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## #1 2006-01-27 09:12:46

God
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### Recursions

Say we define a sequence such that:
a_0 = 1
and for each successive term;
a_n+1 = sin(a_n)

i. The infinite limit of the sequence a_n is 0, right?
ii. Does this infinite series have a finite sum? If so, what is it?
iii. Is there any explicit function ƒ(x) such that ƒ(x+1) = sin( ƒ(x) )?

-thanks

Last edited by God (2006-01-29 02:22:21)

## #2 2006-01-28 18:15:25

ryos
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### Re: Recursions

i. I think so.
ii. It stands to reason that, although the numbers will get so small as to be essentially 0, there will always be some teeny tiny insignificant bit more to add. I wouldn't know how to find out, though.
iii. Sure there is: ƒ(x) = sin(x-1)

El que pega primero pega dos veces.

## #3 2006-01-29 02:22:04

God
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### Re: Recursions

Thanks. I had a typo on part iii though:
I meant such that ƒ(x+1) = sin( ƒ(x) )

## #4 2006-01-29 03:40:13

Ricky
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### Re: Recursions

ii. By the ratio test:

Since
<
for all n:

< 1

Which means that the series converges, although I know of no tests that actually find this number.

And why isn't the latex '<' not working?

Last edited by Ricky (2006-01-29 03:41:37)

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."