Background of Bisection Method of Solving a Nonlinear Equation

In this segment we'll talk about the background of the Bisection method that what the Bisection method is based on so far as solving a nonlinear equation is concerned. So, the theorem goes as follows says if we if we have an equation f(x) = 0, then if f(x) is real and continuous, and we find out that the value of the function at some point x_L, and then some point x_u, if the product of those two values less than zero, then at least one root lies between x_L and x_u. So, that is the theorem on which the bisection method is based on.

So, if we look at from a graphical point of view so, let's suppose somebody gives us f(x) = 0, and we plot f of x as a function of x, let's suppose, and let's suppose the function looks like this, we are seeking this point right here, because that's the root of the equation f(x) = 0. So, in this case if we find out there is some point x_u let's suppose and there is some point x_L then what we are finding out that the value of the function here is different from the value of the function here, or we can say that between these two points, between this point and this point the function is changing sign, then we have at least one root between x_L and x_u. So, although that's a qualitative way of saying it, but if we are writing a program then the way to do it is to calculate the value of the function at one point, and calculate the value of the function of the second point and see if the product of the two functions is negative, because when only when if one is positive, multiplied by another one is negative, do we get a negative number. So, because if we have two negative numbers, then we get a positive number. If we have two positive numbers, we multiplied by another positive number we'll get a positive number, and if we have a negative number and then a positive number then we get a negative number for the product. So, you can very well see that the function in these two cases is changing sign, but in this case in this case the function sign is not changing. So, we can only talk about that when the function is changing sign, which is given by this product here that the product of the two function values will be less than zero, then we can guarantee that there is at least one root which lies between x_L and x_u, provided of course that this function right here is real and is continuous. So, some of the ways you'll find out that this particular theorem which we just talked about is misinterpreted is as follows. So, let's suppose if I have my function my equation like this in my equation f of x equal to 0.

Let’s suppose the function behaves like as shown the figure here. Let's suppose if I choose this as my one value of the function, and this as another value of the function what you'll find out the function is not changing sign. Right? The value of the function here is positive, the value of the function here is positive, so I will get the value of the function here would be of the product of the two function values will be greater than zero. But we have a root here, and a root here, and one might think that that is the violation of that previous theorem. No. The theorem is only telling you that there is at least one root if the function is changing sign. It does not tell you anything if the function is not changing sign. So, if the product is greater than 0, then you know that the function is not changing sign, and there may or may not be a root between those two points. Let's continue with these misconceptions or maybe misperception or perceptions.

So, let's suppose if I have a function here and that is my function as a function of x and my equation is f(x) = 0, let's suppose I choose this as one of the points and I choose this as one of the points, then the function value at x_L, times the value of the function at the upper limit, which is this this point right here, is greater than zero. And in this figure, we don't find any roots between x_L and x_u. But this should not be interpreted as that, hey, if the function does not change sign that there will be no root between these two points. As we found out in the previous example, we had two roots between x_L and x_u when the function was not changing sign so keep that in mind, also, what the theorem is saying. It's only saying something if the function is changing sign. It does not say anything if the function is not changing sign. Let's look at another misconception which people have about the theorem which we just talked about. So, let's suppose if I have the function in this equation which is f(x) = 0, the function which is the left hand side right here, if this is how it behaves, and then I choose my lower limit to be this in the interval in which I say there is a root what you're finding out here is that the value of the function at x_L times the value of the function of x_u is less than zero, and so there will be at least one root between x_L and x_u it turns out that we have three roots between x_L and x_u. So, people somehow think that there should be only one root between those two points. No, the theorem is only saying that if the function is changing sign between the two points, then there will be at least one root. It does not tell you how many of them will be there, but at least one root between x_L and x_u. One more misconception takes place in a function which is like this, for example, here we have f(x) as a function of x, and let's suppose this is for the equation of x = 0, and let's suppose if I choose this to be x_L and this to be x_u, the value of the function at x_L times the value of the function x_u is changing sign, because it is positive here and negative here, but there's no root between x_L and x_u. So, is that a failure of the theorem? No, it is not the failure theorem because there are two other things which the theorem says; that the function has to be real, and it has to be continuous between those two points in order to apply that theorem. So, I’m hoping that it clarifies how we can apply the theorem what the theorem says and what the theorem does not say. And that's the end of this segment.