Contraction function

Question

We wanna show that the sequence $(t_n)_{n=1}^{\infty}$ recursively defined by $t_1=1$, $t_{n+1}=\frac{1}{2+t_n}$ for $\forall n \in \mathbb{N}$ converges to $-1+\sqrt 2$.

NOTE: i have to use the CONTRACTION THEOREM TO SHOW THIS. I have to make the recursively sequence a $f(x)$ and SHOW that it is a contraction.

Now i started on this problem and i used according to the Contraction Theorem that

For every $x\subset$ $I$ where $I=[0,1]$ we have that $f(x)\subset I$

And that there is a number $c$ such that $0\leq c<1$ such that for every $x,y$ in the interval we have $|f(x) - f(y)|\leq c|x-y|$

Now i started of and then after some algebra i got $\leq \frac{|x-y|}{|2+x| |2+y|}$

but now should i take $x=0, y=0$ because that is the minimim of the interval and i will get $c=\frac{1}{4}$? But am left with the fraction $\frac{1}{4} |x-y|$ which is also true according to the definition but i dont understand that why $|x-y|$ in the numerator still can stand after we decided that in the denominator we plugged in $x=0, y=0$

We don't need to actually put $x=y=0$, rather think it this way that $\frac{|x-y|}{|2+x|.|2+y|}\le \frac{|x-y|}4$ for all $x,y\in [0,1]$. — QED, Dec 30 '17 at 01:36
By finding the maximum of $|2+x|.|2+y|$ on the interval $[0,1]$ — QED, Dec 30 '17 at 01:38
@AbishankaSaha why should we focus on the denominator and why should we try to find the max? Also i dont understand the full use of this theorem, why is it needed for what purpose?? — Mark, Dec 30 '17 at 01:40
Sorry, i meant the minimum of the denominator. Because by finding the minimum of the denominator we can find a constant $c$, such that $1/c$ can serve as the contraction factor — QED, Dec 30 '17 at 01:50
@AbishankaSaha but why is 0 the minimum? And should we always plug in the minimum value on the interval? — Mark, Dec 30 '17 at 01:51
Because the function $|2+x|.|2+y|$ is increasing on the interval $[0,1]$ — QED, Dec 30 '17 at 02:21
@AbishankaSaha ok but still i dont get the use of the contraction theorem as a whole. And you are saying that we should not plug in a zero? But if the minimum of the interval [2,infinity] then what would have been the minimum a 4? Should we plug in the numbers x=y=2? In this case — Mark, Dec 30 '17 at 02:28
We wanna show that the sequence ... recursively defined by ... The sentence is unfinished. You want to show that the sequence does what? — dxiv, Dec 30 '17 at 02:29
@Mark What do you mean by the recursive sequence being a contraction function? — dxiv, Dec 30 '17 at 03:26
@dxiv also note that i dont fully understand the use of the contraction theorem i dont know what is does exactly — Mark, Dec 30 '17 at 03:56
For the relationship to contractions, see this previous answer. But you don't have to use contractions in order to prove the convergence of this sequence, so you'll have to decide on which question you mean to ask. — dxiv, Dec 30 '17 at 04:46
If $t$ is the limit then $t=1/(2+t)$, so $t(2+t)=1$, and $t^2+2t-1=0$ with solutions $t=\frac{-2\pm\sqrt8}2=-1\pm\sqrt2$, rule out the negative $-1-\sqrt2$. — Mirko, Dec 30 '17 at 04:52
@Mirko yes but i want to show it according to $f(x)$ a contraction — Mark, Dec 30 '17 at 13:21
@dxiv i dont get the relation $|f(x)-f(y)|<=c|x-y|$ how can that relation show that $f(x)$ will be a contraction please — Mark, Dec 30 '17 at 13:27

rtybase · Accepted Answer · 2017-12-30T18:01:24.527

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Use MVT instead to find $0\leq c<1$, i.e. $\exists \varepsilon \in (x,y) \subset [0,1]$ such that $$\left|f(y)-f(x)\right|=\left|f'(\varepsilon)\right|\cdot\left|y-x\right|$$ where $f(x)=\frac{1}{2+x}$. Or $$\left|f(y)-f(x)\right|=\left|\frac{1}{(2+\varepsilon)^2}\right|\cdot\left|y-x\right|=\frac{1}{(2+\varepsilon)^2}\cdot\left|y-x\right| \tag{1}$$ and since $0\leq \varepsilon \leq 1 \Rightarrow 2\leq 2+\varepsilon \leq 3 \Rightarrow 4\leq (2+\varepsilon)^2 \leq 9 \Rightarrow \frac{1}{4}\geq \frac{1}{(2+\varepsilon)^2} \geq \frac{1}{9}$. As a result, $(1)$ becomes $$\left|f(y)-f(x)\right|\leq\frac{1}{4}\cdot\left|y-x\right| \tag{2}$$ for $\forall x,y \in [0,1]$. So we take $c=\frac{1}{4}$.

Now, let's show that $$\forall x \in [0,1] \Rightarrow f(x) \in [0,1] \tag{3}$$

$0\leq x \leq1 \Rightarrow 2\leq 2+x \leq3 \Rightarrow \frac{1}{2}\geq \frac{1}{2+x} \geq \frac{1}{3} \Rightarrow 1 > \frac{1}{2} \geq f(x) \geq \frac{1}{3}>0$.

From $(2)$ and $(3)$ we have that $f(x)$ a contraction mapping on $[0,1]$. According to Banach fixed-point theorem and the fact that $t_1=1 \in [0,1]$, there $\exists! t^{*} \in [0,1]$ such that $t^{*}=f(t^{*})$ where $\lim\limits_{n\rightarrow\infty}t_n=t^{*}$ or $$t^{*}=\frac{1}{2+t^{*}} \Leftrightarrow (t^{*})^2+2t^{*}-1=0$$ This has 2 solutions $-1+\sqrt{2}$ and $-1-\sqrt{2}$, but only $-1+\sqrt{2} \in [0,1]$

edited Dec 30 '17 at 18:01

answered Dec 30 '17 at 16:42

rtybase

16,907

Why did you take $|f’(x)|$ as c?? Why take the derivative ?? Is the constant $c$ a derivative of f(x)??? And also where did i proved that $f(x)$ lies in the interval [0,1]? – Mark Dec 30 '17 at 16:58
I didn't take $|f'(x)|$ as $c$. I used MVT to maximise $|f'(x)|$ on $[0,1]$. – rtybase Dec 30 '17 at 17:00
ok but it seems you replaced $c$ with the derivative of $f(x)$?and then you maximized the derivative and took that as a c? I really have to use the contraction theorem not the mvt :-( – Mark Dec 30 '17 at 17:04
I really have to understand what $|f(x)-f(y)|<=c|x-y|$ means what does that mean?? How can i sketch that? – Mark Dec 30 '17 at 17:07
Yes, I used max of $|f'(x)|$ on $[0,1]$ as $c$, which fits the purposes. I used MVT to show that there exists such a $c=\frac{1}{4}$ which makes $f(x)$ a contraction mapping. Now, you can use Banach fixed point theorem, more details here – rtybase Dec 30 '17 at 17:15
You also stated "For every $x\subset I$ where $I=[0,1]$ we have that $f(x)\subset I$" ... I assume you proved it. – rtybase Dec 30 '17 at 17:23
no i did not proved that yet. How cAn i prove that? Thanks for the info still somehow not so clear but a bit more now – Mark Dec 30 '17 at 17:55
@Mark I have updated my answer with a lot more details, including extra proofs, let me know if more clarity is still required. Specifically, I believe that by "CONTRACTION THEOREM" you mean "Banach fixed-point theorem" which is a really famous and popular result. – rtybase Dec 30 '17 at 17:59
where were you all the time men!! You really nailed it !!! But i have one more question left what does the statement $|f(x)-f(y)|<=c|x-y|$ really means can you explain it in your own words then i would understand it fully – Mark Dec 30 '17 at 18:25
In the context of contraction mapping and this particular example, it means that "for all $x,y \in [0,1]$ there exists a constant $c$, $0\leq c < 1$, such that $|f(x)-f(y)| \leq c|x-y|$". It is important to remember "for all $x,y$" part. – rtybase Dec 30 '17 at 18:32
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alright thank you alot!!! Its clear to me now!!! – Mark Dec 30 '17 at 20:21

Contraction function

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