CHAPTER 03.04: NEWTON-RAPHSON METHOD: Advantages & Drawbacks for Newton-Raphson Method: Part 2 of 2


You're going to get oscillations near local maxima and minima.  Here is an equation of x squared plus 2, which does not have any real roots, x squared plus 2 equal to 0 does not have real roots, which you can see that if you look at the graph x squared plus 2, it . . . it has . . . it doesn't cross the x-axis anywhere. But you could start with some initial guess, trying to find out the roots of this equation, because you can surely set it up in Newton-Raphson method formula, this particular equation, and so this is what happens, you . . . let's suppose you start with -1, which seems to be a good place to start, you take the tangent, you end up right here, then you go up here, you draw the tangent, again, there to see where it crosses the x-axis, it's crossing right here, you go all the way up to right here, and it crosses the x-axis, the tangent crosses the x-axis right here, and then you go all the way up to there, the tangent crosses the x-axis all the way at here.  Now you might think that it's diverging.  No, actually it's not diverging, it's going to come back and start oscillating, again, back around the local minimum here, which is at x equal to 0, so that can happen when you are using Newton-Raphson method.  Again, you have to understand that this equation does not have any roots between x equal to . . . it does not have any real roots, so but still, if you're going to use Newton-Raphson method algorithm you will find out that it starts oscillating between one point and another, and starts getting closer and closer to x equal to 0, not closer and closer, but it will start oscillating around x equal to 0, as you conduct more and more iterations.  This is another issue which you can have when using Newton-Raphson method on equations.


So if you have equations such as this, x minus 1, whole cubed, plus 0.512 equal to 0.  Now, what is different about this particular equation is that it has an inflection point at x equal to 1, so at x equal to 1, there's an inflection point.  What do we mean by inflection point is that you have . . . you have concavity changing to the left of it and to the right of it.  So here it is concave up, here it is concave down, so that particular point is called an inflection point, and one of the ways to find an inflection point in the function is to see that if the second derivative of the function at that particular point is becoming 0 or does not exist, that does not make it to be an inflection point, just because the second derivative is 0 at a particular point does not make it an inflection point, it makes it a candidate for an inflection point.  Now, since in this case, the second derivative is going to be changing sign to the left of it and the right of it, because it you look at the second derivative of the function will be 6, x minus 1, and you can see that at x equal to just before 1 it's negative, just after 1 it's positive, so since the second derivative is changing sign across the inflection point, we know that this is an inflection point.  And now what happens in Newton-Raphson method formula is that you may start with a good guess of 5, it does not seem to be a bad guess to be finding out the exact root of 0.2.  0.2 is the exact root of this particular equation, and you find out that you start with initial guess of 5 right here, and when you apply the Newton- Raphson method formula, it starts getting closer and closer to 0.2, which is the exact value, but now what you are finding out is that at this particular point at 0.9259 it is getting closer to the inflection point of x equal to 1, and it starts diverging, it goes to -30 as the next initial . . . next iterative value, but again, what happens is that, in this case, it starts, again, to converge, and after twelve more iterations, it converges on the exact root of 0.2. So you are going to have issues with equations which have inflection points which are close to the root, and that's the case here, so here we have the inflection point x equal to 1, the root is at 0.2, this is where the root is, and if you find out that any of the initial guesses or the iterative values turn out to be close to the actual root, you are going to have some divergence problems in the Newton-Raphson method.  And that's the end of this segment.