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# Function as a geometric series

AP.CALC:
LIM‑8 (EU)
,
LIM‑8.E (LO)
,
LIM‑8.E.1 (EK)
,
LIM‑8.F (LO)
,
LIM‑8.F.1 (EK)

## Video transcript

whereas to find a power series for F and they've given us f of X is equal to 6 over 1 plus X to the third power now since they're letting us pick which power series you might see well let me just find the Maclaurin series because the Maclaurin series tends to be the simplest to find because we're centered at 0 and so you might immediately go out and say alright well let me evaluate this function at 0 evaluates derivative at 0 its second derivative 0 so on and so forth and then I can use the formula for the Maclaurin series to just expand it out but very quickly you will run into roadblocks because the first the evaluating this at F at x equals 0 is pretty straightforward evaluating the first derivative is pretty straightforward but then once you start taking the second and third derivatives it gets very hairy very fast you could do a simplification where you could say well let me find the Maclaurin series for F of U is equal to 6 over 1 plus u where u is equal to X to the 3rd so you find this Maclaurin expansion in terms of U and then you substitute for X to the 3rd and actually that makes it a good bit simpler so that is another way to approach it but the simplest way to approach it is to say hey you know this this form right over here this rational expression it looks similar it looks like the sum of a geometric series let's just remind ourselves what the sum of a geometric series looks like if I have a plus a times R so a is my first term R is my common ratio plus I'm going to multiply times R again plus a times R squared plus a times R to the third power and I keep going on and on and on forever we know that this is going to be equal to a our first term over 1 minus our common ratio and this comes from a the sum of a geometric series and notice that what we have here are f of x our definition of f of X and the sum of a geometric series look very very similar if we say that this right over here is a so a is equal to 6 and if negative R is equal to X to the third or we could say let me rewrite this I could write this denominator is 1 minus negative x to the third and so now you can say okay well R could be equal to negative x to the third and just like that we can expand it out well if a is equal to 6 and Nate and R is equal to negative x to the third well then we can just write this out as a geometric series which is very straightforward so let's do that and I will do this in I'll just in this nice pink color so the first term would be 6 plus 6 times our common ratio 6 times negative x to the third and so actually let me just write that as negative 6x to the third and then we're going to multiply by negative x to the third again so that's going to be if I multiply this times negative x to the third that's going to be positive six times X to the sixth power and I'm going to multiply about times negative x to the third again so it's going to be minus 6 times X to the ninth power and I'm going to go on and on and on so it's going to be and then I could keep going on multiply times negative x to the third I will get 6x to the twelfth power now we can go on and on and on and on forever and so the key here was and this is the Maclaurin series expansion for our f of X but the key is to not have to go through all of this business and just to recognize that hey the way this function was defined is it looks a lot like the sum of a geometric series and it can be considered the sum of a geometric series and we can use that to find the power series expansion for our function this is a very very very useful trick
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