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Bartle problem 6 section 3.3

issacnewton

Member
Jan 30, 2012
61
here is the problem.
Let \( a>0\) and \( z_1 > 0\) . define \( z_{n+1}=\sqrt{a+z_n} \) for all \( n\in \mathbb{N} \). Show that \( (z_n) \) converges and find the limit.
I am supposed to use monotone convergence theorem. For that I need to prove that the sequence is bounded and monotone. I can prove that its bounded below by \( 0 \), but I am having trouble about the upper bound. Also sequence can be increasing or decreasing, depending upon the values
of \( a\) and \( z_1 \) . Any hints ?
 

chisigma

Well-known member
Feb 13, 2012
1,704
here is the problem.
Let \( a>0\) and \( z_1 > 0\) . define \( z_{n+1}=\sqrt{a+z_n} \) for all \( n\in \mathbb{N} \). Show that \( (z_n) \) converges and find the limit.
I am supposed to use monotone convergence theorem. For that I need to prove that the sequence is bounded and monotone. I can prove that its bounded below by \( 0 \), but I am having trouble about the upper bound. Also sequence can be increasing or decreasing, depending upon the values
of \( a\) and \( z_1 \) . Any hints ?
Let's suppose that z is real and let's write the recursive relation as...

$\displaystyle \Delta_{n+1}= z_{n+1}-z_{n}= \sqrt{a+z_{n}}-z_{n}= f(z_{n})\ ,\ a>0\ ,\ z_{0}>0$ (1)

In this case f(x) has two 'fixed points' [points where is f(x)=0...], an 'attractive fixed point' at $\displaystyle x_{+}= \frac{1+\sqrt{1+4 a}}{2}$ and a 'repulsive fixed point' at $\displaystyle x_{-}= \frac{1-\sqrt{1+4 a}}{2}$. The combined conditions $a>0$ and $z_{0}>0$ imply that in any case the sequence converges to $x_{+}$, if $0<z_{0}<x_{+}$ the sequence will be 'monotonically increasing', if $z_{0}=x_{+}$ the sequence will be constant, if $z_{0}>x_{+}$ the sequence will be 'monotonically decreasing'...

Kind regards

$\chi$ $\sigma$
 
Last edited:

Alexmahone

Active member
Jan 26, 2012
268
Let us first consider the case when $\displaystyle z_1<\frac{1+\sqrt{1+4a}}{2}$.

Step 1: Assume $\displaystyle z_n<\frac{1+\sqrt{1+4a}}{2}$ and show that $\displaystyle z_{n+1}<\frac{1+\sqrt{1+4a}}{2}$. This will prove (by induction) that $\displaystyle \{z_n\}$ is bounded above.

Step 2: $\displaystyle \frac{z_{n+1}}{z_n}=\sqrt{\frac{a+z_n}{z_n^2}}= \sqrt{\frac{a}{z_n^2}+\frac{1}{z_n}}$

Using $\displaystyle z_n<\frac{1+\sqrt{1+4a}}{2}$ (proved in step 1), show that $\displaystyle \frac{z_{n+1}}{z_n}>1$. This will prove that $\displaystyle \{z_n\}$ is increasing.
 
Last edited:

issacnewton

Member
Jan 30, 2012
61
This is chapter on sequences and the book has not done functions yet. I just want to assume the things which the book assumes at this point.
So with the information given, I think its better to split the problem in 3 cases, depending upon where \( z_1 \) is with respect to \( x_{+} \).
 
Jan 31, 2012
54
Let us first consider the case when $\displaystyle z_1<\frac{1+\sqrt{1+4a}}{2}$.

Step 1: Assume $\displaystyle z_n<\frac{1+\sqrt{1+4a}}{2}$ and show that $\displaystyle z_{n+1}<\frac{1+\sqrt{1+4a}}{2}$. This will prove (by induction) that $\displaystyle \{z_n\}$ is bounded above.

Step 2: $\displaystyle \frac{z_{n+1}}{z_n}=\sqrt{\frac{a+z_n}{z_n^2}}=sqrt\left(\frac{a}{z_n^2}+\frac{1}{z_n}\right)$

Using $\displaystyle z_n<\frac{1+\sqrt{1+4a}}{2}$ (proved in step 1), show that $\displaystyle \frac{z_{n+1}}{z_n}>1$. This will prove that $\displaystyle \{z_n\}$ is increasing.


Step 2*:

True(induction hypothesis) : $z_{n+1}>z_{n}$

Need to prove:

$$ z_{n+2}>z_{n+1}$$


$$z_{n+2}=\sqrt{a+z_{n+1}}>\sqrt{a+z_n}=z_{n+1}$$
 

Alexmahone

Active member
Jan 26, 2012
268

chisigma

Well-known member
Feb 13, 2012
1,704
Let's write again the recursive relation in term of difference equation...

$\displaystyle \Delta_{n}=z_{n+1}-z_{n}= \sqrt{a+z_{n}}-z_{n}= f(z_{n})\ ,\ a>0\ ,\ z_{0}>0$ (1)

In order to have a simple description of the problem the function $\displaystyle f(x)=\sqrt{a+x}-x$ for $a=1$ is represented here...

MHB02.PNG

The f(x) has only one 'attractive fixed point' [a point where is f(x)=0 and the function crosses the x axes with negative slope...] in $\displaystyle x_{+}= \frac{1+\sqrt{5}}{2}$ and it is evident that, because is $\displaystyle |f(x)|< |x-x_{+}|$ [see 'red line'...], any $0<z_{0}< x_{+}$ will generate an increasing sequence with limit $x_{+}$ and any $z_{0}> x_{+}$ will generate a decreasing sequence with limit $x_{+}$...

Kind regards

$\chi$ $\sigma$
 
Last edited:

issacnewton

Member
Jan 30, 2012
61
Thanks all. Solved the problem as suggested.