6.3 Taylor and Maclaurin Series

[STARTS AT 5:31]

Professor: Find some Taylor polynomials. We have the function f of x equals natural log of x, and we are centered at x equals 1. So this will require us to look at several derivatives, in this case, three derivatives. And then evaluate each of those at 1. So f of x is natural log x. So f of 1-- because we are evaluating this one at 1-- is 0. f prime of x is 1 over x, and so f prime of 1 is 1. 

The second derivative-- negative 1 over x squared, and so evaluating the second derivative at 1, we get negative 1. The third derivative of x is 2 over x cubed. So the third derivative evaluated at 1 is 2. All right, so p0, our first-- or our 0-th polynomial-- is just f of a, so 0. 

Our first Taylor polynomial is 0 plus-- and now we have 1 divided by 1 factorial times x minus a, in this case, 1. Now, you're going to notice, that is just x minus 1. That is all the first Taylor polynomial is. 

So the second polynomial, we'll take x minus 1, the last polynomial, plus-- well, in this case minus, because we have a minus 1 here-- minus 1 over 2 factorial times x minus 1 squared, which I will write as x minus 1 minus 1/2 x minus 1 squared. 

Now our third polynomial-- x minus 1 minus 1/2 x minus 1 squared plus 2 over 3 factorial x minus 1 cubed. Well, that means our third polynomial is x minus 1 minus 1/2 x minus 1 squared plus 1/3 x minus 1 cubed. 

Now, if we compare these polynomials, each of them, to the original function, natural log of x, at the value 1, they all have a common value at x equals 1. So as n goes to infinity, as we take the further polynomials as we go, Pn of x appears to converge to f of x-- let's call it f of x-- close to 1. If we're close to 1, our Taylor polynomials are very, very good approximations. And we can use that to our advantage in some other cases. 

[ENDS AT 9:05]