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### 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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# Power series of arctan(2x)

We can represent arctan(2x) with a power series by representing its derivative as a power series and then integrating that series. You have to admit this is pretty neat.

## Want to join the conversation?

- Wouldn't it be easier just to look at the derivative of arctan(2x) as a geometric series with the first term 2 and the ratio -4x^2? We would just write a few terms, then integrate and get to the same result. It seems easier and faster.(76 votes)
- Going through the geometric series route, we'd have to constrain x in such a way that our common ratio falls between -1 and 1 (Remember that a geometric series converges only when |common ratio| < 1)

Going through the Maclaurin Series route, on the other hand does not require us to constrain x in a similar way and hence, gives a more general, less constrained solution.(44 votes)

- Can someone explain why integrating is necessary? I think I missed that.(13 votes)
- we integrated so that we can walk back from the derivative that we expanded. remember, we had to first take the derivative of arctan( ) so that we could expand an equation. we need to go and account for the fact that we took the derivative of the function in our expansion so we take the anti-derivative (integral) of our f(x) function to represent the original (arctan) function.(23 votes)

- Wait, is a power series and a Maclaurin series the same thing?(12 votes)
- Almost. The difference is that Maclaurin series is "centered" at zero. Remember each single term in a power series has the form: a[sub n]*(x-c)^n In a Maclaurin series c=0, and a[sub n] equal to the (nth derivative of f)/n!(20 votes)

- What theorem states we can do this specifically? I think that should have been included in the video just for clarification.(4 votes)
- If you refer to the integration of power series, it essentially follows from the fact that a power series converges uniformly to a continuous function on, say, compact subsets of its interval of convergence.

Suppose`ƒ(x) = ∑ c(n)(x - a)ⁿ`

is a power series about the point`a`

with radius of convergence`R > 0`

, i.e., the series converges on`(a - R, a + R)`

. Then for any`0 < r < R`

, the series converges uniformly to a continuous function on`[a - r, a + r]`

. Since a uniformly convergent series of integrable functions is itself integrable and may be integrated term by term on such intervals, the result follows.(6 votes)

- could he just have used (1)/(1-x) and used a geometric series instead of 1/(1+x)?(4 votes)
- Yes, that would have been a valid path to take. He would have had to define
`f(x) = 2·g(-4x²)`

, and the end result would have been the same.

I guess he didn't do it in order to give another example of the power series.(6 votes)

- Hold on, what am I missing here?

Is it just a given in this video that the series expansion of the derivative of f(x) is the derivative of the series expansion of F(X) (its integral)?

Is that supposed to be obvious to me? It seems a bit glossed over, but perhaps I'm missing some key here.(5 votes)- Your probably missing the first of the series. Try this:

https://www.khanacademy.org/math/integral-calculus/sequences_series_approx_calc/maclaurin_taylor/v/maclauren-and-taylor-series-intuition(2 votes)

- What is the best method for choosing an appropriate g(x) that is close enough to the original f(x)? This problem seems obvious, but I'm not sure I would have come up with that so readily without Sal's help. Just practice?(6 votes)
- In this example, Sal only uses the first 4 non-zero terms to approximate the function. That's close enough to him and to me. If you want to be more accurate, you can use more terms but it's, in some cases, quite nasty. The more term you use, the more accurate it would be(0 votes)

- at6:00when he says that one can interchange the two functions by replacing x with 4x^2, my question is why is this valid when the derivatives will have been different?

for instance if h = (1+x^2 ) ^2 than h' = 4x(1 + x^2), but if i were to let g = (1+x)^2 than g' = 2(1+x) . by replacing x^2 for x on g' != h' . so my question is why is this valid(5 votes)- Because the derivatives are only used to find a Maclaurin approximation for g(x). Once the approximation is computed, he is just evaluating his approximation for g(4x^2). He is not taking any more derivatives, so it does not matter what x is replaced with. That is pretty much the whole trick of the simplification.

Go back just two videos (Maclaurin series based on cos x). We can simply find cos(x^3) is 1 - x^6/2! + x^12/4! - x^18/6! +.... by only knowing the power series of cos(x) and evaluating that alternate, but completely equivalent, representation for x^3. It does not matter what the derivative of x^3 is because we are simply evaluating. Hope this makes sense!(1 vote)

- At6:00- why is it f(x) = 2(g(4x^2))? Why are we multiplying g(4x^2) by 2? I am just having a hard time seeing it... Sorry(3 votes)
- f(x) = 2/(1+4x^2). It's easier to find the Maclaurin polynomial for a simpler series, 1/(1+x). So we found that Maclaurin expansion, and we called it g(x). Now, if g(x) = 1/(1+x), we have to transform function g(x) into f(x) so we know how to transform the expansion for g(x) to get the expansion for f(x). To go from 1/(1+x) to 2/(1+4x^2), we substitute 4x^2 for x, and multiply the function by 2. Thus, we want to find: 2*g(4x^2) to get f(x). That's why f(x) = 2(g(4x^2)). Hope that helps!(4 votes)

- I'm confused. If arcatn(2x) is approximately 2x-8/3*x^3+32/5*x^5-128/7*x^7, why is it that arctan(2) is 1.1 rads and entering 1 into the series yields -12.5, diverging further from the arctan answer as i lengthen the series?(3 votes)
- It's a finite approximation at x = 0, so in a sufficiently small interval centered 0, the approximation would quite nicely match, but when you get further, they would separate drastically. To accurately approximate the function, you need to use an infinite approximation(1 vote)

## Video transcript

- [Voiceover] What I would like us to do in this video is find the
power series representation of or find the power series
approximitation (chuckles) the power series
approximation of arctangent of two x centered at zero and let's just say we want
the first four nonzero terms of the power series
approximation of arctangent of two x centered at zero so it's essentially the
Maclaurin Series of arctangent of two x, the first four
nonzero terms of it. If you feel good about it, I encourage you to pause the video and try to work through it yourself. You might have tried to work through it and you've probably took the first derivative of it. You probably saw that hey, the derivative with respect to x of arctangent of two x is equal to
and this is a refresher if you didn't realize it the first time. It's going to be the
derivative of arctangent of x is one over one plus x squared so this is going to be the derivative of this which is two over one plus this whole thing squared. One plus, I would say
one plus four x squared and then as you tried to
find more of the terms of the Maclaurin Series, you would have taken
more derivatives of this and it would have gotten very hairy, very fast if especially if you're looking for the first four nonzero terms. You probably realize hey, there must be some type of an insight that I hadn't fully appreciated yet when I just tried to just power through or no pun intended, power through finding the power series, the first four nonzero
terms of the power series centered zero of arc tangent of two x. You are right, there
is a key insight here. The key insight here is well, instead of doing it directly, let's see if we could
find the power series representation, the first
four terms of this thing right over here and then we
can take the antiderivative of that to find the power series of arctangent of two x making sure that we get the constant right that it satisfies the fact
that we are being centered at zero, so I know what
you're thinking now. Well that seems like it's getting us to the exact same issue. If I want to find the power
series representation of this, the first four terms of it, I'm still having to take
the derivative of this multiple times it seems just as hard but this is the key insight
I guess you could say. The key insight is let's just say f of x which of course is the derivative of arctangent of two x is two over one plus four x squared. Now, if we had another function that cleans it up a little bit so that we don't get all these hairiness when we take the derivatives. Let's say we had another function g of x, I'm using the color that I have not used. Let's say, that I have g of x is equal to one over one plus x. This is an interesting thing
because it's really easy and this is the same thing as one plus x to the negative one power. g of x is interesting
because it's really easy to take its derivatives. For example, g prime of x is going to be equal to chain rule derivative one plus x is just one so it's going to be equal
to negative one plus x to the negative two power if I want to take the
second derivative of that. g prime prime of x that's
going to be negative two times negative one it's
two times one plus x to the negative three power. If I want to take the
third derivative of that, that's going to be, let's see. Negative three times two its negative six times one plus x to the
negative four power. I know you're saying, "Sal, "aren't we worried about this? "Why are you doing this?" Well just bare with me for a second. Just said quickly, I was able to find the first three derivatives of g of x and it's very easy then to
find the first four terms of its power series representation especially it's Maclaurin, that's Maclaurin series if the power series centered at zero. We just have to evaluate
each of these at zero. g of zero is equal to one, g prime of zero is equal to negative one. g prime, prime of zero, so one plus zero and
then the negative three, that's just one times two is equal to two and then the third
derivative evaluated at zero is equal to negative six. I could write the g of x is approximately equal, so I'm just going to do
the first four terms here is going to be g of zero which is one minus g prime of zero times x. That's minus one times
x, so that's negative x plus g prime, prime of zero two over two factorial times x squared. Well this is just one times x squared so let me just write that. That's just going to be plus x squared and then we have plus g
prime, prime, prime of zero which is negative six over three factorial times x to the third. Well three factorials are six so negative six divided by
six is just negative one. That's going to be
negative x to the third, negative x to the third and
I know what you're thinking. "All right Sal, so you just
started with a hard problem "and you gave yourself
a much easier problem "to find the power series representation. "How is this useful?" Well this is the key insight
that I've been promising throughout this video so far. The key insight, the
long promised key insight is that and I'm finding a
suitable color for a key insight is that we can write f of x. Notice f of x is just two times, f of x is two times g of four x squared. Notice you replace your
x's with a four x squared. You're going to have one
over one plus four x squared and then you multiply that
whole thing times two, you get this thing right over here. If f of x is equal to that then f of x is power series representation its just going to be
taking this power series or at least the first four terms of it and replacing the x's with four x squares and then multiplying the
whole thing times two, so let's do that. We can write that f of x, so we could write the f of x is going to be approximately equal to two times this thing, a value when x is equal to four x squared. It is one minus instead of an x, I'm going to write a four x squared plus x squared but instead of an x, I have a four x squared squared. This is plus four x squared squared. Well that's going to
be 16 x to the fourth, so let me just write that, that's going to be plus 16 x to the fourth and finally minus x to the third but now an x is four x squared so it's minus four x
squared to the third power. Well that's going to be 64 x to the sixth. Let me write that, minus 64 x to the sixth power and then we could say
that this is going to be so f of x is approximately equal to, distribute the two, two
minus eight x squared plus 32 x to the fourth minus 128 x to the sixth. Just like that with a little
bit of a substitution, I was able to reasonably
a non hairy fashion find the first four nonzero terms of the power series of two
over one plus four x squared which is the derivative, which is going to be the derivative of the power series of arctangent of two x. Let's write this down,
so I'm going to write, I'm going to write down, so arctangent of two x, arctangent of two x which is equal to an antiderivative of f of x, dx which is going to be
equal to an antiderivative of this whole, all of this business. It's the antiderivative
of two minus eight, minus eight x squared plus 32 x to the fourth minus 128 x to the sixth. Actually let me make this approximate because now of course
we're doing approximation with the power series. Dx and what is this going to be equal to? We get approximately equal to well I'm going to have a constant there. Let me write the constant first because when we write our power
series or Maclaurin Series, the first term is the constant term. It's our function evaluated to add zero. We're going to have a constant, if I take the antiderivative of two, that's going to be plus two x. The antiderivative of this,
let's see, x to the third divide eight by three so it's negative 8/3 x to the third power then plus 32 x to the fifth over five minus 128 x to the seventh over seven. We're in the home stretch. We've got at least four nonzero terms, if this is nonzero there's going to be five nonzero terms but now let's just make sure that our constant is
appropriate for arctangent of two x. This is essentially
evaluate to what arctangent of when this function,
when x is equal to zero. What is arctangent of zero? Remember this is centered at zero so we better get it right there. That's just the most basic thing if we're doing the Maclaurin
Series representation. We're centering at zero so our approximation evaluated zero better be the same thing as
a function of evaluated zero. Well arctangent of two times zero is just going to be zero and so this thing when
you evaluate it at zero, this gets to C, so this gets to C must be equal to zero. C must be equal to zero if
we want this thing to be zero when X is equal to zero. Just like that we're done. We've been able to figure
out that arctangent of two x is approximately equal to two x minus 8/3 x to the third power plus 32 over five x to the fifth minus 128 over seven x to the seven. If we wanted more terms, we could have gotten more terms by just doing what we just did but doing it for more terms. Hopefully you enjoyed
that fairly hairy problem but as you saw it's not
as hairy as we thought it was going to be.