6.4.3

Instructor: In example 2, we're going to use the table of familiar series to determine the Maclaurin series for these two functions that are composite functions. 

First we have-- in part a, we have cosine of root x. We are going to first, use the fact that we know what the cosine-- the series for cosine is this. So we're going to modify this, and replace-- and actually I'm going to use, instead of cosine root x, I'm going to use cosine of x to the 1/2. That will be helpful here. So if you notice, this is going to be equal to n equals 0 to infinity of negative 1 to the n. And we're going to have x to the 2n. So just 1/2 to the 2n over 2n factorial. Well that numerator becomes x to the n. But we still have the 2n factorial. And there we have the closed form for cosine of root x, making a simple substitution there. 

Let's see if we can apply that same technique to sinh x. Well, there are no hyperbolic functions up here. That's not very nice. But, what we can do is rewrite this as e to the x minus e to the negative x over 2, which is the definition of hyperbolic sin. Now, let's see. Let's write this is e to the x over 2. Or how about this. One half e to the x minus one half to the negative x. 

Now we don't know the specific, like already, we don't know the Taylor series or the Maclaurin series for e to the negative x. But we can use the fact that e to the x, e to the x is the sum n equals 0 to infinity of x to the n over n factorial, which means that e to the negative x is x or negative x. Put that a little bigger. Negative x to the n over n factorial. We can use that fact to go ahead and combine those two. 

Now if you notice, on the right hand side the negative x to the n is-- well it's going to be very similar, actually, to what the x to the n over n is. In fact the even terms, whenever n is even, those two are going to equal 0. Because when n is even we have just n x to the n, an x to the n and we're subtracting those two. So on our even terms, we have equals 0. 

But the odd terms are equal. So we end up-- rather than having a one half and a one half we end up with this. So the two 1/2 terms cancel out. But we need to make sure they only have odd terms, because the even terms are no longer there. So it's going to be the same function, except it's going to be x to the 2n plus 1. So, similarly it's x to the n over n factorial, but instead of just n's it's only the odd ones. So x to the 2n plus 1. And then we'll have 2n plus 1 factorial.