# Square Root

1. Jan 30, 2005

### dduardo

Staff Emeritus
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.

Last edited: Jan 30, 2005
2. Jan 30, 2005

### Gokul43201

Staff Emeritus
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.

3. Jan 30, 2005

### dduardo

Staff Emeritus

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?

4. Jan 30, 2005

### Gokul43201

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

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.

5. Jan 30, 2005

### dduardo

Staff Emeritus
Ok, now I understand. Thanks