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sequence with recursive definition?

skatenerd

Active member
Oct 3, 2012
114
Sorry to spam my problems all over this forum but series have me struggling somewhat. Last problem on my homework is the sequence an defined recursively by:
a1=1
and
an+1= \(\frac{a_n}{2}\) + \(\frac{1}{a_n}\)

First part was the only part i know how to do. it was to find an for n=1 through 5.
However this next part has me stumped. Assume that:
The limit as n approaches infinity = alpha > 0
Obtain the value alpha by taking the limit of both sides of (1) [(1) is the info given in the beginning].
Now I can already tell this limit is going to \(\sqrt{2}\) because it is kind of hinted to in the later parts of this problem. However I am kind of confused as to how to approach doing this limit of both sides of an+1= \(\frac{a_n}{2}\) + \(\frac{1}{a_n}\) . How would I evaluate this limit when the right side of the equation is in terms of a​n?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
One way (though not as directed by the problem) to demonstrate the sequence converges to $\displaystyle \sqrt{2}$ is to observe that if we begin with the function:

$\displaystyle f(x)=x^2-2=0$

which we know has the positive root $\displaystyle \sqrt{2}$ and apply Newton's method:

$\displaystyle f'(x)=2x$

$\displaystyle x_{n+1}=x_n-\frac{x_n^2-2}{2x_n}$

$\displaystyle x_{n+1}=\frac{2x_n^2-x_n^2+2}{2x_n}$

$\displaystyle x_{n+1}=\frac{x_n^2+2}{2x_n}$

$\displaystyle x_{n+1}=\frac{x_n}{2}+\frac{1}{x_n}$

For any positive initial value, the series will converge to the positive root of the defining function.
 

CaptainBlack

Well-known member
Jan 26, 2012
890
Sorry to spam my problems all over this forum but series have me struggling somewhat. Last problem on my homework is the sequence an defined recursively by:
a1=1
and
an+1= \(\frac{a_n}{2}\) + \(\frac{1}{a_n}\)

First part was the only part i know how to do. it was to find an for n=1 through 5.
However this next part has me stumped. Assume that:
The limit as n approaches infinity = alpha > 0
Obtain the value alpha by taking the limit of both sides of (1) [(1) is the info given in the beginning].
Now I can already tell this limit is going to \(\sqrt{2}\) because it is kind of hinted to in the later parts of this problem. However I am kind of confused as to how to approach doing this limit of both sides of an+1= \(\frac{a_n}{2}\) + \(\frac{1}{a_n}\) . How would I evaluate this limit when the right side of the equation is in terms of a​n?
If the limit \(\alpha\) exists and is non-zero then:

\[\alpha=(\alpha/2) + (1/\alpha)\]

(arrived at by taking limits of both sides of the recursion relation)

CB
 

skatenerd

Active member
Oct 3, 2012
114
Oh man thank you to you both. To CaptainBlack, I guess I didn't realize that seeing that if limit of n approaching infinity of an = alpha, then so will the this limit of an+1. Makes sense now! Thanks again.
 

chisigma

Well-known member
Feb 13, 2012
1,704
Sorry to spam my problems all over this forum but series have me struggling somewhat. Last problem on my homework is the sequence an defined recursively by:
a1=1
and
an+1= \(\frac{a_n}{2}\) + \(\frac{1}{a_n}\)

First part was the only part i know how to do. it was to find an for n=1 through 5.
However this next part has me stumped. Assume that:
The limit as n approaches infinity = alpha > 0
Obtain the value alpha by taking the limit of both sides of (1) [(1) is the info given in the beginning].
Now I can already tell this limit is going to \(\sqrt{2}\) because it is kind of hinted to in the later parts of this problem. However I am kind of confused as to how to approach doing this limit of both sides of an+1= \(\frac{a_n}{2}\) + \(\frac{1}{a_n}\) . How would I evaluate this limit when the right side of the equation is in terms of a​n?
The solving procedure for this type of problems is illustrated in...

http://www.mathhelpboards.com/f15/difference-equation-tutorial-draft-part-i-426/

The difference equation can be written as...


$\displaystyle \Delta_{n}= a_{n+1}-a_{n}= \frac{1}{a_{n}}- \frac{a_{n}}{2} = f(a_{n})$ (1)

The function f(x) has two 'attractive fixed points' in $x=\pm \sqrt{2}$ and, because for both the fixed points the theorem 4.2 is satisfied, roughly specking any initial value >0 will generate a sequence which tends to $+\sqrt{2}$ and any initial value <0 will generate a sequence which tends to $-\sqrt{2}$. An important detail in the case $a_{0}>0$ is that the sequence for any n>0 is decreasing...


Kind regards


$\chi$ $\sigma$
 

skatenerd

Active member
Oct 3, 2012
114
My teacher went over this problem recently and showed that it was a funny way of doing a proof of The Newtonian Method. Seeing as it was something we had never learned about before, I guess it was just a little something our teacher threw in to see if we could teach ourselves a concept on our own.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775