# Arithmético-geometrical continuation

A arithmético-geometrical continuation is a continuation with values in a body and defined by recurrence by


\ begin {boxes} u_ {n_0} = U \ \ \ forall N \ geq n_0, \ quad u_ {n+1} = au_n + B \ end {boxes} In general, one works on $\ R$ (body of the real ) or $\ mathbb C$ (body of the complex ).

## Use

The arihmético-geometrical continuation meets in the modeling of certain flows of population (fixed contribution and escape proportional): contribution of 10 and escape of 5%, $u_ \left\{n+1\right\} = u_n+ 10 - 5 \ % \ times u_n$

It also meets in the plane of refunding: a capital C borrowed froma monthly rate T and refunded by monthly payments M led to the development of a plan of refunding. If $R_n$ represents the capital remaining due at the end of N monthly payments, the continuation $\left(R_n\right) \,$ is a arithmético-geometrical continuation of relation of recurrence: $R_ \left\{n+1\right\} = \left(1 + T\right) R_n - M$

One also finds it in a Chaîne of Markov in two states. The stochastic Matrice is then


\begin{pmatrix} has & 1-a \ \ 1-b & B \end{pmatrix}

Relation

$\left(p_ \left\{n+1\right\}, q_ \left\{n+1\right\}\right) = \left(p_n, q_n\right)$
\begin{pmatrix} has & 1-a \ \ 1-b & B \end{pmatrix}

It is deduced that:

$p_ \left\{n+1\right\} = ap_n + \left(1-b\right) q_n \,$.
$q_n = 1-p_n \,$,
while replacing one obtains
$p_ \left\{n+1\right\} = \left(+ B has - 1\right) p_n + 1 - B \,$

## General term

For the commonplace case where = 1 has, one deals with arithmetic Suite.

If $a \ 1$, one seeks by translation to bring back to a geometrical Suite:

One poses $v_n = u_n + c$
One shows that $\left(v_n\right) \,$ is geometrical if $c = - \ frac \left\{B\right\} \left\{1-a\right\}$
One finds whereas $v_n = v_ \left\{n_0\right\} a^ \left\{n-n_0\right\}$
Then, thanks to the relations between $u_n$ and $v_n$, $u_n = \ frac \left\{B\right\} \left\{1-a\right\} + a^ \left\{n-n_0\right\} \left(u_ \left\{n_0\right\} - \ frac \left\{B\right\} \left\{1-a\right\}\right)$

One can as find the general term, by observing as this continuation consists in building the sum of the term of a geometrical continuation. To illustrate it, one can be interested in the case of in the following way defined continuation (definition 2):


\ begin {boxes} u_ {0} &= U \ \ u_ {n+1} &= au_n + B \ end {boxes} ,

One notices whereas

$u_1=aU+b$
$u_2=a^2U+ab + b$
$u_3=a^3U+a^2b + ab + b$
the general term is expressed by
$u_ \left\{N\right\} =a^ \left\{N\right\} u_ \left\{0\right\} + \ sum_ \left\{i=0\right\} ^ \left\{n-1\right\} a^ \left\{I\right\} b$.

With the sum of the first terms of a geometrical continuation, one obtains the following general term:

$u_ \left\{N\right\} =a^ \left\{N\right\} u_ \left\{0\right\} + B \ dfrac \left\{1-a^ \left\{N\right\}\right\} \left\{1-a\right\} = a^ \left\{N\right\} \left(u_ \left\{0\right\} - \ dfrac \left\{B\right\} \left\{1-a\right\}\right) + \ dfrac \left\{B\right\} \left\{1-a\right\}$
By posing $r= \ dfrac \left\{B\right\} \left\{1-a\right\}$, one finds
$u_ \left\{N\right\} =a^ \left\{N\right\} \left(u_ \left\{0\right\} - R\right) +r \,$.

## Summon $n$ first terms

For a continuation defined according to definition 2, one has $\ sum_ \left\{i=0\right\} ^ \left\{n-1\right\} u_ \left\{I\right\} = \left(u_ \left\{0\right\} - R\right) \ dfrac \left\{1-a^ \left\{N\right\}\right\} \left\{1-a\right\} + NR \,$.

## Convergence

The general term and the considerations on the geometrical continuations make it possible to determine the limit of such a continuation according to the values of has and, possibly, the sign of $u_ \left\{n_0\right\} - \ frac \left\{B\right\} \left\{1-a\right\}$.

An interesting remark is to be made if |has| < 1. In this case, the limit of the continuation is $\ frac \left\{B\right\} \left\{1-a\right\}$ whatever the initial value. The limit of a continuation of this type is thus completely independent of the initial conditions . This characteristic is to be put in glance with the continuations at nonlinear recurrence (Suite logistic) which can, they, being very sensitive to the initial conditions. In a Chain of Markov, that proves that the chain converges towards a stationary chain.

Category: Elementary mathematics

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