Re: Array problem

Saysero <saysero@xxxxxxxxx> writes:

On Jan 22, 3:54 pm, s...@xxxxxxxxxxxxxx wrote:
On 22 Jan, 13:15, Saysero <says...@xxxxxxxxx> wrote:

On Jan 22, 1:40 pm, s...@xxxxxxxxxxxxxx wrote:

On 22 Jan, 12:06, Saysero <says...@xxxxxxxxx> wrote:

If a_n in R and y_n = 2*a_n + a_(n-1) and y_n -> y show that a_n ->
a.
Thanks in advance.

Use a_n = y_n/2 -y_{n-1}/4 +y_{n-2}/8-... +/-y_1/2^n -/+a_0/2^n

You will get a = y/3

How?

You have that for any e>0 there exists an integer N such that for
n>=N
|y_n - y| < e

You want to show that for given d>0 there exists an integer M such
that for m>=M
|a_m - y/3| < d

So choose e based on d (e=d/2 will do), which gives you N; then find
an M (based on d, N, the sequence {y_i} for 0<=i<N, and a_0) which
satisfies this.

I don't see how it can be done that way. I am sure it is not that is
not the idea.

That way works fine, but maybe you'd prefer this. Given epsilon > 0,
take N so for all k >= N, |y_k - y| < epsilon/2.
Suppose, say, |a_n - y/3| >= epsilon, for some n much larger than N. Then
|a_(n-1) - y/3| = |y_n - 2 a_n - y/3| >= 2 |a_n - y/3| - |y_n - y|
> (1 + 1/2) epsilon.
Similarly
|a_(n-2) - y/3| >= 2 |a_{n-1} - y/3| - |y_{n-1}-y| > (2 + 1/2) epsilon.
By mathematical induction,
|a_(n-k) - y/3| > (2^(k-1) + 1/2) epsilon as long as n-k >= N.

Take n is large enough that (2^(n-N-1)+1/2) epsilon > |a_N - y/3|, and
this is impossible. Thus for all n sufficiently large,
|a_n - y/3| < epsilon.
--
Robert Israel israel@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
.

Quantcast