a crazy iterated limit.

[@renerpho]

10:41 It's nice of you that you graphed those, but are you planning to show us those graphs, too?

[@Jay-tn5of]

It's clear to see graphically that it converges to 0 as the expression sin(sin(sin...))) can be linearly approximated using y=0

[@ΚωνσταντίνοςΔημητρίου-θ7δ]

@@Jay-tn5of irrelevant to what Daniel said

[@wawawuu]

@@gregorymorse8423 lmao you must be a bitter person.

[@lyrimetacurl0]

It would look like a sine graph slowly being squashed vertically till it was a straight horizontal line at y=0

[@DrWhom]

graph them yourself in Mathematica.

[@goodplacetostop2973]

10:42 Where are the graphs ? 😢 10:55 Good Place To Stop

[@cH3rtzb3rg]

They are on the screen. But the video only shows the blackboard ;)

[@ΕχιΜιμζ]

Video lenght not enough as Fermat said 😆

[@Monolith-yb6yl]

I wanted to see graphs too

[@emmanuellaurens2132]

7:30 or, you know, you can notice that an = -bn because a1 = -b1 and sin (-x) = - sin x. Which will save you time at 10:00 too.

[@goncalofreitas2094]

I thought about it too! sin(x) is odd so the sequences are of opposite signs

[@terryendicott2939]

ODD you should mention this.

[@sinecurve9999]

This problem has a very dynamical systems flavor to it. The graphs you mentioned at the end didn't show up.

[@tomholroyd7519]

It's the circle map. It gets interesting if a*sin(x) = x has more than one fixed point, a > 1

[@joeeeee8738]

A visual representation of this is always helpful. Don't know why it's not included!

[@joeeeee8738]

@@gregorymorse8423 🙄

[@joeeeee8738]

@@gregorymorse8423 🙄

[@nathanielkilmer5022]

@@gregorymorse8423 Yes, close mouth before inserting foot. Brilliant.

[@nathanielkilmer5022]

@@gregorymorse8423 It's pretty hilarious to me that you came into this thread to put down the intelligence of others, but you couldn't even get a simple insult correct. It seems that you... put your foot in your mouth.

[@agrajyadav2951]

@@gregorymorse8423 bruh wtf?

[@websnarf]

sin(-x) = -x. So WLOG, we can pull any negative sign outside, and assume x is always positive. For x > 0, sin(x) < x, and since sin is monotonic on the interval [0, 1], sin(sin(x)) < sin(1) < 1. Thus the sequence sin(x), sin(sin(x)), sin(sin(sin(x))), ... is decreasing, but bounded below, and therefore converges. If it converges to k, then we have k = lim sin(sin( ... (sin(x)) ... )) = sin( lim sin(sin( ... (sin(x)) ... )) = sin(k). k = sin(k) has the sole solution k = 0.

[@thomasdalton1508]

sin(-x)=-sin(x), you mean!

[@leif1075]

You didn't prove why you can pull.sine out and have sine of limit of sine is equal to the sine..I know you can do that with log but you don't know with sine??

[@drillsargentadog]

much easier proof: sine is contractive on [-1,1] with 0 as a unique fixed point. Hence the limit is zero by Banach's theorem.

[@yuseifudo6075]

No fun in this

[@atreidesson]

Well, I appreciated these many graphs that were sliding over the screen in the end, you know, the best feeling is when you hear, "there will be some graphs", and they actually are

[@as-qh1qq]

While it's very intuitive to conclude that the solution must be zero by directly restructuring the original composition to y=sin(y), for those wondering why this video is 10m long, it is Michael's satisfying rigour.

[@albertobarreirohermida3829]

Everyone who ever owned a calculator already knew the answer to this one.

[@edusoto91]

I bet you are talking about the corresponding Cos limit. In the Sine case it takes over 3000000 iterations to drop below 0.001 (starting with 1) while it takes over 50 iterations to achieve 8 correct decimals of the corresonponding Cosine iteration. (It was funny to check this, thanks Michael and Alberto!)

[@albertobarreirohermida3829]

@@edusoto91 Not if you set your calculator to operate in degrees!

[@edusoto91]

@@albertobarreirohermida3829 Ahhhh. That's not fair!

[@wawawuu]

@@gregorymorse8423 This just in, playing around with one's calculator is verboten.

[@sciencysergei]

This is the new math - always trust the calculator! Success of educational reforms. What is 2+5*3? Some calculators give 21. It gotta be true!

[@florisv559]

He could have saved a lot of hard work by noting that for x 0, |sin x| < |x|. So, you have the monotonous sequence and can then go right ahead to point out convergence. But he simply wanted to show how you can use the squeeze theorem in another context than for finding the derivative of the sine function.

[@marc-andredesrosiers523]

A visual complement would have been neat: to build the intuition behind the solution.

[@JonathanMandrake]

We had this in my Numerics class, it converges to 0 using Bouwers Fix Point Theorem

[@theshoulderofgiants]

how would brouwers fixed point theorem hold for f(x) = x+3

[@Arbmosal]

@@theshoulderofgiants it does not apply, because of the boundedness condition

[@JonathanMandrake]

@@theshoulderofgiants 1. Then f(x) isn't a contraction, defined as: |f(b)-f(a)|=L|b-a| where L is a constant

[@YaamFel]

Banach's fixed point theorem*, Brouwer talks about self-maps on contractible sets

[@JonathanMandrake]

@@YaamFel Yes, you're right, I mixed them up

[@Bodyknock]

9:55 This is basically proving a special case of a Fixed Point Convergence Theorem which says that if f is a continuous function then if f(f(f(f(...f(x₀)...)))) converges it must converge to a fixed point x where f(x)=x . In fact clearly you can replace sin() with any continuous function f() in this part of the video and the result would be the same that if it converges it must converge to a fixed point since only the fact that sin is continuous was used in this part. So basically to sum up, sin() isn't all that special here. if you have any monotone, continuous function on a bounded interval to a bounded interval, and you apply that function over and over again starting at some initial point on that interval, then it must converge to a fixed point. Which in the case of sin happens to be 0 is the only fixed point in that interval.

[@xoriun8638]

Isn't this this only true for contractions, i.e. L-lipschitz continuous functions with 0

[@AlcyonEldara]

@@xoriun8638 yes, f(x) = -x is the easiest counter-example.

[@japanada11]

The first paragraph of Doug's comment still holds in general (*if* the sequence converges, then it converges to a fixed point). But yes, the second paragraph is incorrect as stated because it leaves out the crucial "if the sequence converges" assumption.

[@clahey]

I think this is very close to correct. I think you need monotonically increasing and not just monotone, and you could be clearer that all the bounded intervals are the same interval.

[@vladzotov1959]

Would be great, if you did the cos version! It actually converges to an transcidential number

[@Imperial1506]

Just a solution of x=cos(x). Since such equations are called transcendental, no wonder that the solution is also transcendental

[@teambellavsteamalice]

Cool video. I'd probably messed around a bit with approximating sin x with a taylor series which should simplify to f(x)=x as x approaches ±0 (as quadratic and higher terms decrease faster). But that probably lacks the proof f(f(x)) is converging in the first place. p.s. For those interested, the derivatives go from cos x, -sin x, -cos x back to sin x, with the even elements having a factor ±sin 0 = 0 and the odd factors alternate signs. This gives: sin x = 0 + 1x − 0x^2 + − 1/3! x^3 + 0x^4 + · · · sin x = x − (x^3)/3! + (x^5)/5! − (x^7)/7! + · · ·

[@levm.2666]

A simple(r) way: Let sin( sin( ....sin(x)...) = y. Then, sin(y) = y. Also, sin(y) < y (by Taylor) for all y>0 and sin(0) = 0. Therefore y =0, that is, the limit is zero.

[@TheEternalVortex42]

You have to show convergence though

[@jackkalver4644]

Judging from the graph, if y=sin y, y must be 0. Therefore, the answer is 0. Use calculus to make sure. (When x>1, x>1>=sin x. So only test [0,1].)

[@Abhisruta]

Hey Michael, can you sort out Set Theory videos to a playlist?

[@Grassmpl]

Note the convergence is uniform, since it conveges pointwise, and the sequence of absolute differences is uniformy bounded by a sequence approaching 0.

[@roberttelarket4934]

I shouldn’t be watching this as I’m already crazy and don’t want to become crazier!

[@coffeecup1196]

An loosely defined attempt before watching the video: Take f_n(x) = sin(f_n-1(x)) and consider g(x) = x - sin(x). By simple analysis (one could use derivatives to prove this), the function g(x) has precisely one zero on the interval [-1,1], and that is at zero. Everywhere else on the interval, g(x) is an increasing function with no other inflection points existing. After every iteration of f(x), it bounces between the line y = x and sin(x) in a staircase pattern (look up McCabe Thiele plots for the inspiration of this). Since g(x) only has one zero and is increasing everywhere else, every iteration brings it closer to zero. Taking the limit to infinity, the limit is zero. As a note, the convergence is very slow since g(x) quickly approaches zero, but since it never actually hits zero until x = 0, the argument should still hold. Cleaning up edge conditions: Since sin(x) only outputs to the interval [-1,1], any possible x input will be in the valid interval for the above argument after one iteration. Also, if an input is precisely 0 after one iteration, then the limit is obviously 0 because sin(0) = 0. Post-video: It looks like we did basically the same thing except you actually proved g(x) is increasing outside of the point 0 on the interval [-1,1]

[@Monkieteam]

1) Notice that the line y=x is > sin(x) for x>0 and is < sin(x) for x

[@RipleySawzen]

Between -π/2 and π/2, the slope of sin(x) is less than 1 at all points except 0, so the output is going to be closer to 0 than the input* The first sin(x) to be evaluated outputs inside that range All further iterations bring us closer to 0 *I chose this range because it's 1:1, and because it is 1:1, it avoids any possible cycles, though I suppose that's impossible with any function that has slopes less than 1 except at f(x)=0, but that adds lots of complication to an otherwise simple solution. Basically, my mind immediately jumped to "Make x be between -1 and 1" because that is all that matters.

[@martinepstein9826]

Getting closer to 0 doesn't imply the limit is 0. The sequence [1+1/n] is always getting closer to 0 but the limit is 1.

[@RipleySawzen]

@@martinepstein9826 A: That's not a continuous function, like sin B: Its slope greatly exceeds 1, unlike sin C: At what values does repeatedly iterating it bring us closer to 0? And how does doing so make us get to 1? I don't think that it's EVER possible to get closer and closer to 0 and wind up at 1. If the limit approaches 0, the limit is 0. D: Why would you generalize what I'm saying about a specific function to all functions? E: Why are people liking your irrelevant comment more than my intelligent one?

[@martinepstein9826]

​@@RipleySawzen "Why would you generalize what I'm saying about a specific function to all functions?" That's how logic works. If your argument is correct then it will work in different situations. You showed (correctly) that the sequence [x, sin(x), sin(sin(x)), ...] gets closer and closer to 0, and from this you concluded (incorrectly) that the limit is 0. I showed you the sequence [2/1, 3/2, 4/3, 5/4, ...] that gets closer and closer to 0 even though the limit is not 0. That means your argument is incorrect.

[@criskity]

@@martinepstein9826 But on the interval 0 < x

[@martinepstein9826]

@@criskity Well yeah, that's how we know the sequence [x, sin(x), sin(sin(x)),...] gets closer and closer to 0. Now you just need to show that the limit is 0. See my example sequence for why you haven't done this yet, and see the video for how to actually do this.

[@uncreativename9833]

Where are the graphs from Mathematica you wanted to show?

[@yeech]

I guess they are on his computer screen. But this video only shows his blackboard.

[@rossi9366]

Thx professor for this channel. It's a gift.

[@krisbrandenberger544]

Proof that b_n is bounded: 1) We know 0

[@純糖]

A good practice problem of fixed point theorem

[@barbietripping]

Before beginning, I expect: sin(x)=x implies 0 is the answer But I would love to see some unannounced overkill if it happens

[@barbietripping]

Was not dissatisfied, got to see the domain get whittled down by using monotonic boundaries. Visualizing that process is far more appealing than notating it

[@tgx3529]

Our lives are similar. We are gripped by pliers and pushed to zero.Pliers x and -x are ruthless!

[@mrminer071166]

It's an interesting question for beginners: what is the relationship of the LIMIT question (as considered in the video) and the identity question, where does Sin (x) = x? The geometry is clear enough: just graph y=x and y = sin(x), which obviously intersection at 0,0 and one other point in the first quadrant, and its reflection in the 3rd quadrant. But this is a nice problem for beginners because you CAN'T be tempted to "touch" the infinite iteration of fsubn, you are FORCED to think in proper LIMIT terms. And then you get the whole "attractor" situation as well.

[@theprimera387]

ig you can intuitively say that the limit is 0 bc |sin(x)|

[@ulhaque]

It can also be simply analysed by putting the extreme values 1 and -1 . It becomes clear that the value comes closer and closer to zero with each sin .

[@igorstasenko9183]

wouldn't it be easier to prove that |sin(x)| < |x| , for any x in [-1..1] and not 0, which means, that if you apply recursive sin() .. every step in recursion will be closer and closer to 0 , and consequently converges to zero if you apply it infinite number of times?

[@YonatanSetbon]

The same problem but with cosine is much more interesting.. :m

[@albertogarcia4177]

Let be the n-iteraded function F(n,x)=sin(sin(...x)...)) (n times), and assum it has limit L=Lim F(n,x) when n->infinitum, then we also have L=Lim F(n-1,x) for n->infinitum, and since sin x is a continuos function on all real L, and F(n,x)=sin(F(n-1,x)) we get L=Lim F(n,x)=Lim sin(F(n-1,x) =sin Lim F(n-1,x)=sin L, so L=sin L, wich implies L=0 for properties of the sin function

[@jeanefpraxiadis1128]

This exposition looked a lot more rigorous and tedious than it needed to be. The limit function y(x) that we are looking for satisfies y(x) = sin(y(x)) for all x. But only the zero function does this, so y(x) is the zero function.

[@ikirigin]

For x>0, sin(x) < x, and repeated applications approach 0. For xx, and repeated applications approach 0.

[@tredex9107]

If we take f(x) as Y, can we say that Y = Sin(y)? Like those infinite sums of roots

[@Linkga420]

I thought the solution would just be y=sin(sin(...sinx...)) and since this is an infinite sequence you can just say y=sin(y) for which the only solution is y=0. Just like you do it to get the characteristic polynomial for the golden ratio, for example (x=1/(1+x)). I guess, that's not accurate enough?

[@Linkga420]

also the sin is defined for all real numbers and can only become a real number as well.

[@wise_math]

Nice limit

[@matthewbriggs9137]

Where were the promised Mathematica plots?..

[@toddy5241]

What about lim cos(cos(cos(...)))?

[@MCLooyverse]

So this would be looking for a stable fixed point of sin, right? sin x = x => x = 0, so that would be the answer, I think.

[@Ivankarongrafema]

What would happen if instead of sin, we had log?

[@jonastechmanski1592]

Do cosine next, it's different!

[@RizmaYudatama]

I think this kind of problem is quite obvious what the answer is, but it is hard to prove. I know that the answer is 0, but I don't how to prove it.

[@farfa2937]

Same, I saw it and was like ok that's 0 but I wouldn't have known all this procedure.

[@drmathochist06]

Why not just show that x -> sin(x) has a unique attracting fixed point at 0?

[@kenbob1071]

Looked at the thumbnail for the video; took 5 seconds; figured it had to be zero; watched video to see how contorted the proof would be. Bada-bing, bada -boom.

[@elonmusk1270]

I was able to ans just by seeing the question.

[@MrLetter27]

Where do I get such a cool Michael Penn pullover, Professor? Also, what is the backstory for the logo? Perhaps a personal “Quite Easily Done” aka “…and that’s a good place to stop.” Would love to know.

[@iooooooo1]

For the hoodie, it's linked in the video description, teespring.com/stores/michael-penn-math (it redirects to michael-penn-math.creator-spring.com/).

[@shahinjahanlu2199]

Can we use siny=y ?

[@md2perpe]

Yes, but some more is needed, e.g. also showing that the limit exists.

[@shahinjahanlu2199]

@@md2perpe thx

[@felliphegoes57]

Where are the graphs? :((

[@yosefmohamed1591]

We can say sin(x) < x Until x approaches zero Therefore the whole expression approches zero and the sine of zero is zero Because the range narrows around the zero every time we get the sin of the sin of the sin and so on...

[@Misteribel]

So where are the graphs you promised? Your deduction follows intuition, but a graphical presentation would still be nice!

[@rowanpotato9816]

You forgot to show the graphs

[@laojackos]

I really wanted to see the graphs

[@kevinmartin7760]

If x is not real (complex, perhaps) it is not until f_2 that your value is in [-1, 1]

[@lyoukeefen7199]

You see, I could do essentially the same thing for the b_n sequence, ooooorrrrr, I could write similarly b_n approaches 0 QED😎

[@Omar-hm6pu]

Very very nice problem

[@lacryman5541]

Graphs are the first thing to do

[@dushyanthabandarapalipana5492]

Thanks!

[@bossboss-dg6cj]

It was easier to derive that b_n=-a_n rather than working on both sequences.

[@ДмитрийЦымбаленко-п2ц]

you're genious

[@diniaadil6154]

Or you could just notice than sin in an odd function which would save you the trouble of going through both negative and positive cases

[@nournote]

an=-bn

[@delafrog]

it seems like a obvious, because solution of equation x=sin(x) is 0. That problem is much more interesting, but with cos() instead of sin() :)

[@rogerkearns8094]

Just messing about with a calculator soon strongly suggest this!

[@AtariDays80]

Now repeat with cos(x). And then go crazy with co-tangent.

[@kameronpeterson3601]

I did it the naive way by assuming it has a solution n n = sin(sin(sin(sin(sin(...x))))) Notice that, since the amount of sines is infinite, this equation contains itself. That gives us n = sin(n) and the only solution is n = 0.

[@CharlesPanigeo]

It goes to 0 because 0 is an attracting fixed point of sin(x).

[@shohamsen8986]

Maybe do a collab woth baker

[@krampinhofredericoobode3680]

I an a engineer student and I got lost at the first minute 😃

[@janekgroe4304]

If the limt exists it has to be zero since sinx=x has no other solution

[@Cpt.Zenobia]

I think they forgot to show the plot of the seq.

[@xizar0rg]

I guess he left looking at the graphs as homework.

[@relike868p]

I thought it goes like x = sin x and thus x = 0 by drawing graphs and it is done. Next: use cos instead of sin

[@mohamedfarouk9654]

I was wondering if there is a continuous expansion g_t(x) for all real t such that when t = an integer n then g_n(x) = f_n(x). Anyone knows?

[@the_nuwarrior]

Banach fixed-point theorem.

[@AkamiChannel]

Didn't need to do all that. Coulda told you it was 0. I mean, it was a guess, but it turned out right. I'm a genius.

[@Happy_Abe]

Never put the graphs up :(

[@Kualinar]

JUST by looking at the structure of that statement, I can say that the limit must be zero. It could be in degree, radiant, grad or any other system used to measure angles, it don't mater. The only difference is how fast it will converge to zero.

[@jursamaj]

For *any* system of angle measure? Nope. Contrive a measure that has some non-zero value of x such that sin(x)=x, and you'll have more fixed points than just 0. This works for any unit of measure that is larger than a radian (that is, fewer than 2π units per full revolution). Suppose there are sqrt(32) burps per full revolution (5.65685…). Then 45°=sqrt(2) burps. The sine of that is sqrt(2). Boom, fixed point at sqrt(2), doesn't converge to 0. For somebody who likes equilateral triangle, the sextant (60°) fits 6 sextants into a circle. Same problem, altho different specific fixed points. Or more naturally, if your unit is right turns, there are 4 in a circle, and sine(1)=1.

[@Kualinar]

@@jursamaj I must agree with you on that one. I've come up with the «trit» where there are three trits in a full circle after I posted and searched for a system that would invalidate my statement.

[@nikita__222]

When I saw that the video lasts for ~10 minutes, I supposed that the solution of the problem will be based on set theory. I mean that the lower and higher bounds of the range of iterated sine function will be equal and will contain only one element - zero.

[@oscarlama]

The real challenge is cos(cos(cos(cos...cos(x)...))) as the n repetitions tend to infinity

[@ryanye8441]

Can we have a easy way to figure this out? Let's assume that this limit exists (which is N), so we can conclude that sin(N)=N, so N=0 so the limit DOES exists which is 0 any flaw?

[@ezequielangelucci1263]

Flaw: "Let's assume that this limit exists"

[@aaronspeedy7780]

Here's how I solved it. sin(sin(sin(...))) = x Because we can replace sin(sin(...)) with x We can rewrite it like sin(x) = x Which is of course zero

[@soranuareane]

Does the claim that "sin x has only one fixed point, namely sin 0 = 0" require proof? How do we show there isn't something else that sine maps to itself?

[@pwmiles56]

Monotonous sequences, as we used to call them :-))

[@carlosmirandarocha8905]

Where are the graaaaphs bruuhh????

[@jplikesmaths]

Me, who thinks that n corresponds to the n in sine 😂😂😂😂

[@라디안-e9z]

This problem can be solved without the monotonic convergence theorem :)

[@demenion3521]

defining the sequences a_n and b_n is somewhat pointless if you later use the fact that 0 is the only fixed point of sine anyway since the goal limit is also just asking for a value L=sin(L).

[@pawe2148]

sin(sin(sin...sin(x)))=y sin(y)=y y=0

[@md2perpe]

There you have assumed that there is a limit.

[@nombreusering7979]

If he was mentioning that the number of iterations is infinite, then it'll be that quick to solve. But, he mentioned that the number of iterations is n, which is quite a difference.

[@md2perpe]

@@nombreusering7979 If the limit L exists, then, since sin is continuous, we have L = sin(L).

[@nombreusering7979]

@@md2perpe Nope. You're just showing that L=sin(lim(sin(sin...(n))) iterating n-1 times) You can't just rely on the fact that n-> inf and set n=inf. It's a limit. Or perhaps we are not synced

[@md2perpe]

@@nombreusering7979 First note that lim f_n(x) = lim f_(n-1)(x). Assuming that the limit exists we have L(x) := lim f_n(x) = lim sin(f_(n-1)(x)) = { since sin is continuous } = sin(lim f_(n-1)(x)) = sin(L(x)). That the limit exists follows from that if 0 < x then 0 < f(x) < x so that f_n(x) is a decreasing positive sequence. Similarily, if x < 0 then f_n(x) is an increasing negative sequence, and as such has a limit. The case x = 0 is trivial.

[@alexandrosin9101]

Much more interesting task is to find lim √n * sin(sin(sin(....))) -- after solving it (which is quite faster, than in the video) the result of the video becomes trivial.

[@Kapomafioso]

Homework: find the following limits a) lim n-> infty cos(cos(cos(...cos(x)...))) (composed n-times) b) lim n-> infty sin(sin(sin(...sin(x)...)))/sin(sin(...sin(1)...)) (numerator composed n-times, denominator composed n-1 times) c) lim n-> infty sin(sin(sin(...sin(x)...)))/sin(sin(...sin(x)...)) (numerator composed n-times, denominator composed n-1 times) d) lim n-> infty sin((pi/2)*sin((pi/2)*sin(...(pi/2)*sin(x)...))) (composed n-times)

[@沈栋-w9h]

NICE！

[@Veggie13]

The missing graphs, though...

[@hopelessdove]

If you try to do this with a calculator, you will find that the thing converges ridiculously slowly. sin(sin(sin...(1))) (100 times) is still ~0.16

[@cH3rtzb3rg]

In limited precision there will be a fixed point as soon as the difference between sin x and x is less than the relative precision.

[@georgeb8893]

Yes, ridiculously slow. To get it to less than 10^(-10) you would probably need about 10^20 iterations. [I think it converges a lot like sum n=1to inf of n^(-1.5).]

[@kareolaussen819]

@@georgeb8893 My back-of-a-tiny-envelope calculation suggests that the sequence behaves like sqrt(6/n)*sign(sin(x)) for large n.

[@idjles]

Sin x < x for small x, therefore limit=0

[@anshumanagrawal346]

Not quite, a strictly decreasing sequence can be bounded above 0

[@idjles]

@@anshumanagrawal346 I can use an epsilon/delta argument to show that it goes to zero and not asymptote to a non-positive number.

[@anshumanagrawal346]

@@idjles how

[@brian8507]

y = sin y.. only y = 0 solves this. There saved u 10 mins watching video