# TIFR 2013 problem 20 | Sequence limit~Fixed Point

Try this problem 20 from TIFR 2013 based on Sequence Limit - Fixed Point. This problem is an application of the Banach Fixed Point Theorem.

Question: TIFR 2013 problem 20

True/False?

Consider the function $f(x)=ax+b$ with $a,b\in\mathbb{R}$. Then the iteration $x_{n+1}=f(x_n)$; $n\ge0$ for a given $x_0$ converges to $\frac{b}{1-a}$ whenever $0<a<1$.

Hint:

Banach Fixed Point Theorem

Discussion:

Once the existence of limit is guaranteed, we can safely calculate the limit. If limit is $x$ then $x=ax+b$. In other words, $x=\frac{b}{1-a}$.

The question really is does limit exists?

$x_{n+1}-x_{n}=f(x_{n})-f(x_{n-1})$

$=ax_{n}-ax_{n-1}=a^{2}(x_{n-1}-x_{n-2})=...=a^{n}(x_{1}-x_{0})$.

For $n<m$,

$x_m-x_n=x_m-x_{m-1}+x_{m-1}-x_{m-2}+...-x_{n}$

$=(a^{m}+...a^{n})(x_1-x_0)=\frac{a^n}{1-a}(x_1-x_0)$

And the right hand side converges to 0 as n tends to infinity (Here,we are using the fact that $0<a<1$) i.e, the right hand side can be made arbitrarily small using large enough n, so $x_n$ is a Cauchy sequence. We are in $\mathbb{R}$, so by completeness, $x_n$ converges.

Remark:

The above discussion really didn't make use of Banach Fixed Point Theorem. But knowledge is power. And if known that :

"Let $T:X\to X$ be a contraction mapping, X-complete metric space. Then T has precisely one fixed point $u\in X$. Furthermore, for any $x\in X$ the sequence $T^k (x)$ converges and the limit is $u$. " (Banach Fixed Point Theorem).

the answer is immediate. The above solution runs in the line of the proof of this Theorem.