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### Course: Calculus, all content (2017 edition) > Unit 7

Lesson 15: Power series function representation- Function as a geometric series
- Geometric series as a function
- Function as a geometric series
- Geometric series interval of convergence
- Power series of arctan(2x)
- Power series of ln(1+x³)
- Finding function from power series by integrating
- Integrals & derivatives of functions with known power series

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# Function as a geometric series

Expressions of the form a/(1-r) represent the infinite sum of a geometric series whose initial term is a and constant ratio is r, which is written as Σa(r)ⁿ. Since geometric series are a class of power series, we obtained the power series representation of a/(1-r) very quickly.

## Want to join the conversation?

- At1:42, the sum of geometric sequence is a/(1-r), but this is only true for |r|<1, isn't it?

Applying this to f(x) seems like a bit of a stretch to me.(34 votes)- Yes, you are correct. This series converges when |-x³| < 1 ⇔ |x| < 1.(12 votes)

- How can we say that Geometric series are Maclaurin series? or Maclaurin series are Power series? I think there's an information gap there. Can you explain to me?(10 votes)
- I don't like asking these type of questions but What is the point of this ?

what the point of representing a nice rational function with a series that only works when x is between (-1,1)

is this trick used in some proofs or when solving some questions if so can someone give me an example

second question is what if we wanted to have a taylor series and expand it at x=3 for example what would I do ?(8 votes) - How is this a Maclaurin series? Doesn't the Maclaurin series have a factorial denominator in every term?(4 votes)
- Good question. If you take the first, second and third derivative of f(u), you can then evaluate f f' f'' f''' at u=0. That will give you the first four coefficients of the Maclaurin Series. Next, put each of those coefficients to their corresponding terms of the M. series using u as your variable - corresponding terms meaning u^0/0! + u^1/1! + u^2/2! + ... . You can then substitute every u back to x^3. Then simplify. The process will take only a few minutes - using u in place of x^3 - and you will quickly see that it reduces to the same equation as above, in Sal's video.(5 votes)

- How do you get rid of that autoplay button at the end..it doesn't let me rewind the video..so annoying..(5 votes)
- You can push the counter clockwise button in the down left corner to rewatch the video.(2 votes)

- at around0:13, how do we know a Maclaurin series is a power series?(2 votes)
- Well, think about how a power series is defined. If I have a power series centred at c, it'd be the sum from n = 0 (or 1) to infinity of (a_n)(x-c)^n, where a_n is a real number.

See that this is exactly how a Maclaurin series looks like (just that the coefficients themselves need to be found out using a different formula). So, a Maclaurin (and in general, a Taylor) series is a kind of power series.(2 votes)

- When Sal makes this connection with the geometric series formula, how can we be sure it will be the Maclaurin Series? Is there only one unique Maclaurin Series for every function?(2 votes)
- Yes, there is only one unique Maclaurin series for every function. Maclaurin series are always constructed around the function where x=0. To check that this is the Maclaurin series for the function, plug x=0 into any partial sum of the Maclaurin expansion, and you will find that it is equal to the exact function.(1 vote)

- Why is the power series not over n!

Are geometric series a special circumstance that look different from a normal Maclaurin series?(2 votes) - The serie is p(x)=6 -6x^3 +6x^6......

the funtion is f(x)= 6/(1+x^3)

if x=1 f(1)=3 and p(1)=0 or 6

why f and p not equal in a infinite maclaurin's serie the error is zero?(2 votes)- I don't think what you did was allowed I mean isn't this an oscillating sequence that diverges.

I think that you are not allowed to make x=1 in p(x) because we have a constriction that the ratio in a geometric sequence must be less than 1 so we have |x^3| < 1 meaning that

-1<x<1

a lot of contradiction happen when you allow p(1) = 3

I mean notice that I can simply can add p(1) = 6-6+6-6... and it wouldn't change the seires

2p(1)=p(1)

implying that 2=1 which is absurd.

maybe someone who is an expert can shed some more light on this topic.(1 vote)

- at around0:42, f'(x)=-18x^2/(1+x^3)^2.

substitute x=0, we get f'(0)=0.

according to the formula of the Maclaurin series (f^n(0)*x^n/n!), we get the 2nd term of the Maclaurin series to be 0, which is different from -6x^3.

Can anyone point out where I did wrong? Thanks.(1 vote)- You're not wrong. The second term will be 0. However, if you go ahead and find the third derivative, you get -36. So, the third term becomes -36 (x)^(3)/3! which is -6x^3.

Essentially, see that each term of the power series Sal derived has only powers which are multiples of 3. So, even when finding the Maclaurin series with your traditional method, you'll only get a term every third time (You'll get a term on the 3rd, 6th, 9th... derivatives and every other derivative will be 0). So, just to get 5 terms like Sal did, you'll need to take 12 derivatives, which is definitely not something you'd want to do. Hence, he proposed this method.(2 votes)

## 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.