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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)
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.

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  • leaf green style avatar for user Ilya
    At , 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.
    (25 votes)
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  • leaf green style avatar for user Abdo Reda
    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 ?
    (4 votes)
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  • blobby green style avatar for user leifelliott1
    How do you get rid of that autoplay button at the end..it doesn't let me rewind the video..so annoying..
    (3 votes)
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  • piceratops ultimate style avatar for user Andrew Hung
    How is this a Maclaurin series? Doesn't the Maclaurin series have a factorial denominator in every term?
    (2 votes)
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    • blobby green style avatar for user Michael Primavera
      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.
      (3 votes)
  • blobby green style avatar for user 2053936
    Why is the power series not over n!
    Are geometric series a special circumstance that look different from a normal Maclaurin series?
    (2 votes)
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  • male robot donald style avatar for user Anthony Matheus
    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)
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    • leaf green style avatar for user Abdo Reda
      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)
  • blobby green style avatar for user Miray Atar
    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?
    (1 vote)
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  • blobby green style avatar for user Hexuan Sun 8th grade
    What does it mean to find a power series of a function?
    (1 vote)
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  • piceratops ultimate style avatar for user Owen
    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?
    (1 vote)
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    • leaf green style avatar for user Arthur Benson
      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)
  • purple pi teal style avatar for user marry shinota
    why power series have to be a/1-r, aren't a/-r the same?
    (1 vote)
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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.