Contraction principle

doodlenoodle · **Joined:** Sun, 12 Feb 2012 02:16:54 UTC **Posts:** 64

Hello,

Consider the sequence a_n

, where $a_{1}=1$ and $a_{n+1}=1+\frac{1}{1+a_n}$ , for all n in N.

a) is a_n

monotone?
b) Use the contraction principle to show that a_n

converges.
c) Verify that ${a_n}$ converges to radical 2.

a) No not monotone since a1<a2>a3<a4>a5... and so on.
b) $|a_{n+2}-a_{n+1}|<k|a_{n+1}-a_n|$ where k is between 0 and 1.

$|a_{n+2}-a_{n+1}|$ = $\frac{a_{n}+a_{n+1}}{(1+a_{n})(1+a_{n})}$

a_n>1

for all n greater than 1

Since $\frac{1}{1+a_n}<1$ , then $a_{n+1}<2$ for all n

Thus $(1+a_{n})(1+a_{n+1})$ > (2)(2) = 4
So here we take k=1/4 satisfying the contraction principle.

c) Since a_n

converges to some L, we see that $\lim_x\to\infty}a_{n+1}=1+\frac{1}{1+a_n}=L=1+\frac{1}{1+L}= \sqrt{2}$

Does this suffice as an answer?

Shadow · **Posted:** Wed, 21 Mar 2012 01:46:24 UTC

doodlenoodle wrote:

Hello,

Consider the sequence a_n

, where $a_{1}=1$ and $a_{n+1}=1+\frac{1}{1+a_n}$ , for all n in N.

a) is a_n

monotone?
b) Use the contraction principle to show that a_n

converges.
c) Verify that ${a_n}$ converges to radical 2.

a) No not monotone since a1<a2>a3<a4>a5... and so on.
b) $|a_{n+2}-a_{n+1}|<k|a_{n+1}-a_n|$ where k is between 0 and 1.

$|a_{n+2}-a_{n+1}|$ = $\frac{a_{n}+a_{n+1}}{(1+a_{n})(1+a_{n})}$

a_n>1

for all n greater than 1

Since $\frac{1}{1+a_n}<1$ , then $a_{n+1}<2$ for all n

Thus $(1+a_{n})(1+a_{n+1})$ > (2)(2) = 4
So here we take k=1/4 satisfying the contraction principle.

c) Since a_n

converges to some L, we see that $\lim_x\to\infty}a_{n+1}=1+\frac{1}{1+a_n}=L=1+\frac{1}{1+L}= \sqrt{2}$

Does this suffice as an answer?

You need to state that L>0, which is easy, but necessary.

doodlenoodle · **Joined:** Sun, 12 Feb 2012 02:16:54 UTC **Posts:** 64

Otherwise it looks good?

Also, finally getting into the latex format.

Shadow · **Posted:** Wed, 21 Mar 2012 01:55:52 UTC

doodlenoodle wrote:

Otherwise it looks good?

Also, finally getting into the latex format.

Pro tip: when doing limits, start off your code box with a $ it changes $\lim_{x\to a}$ into $$\lim_{x\to a}$

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