Hi, In this lesson we're going to talk, just an intro to numerical analysis. Now what is numerical analysis? Well, it's the way we find a solution to an equation where we can't find the exact solution necessarily. You think you just plug into your calculator? How does your calculator get the answer? Or how does the computer get that answer? It's important for us to know the different ways we can get those answers in the different pros and cons to each approach. We're going to start with talking about how would you find a root of an equation. So for instance, say I want to find an approximation to the square root of 2. If I squared both sides, if I looked at this as my f of x. If I look at f of x is equal to x squared minus 2. What does that look like? It looks something like that. What I want is that point, that is the square root of 2. That is where x squared minus 2 is equal to 0, it's where the height is 0. That's where my f of x is 0. If I want to find that root, what I'm trying to do is to find where this function is 0. This is going to give me an approximation to this number. There's many ways I can do this and these are all called root finding methods. I'm using these methods as an example for you of how numerical analysis is used. One way I could do it is I could use what's called a bisection method. What I could do is, I could say, okay, what is f of 1? f of 1 is negative 1. Here's 1, negative 1. That's that point. F of 2 is 2 squared minus 2, which is 2, so f of 2 is 2. I know that since it's negative here and positive here, that's somewhere in between because it's a continuous function. This had to cross the axis. I know now my root is in the interval between 1 and 2, so I bisect the interval. I look at one-and-half. Let's say, that's f of 3/2, which is 3/2 squared, 9/4 minus 2, minus 8/4, which is 1/4, which is positive. F of 3/2 when I bisect to this interval was positive, it was up here somewhere. I know now my root is in the interval from 1 to 3/2, 1/2 because I got 1 negative and 1 positive. I will bisect that interval. I'll look at f of 1/4. F of 1/4 is 1.5625 minus 2. I don't really care what the actual value is, I just need to know it's negative. When I go to 1/4, which I have bisected this interval I'm at the negative. Now, I know my root is between 1.25 and 1.5. I can keep going like this as many iterations as I want to keep dividing it, dividing it and dividing it till I get as many decimal place accuracy as I want or as my computer or calculator can handle. I will get an approximation. This is never going to give me, particularly in a case of an irrational number, it will never give me an exact answer. It's going to give me an approximation to so many decimal places. Now this is great, but you might think that was slow and really, when I was here, I knew the answer was closer to this number. I could have take something that wasn't quite half the interval. But we're programming this into a computer to do, so we want to be able to do this automatically and find the answer. Now think about this. What you have to do with this method is the pros and cons to this 1 negative is you have to find two points to start the give different signs 1 positive and 1 negative. That can be a problem. You may or may not be able to find those points. You might do a lot of hunting around before you find those points. That's the bad thing. The other bad thing is, it's slow. Those are bad things, those are the cons. The good side is, once you've done this, you will always converge. It always is going to get you the answer. Now, if I had picked this point and this point, it would have taken me to this route, which may not be the route I want. You can imagine if you had a function that looks like this, and you want to find a particular route, then you better start with guesses that are on each side of that root so that you can find a solution. When we come back, we're going to talk about a thing called Newton's method, which has different trade-offs. I'm going to leave these up here and then I'll show you how Newton's method works.
