 This lesson is a continuation of Taylor and McLaurin 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 an 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 one 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 McLaurin 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 one 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 one minus x, then f prime of x is equal to A over one minus x quantity squared. F double prime of x is equal to 2A over one minus x quantity cubed. F triple prime of x is equal to 6A over one 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 one minus x to the nth plus one. Well how does this convert into our Taylor polynomial? We've centered 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 one 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 one plus x. f prime of x is equal to negative A over one plus x quantity squared. f double prime of x is equal to 2A over one plus x quantity cubed. f triple prime of x is equal to negative 6A over one 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 one 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 one plus x to the n plus one. 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 at 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 one 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 one. We have to include the toggling from positive to negative to the nth power A x to the nth power. And in sigma notation, that's equal to sigma from n equals zero to infinity of negative one 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 one to the zero power, which is one. 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 one for n, then I'll have negative one to the first power, which will give a negative one, 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 one minus x is just A plus A x, et cetera. So let's go on. What happens with our binomial series? So if we start off with f of x is equal to one plus x cubed, we can say our function is equal to one plus x quantity cubed. The prime of x is equal to three times one plus x squared. Double prime of x is equal to six times one plus x to the first power. f triple prime of x is equal to six, and f to the fourth prime of x is equal to zero. Scented around zero, we'll have f of zero equals one, f prime of zero equals three, f double prime of zero equals six, f triple prime of zero equals six, and f to the fourth prime of zero, of course, is zero. Putting this into our Taylor polynomial, we will get the expansion. f of x is equal to one plus three x over one factorial, remember that, plus six x squared over two factorial, plus six x cubed over three factorial, plus, of course, a zero. Computing this, we get one plus three x plus three x squared plus x cubed, which is certainly an expansion for one 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 one plus x to the one 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 one plus x to the one half power. f prime of x is equal to one half times one plus x to the negative one half power. f double prime of x is equal to negative one fourth, one plus x to the negative three half power. f triple prime of x is equal to three eighths times one plus x to the negative five half power. If we center this around zero, we will get f of zero is equal to one. f prime of zero is equal to one half. f double prime of zero is equal to negative one fourth. f triple prime of zero is equal to three eighths. Expanding this out into our Taylor polynomial form, we get f of x is equal to one plus one half x minus one fourth x squared over two factorial plus three eighths x cubed over three factorial. And that equals one plus one half x minus one eighth x squared plus one over sixteen x cubed. And if we put it into our binomial formula that we got by working through the general form, which is one plus x to the k power equals one plus kx plus k times k minus one x squared over two factorial plus k times k minus one times k minus two x cubed over three factorial. Since our function is one plus x to the one half power, we will indeed get one plus one half x plus k is one half. So we have one half times negative one half x squared over two factorial plus one half times negative one half times negative three halfs x cubed. Over three factorial and that will give us one plus one half x minus one eighth x squared plus if we multiply everything out, the threes go out here. We have two times two is four times two is eight and then another two down here so we get one over sixteen x cubed. So this creates that binomial theorem.