If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Geometric series as a function

AP.CALC:
LIM‑8 (EU)
,
LIM‑8.E (LO)
,
LIM‑8.E.1 (EK)
,
LIM‑8.F (LO)
,
LIM‑8.F.1 (EK)
Power series of the form Σk(x-a)ⁿ (where k is constant) are a geometric series with initial term k and common ratio (x-a). Since we have an expression for the sum of a geometric series, we can rewrite such power series as a finite expression. Created by Sal Khan.

Want to join the conversation?

  • piceratops ultimate style avatar for user Chester Madrazo
    I'm a little bit confused... Does it mean that the infinite geometric series does not equal the function when x is outside the interval of convergence?
    (5 votes)
    Default Khan Academy avatar avatar for user
    • blobby green style avatar for user Drewqueen13
      If x is outside the interval of convergence, the function will still be a geometric series however it will not converge and the limit does not exist. This means that your geometric series does not add up to an obtainable value as you go towards infinity.
      (20 votes)
  • blobby green style avatar for user gazogang
    I understand that this is the algebraic geometric series centered at 0. How can it be determined centered at another value, c, with the x-term becoming (x-c)? I couldn't find this explanation in this video series or by searching elsewhere. Does it exist in Khan Academy? Thank you.
    (6 votes)
    Default Khan Academy avatar avatar for user
    • blobby green style avatar for user Andrew
      Here we see: ∑ 2*(-4x^2)^n, where the common ratio is (-4x^2). I would THINK that if you can break up your common ratio into the sum of an x-varying part minus a constant part, so it looks like (x-c), then that would show that the series is centered at the constant part.

      Example: ∑ 2*[( 4b^2) - 5]^n. The common ratio is (4b^2)-5. If you made the substitution x = 4b^2. You could write f(x) = ∑ 2*(x-5)^n. I THINK you could reasonably say that this series is centered at c.
      (4 votes)
  • leaf green style avatar for user Harrichael
    Shouldn't Sal be testing the endpoints of the interval due to inconclusive ratio test at 1?
    (2 votes)
    Default Khan Academy avatar avatar for user
  • piceratops seed style avatar for user Cagan Sevencan
    he is confusing me with absolute value thing. he should just take the number out straight up
    (1 vote)
    Default Khan Academy avatar avatar for user
  • piceratops ultimate style avatar for user Sher Gill
    Is it mathematically legal if the starting term 'a' is defined in terms of a variable (i.e. x^2)?
    (1 vote)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user Miray Atar
    Can you express this function graphically as the sum of (∑from n=-infinity to n=-1/2 of f(x)) + 2/(1+4x^2)+ (∑from n=1/2 to n=infinity of f(x))?
    (1 vote)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user Jason Hoenselaar
    let me just understand something here ,i understand the concept of the interval of convergence and all that ,but is it still correct to say that a function is exactly equal to the expression (a/1-r) when r is within the interval of convergence ? what if we have that r is slightly less or more say (r+x) etc .but r+x was still within the radius of convergence .it would be incorrect to say that both [a/1-r ] and [a/1-(r+x)] are exactly equal to our function ,since that would imply they are equal to each other .So isn't the expression [a/1-a] just a really good approximation ?
    (1 vote)
    Default Khan Academy avatar avatar for user
    • piceratops ultimate style avatar for user Matthew Chen
      The misunderstanding you have is that the function where r = r and the function where r = r + x are two different functions. Say the first one if f1, second one is f2. Then f1 = a/(1-r) and f2 = a/(1-(r+x)). The expression a/(1-r) is correct for each function since r is not the same for them.
      (1 vote)
  • leafers ultimate style avatar for user TheModernNinja21
    What does Σ (sigma) mean?
    (I know it's Greek).
    (0 votes)
    Default Khan Academy avatar avatar for user
  • leaf green style avatar for user nicole32021
    why do we raise the common ratio (-4x^2) to the nth power in the sum notation? Wouldn't that change the value when you go to plug in 1 for n, 2 for n, 3 for n, ect...?
    (1 vote)
    Default Khan Academy avatar avatar for user
  • leafers tree style avatar for user Murat
    I don't understand why we didn't test the endpoints at about .
    (1 vote)
    Default Khan Academy avatar avatar for user

Video transcript

- [Instructor] So we have this function that's equal to 2 minus 8x squared plus 32x to the fourth minus 128x to the sixth. And it just keeps going and going. So it's defined as an infinite series. And what I want to explore in this video, is there another way to write this function so it's not expressed as an infinite series? Well, some of you might be thinking, well, this looks like a geometric series on the right hand side, an infinite geometric series, and we know what the sun of an infinite geometric series is if it converges. So maybe that's a way that we can express this. So let's try to do that. So first let's just confirm that this is an infinite geometric series. And in order for it to be a geometric series, each successive term has to be some common ratio times the previous term. So to go from 2 to negative 8x squared, what do you have to multiply by? Well you have to multiply by negative 4x squared. Now let's see if you multiply negative 8x squared times negative 4x squared, what do you get? Well, negative 4 times negative 8 is positive 32. x squared times x squared is x to the fourth. So that works. And then you multiply that times negative 4x squared, and you indeed would get negative 128x to the sixth. So this indeed looks like an infinite geometric series on the right-hand side. In fact, we can rewrite f of x as being equal to the sum from n equals 0 to infinity of, you have your first term, and then you have your common ratio, negative 4x squared to the nth power. Let's confirm that works when n equals 0 this is going to be 1. So 2 times 1 is 2, and that indeed is our first term there. And then to that, you're gonna add it to when n is equal to 1. So that's just going to be two times negative 4X squared, which is indeed this second term right over here. And so this looks like it works. Now what is the sum of an infinite geometric series like this? Well it's going to be a finite value, assuming the absolute value of your common ratio is less than 1. So first of all, let's just think about under what conditions is the absolute value of our common ratio less than 1? And then we could say, okay, that helps us to find a radius of convergence. And then if x is in that zone, or if it's in that interval, then we can figure out a non-infinite geometric series way of expressing this function. So if we just think about under what circumstances will this converge, will it come out to a finite value? That's a situation in which the absolute value of your common ratio is less than 1. And so let's see if we can simplify this a little bit. No matter what x is, it's always going to be not, x squared is always going to be non-negative. And so the only, so this entire expression is always going to be negative. And so if you take the absolute value of it, this is going to be evaluate as 4x squared, which is always going to be positive. So this is equivalent to 4x squared, which needs to be less than 1. Or we could say that x squared needs to be less than 1/4. Or we could say that x needs to be less than 1/2 and greater than negative 1/2. One way to think about it is anywhere in this interval, if you square it, you're going to be less than 1/4. At 1/2, if you square it, it's equal to 1/4. And at negative 1/2, if you square it, it's equal to 1/4. But for lower absolute values, it's going to be less than 1/4. And so that's what this interval right here says. Another way to think about it is the absolute value of x needs to be less than 1/2. And so we've just defined an interval over which this infinite geometric series will converge. You could say this has a radius of convergence of, let me write it this way, radius of convergence, convergence of 1/2, you can go 1/2 above 0 and 1/2 below 0. But now that we've set the conditions under which this would converge, let's rewrite it. So this function is going to be equal to, we know what the sum of an infinite geometric series is. It's going to be equal to the first term over 1 minus your common ratio, 1 minus negative 4x squared. And so we can rewrite our function as f of x is equal to 2 over 1, subtract a negative 1 plus 4x squared for the absolute value of x is less than 1/2. We have the interval over which we converge and there you have it. We are done.