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## Finding Taylor or Maclaurin series for a function

Current time:0:00Total duration:4:01

# 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

- [Instructor] We're asked
to find a power series for f, and they've given us f of
x is equal to six over one plus x to the third power. Now, since they're letting
us pick which power series, you might say, well, let me
just find the Maclaurin series because the Maclaurin series
tends to be the simplest to find 'cause we're centered at zero. And so you might immediately
go out and say all right, well, let me evaluate
this function at zero, evaluate its derivative at zero, its second derivative at
zero, 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, evaluating this at f, at x equals zero 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
six over one plus u, where u is equal to x to the third. So you find this Maclaurin
expansion, in terms of u, and then you substitute
for x to the third. 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 what, 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 gonna multiply it 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 one minus our common ratio, and this just comes from a,
the sum of a geometric series. And notice that what we
have here, our 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 six. And if negative r is
equal to x to the third, or we could say, let me rewrite this. I could write this
denominator as one minus negative x to the third. And so now, you could 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 six, and r is equal to negative
x to the third, well, then we could 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 do
this in this nice pink color. So the first term would be six, plus six times our common ratio, six times negative x to the third. And so actually, let me just write that as negative six x to the third,
and then we're gonna 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 gonna be positive six times x to the sixth power. And then I'm gonna multiply
it by times negative x to the third again, so it's
gonna be minus six times x to the ninth power, and I'm gonna go on, and on, and on. So it's gonna be, and
then I could keep going, I multiply it times
negative x to the third. I will get six x 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.