 This lesson is a continuation of Taylor and McLaren series and we will concentrate on the geometric and binomial series which you had in your lesson. I will be doing a couple more examples for you on your geometric series and your binomial series to help you review this particular lesson. What is a infinite geometric series? Well we already know that an infinite geometric series starts out with a and then we add a r plus a r squared plus a r cubed plus and then we add more to it plus an a r to the n and that's equal to the sum from n equals zero to infinity of a r to the nth power. You learn this in your algebra classes as well as in our review of geometric series. What if we want to find the sum of an infinite geometric series? Again we have a formula for that. So the sum of a geometric series is given as a over 1 minus r. Using Taylor polynomials, how do we get this? How do we go from that summation formula to the series itself? Because when we're thinking of Taylor or McLaren polynomials or series, we think of a function going to a series. So this time let's make f of x equal to a over 1 minus x and use our development of a Taylor polynomial to figure all of this out. So if f of x is equal to a over 1 minus x then f prime of x is equal to a over 1 minus x quantity squared. f double prime of x is equal to 2a over 1 minus x quantity cubed. f triple prime of x is equal to 6a over 1 minus x to the fourth power. And if we continue this on down we get f to the nth prime of x is equal to n factorial a over 1 minus x to the nth plus 1. Well how does this convert into our Taylor polynomial? Well we've sent it around zero. So next thing we need to do is to substitute the zeros in. So f of zero is equal to a. f prime of zero is equal to a again. f double prime of zero is equal to 2a. f triple prime of zero is equal to 6a. And then down to f n prime of zero and that equals n factorial times a. If we want to put this into our Taylor series we'll have f of x is equal to a plus a x plus 2a x squared over 2 factorial plus 6a x cubed over 3 factorial plus dot dot dot plus n factorial a x to the n over n factorial plus dot dot dot. Remember this is infinite therefore we continue on. So simplifying we get a plus a x plus a x squared plus a x cubed plus dot dot dot plus a x to the nth power and in symbols with our sigma we get that equals sigma from n equals zero to infinity a x to the nth power. And that is your geometric series once again shown through the development of a Taylor polynomial. To go on this time we are given f of x is equal to a over 1 plus x. Well the only difference between this one and the last one is that the negative between the one and the x has been changed to a positive and how does this affect our Taylor series centered about zero. So let's go through again and create this one. So we'll have f of x is equal to a over 1 plus x f prime of x is equal to negative a over 1 plus x quantity squared f double prime of x is equal to 2a over 1 plus x quantity cubed f triple prime of x is equal to negative 6a over 1 plus x to the fourth power. And then if we continue on we get f to the nth power of x is equal to and how do we take care of the negatives and positive. We will say negative 1 to the nth power of course if n is even it will be a positive number if n is odd it will be a negative number and then we'll have n factorial a over 1 plus x to the n plus 1. Let's go on and create it around zero centered at zero. So f of zero is equal to a f prime of zero is equal to negative a f double prime zero is equal to 2a f triple prime of zero is equal to negative 6a and then down to f to the nth prime of zero is equal to negative 1 to some nth power times n factorial a. And if we put this into our Taylor series again we will get f of x is equal to a minus a x plus a x squared minus a x cubed plus dot dot dot plus a negative 1 we have to include the toggling from positive to negative to the nth power a x to the nth power and sigma notation that's equal to sigma from n equals zero to infinity of negative 1 to the nth power a x to the nth power. Now let's check to see if indeed this is what we've created. If I put in an n is equal to zero into my formula I'll have negative 1 to the zero power which is 1 a x to the zero power which is a. So I get the proper sign in front of my a. If I put in a 1 for n then I'll have negative 1 to the first power which will give a negative 1 and then it will be a x. So my signs are appropriate so this is the formula that we will use. Again it's just a recreation of your geometric series this time with the plus. So this one actually has the signs changing back and forth whereas the first one with the 1 minus x is just a plus a x etc. So let's go on. What happens with our binomial series? So if we start off with f of x is equal to 1 plus x cubed we can say our function is equal to 1 plus x quantity cubed. The prime of x is equal to 3 times 1 plus x squared. Double prime of x is equal to 6 times 1 plus x to the first power. F triple prime of x is equal to 6 and f to the fourth prime of x is equal to 0. Sent it around 0 we'll have f of 0 equals 1, f prime of 0 equals 3, f double prime of 0 equals 6, f triple prime of 0 equals 6 and f to the fourth prime of 0 of course is 0. Putting this into our Taylor polynomial we will get the expansion f of x is equal to 1 plus 3x over 1 factorial remember that plus 6x squared over 2 factorial plus 6x cubed over 3 factorial plus of course 0. Computing this we get 1 plus 3x plus 3x squared plus x cubed which is certainly an expansion for 1 plus x quantity cubed if we use the binomial theorem. Let's go on and do one more of these. What is the Taylor series for f of x equals 1 plus x to the 1 half power. Again we can expand it through Taylor series or we can use our formula. So if we look at this one we'll have f of x is equal to 1 plus x to the 1 half power. F prime of x is equal to 1 half times 1 plus x to the negative 1 half power. F double prime of x is equal to negative 1 fourth 1 plus x to the negative 3 half power. F triple prime of x is equal to 3 eighths times 1 plus x to the negative 5 half power. If we center this around 0 we will get f of 0 is equal to 1. F prime of 0 is equal to 1 half. F double prime of 0 is equal to negative 1 fourth. F triple prime of 0 is equal to 3 eighths. Expanding this out into our Taylor polynomial form we get f of x is equal to 1 plus 1 half x minus 1 fourth x squared over 2 factorial plus 3 eighths x cubed over 3 factorial and that equals 1 plus 1 half x minus 1 eighth x squared plus 1 over 16 x cubed. And if we put it into our binomial formula that we got by working through the general form which is 1 plus x to the k power equals 1 plus k x plus k times k minus 1 x squared over 2 factorial plus k times k minus 1 times k minus 2 x cubed over 3 factorial since our function is 1 plus x to the 1 half power we will indeed get 1 plus 1 half x plus k is 1 half so we have 1 half times negative 1 half x squared over 2 factorial plus 1 half times negative 1 half times negative 3 halves x cubed over 3 factorial and that will give us 1 plus 1 half x minus 1 eighth x squared plus if we multiply everything out the threes go out here we have 2 times 2 is 4 times 2 is 8 and then another 2 down here so we get 1 over 16 x cubed. So this creates that binomial theorem expansion that may look very very difficult when you are working with it in your book but is very simple to do if you just keep in mind it is a Taylor series or Taylor polynomial depending how far you go out remember if it if it goes towards infinity it's a series if it's just ends at one point it is a Taylor polynomial. This concludes the lesson on geometric series and binomial series.