00:01
Okay, so here we are given a function, f of x is equal to x squared minus a.
00:08
We want to drive newton's recursive formula, so we compute f prime of x.
00:14
So f prime of x is just going to be equal to 2x.
00:18
And we can plug this into newton's formula, so x sub n plus 1 is equal to x of n minus x of n squared minus a over two times x of n.
00:31
And if we can simplify that, we get that x of n plus 1 is equal to, so x of n squared minus a over 2x of n can be written as 2x of n squared minus x squared minus a, all over 2x of n, which is going to be equal to x of n squared plus a over 2x of n, which is equal to x of n plus a over x of n all over two.
01:05
And we've shown then, yes, that x sub n plus 1 is equal to x of n plus a plus a over x of n, all over 2.
01:17
And this is the classic formula for approximating the square root using newton's method.
01:22
And then for the second part, we can compute the iterations with x -0 to be 1, and then a to be equal to 3, and we use x -n plus 1 is going to be equal to x -s -of -n plus 3 over x -n all over 2.
01:41
So if we calculate a few terms, we get x -0 is 1...