Main content
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
© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice
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.
Want to join the conversation?
don't you need to check the limits of the interval to see whether or not it converges to the point?(3 votes)
actually, he didnt check the endpoints of the interval because this is a geometric series. since |r|<1 and not < or = to 1, the endpoints will cause the series to diverge because they are not included.(10 votes)
At3:49, 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)
Let's solve it in the normal way:|-x²/3| < 1↔-1 < -x²/3 < 1↔-3 < -x² < 3↔3 > x² > -3
Becausex² >= 0 > -3 for every xso we just need to take carex² < 3↔-√3 < x < √3.(12 votes)
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)- 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)
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)- Not all series converge everywhere: your series happens to converge nowhere (in the real numbers, that is.). That's all there is to it: not everything necessarily works out.(1 vote)
- 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)
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)
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)
I should understand this, but at1:04how 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)- consider a simpler situation: 1 divided by 12.
1/ (3.4)
= 1/3 (4) = (1/3) /4 ( same as 1/3 divided by 4)(2 votes)
- what is the radius of convergence?(2 votes)
The radius of convergence is the distance from our independent variable (x) to either end of the interval of convergence. In other words, it is half the length of the interval of convergence.
Sal went over these concepts in the previous video in this playlist: https://www.khanacademy.org/math/calculus/sequences_series_approx_calc/power-series-algebra/v/power-series-radius-interval-convergence(2 votes)
- 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)
- 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)
At0:07Sal 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)- 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)
3:49Video 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.