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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 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?
• 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 + ...
• What is the Maclaurin series for tan(x)?
• 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+...
• How would you incorporate imaginary numbers to reconcile e, cos, and sin like Sal alluded to?
• At about Sal 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
• 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.

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! + ...
• Why doesn't Sal talk about the Tan Taylor Series at 0 (Maclaurin) ? Isn't Tan x Sinx/Cos x ?
• 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.
• 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?

• How can we prove that the series Σ (x^n)/n! from n=0 to infinity converges?
• 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.
• 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
• Yes, we can use 0! in Maclaurin series because by definition, 0! = 1.
• what is the general term of e^x/2 and also e ^-x in terms of sumation
• 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!