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I knew it was the golden ratio as soon as I saw the terms of the fibonacci sequence being divided by each other 1=1, 2/1=2, 3/2=1.5, 5/3=1.666, 8/5=1.6, 13/8=1.625, etc.
Thanks, the code I have is starting to improve a little since I can reuse some of my older code and I am starting to find better ways to do things and I finally invested in a new mic. Good to see you liked the changes :D
Solution: Just set this fraction to x. x = 1 + 1/(1 + 1/(1 + 1/(...))) Then, since it is infinite, you can replace the denominator with x as well, as it is just the same repeated. x = 1 + 1/x Now multiply both sides with x x² = x + 1 Now move over x and 1 x² - x - 1 = 0 You can now use several methods to solve for x, but this is a very well known equation, that results in (1 ± √5)/2. But (1 - √5)/2 is negative and therefore can not be the result of the original term. Therefore the only result is (1 + √5)/2 aka φ.
@@Oilup-Tyler-Vitelli2 Not really. All I did, was replacing an infinite loop with a self-reference. If it didn't converge, the resulting equation would result in a contradiction or would not deliver a result. You only need to proof convergence, if you try to add or multiply infinite series. Otherwise you end up with things like the "golden nugget" result of 1+2+3+4+... = -1/12
Without watching the video, it is obvious that the given expression is equal to one plus the inverse of itself, a quadratic expression equivalent to x²-x-1=0. Its positive root is the golden ratio, (1+√5)/2.
Note that this is just the (positive) fixed point of 1/(1+x), at which x=f(x), and the whole thing is done before you're worried about subscripts or infinity or anything. This is the fact handwaved away at 14:20
It doesn't actually break. Even as a divergent series the concept holds true that 1=0 because when n=infinity adding 1 is equivalent to adding nothing. It's conceptually beautiful.
I remember when I was 18 having a question with a continued fraction in a math competition and I didn't know what they were. Was still able to solve it though but it took me time when it's actually quite simple in its simplest form. basically, you pose x = 1/ (1+x) Then you can see that 1 + x = 1 + 1/ (1+x) But the second part is also 1 + 1/(1 + 1/(1+x)) But this second part is also 1 + 1/ (1+ 1/(1+1/1+x))) Etc.. So reversely, the whole infinite fraction is just 1 + x So x = 1/ (1+x) Which give x (1+x) = 1 --> x²+x-1=0 So x = (-1 +- sqrt(5) )/2 But the answer has to be positive, so x = (-1 + sqrt(5) )/2 So the continued fraction's value is 1 + x = (1 + sqrt(5) ) /2 (The golden ratio) Just to be safe, we check that this doesn't give a division by 0 when you expand the whole thing back so it's ok.
Note that if you try to find the first n terms to explore the problem and see if it converges, you can basically define the n terms as you want. By this I mean that you can: - cut by removing the denominator which give 1 + 1 (/...) as a first term - or you can cut by removing the part behind the 1+ in the denominator giving 1 + 1/( 1+ ...) as a first term Both give in fact the same sequence in the long run (shifted by 1 position)
2:57 I would just claim that for the fraction to have a value, it has to converge. By definition of convergence, for any small €>0 there exists an n such that for any u(x) with x >n, we have k-€ < u(x) < k+€ where k is the value towards which the series of terms converges. So for any €>0 no matter how small, there also exist a n such that for any x > n we have u(x-1)-2€ < u(x) < u(x-1)+2€ So lim(x->inf) u(x)) = u(x-1) (I think that this could actually seen as an alternate definition for the convergence of a series) And lim (x->inf) u(x) = 1 / (1+ u(x)) (note: € is easier to type than epsilon)
I actually had to demonstrate how this fraction was equal to phi for a school presentation and I had no idea to start, your video showing up in my feed really saved me!
It makes sense that you can make infinite fractions for any irrational number by doing the method in reverse but can you do it for numbers like pi and e?
13:00 RE. Fixed point of cos x being Irrational, more can be proven. Serge Lang proves in his book Transcendental Numbers, that for non-zero x both x and sin x cannot be algebraic. Also true, if sin x is replaced by cos x, Exp(x) x is allowed to be a complex number as well
Assume convergence and set the continuing fraction equal to a. Due to the nested portion of the expression: a = 1 + 1/a, and multiplying by a yields the quadratic.
These continued fractions are actually easy if we just use our logical brain. If we call the equation 1+(1/1+(1/1+...) as x, then simply we have: x=1+(1/x) and we get the golden ratio, phi!
I am sorry it took so long to make lol but thanks actually teaching this topic to me. I hope you are doing well and I genuinely appreciate the comment. Have a nice day :)
I think its like when you get questions like the lengths or area of a shape being negative due to positive and negative solutions of quadratics but in reality - shapes need to have positive areas and lengths so we just neglect the negative solution. It's a good question though.
Before the video starts, you find this by solving y=1+1/y and after solving the quadratic you find y is phi, the golden ratio this was a great video, starting from nothing and methodically finding the solution. I have a couple things to note. 5:55 it shouldn't invalidate the rules of rationality also because assuming that the result is irrational, and taking into account the denominator equals the result (basically, with some extra things tacked on) that would mean the denominator is also irrational, and so we can see that the definition of rationals, that is being able to write the number as a ratio of whole numbers, does not hold for this infinite fraction, because it would instead be a whole number divided by an irrational number. 11:09 and this approaches a value that I'd bet doesn't have an exact form that be constructed using your regular functions, you'd need to create a more... special one. But it can be easily approximated, and it's about ~0.73909. 13:05 so, impossible to solve? It depends on what functions you allow yourself to write it's closed form solution with. It's similar to e and pi, yes, but at least both of them have a nice quickly converging series. And yeah the solution will be a unending, never repeating string of decimals, but that's not really special nor is it an issue, after all the first equation we solved had phi as the solution, which has that property as well, but you could solve that problem rather easily.
Hey that's actually some really useful feedback. I appreciate you explaining specifically what you thought could be improved since it really does help since I am still quite new to making videos and new to these concepts. I'm glad you like the video and I'll keep these in my mind if I make a video on a similar topic. Have a nice day.
In reasoning I always like to add the 'suppose the sequence converges, then we can assume lim n> inf u(n+1) = u(n)'. Then if we get a result like 0=1 in the example, we know our assumption was wrong
We can solve this without the limit, we can simply let the whole term = x and then the denominator is the same as x therefore it can be written as 1+1/x = x, now we can make quadratic in x and solve this
That is quite a nice way to do it. I used this method to show its applications elsewhere but that is probably better for this specific question actually.
2:49 only if u_n converges. If eg. u_n=(-1)^n, then u_∞≠u_(∞+1), because in general if u_n=1 then u_(n+1)=-1, and the other way around. So whatever u_∞ should equal u_∞=-u_(∞+1)
my solution: Let 'n' be equal to this expression. n= 1 + (1/(1+(1/..)) n= 1+(1/n) n=(n+1)/n n^2=n+1 n^2-n-1=0 Using quadratic equation, we get, n= (1+root(5))/2 OR (1-root(5))/2
Extremely close to the solution. All of it is correct except for the fact that we discount the negative solution since by inspection, we can see the fraction must be positive. It is similar to discounting negative areas and lengths.
I guess the video can be reduced to: If a sequence u_n converges, then lim u_n = lim u_{n+1} (this can be proven using the formal definition of the limit). Now if u_{n+1} = f(u_n) for a continuous function f and the sequence converges, so u = lim u_n exists, we have: u = lim u_n = lim u_{n+1} = lim f(u_n) = f(lim u_n) = f(u), so the limit has to be a fixed point of f.
Boss man has done it once again, brilliant video as usual. Thoughtful and engaging concepts discussed as well. Really like how everytjint is built from the ground up with a no wackadoodle nonsense approach. Much love xx
there is a better way to solve these: let's say 1+1/(1+1/(....))) =x Now, the part 1+(1/ *[ (1+1/(1+...)) ]* ) (indicated with square brackets) is the same as the whole expression So, 1+1/x=x =>1/x=x-1 =>(x-1)*x=1 =>x^2-x-1=0 now solving for x is easy.
Ok before I watch this video, I got both positive and negative solutions for the golden ratio by setting up the equation x = 1 + 1/x, which if that's the solution it's cool af
Aww you are so close. The solutions are the golden ratio but you need to discount the negative solution since you would need to notice that by inspection, there are no negative values in the fraction so the final value cannot be negative. It is similar to discounting values for the negative solutions when calculating lengths and areas. Still good job!
I really liked your video! Explanations are crisp, and the visuals are great. You've sparked my curiosity on finding integer roots with infinite fractions. Could you point me to any resources that would be of interest?
@briancooke4259 Hey I'm glad you liked the video :). I linked some resources in the description although there are some great videos from people like blackpenredpen too. If you want me to link some more, let me know but there are plenty of other great videos on the topic.
Very nice solution although the only issue is that there is no negative solution to this. You can tell by inspection the question only has positive values so the solution must be positive. Its similar to getting negative areas and lengths although aside from that, good job :)
@@academyofuselessideas Quite a while away but maybe?? Who knows? Spooky doesn't describe how cool that would be lol. It looks like you have a lot of videos of your own so you may beat me there xD
@@viks3864 RUclips subscriptions can be modeled by a preferential attachment process... Which is just a fancy way of saying that the jumps in subscriptions have the potential of being exponential... The downside is that the probability of making it big in youtube is very small... but 10K or 100K subscribers is far from out of the question for someone making good videos... Maybe, let's put a pin on this discussion and wait a few years to see how far you make it? P.S: The fact that I have made al of videos actually works as evidence against my channel (you are already closed to 1K subs with only 6 videos!)... But who knows... with preferential attachment processes, anyone can make it big at any moment! In any case, having a community of people creating math educational material makes the whole thing more interesting, and I hope that this also helps to the increment of mathematical knowledge (though to be fair, I am not doing any of this with any altruistic motives... i am here for the fame, the fans, the groupies, and the drugs 😉)
I went straight to sum=1+(1/sum) and solved for sum, choosing the positive root. But I also wondered about the negative root too and what that means! Its not immediately obvious why the negative solution can be discarded, the quadratic is valid so both roots should mean something?! It seems too informal to just hand-wave away the solution that we don't like!
It is similar to when we discard the negative solution when we solve quadratics for lengths or areas of shapes. I mean technically they could be but in reality - they don't exist so we just discard them. Same way in this solution - we generate an extra solution and it is obvious the extra solution is the negative one since there is no negative terms in the fraction although you do bring a good point and it does feel a little unnatural.
I haven't watched the video yet, but I think that we can say that this number is x, and then say that 1+1/x=x,so -x+1+1/x=0, and now multiply -x in both sides: x²-x-1. a=1 b=-1 c=-1 x={-(-1)+-sqrt[(-1)²-4×1×(-1)]}/2×1 x={1+-sqrt[1-(-4)]}/2 x=[1+-sqrt(5)]/2 x1=golden ratio x2=-1/golden ratio
I like that you showed all of your working although the only thing is that we can discount the negative solution since by inspection - this can't be negative. It's similar to when we get negative lengths or areas but good job!
Aww genuinely thanks. I do really appreciate comments like these and I'm glad you liked it. I'm working on the next video now and it hopefully should be out soon - I hope lol. Thanks again.
I think this is the solution: Let's set x = 1 + 1/... X = 1 + 1/x We multiply every thing by x X^2 = x + 1 X^2 -x - 1 = 0 And we solve the quadratic formula a = 1 b = -1 c = -1 Delta = b^2 + 4ac = -3 So there's no real solution but in the complex number: X = (1+√3i)/2 correction: i made a small mistake in the quadratic formula so the result should be : (1 + sqrt(5)i)/2
Oh nice working out but you made one small mistake - your coefficients for the quadratic are correct but it is "b^2 - 4ac" not "b^2 + 4ac" which is where the error occurred. Accounting for that error - you should get the correct answer although you did get incredibly close making only one small error so good job!
oh thanks for correcting me, i'm still a 9th grader and i didn't learn about the quadratic formula in school, just from some videos in youtube, i hope that i'm not going to make such a mistake in the future, anyways thanks again@@viks3864
GERARD!!!! I'm glad you watched the video lol and I'm more glad you liked it :). 1000 subs is still a little while away but maybe eventually?? Thanks for the support :D
The videographer makes this harder than it needs to be. This problem is actually easy, just observe that the solution, x, satisfies x = 1 + 1/x. This leads to x^2 - x - 1 = 0, solved by the quadratic equation.
I see how you can get to the result quite quickly doing that, and it's good that you noticed that, although I wanted to do a more rigorous proof to show its applications elsewhere.
Guys I'll tell you the easiest way. 1+1/(1+1/(1+1/...)). Let's call that x. It goes to infinity and if we cancel one, it won't be changed. So we can say that 1+1/x=x x+1=x² x²-x-1=0 D=(-1)²-4×1×(-1)=5. x=(-(-1)+√5)/2= (1+√5)/2. It can't be (1-√5)/2, because (1-√5)/2 is negative and the solution of that fraction can only be positive. So the answer is (1+√5)/2
I used to do this question at 12 years old my way is to make 1+1/1+1/1+…..= x if 1+1/1+1/1+…..= x then 1+1/x = x now I square both sides - - - > x +1=x^2 --> x^2 - x -1 =0 now I use the quadratics formula and get the answer
That is a really clever way of doing it and a way a lot of people do it. I mainly wanted to mention the formula which I used since it is more general for any converging series but your method is much faster and easier so feel free to use that and good job!
I was originally planning on mentioning it which is why phi is in the thumbnail. Last minute I removed that segment since it seemed to overcomplicate this topic especially for those who haven't seen it before although it's good general knowledge for you to recognise that without it being explicitly mentioned :)
I guess it’s the golden ratio(before watching the video) because it’s the ratio of 2 Fibonacci numbers each time and it’s to infinity so it approaches phi
Doing this before I watch the video let u be 1+1/1+1/……. notice that the fraction’s first denominator is still u u=1+1/u u^2=u+1 u^2-u-1=0 Quadratic equation, use the formula etc etc u=1+-sqrt(5)/2 notice that this is positive u=1+sqrt(5)/2
Yeah I got that too when I was reading into it. The issue is the solution isn't some multiple or square root of e or pi but it has its own value so it look incomplete. It would be similar to solving something like the basel problem as a decimal. Another viewer actually mentioned this number and the solution's constant have a name - 'Dotties Number' although it seemed very niche and had no symbol but was interesting nonetheless.
Sometime I just make an part of this, and at an moment say it's was the previous answer, technicly closer to this fraction everytime I activited my calculator, (; I also already trying to make an number become 0 with cos( and divisions and "prev" ( it just the previous answer) for some raison it worked, also reaching positive gogol, and negative gogol
Rule #1 of youtube math videos : solution is integer and obvious (well it's not the case here, then ...) Rule #2 of youtube math videos : when rule #1 does not apply, solution is golden ratio or related to.
Pretty good set of rules - its weird that a lot of these random maths problems have the golden ratio as a solution anyway. Enjoy the RUclips Maths binge.
I got 0.9998477415310881129598107686798 with repeated cos of a number. Starting with 1, it only took 10 times before the number stopped changing. That makes a lot more sense. That makes a lot more sense because the cos of 5 is .99619.... Form there, it is impossible to get to a lower number because the cos of any number below 1 is going to gravitate towards 1.
That's likely because my solution is correct for radians however yours in degrees which is something I probably should have mentioned in hindsight so sorry for the confusion but nice spot.
@@viks3864 Great, now how long before we stop this archaic foolishness of saying the Limit as n->∞? It serves absolutely no purpose except to waste time. If you are doing this kind of math, you understand this by just saying ∞. I mean really, why go through all that. Your video is a perfect example of yeah, you can't do it an infinite number of times, but if you could....Why pretend we don't know what that means?
@rogerahier4750 I get what you mean but it's just another point I wanted to get across - in a lot of areas of higher maths, especially pure maths - you take limits as expressions tend to something rather than equaling. For example h tends to zero for differentiation by first principles - not being pedantic but it's not zero since that would be illogical; it's the limit of h to zero and in this case, its good introduce a new concept for the those who have not heard it in an intuitive way and for those who do know it - they would lose 10 seconds.
The crazy thing is this question ( 1 + 1/(1 + 1/(1 + ...))) came in one of my 10th class maths questions for entrance to a certain coaching institute 💀
This was very interesting! I was wondering if there was a way to generalise the recurrence relation into a form that allows you to find the value for any n without having to recursively apply the function? This would then allow you to find the value for any n and then when you sub in infinity, you can tell if it diverges or not. For example, turning xn=xn-1 * 2 into xn = 2^n
Oh yeah, I see what you mean and it's something I haven't really considered. I'll have a look into it but give it a try yourself too and let me know if you find anything.
@ronakde6647 @viks3864 Depending on the form of the recursion you can find closed formulas. For linear recursions with constant coefficients, the problem is not too hard. That is, if your recursion is of the form u_n = a_1 u_{n-1} + a_2 u_{n-2} + ... a_m u_{n-m} + b... recursions like that depend on the first initial values (the starting point of your recursion). Engineers usually call those difference equations and they are like discrete analogous of differential equations... Perhaps the easier technique to solve those is using z-transforms but there are other ways of solving them.... The classic example is the formula for the n-th term of the Fibonacci sequence Now, perhaps a more interesting connection here is to link the continued fractions problem with a fixed point problem... when you have an equation of the form x=f(x), what you are attempting to do is to find a fixed point of the map f:R->R. That is a point that does not change when you apply the map... Now, you can ask some interesting questions there like, when do maps have at least one fixed point? what happens when you have several fixed points? and how can you find those fixed points?... In particular, you can ask about the so called stability of a fixed point. That is, assume that x is a solution to x=f(x), and take a small epsilon (because what other letter would you use for something small, right?), does f(x+epsilon) gets closer to f(x)? or does it start diverging? If the fixed point is stable then the algorithm of iterating to get to the fix point will work... otherwise, it will fail... You can do some plots and you kind of realize that the stability of fixed points depend on the slope of the function close to fixed point (and you might be able to calculate as well how fast the algorithm converges to the answer)... Fixed points turned out to be very useful in the practice of mathematics... Amusingly, the same idea of fixed points can be used to generalize the cantor diagonalization argument, and also godel incompleteness theorems! So, they appear on plenty of parts of mathematics!
@@academyofuselessideas Your understanding of the topics is a lot better than mine but that's quite interesting - it's cool to know there are more applications since I never really looked into it.
@@viks3864 The connection between fixed points and cantor diagonalization argument, godel incompleteness theorem, or even the halting problem are not obvious but they are there... All these are connections with logic... but if you want some connections in topology, you might want to check the famous Brouwer fixed point theorem (or some of the other fixed point theorems out there). Some cool consequences of the existence of fixed points are the hairy ball theorem, the determinacy of hex (hex is a famous two player game), the existence of nash equilibrium, the picard-lindelof theorem about the existence and uniqueness of solutions of differential equations, and the list goes on!
Oh that's crazy - I get a surprising number of facts like these in my comments of stuff I had no idea about and it helps me learn a lot of random stuff - Thanks :).
I love these channels. But in my experience at latest at 100k subs quality drops as they try to engage a larger audience and niche people who like it staying complex move on hoping to find a new small channel still going into detail. Nice video would love even more complex and convoluted subjects. Great animations. Im here to at least 100k might jump off if you go broad and less complex.
I'll try my best to keep the quality of content the same :). I'm going to keep experimenting with different formats to see what works but don't worry - I'll try as hard as I can to keep the actual quality the same.
@@viks3864 well your quality is phenomenal. Sure older vids have bad mics but your explanations are great and your visualisations complement your commentary quite well. I've not yet had a point where I had to go out of the video to look up what the hell you're talking about. Unlike the lectures of professors I have where you need to look up quite a bit sometimes. All in all keep it up. Don't drop in quality. Happy I found your channel. Have a nice day!
I got the solutions as + or - the golden ratio but then we discount the negative solution so it is just the positive value for the golden ratio. Your solutions very close but slightly out so consider rechecking your working out.
Yep :) Nice spot. I originally mentioned it alongside the fact that the solution is the golden ratio but the video was long enough as is although good general knowledge and nice spot.
@natanprzybylko7227 aww that's unfortunate. It was just one test though so don't worry. One of my friends got a C all year in chemistry and got an A for his finals and passed so I'm sure you will be fine lol. Good luck.
Quite a close guess actually but it's plus or minus the golden ratio which is around 1.6 although I'm guessing you know the method but may have made an arithmetic error.
End of video: For an infinite string of cosines, I got about 0.739085133215 I can’t exactly pin down how to simplify my answer, but I do note that it’s close to e^2/10
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Nice video bro
Math major from algeria
@amaniaridja2005 oh wow thanks, I'm actually considering doing maths myself too. Glad you enjoyed :)
@@viks3864the continued fraction, is simply (1 + √5)/2, which in turn is the golden ratio.
synce you like to scare people you could show how to iterate roots with the Babylonian method 😅
I knew it was the golden ratio as soon as I saw the terms of the fibonacci sequence being divided by each other
1=1, 2/1=2, 3/2=1.5, 5/3=1.666, 8/5=1.6, 13/8=1.625, etc.
That's actually quite good general knowledge since it's quite convenient to spot a trend like that so early on. Good job.
Same here, that 1 2 3 5 8 13 is just so recognizable and distinct.
same
does this explain why i have recurring nightmares about you benching more than me
That would never happen since you are too wham
yo stay small bozo🤘🏾
@@DhruvPatil364dhruv u are such a good boy, very strong, stronger than freddie
@@DhruvPatil364 You are too wham, I heard you actually beat freddiecoles738 in a fight.
@@srivasavipatil8436This seems like a different person with the same second name as you dhruv!
The animation quality looks way nicer and the mic quality has improved a lot too. Nice video!
Thanks, the code I have is starting to improve a little since I can reuse some of my older code and I am starting to find better ways to do things and I finally invested in a new mic. Good to see you liked the changes :D
I get paralysing headaches whenever i see maths like this but you acc made this make sense.❤️
How could maths give you a headache smh. Glad you liked it though :D
Solution:
Just set this fraction to x.
x = 1 + 1/(1 + 1/(1 + 1/(...)))
Then, since it is infinite, you can replace the denominator with x as well, as it is just the same repeated.
x = 1 + 1/x
Now multiply both sides with x
x² = x + 1
Now move over x and 1
x² - x - 1 = 0
You can now use several methods to solve for x, but this is a very well known equation, that results in (1 ± √5)/2.
But (1 - √5)/2 is negative and therefore can not be the result of the original term.
Therefore the only result is (1 + √5)/2 aka φ.
Perfect - nice working out.
Yes but you need to proof it converges
@@Oilup-Tyler-Vitelli2 Not really. All I did, was replacing an infinite loop with a self-reference. If it didn't converge, the resulting equation would result in a contradiction or would not deliver a result.
You only need to proof convergence, if you try to add or multiply infinite series. Otherwise you end up with things like the "golden nugget" result of 1+2+3+4+... = -1/12
Without watching the video, it is obvious that the given expression is equal to one plus the inverse of itself, a quadratic expression equivalent to x²-x-1=0. Its positive root is the golden ratio, (1+√5)/2.
Yeah good job - nice general knowledge.
Note that this is just the (positive) fixed point of 1/(1+x), at which x=f(x), and the whole thing is done before you're worried about subscripts or infinity or anything. This is the fact handwaved away at 14:20
That was my method of attack too. It's as good a way as any to reach the solution.
It doesn't actually break. Even as a divergent series the concept holds true that 1=0 because when n=infinity adding 1 is equivalent to adding nothing. It's conceptually beautiful.
It's like adding a single grain of salt to the ocean. What meaningful change did I make to the salinity?
@@atoms_dancing That's actually a really nice analogy to explain that idea.
I remember when I was 18 having a question with a continued fraction in a math competition and I didn't know what they were.
Was still able to solve it though but it took me time when it's actually quite simple in its simplest form.
basically, you pose x = 1/ (1+x)
Then you can see that 1 + x = 1 + 1/ (1+x)
But the second part is also 1 + 1/(1 + 1/(1+x))
But this second part is also 1 + 1/ (1+ 1/(1+1/1+x)))
Etc..
So reversely, the whole infinite fraction is just 1 + x
So x = 1/ (1+x)
Which give x (1+x) = 1 --> x²+x-1=0
So x = (-1 +- sqrt(5) )/2
But the answer has to be positive, so x = (-1 + sqrt(5) )/2
So the continued fraction's value is 1 + x = (1 + sqrt(5) ) /2 (The golden ratio)
Just to be safe, we check that this doesn't give a division by 0 when you expand the whole thing back so it's ok.
Note that if you try to find the first n terms to explore the problem and see if it converges, you can basically define the n terms as you want.
By this I mean that you can:
- cut by removing the denominator which give 1 + 1 (/...) as a first term
- or you can cut by removing the part behind the 1+ in the denominator giving 1 + 1/( 1+ ...) as a first term
Both give in fact the same sequence in the long run (shifted by 1 position)
2:57 I would just claim that for the fraction to have a value, it has to converge.
By definition of convergence, for any small €>0 there exists an n such that for any u(x) with x >n, we have k-€ < u(x) < k+€ where k is the value towards which the series of terms converges.
So for any €>0 no matter how small, there also exist a n such that for any x > n we have u(x-1)-2€ < u(x) < u(x-1)+2€
So lim(x->inf) u(x)) = u(x-1) (I think that this could actually seen as an alternate definition for the convergence of a series)
And lim (x->inf) u(x) = 1 / (1+ u(x))
(note: € is easier to type than epsilon)
This one is equal to 1.618… (The Golden Ratio). is so gold!
I actually had to demonstrate how this fraction was equal to phi for a school presentation and I had no idea to start, your video showing up in my feed really saved me!
Wow that's good to hear - glad to help :D
It makes sense that you can make infinite fractions for any irrational number by doing the method in reverse but can you do it for numbers like pi and e?
Yep, its actually discussed in the extra reading link in the description but this works for numbers beyond just square root based irrationals.
Yep I believe you have to replace the first 1 with 3 and the ones before a + with 6 or smth like that
Great video! I’ve seen this problem before and you presented it extremely clearly. Glad to have another math channel to add to the list
Hey thanks :D. I'm glad you liked the video and im working on the next one now so it should be out in the next week give or take 6 months
13:00 RE. Fixed point of cos x being Irrational, more can be proven.
Serge Lang proves in his book Transcendental Numbers, that for non-zero x both x and sin x cannot be algebraic.
Also true, if sin x is replaced by cos x, Exp(x)
x is allowed to be a complex number as well
Assume convergence and set the continuing fraction equal to a. Due to the nested portion of the expression: a = 1 + 1/a, and multiplying by a yields the quadratic.
Yep nice :D
These continued fractions are actually easy if we just use our logical brain.
If we call the equation 1+(1/1+(1/1+...) as x, then simply we have:
x=1+(1/x) and we get the golden ratio, phi!
Great video (as always), Vik!!! I'm very glad the problem I gave you inspired you and I hope to keep learning from your channel myself ;). Best, C.
I am sorry it took so long to make lol but thanks actually teaching this topic to me. I hope you are doing well and I genuinely appreciate the comment. Have a nice day :)
I hadn't given this any thought, but since this series is run to infinity, it must hold that x - 1 = 1/x or x^2 - x - 1 = 0 or x = (1 + √5)/2
I think you just saved my high school maths essay. You're a godsend.
As a person who also did (and probably still does) essays the night before deadlines - I am happy this helped.
Wait why can’t we take the - solution, it doesn’t make sense but all the steps we did to get it still should work
I think its like when you get questions like the lengths or area of a shape being negative due to positive and negative solutions of quadratics but in reality - shapes need to have positive areas and lengths so we just neglect the negative solution. It's a good question though.
@@viks3864 alright thanks
Before the video starts, you find this by solving y=1+1/y and after solving the quadratic you find y is phi, the golden ratio
this was a great video, starting from nothing and methodically finding the solution. I have a couple things to note.
5:55 it shouldn't invalidate the rules of rationality also because assuming that the result is irrational, and taking into account the denominator equals the result (basically, with some extra things tacked on) that would mean the denominator is also irrational, and so we can see that the definition of rationals, that is being able to write the number as a ratio of whole numbers, does not hold for this infinite fraction, because it would instead be a whole number divided by an irrational number.
11:09 and this approaches a value that I'd bet doesn't have an exact form that be constructed using your regular functions, you'd need to create a more... special one. But it can be easily approximated, and it's about ~0.73909.
13:05 so, impossible to solve? It depends on what functions you allow yourself to write it's closed form solution with. It's similar to e and pi, yes, but at least both of them have a nice quickly converging series. And yeah the solution will be a unending, never repeating string of decimals, but that's not really special nor is it an issue, after all the first equation we solved had phi as the solution, which has that property as well, but you could solve that problem rather easily.
Hey that's actually some really useful feedback. I appreciate you explaining specifically what you thought could be improved since it really does help since I am still quite new to making videos and new to these concepts. I'm glad you like the video and I'll keep these in my mind if I make a video on a similar topic. Have a nice day.
In reasoning I always like to add the 'suppose the sequence converges, then we can assume lim n> inf u(n+1) = u(n)'. Then if we get a result like 0=1 in the example, we know our assumption was wrong
Nice video! I liked the applications of this
Thanks lad :)
Audio quality 🔥. Nice vid bossman
Always harishan, enjoy hogwarts
We can solve this without the limit, we can simply let the whole term = x and then the denominator is the same as x therefore it can be written as 1+1/x = x, now we can make quadratic in x and solve this
That is quite a nice way to do it. I used this method to show its applications elsewhere but that is probably better for this specific question actually.
yesssss @@pokejinwwi
so theoretically I can infinitely pull? but also not...
you already do boss man
@@viks3864 yessirr
2:49 only if u_n converges. If eg. u_n=(-1)^n, then u_∞≠u_(∞+1), because in general if u_n=1 then u_(n+1)=-1, and the other way around. So whatever u_∞ should equal u_∞=-u_(∞+1)
Nice - I do mention this slightly later in the video but nice spot.
my solution:
Let 'n' be equal to this expression.
n= 1 + (1/(1+(1/..))
n= 1+(1/n)
n=(n+1)/n
n^2=n+1
n^2-n-1=0
Using quadratic equation, we get,
n= (1+root(5))/2 OR (1-root(5))/2
Extremely close to the solution. All of it is correct except for the fact that we discount the negative solution since by inspection, we can see the fraction must be positive. It is similar to discounting negative areas and lengths.
Yay for induction proofs! 🎉
🎉🎉🎉
I guess the video can be reduced to:
If a sequence u_n converges, then lim u_n = lim u_{n+1} (this can be proven using the formal definition of the limit).
Now if u_{n+1} = f(u_n) for a continuous function f and the sequence converges, so u = lim u_n exists, we have:
u = lim u_n = lim u_{n+1} = lim f(u_n) = f(lim u_n) = f(u), so the limit has to be a fixed point of f.
Pretty concise summary - thanks :).
Boss man has done it once again, brilliant video as usual. Thoughtful and engaging concepts discussed as well. Really like how everytjint is built from the ground up with a no wackadoodle nonsense approach. Much love xx
Wow Mr Aiyush himself, thanks for all the help :)
6:22 TBH the phrase “exact approximation” sounds contradictory 😂
I actually thought that as I was writing it down but hoped nobody would pick up on it lol but fair enough - you proved me wrong.
there is a better way to solve these:
let's say
1+1/(1+1/(....))) =x
Now, the part 1+(1/ *[ (1+1/(1+...)) ]* ) (indicated with square brackets) is the same as the whole expression
So, 1+1/x=x
=>1/x=x-1
=>(x-1)*x=1
=>x^2-x-1=0
now solving for x is easy.
Yep perfect - nice solution and working out.
Ok before I watch this video, I got both positive and negative solutions for the golden ratio by setting up the equation x = 1 + 1/x, which if that's the solution it's cool af
Aww you are so close. The solutions are the golden ratio but you need to discount the negative solution since you would need to notice that by inspection, there are no negative values in the fraction so the final value cannot be negative. It is similar to discounting values for the negative solutions when calculating lengths and areas. Still good job!
@@viks3864 oh nice, I didn't actually factor it all out, just recognized the quadratic. But yea, i guess the negative one would be complex
I really liked your video! Explanations are crisp, and the visuals are great. You've sparked my curiosity on finding integer roots with infinite fractions. Could you point me to any resources that would be of interest?
@briancooke4259 Hey I'm glad you liked the video :). I linked some resources in the description although there are some great videos from people like blackpenredpen too. If you want me to link some more, let me know but there are plenty of other great videos on the topic.
@viks3864 thank you so much! Blackpenredpen had a nice video on continued fractions that did just that! I can now do square roots by hand, yay!
i saw that whatever the result is, 1+1/x=x and something something quadratic formula its (phi,1-phi)
Very nice solution although the only issue is that there is no negative solution to this. You can tell by inspection the question only has positive values so the solution must be positive. Its similar to getting negative areas and lengths although aside from that, good job :)
@@viks3864 interesting! thank you
Road to 1000 subs soon come 👀
Actually may happen... Spooky
@@viks3864 hahaha... of course it will happen... will it be more spooky when you get to 10K, 100K, 1M?
@@academyofuselessideas Quite a while away but maybe?? Who knows? Spooky doesn't describe how cool that would be lol. It looks like you have a lot of videos of your own so you may beat me there xD
@@viks3864 RUclips subscriptions can be modeled by a preferential attachment process... Which is just a fancy way of saying that the jumps in subscriptions have the potential of being exponential... The downside is that the probability of making it big in youtube is very small... but 10K or 100K subscribers is far from out of the question for someone making good videos... Maybe, let's put a pin on this discussion and wait a few years to see how far you make it?
P.S: The fact that I have made al of videos actually works as evidence against my channel (you are already closed to 1K subs with only 6 videos!)... But who knows... with preferential attachment processes, anyone can make it big at any moment! In any case, having a community of people creating math educational material makes the whole thing more interesting, and I hope that this also helps to the increment of mathematical knowledge (though to be fair, I am not doing any of this with any altruistic motives... i am here for the fame, the fans, the groupies, and the drugs 😉)
I went straight to sum=1+(1/sum) and solved for sum, choosing the positive root. But I also wondered about the negative root too and what that means! Its not immediately obvious why the negative solution can be discarded, the quadratic is valid so both roots should mean something?! It seems too informal to just hand-wave away the solution that we don't like!
It is similar to when we discard the negative solution when we solve quadratics for lengths or areas of shapes. I mean technically they could be but in reality - they don't exist so we just discard them. Same way in this solution - we generate an extra solution and it is obvious the extra solution is the negative one since there is no negative terms in the fraction although you do bring a good point and it does feel a little unnatural.
2:46 Thats not always true tho, you should use something like ratio test or generating series or something like that
Yeah it is a slight oversimplification although in a lot of generalised cases, it does suffice.
I haven't watched the video yet, but I think that we can say that this number is x, and then say that
1+1/x=x,so -x+1+1/x=0, and now multiply -x in both sides: x²-x-1.
a=1 b=-1 c=-1
x={-(-1)+-sqrt[(-1)²-4×1×(-1)]}/2×1
x={1+-sqrt[1-(-4)]}/2
x=[1+-sqrt(5)]/2
x1=golden ratio
x2=-1/golden ratio
Now I watched the video, and I forgot that x can't be negative on this problem, so the only solution is x=golden ratio
I like that you showed all of your working although the only thing is that we can discount the negative solution since by inspection - this can't be negative. It's similar to when we get negative lengths or areas but good job!
@@brothersbirolo3223Oh looks like you noticed this yourself but still good job on all the working out.
Incidentally, in degrees, repeating cos of any number is practically 1.0.
Rewatching this video, still so amazing. Theoretical and convergent math is the best math imo; so fascinating and counterintuitive sometimes.
Aww genuinely thanks. I do really appreciate comments like these and I'm glad you liked it. I'm working on the next video now and it hopefully should be out soon - I hope lol. Thanks again.
I think this is the solution:
Let's set x = 1 + 1/...
X = 1 + 1/x
We multiply every thing by x
X^2 = x + 1
X^2 -x - 1 = 0
And we solve the quadratic formula
a = 1 b = -1 c = -1
Delta = b^2 + 4ac = -3
So there's no real solution but in the complex number:
X = (1+√3i)/2
correction: i made a small mistake in the quadratic formula so the result should be : (1 + sqrt(5)i)/2
Oh nice working out but you made one small mistake - your coefficients for the quadratic are correct but it is "b^2 - 4ac" not "b^2 + 4ac" which is where the error occurred. Accounting for that error - you should get the correct answer although you did get incredibly close making only one small error so good job!
oh thanks for correcting me, i'm still a 9th grader and i didn't learn about the quadratic formula in school, just from some videos in youtube, i hope that i'm not going to make such a mistake in the future, anyways thanks again@@viks3864
Nice video!! I love the animations so much. Let's go for the 1000 subs🔥
GERARD!!!! I'm glad you watched the video lol and I'm more glad you liked it :). 1000 subs is still a little while away but maybe eventually?? Thanks for the support :D
@@viks3864 1000 subs was a little while away right? hahahah you should upload more vidios they are amazing🥵
The videographer makes this harder than it needs to be. This problem is actually easy, just observe that the solution, x, satisfies x = 1 + 1/x. This leads to x^2 - x - 1 = 0, solved by the quadratic equation.
I see how you can get to the result quite quickly doing that, and it's good that you noticed that, although I wanted to do a more rigorous proof to show its applications elsewhere.
Guys I'll tell you the easiest way. 1+1/(1+1/(1+1/...)). Let's call that x. It goes to infinity and if we cancel one, it won't be changed. So we can say that 1+1/x=x x+1=x² x²-x-1=0 D=(-1)²-4×1×(-1)=5. x=(-(-1)+√5)/2= (1+√5)/2. It can't be (1-√5)/2, because (1-√5)/2 is negative and the solution of that fraction can only be positive. So the answer is (1+√5)/2
Very nice solution, good job!
I used to do this question at 12 years old my way is to make 1+1/1+1/1+…..= x if 1+1/1+1/1+…..= x then 1+1/x = x now I square both sides - - - > x +1=x^2 --> x^2 - x -1 =0 now I use the quadratics formula and get the answer
Sorry if my English was hard to understand , I’m not a native speaker
That is a really clever way of doing it and a way a lot of people do it. I mainly wanted to mention the formula which I used since it is more general for any converging series but your method is much faster and easier so feel free to use that and good job!
No worries - your English is actually very clear and better than some native speakers I know :)@@oxygenwasters8590
Derivative of cos x is bounded by 1 in absolute value. Thus the fixed point iteration conveges.
Yeah exactly - good job.
Really good video, this feels like it should have 2,000,00 instead of 20,000 views
Honestly 20000 feels like a bit too much lol but thanks :)
ans is φ and -1/φ, but for x>0 x= φ
Nice one - good general knowledge.
Great informative video. I didn’t see you mention that the solution is phi, the golden ratio.
I was originally planning on mentioning it which is why phi is in the thumbnail. Last minute I removed that segment since it seemed to overcomplicate this topic especially for those who haven't seen it before although it's good general knowledge for you to recognise that without it being explicitly mentioned :)
Let the fraction be x.
Then 1+x = 1/x
x^2 + x - 1 = 0
So x is the Golden Ratio, and this expression is the Golden Ratio plus 1.
I guess it’s the golden ratio(before watching the video) because it’s the ratio of 2 Fibonacci numbers each time and it’s to infinity so it approaches phi
Nice general knowledge and good guess. It's useful to know things like this in general so good job.
Let x be the fraction and then we have x=1+1/x, or x^2-x-1=0, which has a positive root of the golden ratio.
Bro this is lit the golden ratio
Yep nice - I was going to mention it but I thought it was slightly off topic and didn't want to scare people who hadn't seen it before.
Doing this before I watch the video
let u be 1+1/1+1/…….
notice that the fraction’s first denominator is still u
u=1+1/u
u^2=u+1
u^2-u-1=0
Quadratic equation, use the formula etc etc
u=1+-sqrt(5)/2
notice that this is positive
u=1+sqrt(5)/2
Very nice working out and perfect solution - good job.
give this man a pay rise
ong he’s actually so good, my teacher uses him sometimes in class
Wow the world renowned UTSAV thinks I should get a pay rise. I wish. Dan Lally deserves a pay raise more than I do lets be honest.
@@dpatil00017WOW ITS DHRUV, I heard you smoked FreddieColes738 in a fight??
14:16 seems rather unsatisfying.
Yeah I got that too when I was reading into it. The issue is the solution isn't some multiple or square root of e or pi but it has its own value so it look incomplete. It would be similar to solving something like the basel problem as a decimal. Another viewer actually mentioned this number and the solution's constant have a name - 'Dotties Number' although it seemed very niche and had no symbol but was interesting nonetheless.
@@viks3864Ah, so it does have a name. Interesting.
Maybe it isn't even as unsatisfying as I thought though. Saying x is about 0.7391 is fine lol.
Sometime I just make an part of this, and at an moment say it's was the previous answer, technicly closer to this fraction everytime I activited my calculator, (;
I also already trying to make an number become 0 with cos( and divisions and "prev" ( it just the previous answer) for some raison it worked, also reaching positive gogol, and negative gogol
1+1/x=x, therefore x= (1±sqrt(5))/2, which is the golden ratio. If you have the right tools you can solve it in an instant.
Rule #1 of youtube math videos : solution is integer and obvious (well it's not the case here, then ...)
Rule #2 of youtube math videos : when rule #1 does not apply, solution is golden ratio or related to.
Pretty good set of rules - its weird that a lot of these random maths problems have the golden ratio as a solution anyway. Enjoy the RUclips Maths binge.
I got 0.9998477415310881129598107686798 with repeated cos of a number. Starting with 1, it only took 10 times before the number stopped changing. That makes a lot more sense. That makes a lot more sense because the cos of 5 is .99619.... Form there, it is impossible to get to a lower number because the cos of any number below 1 is going to gravitate towards 1.
That's likely because my solution is correct for radians however yours in degrees which is something I probably should have mentioned in hindsight so sorry for the confusion but nice spot.
@@viks3864 Great, now how long before we stop this archaic foolishness of saying the Limit as n->∞? It serves absolutely no purpose except to waste time. If you are doing this kind of math, you understand this by just saying ∞. I mean really, why go through all that. Your video is a perfect example of yeah, you can't do it an infinite number of times, but if you could....Why pretend we don't know what that means?
@rogerahier4750 I get what you mean but it's just another point I wanted to get across - in a lot of areas of higher maths, especially pure maths - you take limits as expressions tend to something rather than equaling. For example h tends to zero for differentiation by first principles - not being pedantic but it's not zero since that would be illogical; it's the limit of h to zero and in this case, its good introduce a new concept for the those who have not heard it in an intuitive way and for those who do know it - they would lose 10 seconds.
"golden ratio!" --me, blurting out, at 0:13
Crazy that you got the solution in 13 seconds - good job. Either good general knowledge or insane mental maths.
The crazy thing is this question ( 1 + 1/(1 + 1/(1 + ...))) came in one of my 10th class maths questions for entrance to a certain coaching institute 💀
Oh crazy - hope you managed to solve it during your exam and if not, I hope this was a good explanation on how to solve it :D
@@viks3864 Well unfortunately I couldn't solve it...
But that led me to searching for this video so now I know :)
What a beautiful explanation :)
Thanks, glad you liked it :)
It's very roughly 1 + 1/2 which is of course 1.5. Taking into account further terms in the series, I'd guess the answer is sqrt (2) ....1.414
Good guess with logical reasoning although unfortunately the answer is something else.
Amazing video!
Hey thanks :D I'm glad you enjoyed it
phi, -1/phi are the solutions to x-1-1/x=0 at least. I think -1/phi is an unstable equilibrium whereas phi is stable
y_n would always be >= 1 and so we take phi instead of the other solution since -1/phi is negative.
S=1+1/S -> golden ratio by definition. done!
Quite an efficient proof lol - good job.
It would be fun to explain this through non linear dynamics with cosx as your function trying to find the stable term
My maths is not very advanced lmao, can you explain what that is?
@@viks3864 I dont think i can explain that on a yt comment 😅 anywhere i can reach you and send you some notes on that?
@manolismiliaras4357 Erm you could drop me an email on my contact email address but honestly if its a lot of work don't worry :D.
Really good vid! Keep it up
Hey thanks :) Glad you enjoyed.
This was very interesting! I was wondering if there was a way to generalise the recurrence relation into a form that allows you to find the value for any n without having to recursively apply the function?
This would then allow you to find the value for any n and then when you sub in infinity, you can tell if it diverges or not. For example, turning xn=xn-1 * 2 into xn = 2^n
Oh yeah, I see what you mean and it's something I haven't really considered. I'll have a look into it but give it a try yourself too and let me know if you find anything.
@ronakde6647 @viks3864 Depending on the form of the recursion you can find closed formulas. For linear recursions with constant coefficients, the problem is not too hard. That is, if your recursion is of the form u_n = a_1 u_{n-1} + a_2 u_{n-2} + ... a_m u_{n-m} + b... recursions like that depend on the first initial values (the starting point of your recursion). Engineers usually call those difference equations and they are like discrete analogous of differential equations... Perhaps the easier technique to solve those is using z-transforms but there are other ways of solving them.... The classic example is the formula for the n-th term of the Fibonacci sequence
Now, perhaps a more interesting connection here is to link the continued fractions problem with a fixed point problem... when you have an equation of the form x=f(x), what you are attempting to do is to find a fixed point of the map f:R->R. That is a point that does not change when you apply the map... Now, you can ask some interesting questions there like, when do maps have at least one fixed point? what happens when you have several fixed points? and how can you find those fixed points?... In particular, you can ask about the so called stability of a fixed point. That is, assume that x is a solution to x=f(x), and take a small epsilon (because what other letter would you use for something small, right?), does f(x+epsilon) gets closer to f(x)? or does it start diverging? If the fixed point is stable then the algorithm of iterating to get to the fix point will work... otherwise, it will fail... You can do some plots and you kind of realize that the stability of fixed points depend on the slope of the function close to fixed point (and you might be able to calculate as well how fast the algorithm converges to the answer)...
Fixed points turned out to be very useful in the practice of mathematics... Amusingly, the same idea of fixed points can be used to generalize the cantor diagonalization argument, and also godel incompleteness theorems! So, they appear on plenty of parts of mathematics!
@@academyofuselessideas Your understanding of the topics is a lot better than mine but that's quite interesting - it's cool to know there are more applications since I never really looked into it.
@@viks3864 The connection between fixed points and cantor diagonalization argument, godel incompleteness theorem, or even the halting problem are not obvious but they are there... All these are connections with logic... but if you want some connections in topology, you might want to check the famous Brouwer fixed point theorem (or some of the other fixed point theorems out there). Some cool consequences of the existence of fixed points are the hairy ball theorem, the determinacy of hex (hex is a famous two player game), the existence of nash equilibrium, the picard-lindelof theorem about the existence and uniqueness of solutions of differential equations, and the list goes on!
As soon as I read the title I answered "Golden ratio"
underrated +here before viral
lmao thanks xD. You may have some slightly high hopes but I appreciate it lol.
Me too
@johncorn7905 Lmao if this video gets more than 10k views I'd be surprised.
Nice content. Not sure whether 'trend' is the correct word to describe the pattern of the first few terms.
Thanks - yeah my scripting skills could probably use some improving lol.
Is prove Un=Un+1 when n->inf hard?
The way I explained it in this video is quite an oversimplification but the actual proof isn't too bad and you can find it online if you are curious.
Lim means limes it comes from word used for the roman border
Oh that's crazy - I get a surprising number of facts like these in my comments of stuff I had no idea about and it helps me learn a lot of random stuff - Thanks :).
RUclips dropped dropped an ad in the middle of the sponser's segment... ad-ception.
Lmao I didn't even know that could happen lol
excellent commentary my friend
This was an especially difficult topic for me to wrap around, your vdieo has been a blessing for me :)
@@GeorgeLo-sd6vh Wow what an honour George!
I love these channels. But in my experience at latest at 100k subs quality drops as they try to engage a larger audience and niche people who like it staying complex move on hoping to find a new small channel still going into detail.
Nice video would love even more complex and convoluted subjects. Great animations. Im here to at least 100k might jump off if you go broad and less complex.
I'll try my best to keep the quality of content the same :). I'm going to keep experimenting with different formats to see what works but don't worry - I'll try as hard as I can to keep the actual quality the same.
@@viks3864 well your quality is phenomenal. Sure older vids have bad mics but your explanations are great and your visualisations complement your commentary quite well.
I've not yet had a point where I had to go out of the video to look up what the hell you're talking about.
Unlike the lectures of professors I have where you need to look up quite a bit sometimes.
All in all keep it up. Don't drop in quality. Happy I found your channel.
Have a nice day!
@@YoucanatmeAww genuinely thanks - I'm working on the next one now so I hope it doesn't disappoint.
I never knew I wasn’t the only one who says “as n _tends to_ ♾️”. I always thought I was.
:) I'm glad you found somebody else who is unnecessarily pedantic lol
this is an awesome video
Aw thanks, I'm glad you liked it.
This series leads to Fibonacci numbers which is approximately 0.618 and 1.618 i.e. : x - 1 = 1 / x
I got the solutions as + or - the golden ratio but then we discount the negative solution so it is just the positive value for the golden ratio. Your solutions very close but slightly out so consider rechecking your working out.
0:45 Fibonacci Number omg
Yep :) Nice spot. I originally mentioned it alongside the fact that the solution is the golden ratio but the video was long enough as is although good general knowledge and nice spot.
The golden Ratio
Nice job - good general knowledge :D
La solución de cos(x)=x recibe el nombre de "número de Dottie"
Oh yeah I have never seen that before - it's interesting that it actually has a name, thanks.
it’s the golden ratio, ain’t it
Yep very nice - good general knowledge!
good job
Thanks! Glad you enjoyed the video :D
1 + 1/x = x
x^2 - x - 1 = 0
4x^2 - 4x - 4 = 0
(2x - 1)^2 = 5
2x - 1 = √5
x = (1 + √5)/2
Doing continued fractions for my schools math team in a week or so, instantly clicked on this lol
Lol convenient timing - good luck with the test or competition :D
@@viks3864 Ended up doing pretty bad but the next one is logic & logic gates so maybe I'll do better on that one lol
@natanprzybylko7227 aww that's unfortunate. It was just one test though so don't worry. One of my friends got a C all year in chemistry and got an A for his finals and passed so I'm sure you will be fine lol. Good luck.
Oh golden!
It is the golden ratio :). Weird how it and 'e' seem to show up everywhere.
A number that equals one plus it's reciprocal? Isn't that the golden ratio?
Yep - nice general knowledge.
k=given formula
k=1+1/k
k^2-k-1=0
k>0
k=(1+rt5)/2
Answer is plus or minus sqrt(2)
Quite a close guess actually but it's plus or minus the golden ratio which is around 1.6 although I'm guessing you know the method but may have made an arithmetic error.
i found x^2 - x = 1
this gives x^2 - 1 = x
and i got in mind that phi^2 = phi + 1
Yep perfect - phi is the correct solution and good job noticing that we can use phi rather than the more complicated expression for phi.
End of video: For an infinite string of cosines, I got about 0.739085133215
I can’t exactly pin down how to simplify my answer, but I do note that it’s close to e^2/10
yes i can. i think it's the golden ratio, from memory.
Yep - nice general knowledge :).
I've got Golden ratio after a quick thought.
Nice - Good general knowledge :).