Limit of Sequences: Showing a_n > x^(1/2) & a_n -> x^(1/2)

[crshbr]

Hi there!

Homework Statement

Ok here is my problem concerning a sequence that is bounded and should have a limit.

\Large x\geq0 and \Large a_{0}>\sqrt{x}

The sequence \Large a_{n} is defined by \Large a_{n+1}=\frac{1}{2}(a_{n}+\frac{x}{a_{n}}) where \Large n\geq0

So the first question is to show that \Large a_{n}>\sqrt{x}
then it asks me show that the limit of \Large a_{n}=\sqrt{x} as \Large n\rightarrow\infty

The problem is that I can't seem to get anywhere with the first part without referring to limits. I was told to prove the first part using induction but I don't really know what hypothesis to make from which I could then continue.

As for the second part. I haven't really tried doing it yet, as I want to do the first part before moving on to it. However I can't seem to get the hang of this epsilon-N method. I understand what I am doing and why I am doing it, however I don't really understand how I should choose epsilon. Sometimes I am told to use epsilon=1/N or epsilon/2 or sometimes it's just left as epsilon>0 depending on the question. How would I be able to know instantly what epsilon to use when doing a proof.

Homework Equations

The Attempt at a Solution

Thank you.

 

[LCKurtz]

What you are to prove by induction is: Given x ≥ 0, a₀ ≥ \sqrt x, and

a_{n+1}= (\frac 1 2)(a_n+\frac x {a_n})

prove a_n ≥ \sqrt x for n = 1,2,...

So you need to start with the case n = 0, which is to prove a₁ ≥ \sqrt x, or, using the formula for a₁:

(\frac 1 2)(a_0+\frac x {a_0}) \ge \sqrt x

Do that and you will probably see how to do the induction step too.

 

[crshbr]

Hi,

As a matter of fact I have already done that but I didn't get anything meaningful in my opinion. :(

Here is what I did:

\frac{1}{2}\left(a_{0}+\frac{x}{a_{0}}\right)\geq\sqrt{x}

Let a_{0}=\sqrt{x}+y

a_{1}=\frac{1}{2}\left(\frac{\left(\sqrt{x}+y\right)^{2}+x}{\sqrt{x}+y}\right)

a_{1}=\frac{x+y(\sqrt{x}+y)}{\sqrt{x}+y}

a_{1}=\frac{x}{a_{0}}+y

So how does this really help me forward? This is the point I got stuck on.

Thanks for your help.

 

[LCKurtz]

Man, that is confusing and I don't see what its point is. Try proving the inequality in the last line of my first post directly without introducing new variables. Maybe working backwards using reversible steps will help you.

 

[crshbr]

Not a lot you can do there really though :S

All you end up with is:

a_{0}^{2}+x>2a_{0}\sqrt{x}

which is not necessarily true...seeing as how the inequality only holds up to the point where

a_{0}^{2}>a_{0}\sqrt{x}

 

[LCKurtz]

Put it all on one side and don't cancel anything out. See if you recognize the form...

 

[crshbr]

Hah, that's great!

And for part 2, would this be a correct proof for convergence:

Assuming \left\{a_{n}\right\} is monotonic and decreasing

\forall\epsilon \exists a_{N}\in\left\{a_{n}\right\}

so that

L+\epsilon>a_{N}>L

L+\epsilon>a_{N}>a_{n}>L

L+\epsilon>a_{n}

so \epsilon>|a_{n}-L|

now as \epsilon>0 and L=\sqrt{x}

|a_{n}-\sqrt{x}|<\epsilon

|\sqrt{x}-\sqrt{x}|<\epsilon so therefore the limit is correct

 

[LCKurtz]

A decreasing and bounded below sequence would be convergent.

1. Have you proved the sequence is bounded below?
2. Have you proved the sequence is decreasing?

Your argument above makes about as much sense as your original inequality argument did, which is to say not much. You haven't shown L even exists, let alone that it is \sqrt x.

Once you have shown 1 and 2 you know the limit exists. You can figure out what the limit is by letting n --> oo in the recursion formula. At that point you can use limit theorems instead of epsilon arguments.

 

[LCKurtz]

Also I forgot to ask you, did you do the induction step after you did the n=1 step?