How to Prove the Convergence of a Sequence Defined by a Recursive Function?

[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??

 

[tiny-tim]

Hi transgalactic!

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

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

 

[transgalactic]

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

what else do i need to write??

 

[tiny-tim]

i showed the inequality …

no … you only wrote it

how would you prove it? ​

 

[transgalactic]

i don't 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 can't look at it from - infinity too

how to prove this inequality
??

 

[tiny-tim]

i don't need to prove the inequality its obvious

he he he

it isn't obvious!

no matter what number i will put into x its sinx will be from 1 to -1

irrelevant!

and there is a theorem for which sin x and x have approximately the same value near 0

how does that help?

you need to prove that sinx is smaller than x

 

[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
??

 

[tiny-tim]

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
??

messy, but correct

ok, now it easily follows that a_n is decreasing …

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

 

[transgalactic]

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

i don't know how to prove that its decreasing
i can't use the function method that i used before with derivatives and stuff

i can only say an observation
??