Square Root

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

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

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

Then you have [tex] y = \frac {y}{2} + \frac {x}{2y} [/tex]

This gives you a quadratic in y.

dduardo

Ok, now I understand. Thanks