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## AP®︎/College Calculus BC

### Course: AP®︎/College Calculus BC > Unit 10

Lesson 14: Finding Taylor or Maclaurin series for a function- Function as a geometric series
- Geometric series as a function
- Power series of arctan(2x)
- Power series of ln(1+x³)
- Function as a geometric series
- Maclaurin series of cos(x)
- Maclaurin series of sin(x)
- Maclaurin series of eˣ
- Worked example: power series from cos(x)
- Worked example: cosine function from power series
- Worked example: recognizing function from Taylor series
- Maclaurin series of sin(x), cos(x), and eˣ
- Visualizing Taylor series approximations
- Euler's formula & Euler's identity
- Geometric series interval of convergence

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# Geometric series as a function

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?

- 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?(7 votes)
- 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.(23 votes)

- 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)
- 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)

- Shouldn't Sal be testing the endpoints of the interval due to inconclusive ratio test at 1?(2 votes)
- No, because geometric series don't converge at the endpoints.(1 vote)

- he is confusing me with absolute value thing. he should just take the number out straight up(1 vote)
- It might be useful to review here to get the motivation behind the absolute value thing:

https://www.khanacademy.org/math/integral-calculus/sequences_series_approx_calc/convergence-divergence-tests

In particular, check out the videos on the ratio test and the radius of convergence using the ratio test. The convergence divergence tests section is two chapters before the section with the Functions as Geometric Series video.(2 votes)

- Is it mathematically legal if the starting term 'a' is defined in terms of a variable (i.e. x^2)?(1 vote)
- I would imagine so, since x^2 is actually a constant for a given x. But even if it weren't it would still be a thing that's possible, it just may or may not fit into the definition of a geometric series.(1 vote)

- 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)
- 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)
- 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)

- What does Σ (sigma) mean?

(I know it's Greek).(0 votes)- It means summation, like adding up all the things or numbers.(3 votes)

- 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)
- I think your confusion is over the n starting at 0 rather than 1 as it normally does.
`∞ ∞`

∑ 2*(-4x^2)^n = ∑ 2*(-4x^2)^(n-1)

n=0 n=1(1 vote)

- I don't understand why we didn't test the endpoints at about3:40.(1 vote)

## 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.