# Convergence prove question

1. Mar 11, 2009

### transgalactic

$$a_1=\sin x$$
$$-\infty<x<\infty$$
$$a_{n+1}=\sin a_n$$

prove that $$a_n$$ convergent and find

$$\lim _{n->\infty}a_n=?$$
the solution that i saw is that because
$$a_1=\sin x$$ then its bounded from 1 to -1

so

|$$a_{n+1}$$|<|$$\sin a_n$$|=<|$$a_n$$|

so its non increasing and it goes to 0.

but the teacher says that its not a proof
why its not a proof
how to solve it correctly??

2. Mar 11, 2009

### tiny-tim

Hi transgalactic!
Because you haven't proved it's non increasing …

and even when you do, you'll need to prove it doesn't decrease to a limit > 0

3. Mar 11, 2009

### transgalactic

yes i did
i showed the inequality and i written about the property of sinus

what else do i need to write??

4. Mar 11, 2009

### tiny-tim

no … you only wrote it

how would you prove it?

5. Mar 11, 2009

### transgalactic

i dont need to prove the inequality its obvious

no matter what number i will put into x its sinx will be from 1 to -1
and there is a theorem for which sin x and x have approximately the same value near 0

and i put absolute value because we cant look at it from - infinity too

how to prove this inequality
??

6. Mar 11, 2009

### tiny-tim

he he he :rofl:

it isn't obvious!
irrelevant!
how does that help?

you need to prove that sinx is smaller than x

7. Mar 11, 2009

### transgalactic

ok
i prove that sinx <x
f(x)=sinx-x
f'(x)=cosx-1
so for all x that differs 0 f'(x)<0 so f(x) is decreasing
f(0)=0
so on x=0 its 0 and decreasing
so f(x) <0 everywhere except x=0
so sin<x
what next
??

8. Mar 11, 2009

### tiny-tim

messy, but correct

ok, now it easily follows that an is decreasing …

but you still need to prove that it decreases to 0, and not to some number > 0

9. Mar 11, 2009

### transgalactic

$$a_{n+1}=\sin a_n$$
a1=sin x where -infinity<x<+infinity

i dont know how to prove that its decreasing
i cant use the function method that i used before with derivatives and stuff

i can only say an observation
??