00:01
Okay, so the first thing that we're going to do to approximate this is to find a function.
00:08
So what i'm going to do is i'm going to let x be the square root of 11.
00:13
So then from here, i can square both sides.
00:19
So with that, i'll get that x squared is equal to 11.
00:22
So now if i bring 11 to the left hand side, so i'll have x squared minus 11 is equal to 0.
00:29
So now this becomes my f of x function.
00:33
So now i can erase this.
00:38
And also we'll need another piece of information.
00:42
So we're going to need f prime of x.
00:47
So that's just going to be the derivative of this, which is 2x.
00:53
Okay, so now if we use this formula, the newton's method, so x sub n plus 1 is equal to x sub n minus f of x.
01:07
X sub n over f prime of x sub n.
01:14
Okay, so now the first thing that we need to do is we need to come up with a guess for the square of 11.
01:20
Well, i know 11, the square of 11 is somewhere between the square of 9 and the square of 16.
01:26
So i know it's gonna be around 3.
01:29
So i'm gonna let my guess be 3.
01:35
So with that, i'm gonna say that x of 0 is 3.
01:45
So then from there, because i know this, and i know my function, now i can figure out x1.
01:52
So from here, i'll though i need to find f of x0.
01:58
So that means i need to plug 3 into this function here.
02:02
So this gives me 3 squared minus 11.
02:08
So that'll be negative 2...