Sequence with recursive definition?

[skate_nerd]

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

First part was the only part i know how to do. it was to find a_n 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 a_n+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]

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]

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

First part was the only part i know how to do. it was to find a_n 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 a_n+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

 

[skate_nerd]

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 a_n = alpha, then so will the this limit of a_n+1. Makes sense now! Thanks again.

 

[chisigma]

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

First part was the only part i know how to do. it was to find a_n 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 a_n+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 $\chi$ $\sigma$

 

[skate_nerd]

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 into see if we could teach ourselves a concept on our own.

 

[MarkFL]

I believe the Babylonians are credited with first developing a method for recursively approximating square roots, which is the same process that results from Newton's method.

Methods of computing square roots - Wikipedia, the free encyclopedia