Question

does sin(sin(sin(sin(sin(...sin(x))))))))))))))....) converge to any number?

Answer 1

it should go to zero

Answer 2

0 i expect

Answer 3

right!

Answer 4

because x can be anything and it will go from -1 to 1 and then from -1 to 1 u will get 0 eventually

Answer 5

right

Answer 6

but it may be \(\pm0\) :DD

Answer 7

http://ideone.com/Yrttc

Answer 8

http://ideone.com/TsQ7N

Answer 9

but I'm sure there's a way to show this

Answer 10

analytically

Answer 11

Write out its sum:
\[\sum_{k=0}^{\infty}\frac{(-1)^{2k+1}(\frac{\sum_{k=0}^{\infty} (-1)^{2k+1}(\frac{\sum_{k=0}^{\infty}...}{(2k+1)!})}{(2k+1)!})^{2k+1}}{(2k+1)!}\]
And find the nth partial sum and check for convergence :D

Answer 12

McLaurin series, whatever.

Answer 13

it was question with sin, where did u get all this garbage? :D

Answer 14

that is the mclauron series for sinx

Answer 15

Math Education:\[\sin(x)=\sum_{k=0}^{\infty}\frac{(-1)^{2k+1}x^{2k+1}}{(2k+1)!}\]

Answer 16

iterated many times...
not sure that's solveable.

Answer 17

Turing, just btw, I love your picture. And I've been reading your posts and you seem intelligent so whats up :D When I quit getting on here often you weren't around I think haha

Answer 18

i know only geometric and arithmetic series :/

Answer 19

It depends on how many times you compose sin(x) with itself right? n times or infinite? That changes the problem significantly.

Answer 20

Yeah thanks malevolence.
You definitely know your stuff better than me I think ;)

Answer 21

i would try this. assuming it converges to say "y" then solve
\[y=\sin(y)\]

Answer 22

that gimmick presupposes that it converges to something.

Answer 23

So you assume it converges. But the only thing you can do then is to take the arcsin over and over and over. So you get:
\[\cos^{-1}(\cos^{-1}(...\cos^{-1}(L)))...)\]
Assuming your limit is L.

And Turing, probably not haha. I just try to be creative D:

Answer 24

Those should be sin's***

Answer 25

Oh check one more thing: look at that last post. http://openstudy.com/users/malevolence19#/updates/4ef73f34e4b01ad20b50e6fb

Answer 26

Haha
That's a heck of a rant there malevolence XD

Answer 27

sounds like William S Burroughs...

Answer 28

oh Steve Wright I see.

Answer 29

I was just pulling stuff off the top of the head haha. And that integral didn't yield 0. x.x So apparently I set it up wrong or something. I got 4 pi. The thing is, I did the integral correctly because I got the same result by applying stokes theorem (vector calc). Idk /shrug/ what can I do?

Answer 30

\[\Huge \oint\]

Answer 31

Man, I'll have to pick that apart later, I just got done giving myself a headache trying to figure out what I did wrong on ruhmshalle's post :/

Answer 32

Need help?

Answer 33

It is clear based on satellite's argument that if it converges, then it must converge to zero. Assume x is in the interval [0, pi/2]. We can do this without loss of generality because if x isn't, then +/- sin(x) definitely is, so it quickly becomes irrelevant. Noting that sin(x) < x for all values of x and that y < x implies that sin(y) < sin(x), it is apparent that the sequence defined by
\[a_n = \sin(a_{n-1}), a_0 = x \] is monotonically decreasing and has a liminf of zero. So, it converges to zero.

Answer 34

Very, very slowly though...

Answer 35

I should actually rephrase. It is not immediately clear that the liminf is zero (though I don't know what else it would be) but it is clear that the function is bounded by zero, and all monotonic bounded sequence converge. Satellite's argument takes over from there.

Answer 36

sequences*

Answer 37

I think you're looking at this with the wrong angle, there's a very simple approach using sequences...
You have a sequence in the form
\[U_{n}=f(U_{n-1})\]
With f=sin
This sequence will converge iff there's a point such that x=f(x), and clearly the only point that satisfies this is 0, so it converges on 0, QED.

Answer 38

That is not always the case... if f is x^2, for instance, it actually diverges the other way. Most of us are not terribly familiar with fixed point iterations anyway.

Answer 39

It works because sin is a strictly increasing function in [-pi;pi], and after the first iteration your term will obviously be in this interval.

Answer 40

right. I want to find out what happens when n is infinity with that sequence someone wrote up there.

Answer 41

There are more conditions that are necessary... x^2 is strictly increasing from 0 to infinity, and after the first iteration our term will be in that interval too.

Answer 42

I'm not trying to be belligerent, I'm just trying to say that there are definitely conditions that need to be satisfied for the convergence of a floating point iteration to work out properly, and what I said above about the sequence being monotonically decreasing and bounded is sufficient. Maybe not using the language that you're used to, if you're familiar with F.P.I.'s, but it works.

Answer 43

What are FPIs? Sorry, I'm not familiar with math in english

Answer 44

I just abbreviated Fixed Point Iteration. And oops, I meant fixed point, not floating point o.O

Answer 45

True, anyhow, it can be proved through the method I stated, I didn't type it out rigorously enough though... I think you can also do it with constantly decreasing intervals, since sin goes from R into [-1;1] and sin([-1;1]) will be an even smaller interval, then by iterations your intervals will become smaller and 0 will always be in them, meaning that for all nth iteration there will exist an interval that's a neighborhood of 0, meaning your sequence will tend to 0