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limn->¥sin(cos(sin(cos(sin(cos(......ntimes....(0)))).....)
By Arun Iyer on Saturday, October 13, 2001 - 08:24 pm:

Could anyone please evaluate this limit...

limn->¥sin(cos(sin(cos(sin(cos(......ntimes....(0)))).....)

The last term could be sin(0) or cos(0)
I hope the question is clear.
love arun

By David Loeffler on Saturday, October 20, 2001 - 05:34 pm:

Suppose that the limit does exist and call it L. Then we must have sin(cos(L))=L. We can solve this equation by iteration; it has a unique real root L = 0.6948196907. So if the limit exists it must be this.

To prove that it does exist is a little more irritating. It can be done; have you ever come across "cobweb" diagrams?

David

By Arun Iyer on Sunday, October 21, 2001 - 08:32 pm:

No i have not come across cobweb diagrams...
what are they??can you explain it to me please.

Infact i can't understand the whole thing that you have done here.Can you give me some details on your solution.....

love arun

By David Loeffler on Tuesday, October 23, 2001 - 07:08 pm:

In the first part I have just assumed that the sequence was tending to some (finite) limit.

Let's use some terminology here. Suppose we say that x0=0, and xn+1 = sin(cos(xn). So x1=sin(cos(0)), x2=sin(cos(sin(cos(0)))), etc.

Well, we are assuming that there is a limiting value L. Then if xn is close to L, then xn+1 must be even closer, and so on. But xn+1=sin(cos(xn)). So as xn gets close to L, sin(cos(xn)) must also get close to L; and we must have sin(cos(L))=L. This is an equation we can solve approximately for L, as above.


Now, the hard bit: the existence proof. This is often the case with evaluating limits - once you know there is a limit, it's not difficult to find out what it is, but it's hard to show that it exists at all.

THe cobweb method is one of those things which is very difficult to describe in words, but can be demonstrated very simply by a few diagrams. SO I will have to refer you to another site....

David

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