# Thread: Problem involving an iterative sequence?

1. ## Problem involving an iterative sequence?

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?

2. Hello, mongrel73!

Could you restate the problem?
There seems to be a serious typo . . .

$\displaystyle \text{The sequence is de{f}ined by: }\;x_{n+1} \:=\:\frac{x_{n+2}}{6}\; \text{ where }x_1 = 0$

$\displaystyle \text{and you need to show that: }\;\underbrace{\left|x_{n+1} - x_n\right| \;\leq \; \frac{5}{6}\cdot\left|x_n - x_{n+1}\right|}_{{\color{blue}\text{This can't be true!}}} \;\leq \;\left(\frac{5}{6}\right)^{n-1}$

3. ## sorry no i didn't use latex so i guess it was a little ambiguous

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