6.3 Taylor and Maclaurin Series

[STARTS AT 32:26]

Professor: All right, well, f of x is 1 over x. And we are finding it centered at 1. So f of 1 is 1. f prime of x is negative 1 over x squared. f prime of 1 is negative 1. The second derivative of x is 2 over x cubed. And the second derivative evaluated at 1 is equal to 2. The third derivative is negative 6 over x to the fourth, so evaluating that at 1 is negative 6. 

Now if you notice, these are actually respectively 0 factorial, 1 factorial, 2 factorial, 3 factorial. So once we actually divide-- once we divide by our n factorial on each of those terms, our Taylor series ends up being 1 minus x minus 1 plus x minus 1 squared minus x minus 1 cubed, et cetera. 

So we can write that as, well, it's alternating with terms of x minus 1-- x minus 1 to the n. And it's alternating, so we'll have negative 1 to the n. Now, I'm going to go ahead and write this as-- actually, we'll leave it that way. Let's not rearrange that any. So that is our Taylor series for f of x equals 1 over x. 

So let's see if we can find the interval of convergence. That would first require us to take the ratio test. This would be the absolute value of a n plus 1, so negative 1 n plus 1 x minus 1 n plus 1 divided by negative 1 to the n x minus 1 to the n. Now, we know the absolute value is going to wipe out the negative 1 part of that, negative 1 to the n plus 1, negative 1 to the n. So this is going to be equal to the absolute value of x minus 1. 

Now by the ratio test, for us to-- for this to converge, we need this to be less than 1. Because there's no values of n left, we don't need to take the limit there. So our limit is actually that x minus 1 absolute value. 

So let's take that interval. I'll put that right here. That would be x minus 1 is between 1 and negative 1, which means x is between 2 and 0. So our open interval of convergence is 0 to 2. So we know we have 0 to 2. Now, let's check out what happens at x equals 0. At x equals 0, for this we get the sum equals 0 to infinity of 1 to the n, which quite obviously, diverges, so we don't include 0 in that. 

At x equals 2, we get the sum of negative 1 to the n, which also diverges. So we don't include that. So our interval of convergence is 0 to 2. It's an exclusive interval. And our radius of convergence is 1.

[ENDS AT 36:29]