Video: Solving Problems Involving Taylor Polynomials

Find the fourth-degree polynomial of the function 𝑓(π‘₯) = sin π‘₯ at the point π‘Ž = πœ‹/2.

02:23

Video Transcript

Find the fourth-degree polynomial of the function 𝑓 of π‘₯ equals sin of π‘₯ at the point π‘Ž equals πœ‹ over two.

Let’s start by writing out the general form for a Taylor polynomial which approximates a function 𝑓 of π‘₯ at the point π‘₯ equals π‘Ž. We’ve been asked to find the fourth-degree Taylor polynomial of this function at the point π‘Ž equals πœ‹ over two. So let’s write out our general form for a Taylor polynomial up to the fourth degree and substitute in π‘Ž equals πœ‹ over two.

We can see here that we’re going to need to substitute in values for 𝑓 of πœ‹ over two, the first derivative of 𝑓 at πœ‹ over two, the second derivative of 𝑓 at πœ‹ over two, the third derivative of 𝑓 at πœ‹ over two, and the fourth derivative of 𝑓 at πœ‹ over two.

So let’s start by finding the derivatives that we need. 𝑓 of π‘₯ is the function sin of π‘₯. At this point, we recall this useful cycle, which shows us what each of these functions differentiates to. So we see that sin of π‘₯ differentiates to cos of π‘₯. So 𝑓 prime of π‘₯ equals cos of π‘₯. Then we differentiate the first derivative to get the second derivative. And we see that this is going to be negative sin of π‘₯. We then differentiate again to get the third derivative, which we see is negative cos of π‘₯. And we differentiate once more to get the fourth derivative of 𝑓. And we get sin of π‘₯.

But what we actually need to do is evaluate each of these functions at πœ‹ over two. So we substitute in πœ‹ over two into each of these functions. We can work these out either on a calculator or by using the graphs. So we can see that sin of πœ‹ over two is going to give us one. So negative sin of πœ‹ over two will be negative one. And we can see that cos of πœ‹ over two is zero. So negative cos of πœ‹ over two will also be zero.

So now we can substitute these values back into our working. When we do this, we find that two of our terms are actually zero. From here, we just need to simplify our answer. Remembering that the factorial of a number is the product of that number and the integers below it down to one. So two factorial is two multiplied by one, which is two. And four factorial is four multiplied by three multiplied by two multiplied by one, which is 24. So substituting those values gives us our final answer.

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