Proving Limit of y[n] for Newton-Raphson Method

[dduardo]

Can someone get me going in the right direction.

For the given function:

y[n] = (1/2)(y[n-1] + x[n]/y[n-1])

where x[n] = a * u[n] (u[n] is the unit step function)

and y[-1] = 1

prove that y[n] as n -> infinity is equal to sqrt(a)

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I know that this is the Newton-Raphson Method, but how do I go about analytically proving the above.

I've tried writing out a few terms and seeing if there is a pattern, but couldn't find anything.

The inside looks like a accumulator and tried to do a subsitution, but that didn't work.

Any help would be appreciated.

 

[Gokul43201]

When you see limiting behavior, it means that the differencebetween successive terms gets smaller and smaller. In the infinite limit, the difference between successive terms should be 0.

So, set y[n] = y[n-1] and see what happens.

 

[dduardo]

I think I need a little more information.

So you are saying lim n-> inf (y[n] - y[n-1]) = 0 ? But since I'm dealing with a circular function how would I go about taking the limit?

 

[Gokul43201]

I think I need a little more information.

So you are saying lim n-> inf (y[n] - y[n-1]) = 0 ?

Yes, does it seem reasonable to you that this should be true ?

But since I'm dealing with a circular function how would I go about taking the limit?

There really isn't much to do. Let's call $\lim_{n \rightarrow \infty} y[n] \equiv y$

Then you have $$y = \frac {y}{2} + \frac {x}{2y}$$

This gives you a quadratic in y.

 

[dduardo]

Ok, now I understand. Thanks