Define function $f(x)$ from $\mathbb R^1$ to $\mathbb R^1$ by $f(x) = 1+1/x$. Define $a_n$ inductively by $a_1=1$ and $a_n=f(a_{n-1})$ for $n \geq 2$.
- Prove using the contraction principle that $(a_n)$ is convergent
- Prove that if $\lim_{n\to \infty}$ of $a_n= L$ then $L = F(L)$; hint Q2C
- Determine $\lim_{n\to \infty}$ of $a_n$
My thoughts
I honestly have no idea what the contraction principle is, I have googled it and have found no results. And I also don't know what Q2C means.
The Banach contraction principle (which I state here only for the reals) says that if $f$ is a contraction mapping, that is if there is $q$ with $0\le q<0$ such that $|f(x)-f(y)|\le q |x-y|$ for all $x,y$ then $f$ has a unique fixed point $L$, that is $f(L)=L$, and moreover taking any $a_1$ the sequence $a_1,f(a_1),f(f(a_1)),...$ converges to $L$.
So, $a_1=1$ and $a_2=1+\frac11=2$. We will prove by induction that $\frac32\le a_n \le 2$ whenever $n\ge 2$. Indeed this is true for $n=2$. Assuming it is valid for some $n\ge2$, we have $\frac12\le \frac1{a_n} \le \frac23$, so $1+\frac12=\frac32 \le 1+\frac1{a_n}=a_{n+1} \le 1+\frac23=\frac53<2$, which completes this proof by induction.
So consider the restriction of $f$ to the interval $[\frac32,2]$, we will show that it is a contraction mapping with $q=\frac49$. Indeed, take any $x,y\in[\frac32,2]$. Then $|f(x)-f(y)|=|1+\frac1x-(1+\frac1y)|= |\frac1x-\frac1y|=|\frac{y-x}{x y}|\le|y-x|\cdot(\frac23)^2=\frac49|x-y|$.
So, since $f$ is a contraction mapping on $[\frac32,2]$, the sequence $a_2,f(a_2),f(f(a_2)),...$ converges to some limit $L\in[\frac32,2]$.
To find the value of $L$ use $Q2C$. If you don't know what $Q2C$ means (I do not), then use common sense :) In the equality $a_n=f(a_{n-1})$ take limit as $n\to\infty$, we obtain $L=f(L)=1+\frac1L$ (note that this is exactly what your hint says), so we have the equation $L^2=L+1$, or $L^2-L-1=0$, with roots $L_1=\frac{1-\sqrt{5}}2$ and $L_2=\frac{1+\sqrt{5}}2$. Note that $0>L_1\not\in[\frac32,2]$ hence we must have that $L=L_2=\frac{1+\sqrt{5}}2$.