Video: Approximating Error Using Maclaurin Series

Determine the least degree of the Maclaurin polynomials 𝑛 needed to approximate the value of sin 0.3 with an error less than 0.001 using the Maclaurin series of 𝑓(π‘₯) = sin π‘₯.

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Video Transcript

Determine the least degree of the Maclaurin polynomials 𝑛 needed to approximate the value of sin of 0.3 with an error less than 0.001 using the Maclaurin series of 𝑓 of π‘₯ equals sin of π‘₯.

We begin by recalling that the Lagrange error bound tells us that if the absolute value of the 𝑛 plus oneth derivative of 𝑓 is less than or equal to sum π‘š, then the absolute value of the remainder term 𝑅 sub 𝑛 of π‘₯ is less than or equal to the absolute value of π‘š over 𝑛 plus one factorial times π‘₯ minus π‘Ž to the power of 𝑛 plus one. To answer this question then, we’ll just begin by defining each part. We’re not told the degree of our McLaurin polynomial. So we’ll let 𝑛 be equal to 𝑛? This is what we’re trying to find. We’re told though that 𝑓 of π‘₯ is equal to sin of π‘₯. And we’re finding the Maclaurin series for 𝑓 of π‘₯. That’s the tailor series when π‘Ž is equal to zero.

We’re then going to use this to estimate the value of sin of 0.3. Now, since our function 𝑓 of π‘₯ is equal to sin of π‘₯, we can say that we’re going to let π‘₯ be equal to 0.3. We want to ensure that our error bound is less than 0.001. So we’ll start with this formula. We’re going to replace π‘₯ with 0.3 and π‘Ž with zero. And so we have the absolute value of π‘š over 𝑛 plus one factorial times 0.3 minus zero to the power of 𝑛 plus one. And of course, we want this error to be less than 0.001. Let’s replace 0.3 minus zero with 0.3. And then we know that π‘š is found by maximising the absolute value of the 𝑛 plus oneth derivative on the interval between π‘Ž and π‘₯.

So let’s find an expression for π‘š by considering the 𝑛 plus oneth derivative of our function. We know the 𝑓 of π‘₯ is equal to sin of π‘₯. And when we differentiate sin of π‘₯, we get cos of π‘₯. So the first derivative 𝑓 prime of π‘₯ is cos of π‘₯. Differentiating once more, we get negative sin of π‘₯. And then the third derivative 𝑓 triple prime of π‘₯ is negative cos of π‘₯. We differentiate one more time. And we find that the fourth derivative is sin of π‘₯. And so we see we have a cycle. We want to generalise it. So we’re going to say that, in fact, cos of π‘₯ is equal to sin of π‘₯ plus πœ‹ by two. Remember, that’s simply because the sin and cos graphs are horizontal translations of one another.

We say that negative sin π‘₯ is equal to sin of π‘₯ plus πœ‹, negative cos π‘₯ is sin of π‘₯ plus three πœ‹ by two, and sin of π‘₯ is equal to sin of π‘₯ plus two πœ‹. Now, in fact, if we write πœ‹ as two πœ‹ over two and two πœ‹ as four πœ‹ over two, we see that the 𝑛th derivative of 𝑓 is sin of π‘₯ plus π‘›πœ‹ over two. And so the 𝑛 plus oneth derivative, which is what we’re looking for, is sin of π‘₯ plus 𝑛 plus one πœ‹ over two. We want to maximise the absolute value of sin of π‘₯ plus 𝑛 plus one πœ‹ over two on the closed interval zero to 0.3. Well, the largest value of sin of π‘₯ plus 𝑛 plus one πœ‹ over two is one. So we’re going to let π‘š be equal to one.

Now, note that the maximum value of the trigonometric function sign on the interval between zero and 0.3 might not actually be one. But we do know it must be less than or equal to one. By letting π‘š be equal to one, we know that our error bound might be larger than it has to be, but that’s still allowed. Then our earlier inequation becomes the absolute value of one over 𝑛 plus one factorial times 0.3 to the power of 𝑛 plus one. And this must be less than or equal to 0.001. Now, in fact, this is always positive. So we no longer need the absolute value signs. And unfortunately, there’s no nice way at this stage to solve this inequation. So instead, we’re going to try some values of 𝑛, knowing, of course, that it can take integer values only.

We’ll begin by letting 𝑛 be equal to one. Then we have one over one plus one factorial times 0.3 to the power of one plus one. That gives us 0.045. Now, actually, that’s not less than 0.001, but we’re close. Next, let’s try 𝑛 equals two. We get one over two plus one factorial times 0.3 to the power of two plus one. That’s 0.0045, again not quite less than 0.001, but, again, getting closer. Let’s try 𝑛 equals three. We get one over three plus one factorial times 0.3 to the power of three plus one. That’s 0.0003375, which is indeed less than 0.001. And we can, therefore, say that the value of 𝑛 which ensures that the Maclaurin series approximates sin of 0.3 with an error less than 0.001 is 𝑛 equals three.

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