S.O.S. Mathematics CyberBoard

Hello, first time here @ a math forum, hoping for some help as my class is getting difficult. I am an extreme rookie @ writing proofs prior to this semester and, probably shouldn't be in the class I am, but I kept in it, and got an 84 on my first exam

Alright, Suppose that A is any positive real number. Define the sequence {a_n} recursively by

a_n+1= .5(a_n+A/a_n),

with n in N and a1 is an arbitrary positive number. Prove that {a_n} converges to radical(A) = (A)^.5 Note that if we write a_n+a above as

a_n+1 = a_n - ((a_n)^2 - A)/(2a_n) we obtain the formula for the Newton-Raphson method.

Really do not have a good lead on this problem, I am thinking I want to prove weather it is increasing or decreasing (maybe break it down into cases where A<a1, A>a1, A=1, and see where the sequence is eventually increasing, strictly increasing, or eventually strictly increasing?) and also I think I need to see if the sequence is bounded.

Thanks for any information,
-Doodle

Multiply through by , show it is monotone and bounded so that the limit exists, and take limits on both sides of the infinite recursion that happens.

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I multiplied through by a_n, so I got

a_n+2=.5(an^2+A)

Then I did up to a7, n=5.
a3=.5(a1^2+A)
a4=.5(a2^2+A)
a5=.5(a3^2+A)=.5((.5(an1^2+A)^2+A)=.25a1^4+.25Aa1+.25A^4

I am not sure how to prove weather this is increasing or decreasing. Since A is a constant, as n gets large in an, then that A term will just go away? Thanks for any and all help,

-DoodleNoodle

Eh? I get

So we get



Take limits and letting we see: and solving the quadratic you get what you want. Again, first you need to show the limit exists to pull this trick, but once you do it's easy. Showing you get the positive square root is just a byproduct of checking out and making some easy observations.

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Good point Shadow, thanks a lot for the so-far help.

Can't I just simplify that to

L^2-2L^2+A=0

-1L^2+A=0

A=L^2, then L=square root of A

How do I go about proving that a_n converges?

Monotone + bounded proves convergence, you should know that. Also, you need to make sure you note that your initial guess guarantees you get the positive square root.

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Well,

when A>(a1)^2, then the sequence is increasing, and when A<(a1)^2, the sequence is decreasing. Also, when A=(a1)^2, then the sequence is a constant, a_n=(A)^1/2. A being a positive constant
Since,
a_n+1 = a_n - ((a_n)^2 - A)/(2a_n).

I don't see an obvious bound that I can use to prove

|a_n-L|< epsilon for all n >= N

The whole point is that this is the Newton Rhapson method in disguise. Draw a picture of both cases and you should be able to see why the sequences are bounded and monotone.

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draw a picture for both cases? Meaning the graph for when A>(a1)^2, then the sequence is increasing, and when A<(a1)^2 ?

Just like plot points on a number line, like A = 5, and a1=1, then A=5, a1=10?

Yes, the picture should show you how to proceed with the proof of monotonicity.

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