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Consider the function 𝑓(π‘₯) = cos (π‘₯). Find the Taylor series expansion of 𝑓(π‘₯) = cos (π‘₯) at π‘₯ = πœ‹. Write the Taylor series expansion of 𝑓(π‘₯) in sigma notation.

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Video Transcript

Consider the function 𝑓 of π‘₯ equals cos of π‘₯. Find the Taylor series expansion of 𝑓 of π‘₯ equals cos of π‘₯ at π‘₯ equals πœ‹ and write the Taylor series expansion of 𝑓 of π‘₯ in sigma notation.

And we recall that the Taylor series expansion of a function about π‘₯ equals π‘Ž is given by 𝑓 of π‘₯ equals 𝑓 of π‘Ž plus 𝑓 prime of π‘Ž over one factorial times π‘₯ minus π‘Ž plus 𝑓 double prime of π‘Ž over two factorial times π‘₯ minus π‘Ž all squared and so on. In this question, our function is cos of π‘₯. And we’re looking to find the Taylor series expansion at π‘₯ equals πœ‹. So we’re going to let π‘Ž be equal to πœ‹. And so we see that 𝑓 of π‘₯ here is equal to 𝑓 of πœ‹ plus 𝑓 prime of πœ‹ over one factorial times π‘₯ minus πœ‹ and so on. Now, we can quite easily evaluate 𝑓 of πœ‹. We would substitute πœ‹ into the function cos of π‘₯. But what about 𝑓 prime of πœ‹, 𝑓 double prime of πœ‹, and so on?

Well, we’re going to differentiate our function with respect to π‘₯. We might recall the first derivative of cos of π‘₯ to be negative sin of π‘₯. Then to get the second derivative, we differentiate negative sin π‘₯ with respect to π‘₯ and we get negative cos π‘₯. Differentiating once more, we find that 𝑓 triple prime of π‘₯ is sin π‘₯. And in fact, we’ll repeat this process one more time because when we differentiate sin π‘₯, we get cos π‘₯ again. And you might notice, we have a cycle. The fifth derivative of 𝑓 will be negative sin π‘₯, the sixth derivative will be negative cos π‘₯, and so on. And let’s use all of these to evaluate 𝑓 of πœ‹, 𝑓 prime of πœ‹, 𝑓 double prime of πœ‹, and so on and look for a pattern. 𝑓 of πœ‹ is cos πœ‹, which is negative one. 𝑓 prime of πœ‹ is negative sin πœ‹, which is zero. 𝑓 double prime of πœ‹ is negative cos πœ‹, which is one. And 𝑓 triple prime of πœ‹ is sin πœ‹, which is once again zero.

It, of course, follows that the fourth derivative of πœ‹ will be cos of πœ‹ again, which is negative one. Let’s substitute all of these back into our Taylor series expansion. Now of course, every other term is going to be equal to zero as we’ve seen. And so we find the Taylor series expansion to be negative one plus a half times π‘₯ minus πœ‹ all squared minus one over 24 times π‘₯ minus πœ‹ to the fourth power plus one over 720 times π‘₯ minus πœ‹ to the sixth power and so on. The second part of this question asks us to write the Taylor series expansion of 𝑓 of π‘₯ in sigma notation. And so let’s see if there’s a way we can find a pattern. We saw that 𝑓 prime of πœ‹ was equal to zero, while 𝑓 triple prime of πœ‹ is equal to zero. And extending the pattern, we would see that the fifth derivative, the seventh derivative, and so on would also be equal to zero. And so our derivative alternate between negative one, one, negative one, and so on.

Similarly, our denominators are factorials of ascending even numbers each time this even number is the same as the exponent. So let’s define that even number as two π‘š. Then the denominator is two π‘š factorial and the exponent of π‘₯ minus πœ‹ is two π‘š. To achieve alternating powers of negative one, we write negative one times π‘š plus one. That means when π‘š is zero, negative one to the power of one is negative one, when π‘š is one, negative one to the power of two gives us one, and so on. And so we find that 𝑓 of π‘₯ equals the sum of negative one to the power of π‘š plus one times π‘₯ minus πœ‹ to the power of two π‘š over two π‘š factorial for values of π‘š between zero and infinity.

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