# Sequence limit~Fixed Point (TIFR 2013 problem 20)

Question:

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.