# Solving x = a x + b

Problem: Solving $x = ax +b$ for $x$.

Solution:

Choose $x_0$ arbitrarily, we generate the sequence $\{x_i\}$ recursively from $x_i = a x_{i-1} + b, i=1, 2, ...$:

$x_1= a x_0 + b,$

$x_2 = a x_1 +b =a(a x_0+b)+b =a^2 x_0 +ab + b,$

$x_3 = a x_2 + b = a (a^2 x_0 + ab +b) +b = a^3 x_0 +a^2b +ab +b,$

$\ddots$

$x_n = a x_{n-1} +b = a^n x_0 + b \sum\limits_{i=0}^{n-1} a^i=a^n x_0 + b\cdot\frac{1-a^{n}}{1-a}.$

It follows that for $|a| <1$,

$x = \lim\limits_{n \rightarrow \infty} x_n = \lim\limits_{n\rightarrow \infty}a^n x_0+\frac{b(1-a^n)}{1-a}= \lim\limits_{n\rightarrow \infty}a^n x_0+ \lim\limits_{n \rightarrow \infty}\frac{b(1-a^n)}{1-a}\overset{(\star)}{=}0\cdot x_0 + \frac{b(1-0)}{1-a}.$

i.e.,

$x = \frac{b}{1-a}.$

Fig. 1 shows a CAS -aided solution using Omega CAS Explorer:

Fig. 1

For $|a|> 1$, we rewrite $x=ax +b$ as

$x = \frac{1}{a}x - \frac{b}{a} \overset{A=\frac{1}{a}, B=-\frac{b}{a}}{=} Ax +B.$

Since $|a| >1 \implies \frac{1}{|a|} = |A| <1\implies \lim\limits_{n\rightarrow \infty}A^n = 0$,

$x =\lim\limits_{n \rightarrow \infty}A^n x_0 + B\cdot \frac{1-A^n}{1-A}\frac{}{} = \frac{B}{1-A} = \frac{-\frac{b}{a}}{1-\frac{1}{a}} = \frac{-b}{a-1} = \frac{b}{1-a}.$

We have used fact that

$|a| < 1 \implies \lim\limits_{n \rightarrow \infty} a^n = 0.\quad\quad\quad(\star)$

A proof is as follows:

Since

$|a^n| = |\underbrace{a\cdot a\cdot a ... a}_{n\;a's}| =\overbrace{|a||a|...|a|}^{n\;|a|'s} = |a|^n,$

if $a=0$ then $a^n = 0\cdot 0 ... 0 = 0.$ It means

$(\forall \epsilon >0, \forall n \in N^+, |a^n-0|<\epsilon)\implies \lim\limits_{n \rightarrow \infty} a^n = 0.$

Otherwise ($a \ne 0$),

$\forall \epsilon >0, |a^n-0| =|a|^n < \epsilon \implies n \log(|a|)<\log(\epsilon).$

For $|a|<1,$

$n\log(|a|)<\log(\epsilon) \overset{\log(|a|) < 0}{\implies} n > \frac{\log(\epsilon)}{\log(|a|)}.$

And so,

$(\forall \epsilon > 0, \exists n^* = \lceil\frac{\log(\epsilon)}{\log(|a|)}\rceil \ni n > n^*, |a^n-0| < \epsilon) \implies \lim\limits_{n \rightarrow \infty}a^n = 0.$

Exercise-1 Prove by mathematical induction:

$(x_{k} = a x_{k-1} + b, k=1,2,...,n ) \implies x_n = a^n x_0 + b\sum\limits_{i=0}^{n-1}a^i.$