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## Integral Calculus

### Course: Integral Calculus > Unit 5

Lesson 15: Maclaurin series of eˣ, sin(x), and cos(x)- 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

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# Maclaurin series of eˣ

AP.CALC:

LIM‑8 (EU)

, LIM‑8.E (LO)

, LIM‑8.E.1 (EK)

, LIM‑8.F (LO)

, LIM‑8.F.2 (EK)

Approximating eˣ with a Maclaurin series (which is like a Taylor polynomial centered at x=0 with infinitely many terms). It turns out that this series is exactly the same as the function itself! Created by Sal Khan.

## Want to join the conversation?

- how would you make the maclaurin series for number pi?(35 votes)
- In approximating pi using a series, here is one way that uses an inverse trig function. (If you're interested in the derivation I'm sure you can find it online.)

arctan(x) = x - (x^3)/3 + (x^5)/5 - (x^7)/7 + ...

We know that arctan(1) = pi/4, so let x = 1 to get

arctan(1) = pi/4 = 1 - 1/3 + 1/5 - 1/7 + ...

So multiply both sides by 4:

pi = 4 - 4/3 + 4/5 - 4/7 + ...(97 votes)

- What is the Maclaurin series for tan(x)?(7 votes)
- Quite complex ! Only odd powers of x and the coefficient for x^(2k-1) is equal to 2*(4^k-1)*Z(2k)/pi^(2k) where Z is the Riemann Zeta function. It starts as :

x+(1/3)x^3+(2/15)x^5+(17/315)x^7+...(10 votes)

- How would you incorporate imaginary numbers to reconcile e, cos, and sin like Sal alluded to?(5 votes)
- At about3:30Sal derives an equation for a numerical value of e,

where e = 2 + 1/2! + 1/3! +1/4!...

Can anyone explain how to get from this to the other accepted formula for e,

where e = (1 + 1/n)^n as n approaches infinity.

Thx(2 votes)- Alright, this is a bit complicated.

We first want to prove that d/dx e^x = e^x, only given the limit definition of e: lim {n→infinity} (1 + 1/n)^n.

Start with the definition of the derivative:

lim_{h→0} (e^(x+h) - e^x) / h

You can factor out an e^x, since it doesn't depend on h:

e^x * lim_{h→0} (e^h - 1) / h

Now we have to prove that the limit above is 1. It's rather involved, but here's a proof: http://www.proofwiki.org/wiki/Derivative_of_Exponential_at_Zero/Proof_2

Now that we've proven that d/dx e^x = e^x, we can construct the Maclaurin series for e^x, as Sal did in the video.

e^x =Σ{n=0 to infinity} x^n / n!

Now we just have to plug in x = 1:

e^1 = e = Σ{n=0 to infinity} 1^n / n! = 1/0! + 1/1! + 1/2! + 1/3! + 1/4! + ...

= 2 + 1/2! + 1/3! + 1/4! + ...(5 votes)

- Why doesn't Sal talk about the Tan Taylor Series at 0 (Maclaurin) ? Isn't Tan x Sinx/Cos x ?(2 votes)
- It doesn't have a "nice" Maclaurin series expansion (or at least not as nice as sine or cosine). Yes, tan x = sin(x)/cos(x), but it's generally difficult to divide power series. However, arctan x has a "nice" easy Maclaurin expansion.(4 votes)

- At the end of this video, Sal says that he will show how is the power series of e connected to the power series of cos(x) and sin(x). But I don't see it in the next video? Where is it?

Thanks in advance(3 votes)- I managed to find it later on in the playlist.

https://www.khanacademy.org/math/ap-calculus-bc/bc-series-new/bc-10-14/v/euler-s-formula-and-euler-s-identity(2 votes)

- How can we prove that the series Σ (x^n)/n! from n=0 to infinity converges?(2 votes)
- Use the ratio test - it is a very straightforward (no algebraic tricks needed) proof that clearly shows that as n tends to infinity, the limit of the sum tends to 0 regardless of the value of x.(2 votes)

- To the maclaurin series is correct to put 0!=since it is 1. Like, P(x)=f(0).(x^0 /0!)+f'(0).(x^1 /1!)+...

I hope that I ask my question correct. Thanks(2 votes)- Yes, we can use 0! in Maclaurin series because by definition, 0! = 1.(2 votes)

- what is the general term of e^x/2 and also e ^-x in terms of sumation(2 votes)
- Write out the Taylor expansion for e^x = 1 + x + x^2/2! + x^3/3! + ... + x^n/n! , then for x substitute x/2. e^x/2 = 1 + x/2 + (x/2)^2/2! + (x/2)^3/3! + ... + (x/2)^n/n! . This simplifies to e^x/2 = 1 + x/2 + x^2/(2^2)2! + x^3/(2^3)3! + ... + x^n/(2^n)n! which can be written as summation x^n/(2^n)n!

Or solve directly by substituting x/2 for x. Knowing e^x = sum x^n/n! , e^(x/2) = sum (x/2)^n/n! which simplifies to x^n/2^n divided by n! or x^n/2^n n!

Similarly e^(-x) is simply sum (-x)^n/n!(2 votes)

- How would the Maclaurin series for [(1+e to the x) squared] be found?(1 vote)
- Is this what you mean?

(1 + e^x)^2 = 1 + 2e^x + e^(2x)

This should allow you to write two different series (one for e^x and the other for e^(2x)). The manipulations from there should be trivial.(2 votes)

## Video transcript

Now let's do something
pretty interesting. And this will, to some degree,
be one of the easiest functions to find the Maclaurin
series representation of. But let's try to
approximate e to the x. f of x is equal to e to the x. And what makes
this really simple is, when you take the
derivative-- and this is, frankly, one of the amazing
things about the number e-- is that when you take the
derivative of e to the x, you get e to the x. So this is equal
to f prime of x. This is equal to f, the
second derivative of x. This is equal to the
third derivative of x. This is equal to the
n-th derivative of x. It's always equal to e to the x. That's kind of the
first mind blowing thing about the number e. It's just, you could keep
taking its derivative. The slope at any
point on that curve is the same as the value of
that point on that curve. That's kind of crazy. Anyway, with that said,
let's take its Maclaurin representation. So we have to find f
of 0, f prime of 0, the second derivative at 0. Well, when you take e to the 0,
e to the 0 is just equal to 1. And so this is going
to be equal to f of 0. This is going to be
equal to f prime of 0. It's going to be equal to any of
the derivatives evaluated at 0. The n-th derivative
evaluated at 0. And that's why it makes
applying the Maclaurin series formula fairly straightforward. If I wanted to
approximate e to the x using a Maclaurin series--
so e to the x-- and I'll put a little
approximately over here. And we'll get closer and
closer to the real e to the x as we keep adding
more and more terms. And especially if we had an
infinite number of terms, it would look like
this. f of 0-- let me do it in-- what colors
did I use for cosine and sine? So I used pink and I used green. So let me use a
non-pink, non-green. I'll use the yellow here. So f of 0 is 1 plus f prime
of 0 times x. f prime of 0 is also 1. So plus x plus, this
is also 1, so it's going to be x squared
over 2 factorial. So plus x squared
over 2 factorial. All of these things
are going to be 1. This is 1, this is 1, when
we're talking about e to the x. So you go to the third term. This is 1. You just have x to the
third over 3 factorial. Plus x to the third
over 3 factorial. And I think you see
the pattern here. We just keep adding terms.
x to the fourth over 4 factorial plus x
to the fifth over 5 factorial plus x to the
sixth over 6 factorial. And something pretty neat
is starting to emerge. Is that e to x, 1-- this is just
really cool-- that e to the x can be approximated by
1 plus x plus x squared over 2 factorial plus x to
the third over 3 factorial. Once again, e to
the x is starting to look like a
pretty cool thing. This also leads to other
interesting results. That if you wanted
to approximate e, you just evaluate this
at x is equal to 1. So if you wanted to
approximate e, you'd say e is approximate to-- well,
e is e to the first power. And that's going
to be approximately equal to this polynomial
evaluated at 1. If x is 1 here, we
make x 1 over here. So it'll be 1 plus 1. So it'll be 1 plus 1
plus 1 over 2 factorial, plus 1 over 3 factorial,
plus 1 over 4 factorial. So on and so forth, all
the way into infinity. And you could also view this
as 1 over 1 factorial as well. 1 over 1 factorial. But what's really cool is this
is another really neat way to represent e. It shows that e once again shows
up in this kind of neat thing. It's kind of 2 plus
1/2, plus 1/6, plus-- if you just keep doing this,
you get close to the number e. But that by itself isn't
the only fascinating thing. If we look back at our
Maclaurin representations of these other
functions, cosine of x-- let me copy and
paste cosine of x. So cosine of x right up here. So let me do my best to copy
and paste the whole thing. So copy and paste. Copy and paste. So that is cosine of x. And let's do the same
thing for the sine of x that we did last video. So the sine of x. Let me copy and paste that. Copy, and then
let me paste that. Edit, paste. So do we see any relationship
between these approximations? So before, you probably
would have guessed maybe there's some relationship
between cosine and sine, but what about e to the x? And what you see
here is that cosine of x looks a lot like
this term plus this term, although we would want to put
a negative out front here. So it's a negative version
of this term right here, plus this term right here,
plus a negative version of this term right over here. And sine of x looks
just like this term plus a negative version of
this term, plus this term, plus a negative version
of the next term. So if we could somehow
reconcile the negatives in some interesting
way, it looks like e to the x is
somehow-- or at least its polynomial representation
of e to the x-- is somehow related
to a combination of the polynomial
representations of cosine of x and sine of x. So this is starting to get
really, really, really cool. We're starting to see a
connection between something related to compound interest or
a function whose derivative is always equal to that function. And these things that
come out of the unit circle in oscillatory motion
and all of those things, there starts to seem some type of
pure connectedness there. But I'll leave you
there in that video. And in the next
video, I'll show you how we can actually
reconcile these three fascinating functions.