Hidden Fibonacci and Golden Ratio: MIT Integration Bee (19)

[@speedspeed121]

Honestly, every math and physics student should love watching these videos. Although they might be too difficult for us to answer, the enjoyment comes with learning, obviously, but the competitive spirit he has when he has a challenge that needs to be overcome.

[@69erthx1138]

It's just the yearning for his students to teach him.

[@gold9994]

it has some easy way though, he explains it rigorously, which is useless if being implemented to the bee, but it's always good to know about it

[@ryanlian7656]

Couldnt you do y=that infinite thing , 1/1+y=y and solve for y, which becomes a constant(that number)

[@LetsSolveMathProblems]

That is an excellent approach also! I probably should have mentioned your standard technique of evaluating continued fraction, but I wanted to illustrate the fundamental reason for why the answer came to be one over the golden ratio. Thank you for sharing your solution! =)

[@alexanderterry187]

Pretty much what I did. Differentiate both sides so you get dI/dx = 1/(1+dI/dx) and go from there.

[@ekadria-bo4962]

so its mean we can evaluated any continued fraction with recursion?

[@Vivenk88]

This is what I was about to comment too.

[@guillaumebignon7004]

You have to prove the convergence first to do this. The method in the video is more rigorous

[@zeeshanmehmood4522]

I saw the golden ratio in the continued fraction, but I didn't think to simplify the fraction to find the actual Fibonacci sequence. This is why you're videos are so great, you always find something new and cool to show that I wouldn't think of on my own.

[@rajshekhardubey8371]

dear respected sir. I have no words for this great explanation. You are doing really a great job. You are brilliant. Such a deep analysis of such a tough mathematical calculus problem. Students referring to your video will definitely fall in love with calculus sir. May God give you long life to keep on spreading this great knowledge throughout the world. Really I respect you sir.

[@LetsSolveMathProblems]

I appreciate your compliment, although I admit I still have much to learn and improve on. I am glad you enjoyed the video! =)

[@mohammadalkousa2856]

Recently it was published a book about MIT integration bee, under the title " MIT Integration Bee, Solutions of Qualifying Tests from 2010 to 2023" You can simply find it!

[@Czeckie]

the discussion why you can interchange limit and integral would be interesting. Maybe next (more theoretical) video?

[@anonymous10567]

You can apply the dominated convergence theorem (with g = 1 on [0, 1]) - this seems like something an integration-bee taker could reasonably be expected to know.

[@LetsSolveMathProblems]

I am not an expert in analysis, so I will not make a video on the formal proof (at least anytime soon). However, I will provide a quick summary of a possible approach in this comment. As anonymous10567 mentioned, you may utilize the dominated convergence theorem, which, when its constraints are satisfied, verify the equality between limit of integral of a sequence and the integral of the function the sequence pointwise converges to. To do so, a few steps are required. First, we have to show that the sequence of continued fraction pointwise converges to 1/(golden ratio)--that is, the limit at infinity of the sequence is 1/(golden ratio). This can be done by proving that the limit of successive terms in fibonacci sequence exists (perhaps using alternating series representation of the ratio), then following the steps in this video to prove the limit as golden ratio. Finally, we have to find a function g that is always greater than or equal to the magnitude of each term of the sequence: I believe that g = 1 (as anonymous10567 mentioned) does the trick due to our fibonacci sequence increasing in magnitude (except for the case x = 0: in that case, we have 1, 1, 1/2, 2/3 ..., but the ratio of one term and the immediate next term is still always less than or equal to 1). Unless I am mistaken, the preceding steps (if done formally) prove the validity of the "swapping of limit and integral" by the dominated convergence theorem. Hopefully this helps.

[@h-a-y-k4149]

@@LetsSolveMathProblems I have thought this, even though, I think, I might be wrong. So as the sequence converges to 1/phi then we can rewrite the integral as integral from 0 to 1 (1/phi + a_n)dx where a_n is an infinitely small sequence. On the other hand we can rewrite this using the integral definition. lim n -> infinity sum k = 1 to n ( 1/n (1/phi + a_n)) = lim n -> inf (1/phi + a_n) = 1/phi.

[@arthurgames9610]

It was easy. I done this in 2 minutes. You have to only percive that, when n➡️♾, In goes to an integral of an constant, who is 1/(1+(1/1+(1/...))). And, when you solve for 1+(1/1+(1/...)), you find that its fi, so its the integral of 0 to 1 of (1/fi), that's obviusly equals to fi Ps.: sorry if my inglish isnt very good, its because im Braziliam.

[@phychemat5034]

Atleast spell ur nation correct 🤦‍♂️

[@angelmendez-rivera351]

Notice that 1/[1 + 1/(1 + SqRt x)] = (1 + SqRt x)/(2 + SqRt x) & 1/(1 + 1/[1 + 1/(1 + SqRt x)]) = 1/[1 + (1 + SqRt x)/(2 + SqRt x)] = (2 + SqRt x)/(3 + SqRt x). You can use the principle of mathematical induction to prove that for n > 1, I(n) = Integral 0 to 1 of [(n - 1 + SqRt x)/(n + SqRt x)] dx. The integrand is simply equal to 1 - 1/[n + SqRt(x)]. This easy to integrate. Use linearity to concert this into a difference of integrals. The integral of 1 from 0 to 1 is 1, and the integral of 1/(n + SqRt(x)) can be obtained by substituting. Note that u = n + SqRt(x) => du = 1/[2 SqRt(x)] dx = 1/[2(u - n)] dx => dx = 2(u - n) du. x = 0 => u = n, x = 1 => u = n + 1. The integrand is simply 2(u - n)/u = 2 - 2n/u. Now we can use linearity again to convert this into a difference of integrals. The integral of 2 from n to n + 1 is 2. So far the total integral is 1 - (2 - Integral from n to n + 1 of 2n/u du) = -1 + Integral from n to n + 1 of 2n/u. The remaining integral is trivial, and it equals 2n log(1 + 1/n) = 2 log[(1 + 1/n)^n]. Hence I(n) = 2 log[(1 + 1/n)^n] - 1. As n -> ♾, (1 + 1/n)^n -> e, so log[(1 + 1/n)^n] -> 1, so I(n) = 2 - 1 = 1. Now, I calculated this before even watching the video, but I am fairly confident in this to some extent for some reason. Edit: I realized I made a very basic arithmetic mistake, so in fact, the pattern I presented is false. Of course it could not be so simple. Also, you said proving the limit exists is cumbersome, but in reality you can prove that the limit exists and that the limit is the golden ratio all within one proof, simply by obtaining the limit of the explicit expression for the Fibonacci sequence, which can be found relatively easy. Consider the equation a(n) + a(n + 1) = a(n + 2). This is equivalent to a(n + 2) - a(n + 1) - a(n) = 0. Now suppose there is a constant λ such that a(n + 1) = λ·a(n). Then λ^2·a(n) - λ·a(n) - a(n) = (λ^2 - λ - 1)a(n) = 0, which implies λ^2 - λ - 1 = 0. This is solved with λ = φ & λ = -1/φ. This means a(n) = Aφ^n + B(-1/φ)^n. Let n = 0, so a(0) = A + B, and n = 1 so a(1) = Aφ - B/φ. B = a(0) - A, so a(1) = Aφ - [a(0) - A]/φ = A(φ + 1/φ) - a(0)/φ. φa(1) + a(0) = A(φ^2 + 1), So A = a(1)[φ/(φ^2 + 1)] + a(0)/(φ^2 + 1). Therefore, B = a(0) - A = a(0)[φ^2/(φ^2 + 1)] - a(1)[φ/(φ^2 + 1)]. From this formula, it is obvious that (-1/φ)^n -> 0, and a(n + 1)/a(n) ~ φ. Also, notice that φ^(n + 2) = φ(n + 1) + φ^n for all integers n, from which you should learn 1/φ = φ - 1.

[@Azelide]

1 plus 1 over 1 plus 1 over 1 plus 1 over 1 plus 1 over 1 :)

[@C204unit]

I wanted to point out that the various sequences mentioned with different starting values are actually called Lucas sequences. They are closely related to the Fibonacci sequence and a generalized form of their respective starting numbers.

[@angelmendez-rivera351]

The pattern I noticed is that I(n) is equal to the integral on 0 < x < 1 of (f^n)[sqrt(x)], where f(y) = 1/(1 + y). One can let y = sqrt(x), so that 2y·dy = dx, and I(n) is equal to the integral on 0 < y < 1 of 2y·(f^n)(y). f(y) = 1/(1 + y) = (1 + 0y)/(1 + y) (f^2)(y) = 1/[1 + 1/(1 + y)] = (1 + y)/(2 + y) = 1 - 1/(2 + y) = 1 - f(y + 1) (f^3)(y) = 1/[1 + (1 + y)/(2 + y)] = (2 + y)/(3 + 2y). From this exploration, it is suggested that (f^n)(y) = [F(n) + F(n - 1)·y]/[F(n + 1) + F(n)·y], where F(n) is the nth Fibonacci number, with F(0) = 0 and F(1) = 1. This can be proven with induction. If n = 1, then f(y) = (1 + 0y)/(1 + y) = [F(1) + F(0)·y]/[F(2) + F(1)·y]. Assume (f^m)(y) = [F(m) + F(m - 1)·y]/[F(m + 1) + F(m)·y]. Thus [f^(m + 1)](y) = f[(f^m)(y)] = 1/[1 + (f^m)(y)] = 1/(1 + [F(m) + F(m - 1)·y]/[F(m + 1) + F(m)·y]) = [F(m + 1) + F(m)·y]/[F(m + 1) + F(m) + F(m)·y + F(m - 1)·y] = [F(m + 1) + F(m)·y]/[F(m + 2) + F(m + 1)·y], as conjectured. This concludes this part of the proof. Therefore, I(n) is equal to the integral on 0 < y < 1 of 2y·[F(n) + F(n - 1)·y]/[F(n + 1) + F(n)·y] = 2·F(n - 1)·y·[F(n)/F(n - 1) + y]/[F(n + 1) + F(n)·y] = 2·F(n - 1)/F(n)·y·[F(n)^2/F(n - 1) + F(n)·y]/[F(n + 1) + F(n)·y] = 2·F(n - 1)/F(n)·y + 2·F(n - 1)/F(n)·y·[F(n)^2/F(n - 1) - F(n + 1)]/[F(n + 1) - F(n)·y] = 2·F(n - 1)/F(n)·y - 2/F(n)·y·(-1)^n/[F(n + 1) + F(n)·y] = 2·F(n - 1)/F(n)·y - 2·(-1)^n/F(n)^2·y/[F(n + 1)/F(n) + y] = 2·F(n - 1)/F(n)·y - 2·(-1)^n/F(n)^2 + 2·(-1)^n/F(n)^2·[F(n + 1)/F(n)]/[F(n + 1)/F(n) + y]. Integrating this on 0 < y < 1 results in F(n - 1)/F(n) - 2·(-1)^n/F(n)^2 + 2·(-1)^n·F(n + 1)/F(n)^3·ln([F(n + 1)/F(n) + 1]/[F(n + 1)/F(n)]) = F(n - 1)/F(n) - 2·(-1)^n/F(n)^2 + 2·(-1)^n·F(n + 1)/F(n)^3·ln[F(n + 2)/F(n + 1)]. As n -> ♾, 2·(-1)^n/F(n)^2 + 2·(-1)^n·F(n + 1)/F(n)^3·ln[F(n + 2)/F(n + 1)] -> 0, so I(n) -> 1/φ. This is how one would answer the question without using uniform convergence.

[@liberalaccidental]

Love the video, but think you should have discussed more the conditions for limit of the integral to be equal of integral of the limit (uniform convergence)

[@antonioskanitsakis7374]

A somewhat more rigorous proof would be this: Integrate from x=0 to x=1 of (F(n+1) + Fn*sqrtx)/(F(n+2)+F(n+1)*sqrtx). As you shown, this is the form of the continued fraction after n iterations (Fn= the nth fibonacci number). This integral results in a somewhat ugly expression but taking the limit as n approaches infinity (therefore F(n+2)/F(n+1)=F(n+1)/Fn=Φ) you get the correct result 1/Φ. This way you don't have to interchange limit and integral! Hope this helps.

[@mohammadowaiskhokhar2318]

I see it as a notebook case of recursion. Let the quantity be y. Then by recursion you have 1/(1+y)=y Solve for y which is y=(-1+✔️5)/2 Integrate both sides. And you have your answer(which is same as y)

[@easymathematik]

At 01:30 u say: "U may ask what am I doing with Fibonacci." I wanna answer it for u. :D If one has experience with Fib and golden ratio, then one knows about its continious fraction. The golden ratio has the fraction "1 over + 1 over + 1 over ...." This explains your idea with Fib. :) If one knows this, the solution is no surprise. By the way: great channel! I like your stuff.

[@stanrocks123]

Another method is that you know that the integral from 0 to 1 of a constant will give you a constant (this fraction is a constant as it goes infinitely far). And then just figure out the rest of the golden ration stuff

[@rohandhanju5843]

Wow. I loved your incredible explanation sir! I understood every bit of the problem.

[@Chalisque]

And of course sqrt(x) can be replaced by any function of x (subject to the integral existing).

[@fengshengqin6993]

If n go the infinity ,then the integral willchange , change from a function to a constant integral. This equal to ask you this infinity constant ,integral from 0 to 1 , So the result is (1+sqr5)/2.

[@fengshengqin6993]

sorry ,it should be the (sqr5-1)/2.

[@moonlightcocktail]

But... this does not depend on the fact that we were integrating sqrt x, right? So we could put some unintegrable thing there like (1/ln x) or (sin x / x ) and it would still we or the same...

[@Ben-kh2rh]

Love the way you explain things. Keep more videos coming!

[@nicolascamargo8339]

Excelente explicación

[@moritzhertler]

This is beautiful!

[@johnniegilkerson4724]

@10;38 (1+(1/(1+sqrt(x)) times(1+ sqrt(x)) is 2, where di the sqrt (x) come from?

[@tsunningwah3471]

Hi, may I ask how do you know the ratio of an , an+1 is phi.

[@rakshithgowda1606]

Your videos are very useful. Please don't stop making more.

[@eric3813]

Incredible! Great video!😄

[@alimghazzawi3700]

I honestly solved it in much simpler way in two min but I wasn’t sure about the solution. So what I did consider we want to find y in a nicer way so let y= the whole miss and realized that u can write it as 1/(1+y) then y=1/(1+y) and solved for y and appeared to be 2 soloutins one pos and another neg which are constants then when we integrate both of 2 cases from 0 to 1 it would became it self so I lift with choosing which result is right ? Then I thought it has to be the positive one because the graph it has to be up the x axis because of 2 reasons first square root of x It has to be pos and secondly the graph it seem to up not down so (sqrt(5)-1)2 seems to have more sense to me

[@alimghazzawi3700]

I really want u to read it

[@soumyadipdas6636]

Yeah I have also done the problem this way

[@hyperreed]

If you replace sqrt(x) with just x do you get the same answer?

[@mjnyc8655]

Wow! This seems the sort of problem Ramanujan would have loved to tackle.

[@awanitranjan9730]

Excellent soln

[@meetturtle2914]

Woah....this is so tedious ,but cool too ...I will love to try one .

[@adamfattal9602]

All I want to know is what was going through the minds of the participants of the MIT Integration Bee while seeing this integral. And whomever solved it is a fucking genius.

[@easymathematik]

The connection to Fibonacci u can see if u know that the golden ratio is the "most irrational" number in a specific sense which is: the chain fraction of phi is "1 over 1 over 1 over ...". Every mathematician knows this. Also he knows the connection between Fib. and phi. So his start to solve this in the way he showed makes it "quite natural" in the sense "what do I have? 1 over 1 over 1 over .... Ha! Golden ratio". :) I had the same idea immediately I saw the integral. No offense to the video, oh my gosh, ... I love watching his videos. But things become "trivial" if you know a lot about certain stuff. Remember his first questiin in the video. He said: "U may ask what Fibonacci has to do here?" This is exactly what I want to say: His experience with Fib and golden ratio makes this idea trivial.

[@adamfattal9602]

Hahahahhahah Great reply!@@easymathematik

[@Grassmpl]

You can't just switch the limit and integral. You need the Lebesgue dominated convergence theorem.

[@spaceexplorer5481]

I am in 8 th class and still I loved watching this

[@okitobing5]

This guy is insane

[@saumyaranjan9256]

Fun fact: Fibonacci sequence was not discovered by fibonacci rather an Indian mathematician. Great video.

[@anshulagrawal3859]

Golden ratio is limit without integral.How can we apply concept of golden inside integral

[@indiablackwell]

this is beautiful

[@ShadowCooper77]

Imagine you get to the final integral and then you mess it up, that would be devastating 😂😂 Anyway great video 👍

[@KelfranGt]

Simply amazing 👏👏👏

[@FaerieDragonZook]

But what if your sequence starts a_1 = phi, a_2 = -1?

[@Mihau_desu]

Then it approches (1 - sqrt (5))/2 But it's just one case when it happens.

[@RedefiningtheConcepts]

Hi bro I am your fan.But Please Please clear my doubt. Riemann integral definite integral and integral as Limit of sum all these are same?.And How many are type of integrals.

[@LetsSolveMathProblems]

Most introductory calculus texts *define* the Riemann integral (which is the most common type of definite integral) to be the limit of the sum of rectangular areas. There do exist other types of integrals, such as the Riemann-Stieltjes integral and the Lebesgues integral, the latter of which is widely used in graduate level mathematics. (It is possible to define other types of integrals, but the three I mentioned are the most well-known.) However, in most math problems at undergrad level or lower (like in the video), the integral is almost always Riemann. =)

[@williamwen7190]

Golden Ratio. OMG.

[@LetsSolveMathProblems]

I know! =)

[@iiimusika]

ooh woow

[@iustintaga]

I love this

[@Andrei-rp3dz]

I thought this was Integration Bee 2017 not 2019

[@LetsSolveMathProblems]

The "19" in the title is meant to indicate that this is the 19th video in my "MIT Integration Bee" playlist. I include the link to the playlist below, in case you are interested in exploring a few more integrals. =) kzbin.info/aero/PLpoKXj-PWCbYGk3e4TohJ3L0EoZXfglxK

[@pablowo99]

beautiful

[@KwongBaby]

that is really cool❤❤❤

[@neolokis]

Wow

[@KrisEditz29]

Just great

[@manosxa]

The Fibonacci sequence is NOT fibonacci sequence, Pythagoras and Eucleades and many other Greek mathematicians first developt the concept of φ and that is why we call it golden ration which is how the Greeks called it. Pythagorians also found the assimptotic properties of the so called wrongly fibonacci sequence milleniums before and is very unfair that people call febonacci an analogy that is everywhere in Greece like the parthenon and hundrents of other statues and monuments.

[@angelmendez-rivera351]

manos Much of what you said is stated nonsensically, but also, you said “The Fibonacci sequence is NOT the Fibonacci sequence”, which is a false statement by definition, because it is a contradiction.

[@mahiramerulislam3382]

good.

[@C_A_LLl_A]

"Сквырлдобэкс" втф????

[@remysanlaville3085]

#squaradovex

[@shikhanshu]

MOTHER OF GOD

[@kartiksharma7166]

Grt

[@srpenguinbr]

INtegral vs inTEgral

[@plekkchand]

Yes, exactly, please speak English. And it's not the Gordon ratio, either. But it's an excellent video.

[@udbhavkharad8581]

plekkchand he is speaking English

[@syemtous4510]

Ur fucking genius

[@Koj4]

Holy shit

[@JensenPlaysMC]

Is it just me or did anyone else do this in their head in about 20 seconds?

[@joelsagflaatholmberg3922]

Yeah, it's an easy integral.

[@jethrosu1533]

This is really a bad question. It has nothing to do with integration. There is really no point of putting an integration sign around a constant. It looks confusing.

[@fidgetspinner1050]

OMFG

[@chaos4785]

i can't believe i solved it in a minute😅

[@franzschubert4480]

Just from looking at it I was pretty sure it has to be 1/phi within 10 seconds.

[@connorhorman]

same. As we take the limit, we just get dx/1+1/... which is dx/phi. Easiest question ever.

[@soumyadipdas6636]

I think I can solve it more easily then this

[@subnow4862]

Nice integral!

[@khalidjamhour945]

Wow