant by it) 😁 I'll be looking in on you soon JIM
+Argentarii Homini that wasn't the method he should to approximate them infinite fractions. You included the final -1 or -2 which drastically changes the answer. By leaving out the last number, the limit is more meaningful in a way.
he said the limit of the series of partial fraction must always converge to the number you say it is equal to. So, in this case, it does not converge to 2. I don't think this is relavent to speak about the last number in an infinite serie.
the fraction is meaningless because it converges to 1, but since the fraction is infinite there's always going to be a 2 above the 1, which will result in a quotient of 2
Others seem to have said the same thing, but this is what I came up with. So long as you terminate the infinite fractions with either a 1 or 2, the expression is valid. The problem is that you are treating the continued fraction of a rational number like that of an irrational number, allowing it to iterate to infinity. If you stop after a finite number of iterations, the identity still holds -- meaning that the problem is trying to continue to infinity. I mean, I have no idea why exactly what you did is wrong (I don't see the flaw in the algebraic logic -- somebody please point it out!), but I just understand that the error lies in treating the rational number like it's irrational and continuing the fraction forever. By the way, this is the first I've heard of this channel. This was probably one of the most enjoyable maths videos I've ever watched. Do you have a donation link somewhere?
It depends on what you end your fraction with. If 1, the whole thing equals 1, if 2, it equals 2. This answer does not feel satisfying because it does not seem to apply to the "infinitly long fraction", but it actually applies to the infinite case too! Think about it: it depends on what is your procedure to generate the fraction. One procedure is to replace the 1 in the denominator with 2/(3-1), while the other is to replace the 2 in the denominator with 2/(3-2). Since you cannot hand me the infinite fraction, you must hand me part of it and then exactly define what the three dots mean. The three dots mean different things in the two cases, and that is: which procedure are you using?
Very informative video, but is there some way to define, determine, or identify the characteristics of whichever irrational number would be the LEAST irrational? I'd suspect the answer may rely on your definition, but I'd be curious if this has been thought out too.
This is effectively an iteration because you have constructed a chain of fractions, and you have to compute the result by starting at the bottom and working upwards. If you start way down in the fraction, and enter x, the next level up in the fraction is 2/(3-x). Treat this as an iteration, i.e. x_new = 2/(3-x_old). It turns out that this iteration has two(-ish) behaviours. BEHAVIOUR 1:- If you put in 1+delta, delta *HALVES* each time, so it will converge towards 1. BEHAVIOUR 2a:- If you put in 2, it is stable... ... but... BEHAVIOUR 2b:- If you put in 2+delta where delta is VERY SMALL, delta will *DOUBLE* on each extra level up the fraction, until eventually the number flips sign and it converges towards 1. In other words, you have constructed an iteration which has two fixed points. One of those fixed points (i.e. 1) is stable and the other (i.e. 2) is unstable. Therefore the iteration (continued fraction) will converge towards 1, from almost any number at all, i.e. its radius of convergence is infinity. However, the fixed point 2 has a radius of convergence of ZERO, so 2 will remain 2, but anything slightly away from 2 will move farther away from 2 with each iteration.
If you start with 1, the sequence of partial fractions is constantly 1, thus converges to 1. If you start with 2, same thing just with 2. If you just look at the infinite fraction, you get a quadratic equation by calculating its value. This equation has two solutions: 1 and 2. So the infinite fractions are basically the same, but their values depend on the value you start with.
Indeed 🎯.
Many thanks! I learned a great deal from this, and enjoyed it immensely!
That's great :)
Define a series: a[1]=2, a[n+1]=2/(3-2/a[n]) for n=1,2,..
For example, three following elements are:
a[2]=2/(3-2/a[1]) --> a[2]=2/(3-2/2)=1,
a[3]=2/(3-2/a[2]) --> a[3]=2/(3-2/1)=2
a[4]=2/(3-2/a[3]) --> a[4]=2/(3-2/2)=1
It's easy to prove using induction that a[n]=2 for odd n and a[n]=1 for even n.
Element a[n] in this series can represent the n-th step in the finite construction of the infinite fraction.
The series does not converge and the question "what it converges to" has no meaning.
I consider this as a sequence: start from a_0, cook up a_1 as a_1= 2/(3-a_0), and in general, a_{n+1} = 2/(3-a_n). This sequence has the general term
a_{n}
= ( 2^{n+1} - 2 - (2^n - 2) a_0 ) / ( 2^{n+1} - 1 - (2^n - 1) a_0 )
= ( 2^n (2 - a_0) + 2 a_0 - 2 ) / ( 2^n (2 - a_0) + a_0 - 1 ).
So, when a_0 != 2, the sequence converges to 1, and when a_0 = 2, the sequence converges to 2.
The 'usual' way to chop off a continuous fraction is to take a_0 = 0, and consider the continuous fraction as the limit of the above-mentioned sequence. So the 'correct' value of the continuous fraction is lim_{n->\infty} a_n (with a_0 = 0), which is 1.
Sorry if it appears twice, I cannot see my initial post.
I think property holds only for + operation not for negative operation and may be there is quadratic equation associated with for two different values.....
The equalities shown at the beginning (i.e. "1=..." and "2=...") were correct until the fractions were made infinite, at which point the value of the expression itself changed. 1 does not equal that infinite fraction, and neither does 2. Therefore it does not follow that 1=2.
A comment passing by. I think the 1=2 is about importance of notation, like the 1 / 0.414213... is at the bottom of the partial fraction at 7:39.
Obviously for irrational numbers, the decimal place continues forever, so when you write them in the partial draction form, I think it's "correct" to write 1 / 0.few_decemal_points_of_the_irrational_number to make it less CONFUSING.
At 7:39, sqrt(2) is represented by partial fraction with numbers 1, 2, 2, 2, 2. And I could just make up some irrational number now whose partial fraction representation starts with 1, 2, 2, 2, 2, which equal to first five of sqrt(2) then it can be some random sequence of infinitely many numbers.
Both the sqrt(2) and the random irrational number I made share same first five "coefficients" so when they are abbreviated to partial fraction with five coefficients and dot dot dot, they both have the notation but they obviously are different numbers. So to reduce confusion, you'd write the 1 / 0.decimal place for the irrational numbet.
Just like this, 1 = 2 / (3 - 1 / (3 - 1 / ...)) and 2 = 2 / (3 - 1 / (3 - 1 / ...)) should be notated as
1 = 2 / (3 - 1 / (3 - 1 / (... / 3 - 1) )) and
2 = 2 / (3 - 1 / (3 - 1 / (... / 3 - 2) ))
So according to my theory on the paradox is that the notations for 1 and 2 is that the notation missed on important terms that made two different numbers look the same, just like how 10 and 1000 look the same without "0"s.
So, basically it's like saying
10 = 1 * 10 and if you replace 10 with 1 * 10 few times, it's 10 = 1 * 1 * 1 * 1 * 10. And if you do it infinitely, it's 10 = 1 * 1 * 1 * 1 * ....
And for 100 = 1 * 100, 100 = 1 * 1 * 1 * 1 * 100 becomes 1 * 1 * 1 * 1 * ...
Oh, wait, both 10 and 100 have same 1 * 1 * 1 * ... part, so they must be equal???
What I'm saying is that, if you want to do that, you should 10 = 1 * 1 * 1 * ... * 10 and 100 = 1 * 1 * 1 * ... * 100.
You showed how there is a clear pattern in the integer parts of the infinite fraction representation of e. Is there such a pattern in pi as well, or is it as random as the decimals are?
Pi 's simple continued fraction is "random". However, there are other continued fraction representation of pi that exhibit nice patterns, e.g. functions.wolfram.com/Constants/Pi/10/0002/MainEq1.gif
A quadratic approach...
Let, x=2/3-2/3-2/3-.....
Therefore it can be written as
X=2/3-x
3x-x^2=2
Solving out,
We get,
(X-1)(x-2)=0
X=1 or x=2..
Inference: That continued fraction will be equal 1 OR 2...if we solve out a quadratic problem we get two roots but that doesn't mean we compare both roots..and come to a contradiction..the same is with the case of this continued fraction....
Please comment about my proof..and if there is something wrong.Please be kind enough to point out the mistake..
Thank you
The sqwr root 2 trick works because the silver ratio (the solution x to x^2-x-2=0) is sqwr_root_2 +1.
Well. 1 isnt 2.
You solved the goddamn puzzle
Good
Is the mistake because the infinite fraction representing 2 cannot actually be written in the way you did, because there is a final step that is not taken into account with the . . . ? In other words, the . . . represents you performing a function that returns itself an infinite number of times in the middle of an existing expression, rather than actually representing the entire expression?
Like how if you started with 1 = 1* (1) and expanded that to 1 = 1* (1* (1* ( . . . and also had 2 = 1* (2) and expanded that to 2 = 1* (1* (1* (1* . . . the two expansions would be written the same way, but wouldn't be equal, since the latter is correctly written as 2 = 1* (1* (1* . . . (2).
The sequence of partial fractions does not converge to any specific number. This is because 1 and 2 are rational, thus their fractional representations should be finite.
MAYBE 1 REALLY DOES EQUAL 2
Being familiar with post-modernism, that sounds totally reasonable.
+Birric Forcella I wouldn't say something that is post-modernistic is reasonable. More like it's not reasonable but who cares it doesn't matter, if I feel it's right then it is right in some form that no one would even bother
Gee, haven't you ever heard of something called humor or sarcasm? (Facepalm)
Birric Forcella it wasn't even sarcasm nor it was funny
Do you care? Math is broken anyway
My answer to the riddle: Is it = 1 or = 2?
In that continued fraction, as you write it, you alternate between ending with a numerator and ending with a denominator.
If you stop to evaluate it, ending always with a numerator, that sequence converges to 2 - in fact, it IS just 2, every time!
But if you always evaluate it ending with a denominator, the sequence converges to 1.
Of course, if you stop at every numerator and every denominator, the sequence fails to converge.
So the meta-answer is, you have to define how you mean to interpret the CF; the usual way being the 2nd - always stop at a denominator. That definition makes the answer, 1.
Great answer :)
Thanks!
On further contemplation, it seems that you will run into this conundrum when and only when you allow numerators to exceed denominators, in a non-vanishing fraction of the terms. Or some condition close to that.
Which may be one reason that the 'all numerators must = 1' case is of special interest.
Basically, no matter how far you carry the calculation, whether you stop at a 2 or a 3 radically affects the end value. This is basically saying that the fraction doesn’t converge, which is what Mathologer said is part of recognizing that the fraction is not meaningful, since the truly infinite version does not stop at any point.
So basically -1/12?
Hey @@ffggddss , great answer above. Although I think the theory of numerators exceeding denominator might not be right. Consider (1/2)/((3/2)-((1/2)/(3/2)-((1/2)/(3/2)... ( This was the best font I could come up with in RUclips comments). This above sequence converges to either 1 or 1/2 depending on your theory
you never told us what the continued fraction in your thumbnail evaluates to...that one's the coolest one...
I am so curious
That can be regarded as the most rational irrational number, since it can be approximated better and better by rationals the deeper you go.
It equals I0(2)/ I1(2) = 1.4331274..., where the In are modified Bessel functions of the first kind.
@@landsgevaer Hi, I’m not good at math and I know I’m a month late, but would you mind explaining it in a way where I wouldn’t need to know Bessel function is? Or is there a way of expressing this number as a radical?
@@landsgevaer What about
2 + (1/(4 + 1/(8 + 1... with powers of 2? or powers of 3?
@@idklol2566 Although I knew the Bessel functions, I only found that result somewhere, so I have no proof or derivation.
Let the infinite fraction from the first problem equal x.
Now replace the "infinite part" with x to get x=2/(3-x)
Rearrange ==>. 3x-x^2=2
x^2-3x+2=0
(x-2)(x-1)=0
Therefore there are two separate solutions x=1 and x=2 but since they are separate, 1 does not equal 2! Maths is not broken! Rejoice!
x=1 and x=2... still doesnt answer the question
+Eden Lumbroso no, because a polinomial of nth degree has n solutions(zeros). Damien ended up with a polinomial of the 2nd degree with two solutions x1 = 1, x2 = 2.
+My Name is Chef still doesn't answer the question how come 1 and 2 both equal the same thing (the fraction. not x).
a guy wrote 2 is not a valid answer because the fraction does not get closer and closer to it like it does with 1.
+Eden Lumbroso it does, substitute x=1 and x=2 into the original equation and they both work but not at the same time, it is either one or the other. This is due to the nature of quadratics and polynomials in that there will be n roots where n is the highest order of the polynomial (n=2 for quadratics) but those roots are not equal to each other and with the rearrangement I did, I proved the the infinite fraction is just a quadratic in disguise and will follow the same rules as polynomials.
It does; it shows that for any single-repeating-digit continued fraction there are two valid solutions.
In general for q = 1/(a+(1/a+(1/a + ...))) we see that:
1/q - a = q
so 1 - qa = q^2
q^2 + aq -1 = 0
Without factorising, we see that it has two solutions because it's order two.
Step 1: Turn into x = 2/(3-x)
Step 2: Solve for x to get x = 1 or x = 2
Step 3: Cry because this simple method that you taught me doesn't work and I suck at maths
the funny thing is it actually is theright answer, i think :]
Think about how you solved it (don't say calculator) the answer is in the method
Both 1 and 2 work for that equation (and that infinite fraction above), but that doesn't mean they are equivalent. x=1 OR x=2. Not 1=x=2 => 1=2. I believe that is the problem answer.
What you've done in step 1 and 2 is find the invariant point(s) of the sequence defined by u_{n+1} = 2/(3-u_{n})
I think you nailed it: just because there are two solutions to the equation doesn't mean the two solutions are identical.
on 9:30 I noticed that after 333/106 comes 355/113 witch is (333 + 22)/(106 + 7)! Wow, weird!! [22/7 is the first one]
That's an interesting observation. I should add that if you've got three consecutive best approximations they are always related like this a/b, c/d, (a+kc)/(b+kd) for some k. Try it :)
That’s because the last number in the finite fraction is a 1. For example, phi has a continued fraction of all 1s, and its finite approximations are ratios of consecutive Fibonacci numbers, which always have that property. In fact, this works for any pair of approximations, except you have to multiply top and bottom of one of them by the last number of the fraction before adding things up; it’s just more visible when that last number is 1.
It's kinda easy to see empirically, approximation fractions usually consist of numbers that don't share factors
@@Mathologer you should definitely make a vid. explaining this
very circular
"22 over 7"
BOOM dead giveaway
after 3Blue1Brown ,anyone?
Yeeep
Mhm
Yep XD
Yepp
Me too, and I got REALLY disappointed. First, I basically knew everything he said, and second, the explanation of why phi is the most irrational number was just wrong. I explained it in this comment:
ruclips.net/video/CaasbfdJdJg/видео.html&lc=Ugxik2-iu5Y5EYQQj5R4AaABAg
A couple of interesting answers to the puzzle already, but I'll refrain from making any comments myself until tomorrow (here in Australia). Have fun until then :)
Also, thank you to Zacháry Dorris for contributing English subtitles to the last video (Riemann's paradox), Rodrigo Naranjo for contributing Spanish subtitles and Étienne Leb for his French subtitles!
Added the next day (in Australia). Quite a few answers that are spot on. Before I give my own answer to the puzzle just two hints: 1) The continued fractions in the puzzle are a bit different from the simple continued fractions that I discuss in the rest of the video. 2) What I say in the interlude is crucial for getting to the bottom of what is going on here.
I think that the simplest answer is that a rational number can't be expressed as an infinite fraction, only irrational numbers can be expressed that way, so I think you did say the answer in the video, you tried to got us! 😆
I don't think so. If you use the method explained in the video in order to get a continued fraction of either 1 or 2 the continued fraction will be finite. I don't think it means you cannot get an infinite fraction of a rational number.
+yotam vaknin But he himself says that for a rational number, there's no infinite fraction, it can be expressed as a fraction, but not an infinite one.
+Ⲇⲁⲛⲓ Φi No he said that using the methode to create a fraction will end, not that any given fraction representing that number has to end (aus far as I understood)
The partial fractions do not converge.
I believe the answer to the puzzle is in the interlude. You can only place the equals sign if the partial sums converge to the answer.
But in the expansion for 2 they don't, so you cannot say 2 = (the written infinite fraction) = 1, because it only converges to 1.
that fact depends on how you write the equation.
One sequence follows: {2/3, 6/7, 14/15, ... k/(k+1)...}.
Another: {2, 2, 2, ...}
So both converge.
Michael Esplin The partial sums were defined to include the entire fractions. I'm just using Mathologer's definitions.
+mrBorkD Fair point.
Then why doesn't the substitution for two work? Obviously 2 does equal 2/(3-2), so anywhere you see a two, you can replace it with 2/(3-2).
Both identities and infinite fractions are, in fact, correct. The nuance here is at 1:23 when he states the right sides are both identical; they are not, because in the case of 1 we have been replacing a 1 the entire time, in the case of 2, a 2. The "last digit" of this series is, in fact, quite important, because it is the number we have been substituting for this entire time. The "dot dot dot"s are not the same at all, as one has a value 1, the other, 2.
Here , we can write
x=2÷(3-x)
Now , quadratic equation form...
x^2 - 3x + 2 = 0
(×-2)(×-1)=0
×=1,2
So, x=2÷(3-×) satisfies for both .
You have to remember that in one series, 3-2=1 and in the other 3-1=2. They both are technically correct in and of themselves as they go on towards infinity, but the part that is being replaced is different even though they appear the same at a higher level. Also, keeping the above in mind, you can end either series at any point and so they are not an irrational infinite series, more of a faked one.
The author just uses dot dot dot to replace the series, but the dot dot dot in each series are not the same. You explained it in words much better of course.
this channel is incredible
Oh man, your videos are so good, they must take forever to make, but I love them. I wached them all. keep up the good work.👍
herrreinsch why the fuck am I seeing you here lmao
The partial fraction does converge to both 1 and 2 but it depends on the way one decides to evaluate the partial fraction. If we evaluate it by just considering the fraction before the negative sign then it will converge to 1, but if we were to evaluate it by removing everything below the denominator then the fraction will always be 2!
I just ate some 22/7
The infinite fraction 2/(3-2/(3-2/... is divergent. The sequence of partial sums alternates between numbers approaching 2 and numbers approaching 1.
Actually that's not the case.
2/3 = 2/3
2/(3-2/3) = 6/7
2/(3-2/(3-2/3)) = 2/(3-6/7) = 14/15
2/(3-2/(3-2/(3-2/3))) = 2/(3-14/15) = 30/31
62/63...124/125...etc
+Not Broihon I think that if you use the partial fractions for the "=1" part you have to end with 1 and for the other one you have to end with 2
I will give try when I arrive at home
2/3 = 2/3
2/(3-2) = 2
2/(3-2/3) = 6/7
2/(3-2/(3-2))= 2
etc
Depending on how you take the partial sums, it'll either converge to 1 or always be 2.
EDIT, I believe the way to take the partial sums that lead to 2 are false.
+Not Brohion I did the math again and you're correct. Don't know how I messed that up.
There are no partial sums but fractions ;). He also mentions that you chop the seq. before the further addition for partial fractions, which only makes sense as you can´t add the -2 without aknowledging that it functions as a numerator thus cant be added without adding the denominator, as mult. has precedence over add.
10:10 turn on captions
Bruh! Plot Twist hahaha
yeah right? these captionmakers are so annoying
I think I've got it. Mathologer, I'd love to know if I am right or not.
The interlude is really important because you explain what the "equals" means here. The infinite series has to converge to the number in question for us to label it as equals.
If we start taking the partial sums by cutting off at the minus signs, we first get 2/3 (0.6667).
The next term is 2/(3-2/3) which is 6/7 (0.8571)
Third term is 2/(3-2/(3-2/3)), which is 14/15 (0.9333)
The fourth term is really nasty to write out, but it gives us 30/31 (0.9677).
The pattern emerges to show that each term can be determined by its preceding term. The numerator of any term n will be double the denominator of term n-1, and the denominator of term n will be 1 greater than that. While we see this series does converge to 1, we can prove it using calculus.
This series is just a faster converging series of the form n/(n+1). The nth term test gives us infinity/infinity, but L'hopital's rule says we can take the derivative of the top and bottom, perform the nth term test again, and find out what the series converges to. The derivative of n is 1, and the derivative of n+1 is 1. The limit of 1/1 as n approaches infinity is 1, so the series n/(n+1) also converges to 1. We can use this information to also conclude that the infinite fraction also converges to 1.
Thanks for leaving us this problem to work out. Great video!
It is a recurrent infinite sequence of the form 2/(3 - X), where X
1:19 You forgot to change the 1 into a 2! Just because it's small doesn't mean we won't notice.
I think that the last value in "..." should be 1 to equal 1 and 2 to equal 2. To illustrate this, let's consider the following recursive functions:
f(0) = 1
f(k) = 2 / (3 - f(k-1))
g(0) = 2
g(k) = 2 / (3 - g(k-1))
For any n>=0, we have the following identities:
f(n) = 1
g(n) = 2
I think that the problem occurs when you transform the fractions into infinite fractions. You took a really important part away. The fraction with 1 always had a 1 to be substituted again. And the fraction of 2 always had a 2 to be substituted. So the fraction of 1 has a 1 in the end (that is hidden by the elipses) and the fraction of 2 has a 2 in the end (which is also hidden by the ellipses). And with that in mind we can say that they aren't really equal.
there isnt "at the end"
Well, right. The issue is that the "1" version needs a 1 at the end to maintain the identity. Filling it in with the ellipsis actually breaks the identity by changing it to a sequence of partial fractions which converges to 2, not 1.
If we abstract the infinite fraction: x=2/(3-x); -x²+3x-2=0; Using the cuadratic formula it has two results: x=2 and x=1
(sorry for my bad english)
not sure if i'd agree that phi is "more irrational" than pi, although I guess it depends on the context of what you're asking. phi after all is algebraic and pi is transcendental so i would have thought that pi would have been "more irrational" in that sense.
imagine if you will two line graphs with a gradient of the natural number you want to test
y = phi*x which can be compared to another, your example includes pi so
y = pi*x
the closer either line graph is to the cardinal points, the "more" rational it is. This happens when their convergent fractions come very close to approximating their value. ie in each convergent fractions second iteration ( pi ~ 22/7 as opposed to phi ~ 3/2 )
Absolutely correct. More formally, the Liouville-Roth irrationality measure of the golden ratio is 2, while that of any Liouville number would be infinite.
en.m.wikipedia.org/wiki/Liouville_number#Irrationality_measure
@@dbodow Thank you!!
This was hugely informative. Thank you so much for your videos Mathologer. I love you!
The problem is that the notation of the first example with three dots isn't even rigorously defined at all. We should do:
Lets define f(n)=2/(3-n). {a(n)} and {b(n)} are 2 recurrent sequences defined by a(n+1)=h(a(n)) and b(n+1)=h(b(n)), but a(0)=1 and b(0)=2, which means the 2 sequences aren't the same one. That implies that their limits aren't necessarily the same. An expression of the form in the video doesn't give us an clue of the original term/seed of the sequence in question and isn't well defined because of that.
The notation works for the rest of the video because the sum (and so the limit of the sequence) is actually done in the other way: from left to right. So, it actually has a very well defined first term.
Thanks for the interesting video!
I've also been watching videos about "distributed mode loudspeakers" which are essentially a flat panel (of wood, foam, etc.) with a small speaker driver attached to the back.
One of the recurring problems with DML construction is resonance - depending on the dimensions of the panel and the position of the driver certain wavelengths will cause the panel to resonate. For example, with a square 1m x 1m panel, and the driver placed in the middle, there will be resonance at frequencies that correspond to a wavelength of 50cm, as the wave will reflect off the edges of the panel and constructively interfere with itself. But it's worse than that, as as well as all these simple reflections along a single axis you also get more complex 2D resonances - all the different patterns you that Chladni plates show are visualizations of all the different resonances that we want to minimize.
A rule of thumb I've heard from DML designers is to use a panel whose side lengths are not a simple ratio of each other, and to place the driver roughly ⅖ from two adjacent edges and ⅗ from the other two edges. I wonder if the golden ratio has applications here? Specifically, I hypothesize that the amplitude of resonance peaks may be minimised by a panel whose side lengths have the golden ratio to each other, and the driver is positioned according to the golden ratio - I think (1/φ) from two adjacent edges and (1 - 1/φ) from the others?
It should be possible to simulate this but I can't find any existing Chladni plate simulators that allow for non-square plates, and I don't think they model the position of a 'driver' either.
If 1 equals 2 that means I’ll never be alone cuz I can date myself :)
I want to write a report on this video if someone having data please reply me 🙂❤️
The identity at the start of the video only holds for finite expansions of 1 and 2. If we do not end it with a 1 or 2, then we are essentially breaking the equal sign. On top of that, if you compute the sequence of partial fractions for these two sequences, you will find it converging to 1. For example: try 2/(3 - 2/(3 - 2/(3 -2/(3 - 2/3)))).
if 2=(2/(3-2)), then why don't you substitute the right side of the equation to both numerator and denominator?
In 1=2 riddle, it depends wether we end the continued fraction with 2 or 1. If continued fraction ends with 2 in its tail, it collapses to 2 while if continued fraction ends with 1 in its tail, it collapses. Since, the 3 dots doesn't tells us anything about term on tail, the continued fraction doesn't have a particular value.
This is similar to infinite series 1-1+1-1+1... whose sum is either 1 or 0 depending on the last term and since this doesn't conclude that 0=1, our ICF paradox doesn't conclude 1=2.
Binomials expansion gives 1-1+1-1+… equals 0.5
lol!
@@nuklearboysymbiote will you please elaborate? Expansion of which bi. expression?
@@1paper1pen63 (1 + x)^-1⠀⠀when x = 1 (which of course wouldn't be a "valid" expansion but it does work out to 0.5 if you just evaluate (1 + 1)^-1, but the expansion becomes 1-1+1-1+…)
as other people have noticed, x=2/(3-x) has 2 solutions: x=1,2. however, for the iterative process x_{n+1}=2/(3-x_n) to converge to either 2 or 1, we need to look at the derivative of the iteration. the derivative of 2/(3-x) is 2/(3-x)^2, which is 0.51 for x=2. That is why we approach 1 with the continued fractions (derivative less than 1).
if we play with the equation a bit and rewrite it as x=3-2/x, then the iterations (not similar to continued fractions so much anymore) would actually converge to 2: 3, 7/3, 15/7, 31/15,...
this is because the derivative of 3-2/x is 2/x^2, which is 2(>1) for x=1 and 0.5(
11:40 I used to think you could 1-up φ, by choosing 0’s in all the denominators (because there’s still the infinite ”+(1/[stuff])” -tail, also, in each denominator; it’s not division by 0). You *_CAN_* do that. However; it turns out to be anything but irrational:
0+(1/(0+(1/(0+(1/(0+(1/(0+…)))))))) =
1/(1/(1/(1/(1/(1/(1/…))))));
which, actually, collapses to just 1 (because it’s just: ”1 over 1 over 1 …”, over and over again, it stays, at 1, through all the infinitely many truncations, we get out of it. So, trying to get too greedy, results in the complete opposite of an irrational number: An Integer; and a basic one, at that: *1.*
😅
My take on the puzzle:
If you consider the function f(x) = 2/(3-x), it has two fixed points where f(x) = x: x=1 and x=2.
However, x=1 is an attracting fixed point of the function, and x=2 is a repelling fixed point. That is, if you take some value close to x=1 and then iterate the function, calculating f(x), f(f(x)), f(f(f(x))) and so on, the values will get closer to 1; but if you start with some value close to 2, the values will run away from 2. This is because the function's slope is shallower than 1 at x=1, but greater than 1 at x=2.
I am thinking this is a thing to be wary of when working with continued fractions.
...and the Mathologer's "interlude" about partial fractions about halfway through really gives the key. An infinite fraction expansion only really makes sense if the sequence of partial fractions converges.
no matter what value you pick the iterations converge to 1.
Similar to what 3Blue1Brown said in his “Things They Won’t Teach You in Calculus” video (ruclips.net/video/CfW845LNObM/видео.html ) about why the proper solution for 1+1/1+1/1+1/1+1/.... is phi, rather than its negative counterpart, even if that also satisfies x=1+1/x.
Interesting perspective by modeling it with a dynamical system.
@@roberttheiss6377 unless you pick 2
For the 1=2 thing, you can try to solve for what 2/(3-2/(3-... equals by setting it equal to x. So you get x=2/(3-x), or x(3-x)=2, and you get the quadratic x^2-3x+2=0, and when you solve for x you get x=1 or x=2, so both answers are correct. They differ depending on what you interpret the dot dot dot as
So that's not an eyelash on my screen, it is you?
I think the problem is that 1 is the loneliest number, and since 2 can OBVIOUSLY be as bad as 1 (because its the loneliest number since number 1), there is no way to say they are completely equal. Not to mention the that fact that 1 + 2 = buckle my shoe, and shoes don't have buckles anymore, so it's clear that something weird is going on. Either way, it's been 1 week since you looked at me, and even though I TOLD YOU I had 2 tickets to paradise, you INSIST that I start back at 1, just to make your dreams come true.
hahahaha omg best response ever.
lol I thought it was because they were both rational numbers that couldn't be applied to an infinite fraction, but after reading your post I realize that you were right all along. It just makes so much sense!
Can you support this logic with a few more son.. I mean, uh... historical mathematical... pieces?
You should have listened to your heart.
Well, you are right, the heart will go on. I would say it's a shot to the heart, and that you're just a bit too late, but I would just be left with an akey-breaky heart, which is never a good thing. But if you could just look into your heart shaped box and see that a total eclipse of the heart is bound to happen when you check in at the heartbreak hotel, then you would see that you're terrin up my heart when I'm with you, and that sgt pepper and his lonely band just want to have fun.
You can't ... it since there is - not a +. This means you don't get smaller more meaningless numbers out. Only number you want to know is the last number 1 or 2.
Well 2/3-2/3-2... continued fraction.
Let it be x, so x= 2/3-x
or x(3-x)=2
or x²-3x+2=0
or (x-1)(x-2)=0
Which means this quadratic equation has 2 solutions 1 and 2.
That explains that 1≠2 rather x=1,2. Or we can say that our continued fraction has 2 solutions 1 and 2.
Thanks for the video!
One nitpick - Everytime I heard "square of 2" instead of "square root of 2", a little piece of me died inside. :(
Same here 😢.
Maybe the problem is that you created an infinit fraction for a rational number, and that only works with irrationals?
That's what I'm thinking.
Using the simple form: 1's in the numerator and addition of positives in the denominator --ur assertion is true. That is, a simple continued fraction terminates if and only if it is an expansion of a rational number.
However the form here is not simple. And so has nothing to do with that fact.
That seems legit to me. Also, choosing 2 over (3-1) seems somewhat cheaty to me. Like, you could possibly do this with any number and get an identity. Like in the video, 3 = 3 over (4-3) having the 3 in the denominator again, thus also the infinite fraction.
aa Maybe if you cared to explain why it is dead wrong instead of simply complaining, then that wouldn't be a problem.
aa Well, that was sad. Just go around like Homer and proclaim, "I am so smart! I am so smart! S-M-R-T!"
Okay, I'm gonna try and answer this now. Here's the problem - the way we represent continued fractions and the numbers used obscure the problem a bit. The thing is, infinite fractions don't automatically have any sort of meaning, we need to define what they mean. Otherwise we can just throw whatever numbers we want in and have no idea whether it actually represents a real number or not. That's why we use convergence as our definition - in the fraction given the sequence we get is 2/3, 2/(3-(2/3)), 2/(3-(2/(3-(2/3)))) and so on, the nth term of the sequence evaluating to (2^(n+1)-2)/(2^(n+1)-1), which tends to 1 as n tends to ∞. Notice, however, that we took only the integer part of each denominator in our partial terms - why? It's purely definition. We take the integer parts as they're what defines a continued fraction, at least in the case that the numerator is only allowed to be 1, so we continue with this definition when extending to more general fractions. Furthermore, the (cont)
fractional part should be overwhelmed by the integer part - if instead of just including the integer part, we include the numerator of the fractional part too, the terms we get aren't really representative of the actual continued fraction. In an infinite continued fraction with numerators set to 1, this shouldn't really matter, but when we're working with fractions which don't have this regular behaviour and you start treating what should be a fraction as an integer when looking at convergence? That doesn't really make sense. However, if we take this approach when looking at the given fraction we end up with the sequence 2/(3-1), 2/(3-(2/(3-1))), 2/(3-(2/(3-(2/(3-1))))) and so on, which evaluates to an sequence of 2's. This is pretty obvious as in this case the numerator of the fraction in question actually is equal to the fraction as a whole! (That is, 2=2/(3-1)). I think that for this reason the definition of what counts as a continued fraction is somewhat obscured, but it (cont)
+Hwd405 could explain why two different values emerge. It's all about definition and which way we approach a continued fraction, but usually we take the former approach and the result is convergence to 1. I'm not sure either result is "more correct" than the other though, but it's important to choose a mode of convergence and stick with it so that we don't end up with concepts that aren't well defined.
I see what you're saying, of course you can define things however you like, although you don't have to add any definitions. Sometimes it's alright for things to equal two different values (example bellow) without it breaking maths.
Both values of the continued fraction can be found without looking at convergence of partial sums (but I go into that more in my main reply to the video)
y = f(x) = √x
for x = 4
y = 2 or -2
f(2) = 2
f(2) = -2
f(2) = f(2)
-2 = 2
Hopefully thats an illustration that the problem here isn't that the continued fraction doesn't equal 1 and it isn't that the continued fraction doesn't equal 2. The problem is in the fallacy that because it can equal 1 and 2, means that 1 equals 2.
+oliver maclean that's not a valid example, a function can only have one output given any input. f(2) must take the value of 2 OR -2, not both. Indeed, we usually define the square root symbol to mean the principal root (the positive one). Furthermore, the repeated fraction notation isn't a function - it's notation. If it's consistent well defined notation it shouldn't be equal to two different numbers, otherwise that breaks the transitivity of the equals relation.
+Hwd405 sorry, I meant to say f(4), sorry for any confusion
the answer in my opinion is that the fraction should always equal numbers between 0 and 1 and to get other number you add whole numbers
like you took the whole part of pi and the golden ratio
and you always take the entire fraction out so it would be
x=
x=2/3
x=2/(3-2/3)
x=2(3-2/(3-2/3))
so, x just gets closer to 1
until it becomes 1
so x=1
i am still in school and i my "natural" laguage is not english so i don't understand everything and am probably wrong
The answer is this: The result of the infinite operation depends on the definition of "partial continued fraction." Such a fraction as written means nothing more than "the sequence of partial fractions of this form (form is to be understood after a certain definition that's not been made precise here) converges to such and such a number." The devil is in the precise definition of how to extract the partial fractions from the general formula with "...". Of course, "..." means "and so on" but this is not a precise mathematical phrase. The fact that we often write "..." assumes that we KNOW what we mean by this. Here, it's not clear what it means. It could mean a sequence 2/3, 2/(3-2/3), 2/(3-2/(3-2/3))... or it could mean 2, 2/(3-2), 2/(3-(2/(3-2))... The first sequence will converge to 1, the second converges to 2 (is in fact 2 all the time). The answer is therefore a question of definition of the partial operations. It can (most likely) be proved that the first sequence is always < 1 and strictly monotonic. I say "most likely" because I have not (yet at least) made any proof :)
That shirt is amazing.
@Mathologer:
Very neat! Continued fractions are also an interest of mine, and here's another twist on "most irrational."
What you've shown us is the traditional way to do CF's with unit numerators.
But notice that when a "1" occurs in the string of denominators, this is a signal that the CF up to (just before) that point, is pretty crummy. (This can be quantified by defining a figure-of-merit (FOM), that compares the fractional error
|x - (n/d)| / x
with the "size" of both num. and denom. - actually, with the reciprocal of the product of the two, 1/[nd] ). The rationale here is that the product, nd, is the "outlay" you make to get the fractional precision you obtain.
FOM = (1/[nd]) / (|x - (n/d)| / x) = x / (n |dx - n|)
Those "crummy" approximations can be eliminated by allowing negative denominators, d, but restricting them to |d| ≥ 2, with the additional proviso that when d = ±2, the next d must have the same sign.
On a calculator, e.g., these denominators are found iteratively:
• subtract the *nearest* integer to the current result [that integer is the current denominator]
• then invert that difference to get the next result
This technique gives, for π,
{3; 7, 16, -294, 3, -4, 5, ...} rather than {3; 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, ...}
- these are given to the same rational value: 5419351/1725033;
integer + 6 fraction coefficients vs integer + 11 fraction coefficients, for the same precision.
For e (allowing a little exception to the rules for the first couple terms for the sake of mathematical beauty), it gives:
{1; 1, -2, -3, 2, 5, -2, -7, 2, 9, -2, -11, ...} rather than {2; 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, 1, 1, 10, 1, ...}
- these, too, are shown out to the same fraction: 566827/208524
- The sequence without the exception is
{3, -4, 2, 5, -2, -7, 2, 9, -2, -11, ...}
integer + 9 fraction coefficients vs integer + 15 fraction coefficients, for the same precision.
For φ, it gives:
{2, -3, 3, ...}, with [-3, 3] being the infinitely repeating part. This makes φ seem not so very irrational.
But for √2, it gives (as does the traditional method):
{1; 2, 2, 2, 2, ...}, and now √2 takes over the title of "most irrational," because under the new rules, this is the "crummiest" possible convergent sequence.
This can be verified using the FOM, which actually just gives a value very close to the (abs value of the) next result in the iterative procedure described above ("On a calculator, ...").
Any thoughts?
Addendum:
Under the "new rules," the FOM for φ is φ+1 = φ² = 2.618...
And for √2, the FOM is 1 + √2 = 2.414...
So the contest for "most irrational" is actually pretty close!
Also worth noting, is that the FOM for a general irrational number, bounces all over the place as you proceed down the CF.
Heck, just look at π, e.g.!
You do such great work here!Have you ever considered doing something about Turing's halting problem and the associated Busy Beaver numbers?Just a thought- I eagerly await you next video, whatever the topic may be.Thanks again!
Gotta love, how Mathologer keeps referring to sqrt(2) as: ”the square of 2” (which, obviously, would be 2² = 4) 😅.
You should have done rationalising the denominator for The sqrt2 Continued fraction because it’s not clear why the decimals are the same (sqrt2-1)(1+sqrt2)=1
*Mathologer*, thank you so much for that introduction into *infinite fractions*.
My answer to your puzzle is that *infinite fractions* need a _mathematically strict_ definition (just as *infinite sums* do).
In the sense of definition you gave at 6:50 the fraction being considered converges to 1 and to 1 only. Again as one did with *infinite sums* he may define it as a limit of _sequence of numbers_ -- *partial fractions* (compare to *partial sums*).
When you try to mess around with such objects only relying on , you may face (like with conditionally convergent series).
However one may come up with another definition. When you calculate _N_ th partial fraction you may replace _diagonal dots_ with some particular number _d_ instead of zero.
If you choose _d_ = 2 it converges to 2.
For all other values of _d_ but 3, 7/3, ... when division by zero takes place (see comments below) it converges to 1.
One may also notice that if such fraction has a value it _should be_ the root of x(3-x)=2. However such trick only works for recursive fraction expansions (e.g. not for pi).
P.S. I'm looking forward to watching your video on Phyllotaxis phenomenon since I'm only familiar with Physicists' explanation of it. I would like to share it with you. Here are the lecture slides www.mit.edu/~levitov/FibonacciPhyllotaxis.pdf and here is the experimental proof arxiv.org/pdf/1002.0622.pdf.
Great spot-on comment and thank you very much for the links. There was also a great feature article on the American math society site on all all this. Just remembered that I want one of those magnetic cacti !
Small clarification:
If you choose d = 2 it converges to 2.
If you choose d = (2^(n+1)-1)/(2^n-1), that is { 3; 7/3; 15/7; 31/15 ... } it is division by 0. (If you define 2/0 = infinity and 2/infinity = 0, it still converges to 1)
For all other d it converges to 1.
+David Saykin Are you sure it diverges for d > 2? What about 1000? The resulting term of the first iteration is very close to 0, which leads to a term very close to 2/3, and for d < 2 it converges to 1, right?
Also try d = 3 or d = 7/3 or d = 15 / 7 etc. :-)
My favorite response to this video is the one from some guy who compared this proof to the following one:
1 = 0 + 1
substitue 1 by 0 + 1:
1 = 0 + 0 + 1
substitute endlessly:
1 = 0 + 0 + 0 +..... = 0
+Kamil Musiał thank you. I wasn't careful enough.
+earthbind83 I made a mistake. You're right. Your example shows how trick was done pretty well, I myself also find it similar to infinite sums.
+TheExaltedPheonix that's how Mathologer calls them.
The problem with the identity is that the second part of the identity is false. 2 is not equal to 2/(3-2/(3-... If we look at the partial fractures, 2/3, 2/(3+2/3), 2/(3+2/(3+2/3), ... , we see them converge to 1. The whole fracture only becomes 2 if you put a 2 at the end. In the infinite fracture, there is no 2 at the end, because there is no end, so the infinite fracture converges to 1. So the second identity claimed is false.
Good answer :)
Mathologer, once again great video. But can you explain how you get from "partial fractures" to partial sums?
Hi, Mathologer.
Why do you say: "Good answer"?
Yes, it is a very good answer, but I think that you say this because this answer, as thoughtful as it is, does not reveal the real answer.
And you want the secret, the real answer, kept hidden and the discussion to keep going.
So you say "Good answer" because this answer hints at the answer and does not really reveal it.
And you want that. And I do not like that you want that (if, of course, I am correct).
I guess that others that talked about attractors, basins, attracting and repelling fixed points, recursive calculations, nonlinear recursion x(n+1) = 2/(3-x(n)), f(f(f(x))), etc. really have the correct answer.
The correct answer as to why Mathematicians choose to equate this to 1, instead of:
a) 2
b) either 1 or 2
c) not convergent, because it alternates between 1 and 2
d) something else, like the average of 1 and 2.
So, am I correct or have I misjudged you?
Can’t the square root of 5 be approximated by finding two perfect squares whose ratio is approximately 5 (e.g. 81 and 16)? Based on that, isn’t 9/4 a fairly good approximation of sqrt(5), and thus 13/8 a good approximation of phi?
You can do that trick with any √n, where n > 0 is a non-square. It's a Diophantine problem (solving algebraic equations in integers) called Pell's Equation:
b² = na² + 1
Every +ve non-square n has a "primitive" solution (a₁, b₁), from which an infinite number of larger solutions can be generated by the recursion
a₊ = b₁a + a₁b
b₊ = na₁a + b₁b
For n = 5, (a₁, b₁) = (4, 9); the one you quote. The next couple of solutions are (72, 161), (1292, 2889).
Of some interest, you can get more "best fractions" for n=5 because it's one of the n's that also has a solution to the "Associated Pell's Equation":
b² = na² - 1
namely, (1, 2), which can be used to generate more "best fractions," every other one being in the above list, so that they alternate between satisfying P.E. and the A.P.E. :
(1, 2), (4, 9), (17, 38), (72, 161), (305, 682), (1292, 2889), ...
For general n, the primitive solution, (a₁, b₁) jumps around erratically. Some are easy to find by trial-and-error; some are next to impossible that way.
n = 13 is somewhat challenging; n = 61 is very hard. Luckily, there's a procedure involving - wait for it -
Continued Fractions!
- that will always yield the first solution.
Fred
It makes more sense to me to call phi the most *rational* irrational number. Think e and Liouville's constant. These are quintessential transcendental numbers, and the reason we can prove they are transcendental is that they are well approximated by rationals. Does this make them similar or dissimilar to rational numbers? Well, are rationals well approximated by rationals? If we can use the number itself as an "approximation" then yes, but if we can't use the number itself then rationals are the worst to approximate by (other) rationals. So if we don't count numbers as approximations of themselves we get rationals on one end of the spectrum and e and Liouville's constant on the other end, as it should be. From this point of view phi is the most rational irrational number.
IT IS ACTUALLY NOT THE SAME BECAUSE "THE 3 DOTS" IN THE "FRACTION OF 2" EQUALS 2 AND "THE 3 DOTS" IN THE "FRACTION OF 1" EQUALS 1. THOSE ARE NOT THE SAME 3 DOTS AND THEY DON'T EVER CHANGE. THEY DON'T CONVERGE TOWARDS 0.
Well, the problem is that you can't really keep the equal sign when you do that with two because the sum doesn't converge to 2, but to 1 [2/3;6/7;14/15;30/31;...].
If you call the infinite fraction x, you can get this equation because of the pattern: x=2/(3-x) and the equation roots are 1 and 2. But, even though 2 solves the equation, it breaks the condition of the partial fractions converging to the number, so it have to be discarded.
The necessity of discarding one of the possible solutions becomes more obvious if you try to solve the square root of two equation, because you'll get sqrt(2) and -sqrt(2) and the negative answer is clearly false because you are summing only positive numbers.
Here is how I solved the sqrt(2) equation:
x = 1+1/2+1/2+1/2+...
x-1 = 1/2+1/2+1/2+...
x-1 = 1/2+(x-1)
x-1 = 1/x+1
x^2 - 1 = 1
x^2 = 2
x = +-sqrt(2)
Assume that the infinite fraction converges. Then there exists a limit a. Then a = 2/(3-a). Solve for a to get a1=1,a2=2. This contradicts that the limit is unique and therefore the infinite fraction diverges.
wrong
It would be interesting to see continued partial fractions of complex numbers. Awesome video btw, too bad our brain is used to simple summation, and not fractional one. I have read about a guy who could do them as easily as we normal people do ordinary summation. Incredible!
Infinity is not a number, it is not included in the set of real numbers or any number set. It is the limit of the real numbers set but not part of it. That is why we have to use the lim operation whenever we have something that "continues to infinity". Not every infinite limit converges.
If you are given that infinite fraction without being told how it is produced, then you cannot find the limit because it diverges.
Sorry if that doesn't make much sens e, English is not my first language
X=2/(3-X)
3X-X^2=2
(x^2)-3X+2=0
Factorise for:
(X-2)(X-1)=0
Therefore X=2,1
That was a fun one! Hidden polynomials are great puzzlers ;)
ohhh, so it's not that 1=2, it's that both of them are the correct answer?
My answer was exactly the same (just, via the quadratic formula; rather, than factorization). 🎯👍🏻
@@nicreven Exactly 🎯!
one and two can written as rational numbers; therefore the fractional expansions are finite
But he is not doing continued fractions in the form of a_0 + 1/(a_1 + 1/(a_2 + ...)
same approach: 1=0+1=0+0+1=0+0+0+1=0+0+0+...=0
This one is my favorite answer! :-)
I hope Mathologer makes a video about the most interesting answers given here, instead of just commenting.
The elipses represent different values which propagate up the expression to produce different results.
Hey Mathologer ! Nice video, as usual. I was wondering if there's a "true definition" (conventional definition, rather) of "more irrational" ? If not, don't you think the definition we invent should be such that phi is less irrational than pi, since the latter is a transcendental number, whereas phi is one of the "most algebraic" numbers there are, so in this regard, phi is "more rational", than pi ?
Pi is both trascendental and irrational; if it's trascendental it needs to be irrational, but it doesn't mean the trascendentality makes it more irrational (trascendentality only speaks of our inability to express that number algebraically, this of course means we can't express it as a finite fraction, but it says nothing about how badly we can approximate it). If we extrapolate the definition of irrationality to be different depending on which irrational number you chose, you'd have the definition seen in the video: an irrational number m is more irrational than n if it takes more digits in the denominator to approximate the same number of digits in m as in n. This makes Φ the most irrational number (over any other irrational *and* trascendental number).
Also, here's an excerpt from the description: "If you are reasonably clued up mathwise have a look at the following VERY nice textbook chapter on infinite fractions by Professor Paul Loya from Binghampton University: [LINK] In particular, check out section 7.5.1. The mystery of π and good and best approximations. *I use the definition of "best rational approximation" given there.*"
Because 1 and 2 are rationals, so the diversion is finite..? - which means we can choose if to 2 or 1 at the end of it.
Is that the answer?
This argument only holds if the continued fraction only has 1's as numerators.
I miss the logic in it. divide or multiply to change it around.
Okay, (x-2)(x-0.5) = 0 -> x = 1/(2.5 - x) is true for x element of {2,1/2}
Some of the most educational videos are on this channel. What a wonderful way of making sometimes complex topics so accessible. Simply amazing.
The answer to the puzzle lies in the jump from finite to infinite.
You could replace that fraction 10^10^10^10^10 times (just to mention a staggering number) and it would still equal 1 or 2, depending on the innermost digit. But as soon as you jump to an infinite sequence, it becomes an irrational number, which cannot be either 1 or 2.
This reminds me of geometrical paradoxes. Take a square of side x and draw a stepped diagonal through it, like a staircase from one corner to its opposite. No matter how many steps-and how small-you draw, its length will always be 2x.
Now jump to infinite steps, each infinitely small, and you get the usual diagonal, whose length is, unsurprisingly, an irrational number: x√2
Because that's what irrational (computable) numbers are: they are "placeholders" for what you would get it you could run a given algorithm forever. π, e, √2 are all such objects.
In this sense irrational numbers don't really exist in the real world, because you can't run computations forever. They also have intrinsic fallacies. For example you can't always compare two algorithms for equality or ordering (it's undecidable in the general case) so it follows that equality and ordering aren't always defined for irrational numbers as well.
But wait, there's more. According to current orthodox math, 100% of "real" numbers are not only irrational, they aren't even computable. So you can't even give them a name or identify them, except in a few rare cases (such as Ω, defined over the halting problem, which is uncomputable.)
Conclusion: don't believe what math books tell you about numbers and only use fractions!
What a strange conclusion. Maybe have another look at the interlude ruclips.net/video/CaasbfdJdJg/видео.htmlm11s
Mathologer Yes, the conclusion was a bit of a hyperbole. But I'm not the only one to thing the current view of real numbers is nonsense.
See for example Prof. N. J. Wildberger here: njwildberger.com/2016/01/01/uncomputable-decimals-and-measure-theory-is-it-nonsense/
«This is modern pure mathematics going beyond parody. Future generations are going to shake their heads in disbelief that we happily swallowed this kind of thing without even a trace of resistance, or at least disbelief.»
Very smart guy and definitely a real mathematician. However, I think it is important to stress that he is really only speaking for himself when he criticises the foundations of mathematics and in particular the real numbers. He is a professor of mathematics and so am I and neither I nor any other professional mathematician I know agrees with him in this respect. Having said that, I find a lot of what he has to say very interesting and thought provoking. As well, his videos are a great resource for learning a lot of great mathematics. So, despite of me disagreeing with him in many way I consider him to be one of the "good guys" :)
Fair enough.
In any case I was wrong: the "infinite" fraction 2/(3 - 2/(3 - ...)) is nothing but a fancy way to write the equation x = 2/(3 - x) which has exactly two solutions, 1 and 2. Nothing irrational here.
Well, if we replace the "solutions" with x, we can find _a_ reason using algebra:
x=2/(3-x) → 3x-x^2=2 → -x^2+3x-2=0 → x^2-3x+2=0 → (x-2)(x-1)=0
Very clever
amazing!
x is 1 OR 2
so 1=2 is wrong
Exactly! It even "works" the same if you change the numbers:
(x-3)(x-4) → x^2-7x+12 → x=12/(7-x) ∴ 3=4 lol
Well, no... x = 1 or x = 2, in the context of the above equation. Period. No other implication.
Evan490BC but if you noticed anything past "numbers", you'd see that you can make any version of this "problem" with any numbers you want, like I did where I changed it from 1 and 2 to 3 and 4, making it ?=12/(7-?)
I want to write a report on this video can someone help me pls 🙂
Tera hone ke baad mujhe send kar
@@ajaymane3572 😂😂
What help you want , i can ...
Mereko bhi send kr
@@veronx2005 Write a report on this video and send me ASAP XD XD
I am unaware if you are still looking at the comments, but with the 1=2 madness in the beginning, we can use the partial sums like you say in the interlude. If we do this, we get 2/3, 6/7, 14/15, 30/31, 62/63, so on so forth. This seems to converge to 1, so I believe the final answer is 1
8:06 As an engineer, I knew it was pi the moment I saw "3", let alone 22/7.
7:33 "well that's only justified if we pin down what we mean for it to be equal at that point."
Can someone explain to me what this means
it's the argument wether a converging infinite Series/Sequence etc is really equal to the Number it converges to, basically if you can really treat that infinite series as a number itself.
recryed thank you
@@ryan2-518 sure mate
glad you upload it dude. appreciate it
:)
i am following this channel for quite sometime...
it is really fun to dig into this mystries of number...
Glad you like what I am doing here and thank you for saying so :)
+Mathologer sir, i would also like to know the prime number theorem of ramanujan of what proof he did... also the partition theorem .. please post a video on these topic sir...
eagerly waiting..
Well, there should be some more of Ramanujan's math in a video in the next couple of month, stay tuned :)
thank you sir... there was no direct proof of his work in the internet.. so, number theory lovers are finding it hard to know about this great man.. please do it sir..
thank you
Burkard Polster. I studied Philosophy and Math at University in the 70's and look at at least one math video every day. I just recently discovered your channel and have subscribed and started to put thumbs up on your shows (although I suspect this is the reverse of what the ancient Romans meant by it) 😁 I'll be looking in on you soon JIM
Very fluent and intuitive coverage of a topic which can be very algebraic and opaque. A lot of ground is covered in a very short talk.
x=2/(3-x)
x(3-x)=2
3x-x^2=2
0=x^2-3x+2
0=(x-2)(x-1)
x=2,1
x has 2 possible solutions
I just checked. The serie doesn t converge to 2, it converge to one. It goes like this 2/3, 6/7, 14/15, 30/31...
+Argentarii Homini that wasn't the method he should to approximate them infinite fractions. You included the final -1 or -2 which drastically changes the answer. By leaving out the last number, the limit is more meaningful in a way.
+Argentarii Homini 2/(3), 2/(3-2/(3)), 2(3-2/(3-2/(3)))...
he said the limit of the series of partial fraction must always converge to the number you say it is equal to. So, in this case, it does not converge to 2. I don't think this is relavent to speak about the last number in an infinite serie.
Yes exactly
+Argentarii Homini this is an infinite serie, there is no last digit or last part, or else it would mean this is finite.
the fraction is meaningless because it converges to 1, but since the fraction is infinite there's always going to be a 2 above the 1, which will result in a quotient of 2
i'm gonna stake my imaginary bets on this being right
hahahasan lol
Others seem to have said the same thing, but this is what I came up with.
So long as you terminate the infinite fractions with either a 1 or 2, the expression is valid. The problem is that you are treating the continued fraction of a rational number like that of an irrational number, allowing it to iterate to infinity. If you stop after a finite number of iterations, the identity still holds -- meaning that the problem is trying to continue to infinity.
I mean, I have no idea why exactly what you did is wrong (I don't see the flaw in the algebraic logic -- somebody please point it out!), but I just understand that the error lies in treating the rational number like it's irrational and continuing the fraction forever.
By the way, this is the first I've heard of this channel. This was probably one of the most enjoyable maths videos I've ever watched. Do you have a donation link somewhere?
It depends on what you end your fraction with. If 1, the whole thing equals 1, if 2, it equals 2. This answer does not feel satisfying because it does not seem to apply to the "infinitly long fraction", but it actually applies to the infinite case too!
Think about it: it depends on what is your procedure to generate the fraction. One procedure is to replace the 1 in the denominator with 2/(3-1), while the other is to replace the 2 in the denominator with 2/(3-2). Since you cannot hand me the infinite fraction, you must hand me part of it and then exactly define what the three dots mean. The three dots mean different things in the two cases, and that is: which procedure are you using?
This was essentially my answer, 4 years ago. The ellipsis hasn't been precisely defined. Therein hides the ambiguity.
Fred
1.189723917041748936076897214983267846937217489273489748923643264782367846327846327864278309187482197283987401374836487254309878901742484423416524556244122334564356613556443456455364455346155634453486542534156234896743896457485645600100061068748947789361802165234455045624548675864250256515464556135055634544356504356541436554543650346557897480289723270492384888888888882908490247802785692575278942524325035224897590801978403714134785432205354984713904870913725816347826575892132022346324562434684325...
1:50 That infinite fraction is convertible into a quadratic equation, whose roots/solutions are: x = 1 *_OR_* x = 2.
Namely: -x² + 3x - 2 = 0 *_OR_*
x² - 3x + 2 = 0
como demostraron que para phi la mejor aproximación es con 1's
Very informative video, but is there some way to define, determine, or identify the characteristics of whichever irrational number would be the LEAST irrational? I'd suspect the answer may rely on your definition, but I'd be curious if this has been thought out too.
This is effectively an iteration because you have constructed a chain of fractions, and you have to compute the result by starting at the bottom and working upwards.
If you start way down in the fraction, and enter x, the next level up in the fraction is 2/(3-x). Treat this as an iteration, i.e. x_new = 2/(3-x_old). It turns out that this iteration has two(-ish) behaviours.
BEHAVIOUR 1:- If you put in 1+delta, delta *HALVES* each time, so it will converge towards 1.
BEHAVIOUR 2a:- If you put in 2, it is stable... ... but...
BEHAVIOUR 2b:- If you put in 2+delta where delta is VERY SMALL, delta will *DOUBLE* on each extra level up the fraction, until eventually the number flips sign and it converges towards 1.
In other words, you have constructed an iteration which has two fixed points. One of those fixed points (i.e. 1) is stable and the other (i.e. 2) is unstable. Therefore the iteration (continued fraction) will converge towards 1, from almost any number at all, i.e. its radius of convergence is infinity. However, the fixed point 2 has a radius of convergence of ZERO, so 2 will remain 2, but anything slightly away from 2 will move farther away from 2 with each iteration.
In 4:53 can't you replace 3 by its continued fraction