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Geometric series interval of convergence

When a power series is a geometric series, we can find its interval of convergence without using the ratio test! Created by Sal Khan.

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  • aqualine sapling style avatar for user Thomas Turkington
    At , how is the abs(-x^2/3) the same as abs(x^2/3) ? Wouldn't the negative sign have to be like this: abs{(-x^2)/3}, for the negative sign to disappear since -x times -x is positive x^2? I know this is very basic, but it's confusing me...
    (6 votes)
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  • male robot donald style avatar for user harry park
    Someone help me understand:
    So the point of this video is to demonstrate how if we take a function, and then manipulate it in a way that it looks like a/1-x, and if we take the infinite sum of that function, then sum formula will be the same as the original function we had in the beginning?
    (6 votes)
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    • ohnoes default style avatar for user Anne
      Yes - sometimes it's best to solve a problem by manipulating the function to look like a well-known model function that is easy to work with, such as a/1-r. The original function and the new one are equivalent. This technique is also used, for example, to integrate inverse trig functions such as arctan x.
      (6 votes)
  • leaf green style avatar for user mzhao8
    don't you need to check the limits of the interval to see whether or not it converges to the point?
    (3 votes)
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  • mr pants teal style avatar for user arpithaprasad
    Can't we also define the function as 1 / [ 1- (-2 - x^2)] ? In which case the interval of convergence would be |2 + x^2| < 1. How do you reduce this? This doesn't seem correct cuz 2 + x^2 is always positive and 2 + x^2 < 1 which implies x^2 < -1 ! So this first representation is wrong because I should not have any constant terms in the common ratio? Can someone please explain this?
    (4 votes)
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  • leaf green style avatar for user Jacopo N. Silvestri
    Is it correct to say that for functions such as the ones shown in the video the interval of convergence is the domain of the function ? Thanks for the patience :)
    (2 votes)
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    • male robot hal style avatar for user Sid
      Let's see.
      f(x) = 1 / (1 - x)
      That converges for -1 < x < 1.
      Is f(x) defined for |x| > 1 ? It seems to work for any x except 1, so I'd say yes. Hence, the domain is not restricted to the interval of convergence.
      (4 votes)
  • starky sapling style avatar for user 20leunge
    If x = 0, then the sum's first term will be (1/3)*0^0 power. 0^0 is undefined, so shouldn't we exclude that from the interval of convergence?
    (3 votes)
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  • blobby green style avatar for user Joy Wang
    what is the radius of convergence?
    (2 votes)
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  • female robot ada style avatar for user Lauren
    I should understand this, but at how does Sal get [1/3] / [ 1 + x ] ?? More specifically I don't understand how he got the 1/3 on top. Any help would be great!
    (2 votes)
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  • leaf green style avatar for user nicole32021
    In (1+(x^2/3)) why did the positive x^2/3 change to a negative? Wouldn't it stay positive? The formula a/1-x is saying to subtract the x value and 2/3 is positive. So would't it just change to (1-(x^2/3)) instead of (1- -(x^2/3))?
    (1 vote)
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    • male robot hal style avatar for user Yamanqui García Rosales
      You are just rearranging the function h(x), to have it look as similar as possible to the formula of the sum of the geometric series. Since you are only rearranging, you cannot do anything that would change the value of the function, that is why you have to put the 2 minus signs, the first one is to make the function look like the formula, and the second one is to ensure that the value of the function doesn't change.
      (2 votes)
  • starky seed style avatar for user James Maiorana
    At Sal says assuming that the absolute value of the common ration is less than 1 then the series converges. We are assuming because we do not have actual values for x in these examples to know that the absolute value of the common ratio is less than 1? So when there are values, we can first figure out the common ratio and therefore conclude the series converges?
    (1 vote)
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    • blobby green style avatar for user Creeksider
      What we're getting at here is the notion that we have an alternate way of representing a function -- which will turn out to be useful in various ways -- but we have to be careful because this alternate way of representing the function is valid only within a particular range of values known as the interval of convergence. Certain aspects of the function may become easier to deal with when we change its representation into a series, and we're preparing for that possibility by learning how to create that representation and also how to determine the range of values for which it is valid.
      (2 votes)

Video transcript

So as we talked about in the last video, we've seen many examples of starting with a geometric series expanded out, and then assuming that its common ratio, that the absolute value of the common ratio is less than 1, finding what the sum of that might be. We've proven with this formula in previous videos. But now let's go the other way around. Let's try to take some function-- let's say h of x being equal to 1 over 3 plus x squared-- and let's try to put it in this form. And then once we put it in that form, we can think about what a and our common ratio is. And then try to represent it as an actual geometric series. So I encourage you to pause the video and try to do that right now. So let's see, the first thing that you might notice is we have a 1 here instead of a 3. So let's try to factor out a 3. So this is equal to 1 over 3 times 1 plus x squared over 3. And now, since we don't want that 3 in the denominator, we can think about this as 1 over 3. So we could say this is 1/3 over-- let me do it in that purple color. 1/3 over 1. And we don't want to just add something, we want to subtract our common ratio. So 1 minus-- and let me write our common ratio here in yellow. 1 minus negative x squared over 3. So now we've written this in that form. And so now we could say that the sum-- let me write it here in-- let me do it in a new color. So let me do it in blue. So now we could say that the sum from n equals 0 to infinity of-- let's see, our first term is 1/3. 1/3 times our common ratio to the n-th power. Common ratio is negative x squared over 3. And if we wanted to expand this out, this would be equal to-- so the first term is 1/3 times all of this to the 0-th power. So it's just going to be 1/3. And so each successive term is just going to be the previous term times our common ratio. So 1/3 times negative x squared over 3 is going to be negative 1/9 x squared. To go from that to that, you have to multiply by-- let's see, 1/3 to negative 1/3, you have to multiply it by negative 1/3. And we multiplied by x squared as well. Now in our next term, we're going to multiply by negative x squared over 3 again. So it's going to be plus-- a negative times a negative is a positive-- plus 1/27 x to the fourth. x squared times x squared, x to the fourth power. And we just keep going on and on and on. And when this converges, so over the interval of convergence, this is going to converge to h of x. Now, what is the interval of convergence here? And I encourage you to pause the video and think about it. Well, the interval of convergence is the interval over which your common ratio, the absolute value of your common ratio, is less than 1. So let me write this right over here. So our absolute value of negative x squared over 3 has to be less than 1. Well, the absolute value, this is going to be a negative number. This is the same thing as saying-- let me scroll down a little bit. This is the same thing as saying that the absolute value of x squared over 3 has to be less than 1. And this is another way of saying-- well, one thing that might jump out at you is that x squared, this is going to be positive no matter what. Or I guess I should say, this is going to be non-negative no matter what. So this is another way of saying that x squared over 3 has to be less than 1. Right? I don't want to confuse you in this step right over here. But the absolute value of x squared over 3 is just going to be x squared over 3, because this is never going to take on a negative value. And so we can multiply both sides by 3. I'll go up here now to do it. Multiply both sides by 3 to say that x squared needs to be less than 3. And so that means that the absolute value of x needs to be less than the square root of 3. Or we could say that x is greater than the negative square root of 3, and it is less than the square root of 3. So this is the interval of convergence. This is the interval of convergence for this series, for this power series. It's a geometric series, which is a special case of a power series. And over the interval of convergence, that is going to be equal to 1 over 3 plus x squared. So as long as x is in this interval, it's going to take on the same values as our original function, which is a pretty neat idea.