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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.

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  • leaf green style avatar for user Alejandro Artiles
    how would you make the maclaurin series for number pi?
    (35 votes)
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    • blobby green style avatar for user Jesse.T.Robinson
      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 + ...
      (96 votes)
  • blobby green style avatar for user conleygibbs
    What is the Maclaurin series for tan(x)?
    (7 votes)
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  • leafers tree style avatar for user Phil Oats
    How would you incorporate imaginary numbers to reconcile e, cos, and sin like Sal alluded to?
    (5 votes)
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  • piceratops ultimate style avatar for user Mark
    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
    (2 votes)
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    • leaf green style avatar for user ArDeeJ
      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)
  • leaf green style avatar for user fthanedar
    Why doesn't Sal talk about the Tan Taylor Series at 0 (Maclaurin) ? Isn't Tan x Sinx/Cos x ?
    (2 votes)
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    • blobby green style avatar for user LHS
      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)
  • leaf green style avatar for user Aaron Lin
    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)
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  • duskpin ultimate style avatar for user Kostas Giotis
    How can we prove that the series Σ (x^n)/n! from n=0 to infinity converges?
    (2 votes)
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  • male robot johnny style avatar for user berat han
    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)
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  • blobby green style avatar for user yasmeen yehia
    what is the general term of e^x/2 and also e ^-x in terms of sumation
    (2 votes)
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    • blobby green style avatar for user Slacks
      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)
  • blobby green style avatar for user TamimHerati
    What is e^x? And how did they discover e^x. I think its something like pi, which you see in a circle. But what is the e(^x)?
    (2 votes)
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    • male robot hal style avatar for user Andrew M
      e is an irrational number that, like pi, is very important in math, physics, and engineering. It is named after the mathematician Euler, who was not the first to discover it but was the first to make good use of it in the study of logarithms.
      (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.