CHAPTER 01.07: TAYLOR SERIES REVISITED: Taylor Series: Example
In this segment we're going to take an example of Taylor series, and see how we can apply Taylor series to be able to get the value of the function at some other point. What I mean to say is that, let me write down the Taylor series again. So the general form of the Taylor series, or Taylor's Theorem, is that the value of the function can be calculated at some point x plus h, if we know the value of the function at x, if we know the value of the derivative of the function at x, we know the value of the second derivative of the function at x, and we know the third derivative of the value of the function at x, and so on and so forth, then I can give you the value of the function at some other point.
So in this example, let's suppose somebody tells you the following, that, hey, says you are given the value of the function at 4, some function, is given as 125, the value of the first derivative of the function is given as 74, the value of the second derivative at 4, same point, is given 30, the value of the third derivative at 4 is given as 6, and all other derivatives are 0. So you're given the value of the function at 4, the first derivative at 4, the second derivative at 4, and third derivative at 4, and then all the other derivatives are given as 0, that means the fourth derivative at 4, the fifth derivative at 4, the sixth derivative at 4, and all those other derivatives are all given to be 0. So those are all 0, and what is asked for you is to find the value of the function at 6. Of course, it is already given to you, the fine print is already given to you that all these derivatives are continuous and exist between the point x equal to 4 to x equal to 6, that's the only way you can apply Taylor's Theorem. So let's go ahead and see that, how we can find the value of the function at 6. Now, again, I want to point out here is that nowhere are you given what the function itself is, because if you were given what the function itself is, then you could just substitute the value of x equal to 6, and you would get what the value of the function at 6 is. So this is the beauty of Taylor's Theorem, that you are only given the value of the function at a particular point and all its derivatives at that particular point, and you are asked to find out what the value of the function at some other point is, in this case it is the value of x equal to 6. So if I'm going to apply my Taylor series, my x is 4, my h is 2, because x plus h will then give me the value of 6, and that's where I'm interested in finding out the value of the function, right? So what I'll get is that f of 6 is exactly equal to the value of the function at 4, plus the derivative of the function at 4 times h, which is 2, plus the second derivative of the function at 4 times h squared divided by factorial 2, h I have 2, plus the third derivative of the function at 4 divided by factorial 3 times h cubed, which is 2 cubed. And then the fourth derivative and so on and so forth, which are all 0. All the . . . all the fourth derivative, fifth derivative, sixth derivative, and so on and so forth, they're all 0, so I will not get any contribution from those terms. So this value of the function which I'm going to calculate at 6 will be an exact value, there is no approximation involved in it, because all the derivatives are given to me, and also that the derivatives which are after the fourth derivative, they're all 0. So f of 4, which was given to me as 125, plus f prime of 4, which is given to me as 74, plus f double prime of 4, which is given to me as 30, plus f triple prime of 4, which is the third derivative, which is given as 6. And this value here turns out to be 341. In fact, what I'm going to do is I'm going to write down the individual contributions, I get 125 from here, 148 from here, 60 from here, and 8 from there. So, from the four terms which are available for the Taylor series, because the other ones are 0, I get 125 contribution from here, 148 from here, 60 from here, and 8 from there. So you can see that the contribution will decrease as you add more terms. It doesn't mean that it's always going to be decreasing, it could be increasing in certain cases, but if you choose h to be small enough, it's going to continue to decrease. Here you are finding out the second term is contributing more than the first term, so please don't think that as you keep on using more terms in the Taylor series, that their contribution is smaller and smaller, term by term, and the value turns out to be 341, and this is the value of the function at 6. And you were able to calculate the value of the function at 6 without knowing what the expression for the function is, but from the information that you knew what the initial value at x equal to 4 is, what the derivative of the function at 4 is, what the second derivative of the function at 4 is, the third derivative of the function at 4 is, and all the other derivatives, you're able to calculate what the value of the function at 6 is, in this case being 341, okay? And that's the end of this segment. |