Proof of convergence for a recursive sequences

[simo...@yahoo.com]

Hi, Everyone!

Underscore sign designates subscript. Thus A_(n+1) means A sub n+1, not
A_n + 1. We have two constants k1 =10^-2 and k2=10^-3. Initially we
have A_n = 1 and B_n = 0.

A_(n+1) = A_n - k1*A_n
B_(n+1) = B_n + k1*A_n
B_(n+2) = B_(n+1) - k2*B_(n+1)
A_(n+2) = A_(n+1) + k2*B_(n+1)

Proof that k1*A_(n+R) where R -> infinity will equal k2*B_(n+R) where R
goes to infinity.

Thank You Very Much.

Truly Yours, Simon Dexter

[Robert Israel]

In article <1126889362.7...@z14g2000cwz.googlegroups.com>,

<simo...@yahoo.com> wrote:
>Hi, Everyone!
>
>Underscore sign designates subscript. Thus A_(n+1) means A sub n+1, not
>A_n + 1. We have two constants k1 =10^-2 and k2=10^-3. Initially we
>have A_n = 1 and B_n = 0.

i.e. for n=0?

>A_(n+1) = A_n - k1*A_n
>B_(n+1) = B_n + k1*A_n
>B_(n+2) = B_(n+1) - k2*B_(n+1)
>A_(n+2) = A_(n+1) + k2*B_(n+1)
>
>Proof that k1*A_(n+R) where R -> infinity will equal k2*B_(n+R) where R
>goes to infinity.

It's not true, but close.

If x_j is the vector
[ A_{2j} ]
[ B_{2j} ]
the recurrence can be written x_{j+1} = M x_j where M is the matrix
[ 1 - k1 + k1 k2 k2 ] [ .99001 .001 ]
[ k1 - k1 k2 1 - k2 ] = [ .00999 .999 ]
This has eigenvalues 1 and .98901, a normalized eigenvector for 1 being
[ 100 ]
u_1 = [ 999 ]/sqrt(1008001)
As n -> infinity, M^n will have the limit
[ 10000 99900 ]
u_1 u_1^T = [ 99900 998001 ]/1008001
[ 1 ] [ 10000/1008001 ]
With x_0 = [ 0 ], x_n has the limit [ 99900/1008001 ]
So for even n, A_n/B_n approaches 100/999, which is not
k2/k1 = 1/10.

Robert Israel isr...@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada