CHAPTER 03.05: SECANT METHOD: Derivation of Secant Method: Approach 2 of 2
In this segment, we are going to derive secant method of finding roots of a nonlinear equation. So, if somebody gives you a nonlinear equation like this, f of x equal to 0, which basically implies that if I draw it graphically that you're going to plot the function f of x of that equation f of x equal to 0, and you're going to draw it like this, then what you are trying to do is you're trying to find out where does the function become 0, and that becomes the root of f of x equal to 0, or some people might say it is the zero of the function f of x, which is the same thing. So that's what we want to be able to do, and let's go ahead and see how we can derive secant method and what it is based on. So, let me go ahead and draw . . . draw this . . . this plot again, so that we can show what secant method is all about. So, let's suppose somebody gives you a function like this, and says . . . of the function . . . of the equation f of x equal to 0, and wants you to find out the root of this equation which is right here. So, what secant method does is that it says, hey, let me go ahead and take some initial guess, let's suppose xi, let me go ahead and take another initial guess, let's suppose x-sub-i-minus-1 here, let's suppose, and what I'm going to do is I'm going to draw a secant line, I'm going to draw a secant line between the two points, so this is one point and this is another point. So this point here is nothing but xi, and the value of y is the value of the function at that particular point. This here, again, is x-sub-i-minus-1 and the value of the function at x-sub-i-minus-1. So what you want to do is you take the secant line and you continue this secant line now, and see where it crosses the x-axis. So that's what you are doing, you're taking this secant line which you have, so you're taking this secant line which you have, it's not looking exactly straight, but it's a one secant straight line, and you find out where it crosses the x-axis, so that's where it's crossing the x-axis, so it's at x-sub-i-plus-1, and that's going to become your estimate of the root of the equation. So this becomes the estimate of the second . . . next estimate of the root of the equation, starting with this estimate and this estimate here, and then you're going to do the same thing, you're going to draw a secant line between this point and this point here, and, again, see where the secant crosses the x-axis, and that becomes your root. So, let's suppose if I call this to be point A, I call this to be B, I call this to be C, I call this to be D, and I call this to be E. Now we can very well see that triangle ABC and triangle ADE are similar. Now what does that mean is that you can very well see that they are similar, this triangle ABC and triangle ADE are similar, because all the angles are the same. So, in similar triangles, what you can very well say is that then this height here divided by this height will be equal to this width divided by this width right here. So what I'm trying to say is that DE, so if I take this height, DE, and I divide it by this height, which is BC, I'll put it equal to then this length of AD, and then divided by this length, which is AB. So we've got, from similar triangles, that's what we get from similar triangles, that the two heights are proportional to the two widths which we are seeing right here. So, but what is . . . what is DE? DE is nothing but the value, or the length of DE is nothing but the value of the function at xi. And what is BC? BC is nothing but the value of the function at x-sub-i-minus-1. Now, what is AD, what is the length of AD? The length of AD is nothing but xi minus x-sub-i-plus-1, because it is this point here, D, and A is x-sub-i-plus-1, so you've xi and x-sub-i-plus-1, so it's the difference between the two, and same thing, AB, which is this length right here, is the difference between this point and this point, which is, again, x-sub-i-minus-1 minus x-sub-i-plus-1, so that's what we're getting from there. So you can very well see that, now, in this case, if I am choosing x-sub-i-minus-1 to be one of the initial guesses, xi to be the other initial guess, then the only unknown which I have is x-sub-i-plus-1. So, bottom line is to be able to solve this simple linear equation for x-sub-i-plus-1, and this is what I'm going to get, if I take x-sub-i-plus-1, I'm going to get x-sub-i-plus-1, after a little bit of algebra. So that's what I get as the simplification by writing down the unknown, x-sub-i-plus-1, in terms of the other two . . . the two guesses, xi and x-sub-i-minus-1, and the values of the function at those particular points. And this is nothing but the formula for the secant method. So, secant method has been derived here by taking two initial guesses, x-sub-i-minus-1 and xi, and you are able to get x-sub-i-plus-1, so you have to start with two initial guesses. Again, keep in mind that although you're going to start with two initial guesses, you still . . . this stall falls under the category of the open method, just like other methods, such as Newton-Raphson method, and the reason why that is so is because these two initial guesses, or any of the estimates, two estimates which you're going to use for the roots or root of the equation f of x equal to 0, they don't necessarily have to bracket the root, they can be outside of the bracket, they can be . . . they can be bracketing the root, so that's why it's still called the open method. And that's the end of this segment. |