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_ {n+1} = 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 $(R_n) \,$ is a arithmético-geometrical continuation of relation of recurrence: $R_ {n+1} = (1 + T) R_n - M$

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

(p_ {n+1}, q_ {n+1}) = (p_n, q_n)

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

It is deduced that:

p_ {n+1} = ap_n + (1-b) q_n \,

Like in addition,

q_n = 1-p_n \,

while replacing one obtains

p_ {n+1} = (+ B has - 1) 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

(v_n) \,

is geometrical if

c = - \ frac {B} {1-a}

One finds whereas

v_n = v_ {n_0} a^ {n-n_0}

Then, thanks to the relations between

u_n

and

v_n

u_n = \ frac {B} {1-a} + a^ {n-n_0} (u_ {n_0} - \ frac {B} {1-a})

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_ {N} =a^ {N} u_ {0} + \ sum_ {i=0} ^ {n-1} a^ {I} b

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

u_ {N} =a^ {N} u_ {0} + B \ dfrac {1-a^ {N}} {1-a} = a^ {N} (u_ {0} - \ dfrac {B} {1-a}) + \ dfrac {B} {1-a}

By posing

r= \ dfrac {B} {1-a}

, one finds

u_ {N} =a^ {N} (u_ {0} - R) +r \,

Summon $n$ first terms

For a continuation defined according to definition 2, one has

\ sum_ {i=0} ^ {n-1} u_ {I} = (u_ {0} - R) \ dfrac {1-a^ {N}} {1-a} + 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_ {n_0} - \ frac {B} {1-a}

An interesting remark is to be made if |has| < 1. In this case, the limit of the continuation is $\ frac {B} {1-a}$ 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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Arithmético-geometrical continuation

Use

General term

Summon n first terms

Convergence

Summon $n$ first terms