00:05
So this problem, we're asked to use newton's method to approximate the solutions to x squared equal 10 to four decimal places.
00:15
So according to newton's formula, the solution is going to be an iterative solution.
00:21
So you're going to have an initial guess.
00:24
So it's going to be my initial guess and then minus f of x0 over f prime of x0.
00:36
And then i can just iterate for let me to try this over and over.
00:38
Okay, you know that 3 squared is equal to 9.
00:41
So let's start with x0 equal 3.
00:46
Then what i'm going to come up with is my guess is going to be x is equal to 3 minus, okay? and if you look at it's the function x squared minus 10.
00:58
And so f of 3 is going to be 9 minus 10, which is negative 1.
01:04
And so if this is f of x, then you, you're going to be 9 minus 10, which is negative 1.
01:09
You know that f prime of x is going to be 2x so this is two times three this is three plus one six so this is what 19 over six and then if i get an approximation of that so 19 divided by six is 3 .1 6667 so 3 .1 16667 so 3 .1 1666 seven, okay? and now i just want to iterate until i get four decimal places of accuracy.
01:48
So let's just use our calculator for this.
01:51
Okay.
01:52
So x squared minus 10.
01:55
I'm going to store that as my original function.
02:01
And then the derivative of that is going to be 2x.
02:05
So i'm going to store that as df.
02:08
And let's just pick a value of a for my iteration.
02:11
So the next value i'm going to pick is 196.
02:15
So 19 over 6...