Hello, mongrel73!
Could you restate the problem?
There seems to be a serious typo . . .
I'm struggling on this but i get the feeling it's easier than i think.
The sequence is defined by:
xn+1 = (xn + 2) / 6
where the initial value is 0.
and you need to show that
| xn+1 - xn | </= (5/6) * |xn - xn+1 | </= (1/3) * (5/6)^n-1
after having shown that 0< xn <1 (which i was able to do).
( by the way </= means 'less that or equal to", i didn't know how to put it on the computer though.)
can anyone help?
that wasn't what the definition of the iteration was. i'll try it in latex:
ok i tried in latex it didn't work so i'll just try to be clearer:
xn+1 = [ ( (xn)^5 ) + 2 ] / 6
is the correct formula
as opposed to
xn+1 = [ xn+2 ] / 6
which is where the misunderstanding came from i think.
also the right-most term of the double inequality should be:
(1/3) * [ (5/6)^(n-1) ]
rather than just:
(5/6)^(n-1)
I think i just completely forgot to put in the ^5 part!
Then I made another typo with the actual problem:
| xn+1 - xn | </= (5/6) * |xn - xn-1 | </= (1/3) * (5/6)^n-1
Thanks, please help!