a crazy iterated limit.

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[sinecurve]

This problem has a very dynamical systems flavor to it.

The graphs you mentioned at the end didn't show up.

[renerpho]

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

[emmanuellaurens]

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.

[goodplacetostop]

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

[joeeeee]

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

[lyrimetacurl]

The easy way:
y=sin(sin...))
y=sin(y)
take the inverse;
arcsin(y) = y
Solution exists only when arcsin(y) = sin(y) which is when y=0

[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

[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.

[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.

[albertobarreirohermida]

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

[florisv]

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.

[buhlaigah]

As an engineer, I'd recognize that this is a fixed point iteration, solve sin(x)=x, and call it a day lol

[as-qhqq]

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.

[luckycandy]

From the fixed point convergence theorem, for all real x, if x≠pi*k for some integer k, |d/dx(sin(x)|=|cos(x)|<1 and thus the series converges to the fixed point x=sin(x) which is of course 0 (can be proven in numerous ways),
and if x=pi*k for some integer k, sin(x)=0 and thus the whole sequence is zero from n=2 and so converges to 0 as well.

Very nice proof though, I love seeing ground level proofs from time to time.

[samosamo]

Ground level proof. Great professor. In this way we make Maths easy for all students. By the way one can also see that convergence is uniform.

[marc-andredesrosiers]

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

[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.

[JonathanMandrake]

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

[davidmacfarlane]

Goodness, How does he write so fast with chalk? I can hardly write a fraction of that speed with my own pen and paper, and with less neatness than him!

[markusvohburger]

Although I have no use for higher mathematics in my everyday live, I really enjoy watching your new videos every day. Thank you very much Professor!