CHAPTER 01.03: SOURCES OF ERROR : Truncation Error: Series
In this segment we're going to look at an example of truncation error. And this example is going to relate to a Maclaurin series. So, for example, if we have e to the power x, the Maclaurin series for that is given as 1 plus x plus x squared by factorial 2 plus x cubed by factorial 3, and so on and so forth. And if somebody told me that, hey, I'm going to only use the first three terms of this Maclaurin series, and I'm going to forget about the rest of them, this part here turns out to be your truncation error, that part is your truncation error. So no matter how many terms you take, you might say, hey, maybe this seems to be too big a truncation error to take, what I will do is I will take a million terms of this Maclaurin series to calculate e to the power x, but still, the million-and-oneth term which you'll have, and any term after that will still be part of the truncation error.
So let's go ahead and take an example here for putting the value of x to be 1.2. So let's suppose somebody's using this particular series here to calculate e to the power 1.2. So in that case, the series would become this, divided by 2 factorial plus 1.2 cubed cubed divided by 3 factorial, and so on and so forth. So, again here, we'll have infinite number of terms in the series, and what we need to do is, we need to be able to see that how many terms do we really need to be able to . . . of calculating e to the power 1.2. But there will be always truncation error associated with the calculation of e to the power 1.2, because of the fact that there are infinite number of terms in the series. So I'm going to draw a table here, and let's go ahead and see what kind of numbers do we get for e to the power 1.2 as we keep on increasing the number of terms. So in this particular table, I'm going to use n, which is the number of terms in the series. I'm going to use e to the power 1.2. What approximate value do I get for e to the power 1.2 by taking so many terms in the series, what is the approximate error, and what is the absolute relative approximate error, in terms of percentages? Let me go ahead and, for completion purposes, let me write down the series again. The series is 1 plus 1.2 plus 1.2 squared by factorial 2 plus 1.2 cubed by factorial 3, and so on and so forth. So if I was going to take one term in the series, e to the power 1.2 will be just 1. And in this case I cannot calculate my approximate error, because I don't have a previous approximation, and the relative approximate error also cannot be calculated, because I don't have a previous approximation. If I take two terms in the series, what value of e to the power 1.2 I'm going to get? I'm going to get 1 plus 1.2, which is 2.2. In this case, the approximate error will be the difference between the present approximation and the previous approximation, which will be 1.2, and the relative . . . absolute relative approximate will be 1.2 divided by the present approximation times 100, and that number will turn out to be 54.5 percent. That's what I'm going to get as my absolute relative approximate error. If I choose three terms, I'll be choosing 1, 1.2, and 1.2 squared divided by factorial 2, and that number turns out to be 2.92, my approximate error is 0.72, and my relative approximate error is 24.658. And you can see that, as I keep on increasing the number of terms, so it might be a good idea for you to fill this table up at home, as homework. Now I'm going to show you what do I get for six, so you can go ahead and calculate the value at six at home, and see whether you match what I am getting. I get 3.3151, the approximate error is 0.020736, and this is 0.62255 percent. So one of the things which you are seeing here is that as you keep on increasing the number of terms which you are using in your Maclaurin series, that your value for e to the power of 1.2 is asymptotically getting to the value . . . exact value e to the power 1.2. Your approximate error is decreasing, and also your relative approximate error is decreasing, it went from 54 to 24 to 0.62255, which, in this case you are finding out that this number here is less than or equal to 5 percent. So what that means is that, it's not less than 0.5 percent, but it's less than 5 percent, what that means is that one of the . . . at least one of the significant digits are correct in this solution which you just obtained right here. So you have 3, which you got there, so you can say that this particular digit here is at least correct.
Now as you keep on increasing the number of terms, what's going to happen is that this absolute relative approximate error is also going to keep on decreasing, hence you will have more confidence, you can say more of the number of significant digits are at least correct. Because as soon as this number here, the relative approximate turns out to be less that 0.5, then you have two significant digits to be at least correct, if it turns out to be 0.05 percent, and at least three significant will be at least correct. So that's how you overcome the problem of the truncation error, that you are going to have truncation error, but what you want to try to do is that you want to minimize the amount of truncation error, and that's something which you do have some control of. You cannot totally get rid of truncation error, but you can surely minimize it. And that's the end of this segment. |