Sal's old Maclaurin and Taylor series tutorial
Polynomial approximations of functions (part 5) MacLaurin representation of sin x
Polynomial approximations of functions (part 5)
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- Welcome back.
- Well, in the last video we took the Maclaurin series for
- cosine of x, the Maclaurin series representation.
- So I guess we might as well do the same for sine of x.
- And I'm sounding very nonchalant, but I have a reason
- behind why I'm doing this, and you'll see in a few videos from
- now, when we come up with the grand conclusion.
- So anyway, let's just set up f of x.
- And you might just want to do this yourself instead of
- watching me do it, because it should be pretty
- self-explanatory now that you saw cosine of x.
- And then you could check your work.
- You can pause right now, and then you can check to see
- that we got the same answer.
- You'll probably be right and I probably made
- a careless mistake.
- So f of x is equal to sine of x.
- First of all, let's just figure out all of the
- derivatives of sine of x.
- We can already guess that it probably cycles similar to
- cosine of x, with slight variation.
- So what's the derivative of sine of x?
- Well, that's just cosine of x.
- What's the second derivative?
- I'm just going to stop putting parentheses
- around these numbers.
- The second derivative, well, that's just derivative of that.
- It's minus sine of x.
- The third derivative of x.
- Well, I guess I'll put the parentheses so you don't think
- it's f to the third times x.
- The third derivative, well, that's just going to be the
- minus cosine, rightu0008?
- Because the derivative of sine is cosine, but then we have
- that minus sine there.
- And the fourth derivative.
- The derivative of cosine is minus sine, but we have a minus
- there, so we get back to sine.
- And then the cycle continues.
- So the fifth derivative is just going to be cosine of x, or
- just the derivative of the sine of x.
- And then the cycle continues, right?
- All right.
- So we know the derivatives, and we can just keep going.
- So let's evaluate the derivatives of our function
- of sine of x at x equals 0.
- f of 0.
- Well, that's sine of zero.
- What's sine of 0?
- Well, sine of 0 is 0.
- f prime of 0 is equal to cosine of 0.
- That's equal to 1.
- The second derivative at 0, that's minus sine of 0.
- Well, sine of 0 is still 0.
- So that's 0.
- You can see it's a very similar pattern to what we saw in the
- Maclaurin series for cosine of x.
- And then the third derivative evaluated at 0.
- That's cosine of 0 is 1.
- But we have a minus sine, so that's minus 1.
- This you already know.
- The fourth derivative is 0.
- Sine of 0 is 0.
- And then it starts to cycle again.
- The fifth derivative, 0 equal to 1.
- So we start with 0, then positive 1, then 0, then minus
- 1, then 0, then positive 1.
- Every other number is a 0.
- Every other, I guess you could say, coefficient in the
- Maclaurin series is a 0.
- The coefficient when you don't include the factorial term.
- And then the ones in between oscillate between positive
- 1 and negative 1.
- So what would be the Maclaurin series for sine of x?
- The Maclaurin series representation?
- So we could say that sine of x -- And remember, I haven't
- proven to you that the Maclaurin series representation
- of sine of x or cosine of x or e to the x really is equal to
- those functions over the entire domain.
- I might do that later.
- Frankly, I've been thinking about the proof myself.
- It hasn't been completely intuitive how to do that proof.
- Although if you test them out, it does seem to
- make a lot of sense.
- But you shouldn't just take my word for it.
- I'm going to look up the proof and I will prove
- it to you, eventually.
- But for now, you just have to take it as a bit of a leap of
- faith that the Maclaurin series representations just don't
- approximate those functions around 0, that when you take
- the infinite series, it actually equals the function.
- So sine of x.
- The Maclaurin series representation is going to be
- equal to -- well, f of 0, that's 0 plus f of -- so the
- first derivative, it's going to be 1 times x to the 1 over 1
- factorial, which is just 1.
- This is just 1.
- And then we have -- this is just plus 0 -- and then we have
- minus -- and now this is the third derivative -- so, minus 1
- times x to the third over 3 factorial.
- And then you have a 0.
- Then we have a plus 1.
- And this is now the fifth derivative, so x to the
- fifth over five factorial.
- And we'll just keep going, but I think you see the pattern
- as we write and rewrite it.
- Sine of x is equal to 0, so we get x to the first, so that's
- just x minus x to the third over 3 factorial plus x to
- the fifth over 5 factorial.
- And then you can imagine the pattern.
- We'd go minus -- we're just taking the odd numbers -- x to
- the seventh over 7 factorial plus x to the ninth over 9
- factorial minus x to the eleventh over 11 factorial,
- and we'll just keep going.
- And so we'll just keep oscillating in sine -- that's
- a bit of a pun -- and we use all of the odd exponents.
- So if I were to write that in sigma notation, and sigma
- notation often is the hard part.
- Well once again, the first term, when the term is 0 --
- this is the first term, right -- we get a positive sine.
- Because we're going to oscillate in sine, we're
- probably going to take negative 1 to some power.
- It'll be negative 1 to the n plus 1.
- So let's see if that works.
- If this is the first term, this will be a -- no,
- no, no, that won't work.
- It'll be to the 2n plus 1.
- Actually, I think I should have done that in the
- previous video, too.
- I think it should have been negative 1 to the 2n,
- not negative 1 to the n.
- I'm sorry for that mistake.
- So it's negative 1 to the 2n plus 1 times x to the 2n.
- Oh, no, no, sorry, I was right in the previous video.
- See, I'm confusing myself, because I don't
- count the 0 terms.
- It would probably help me to write the sigma down first.
- So this is equal to -- as you can see I do all of this in
- real time -- infinity from n is equal to 0.
- And so, the first term is positive, so it'll be
- negative 1 to the n, right?
- Because negative 1 to the 0 power is 1, right?
- So that's positive and then the second term is negative, then
- positive, negative, right?
- And so, the zeroth term is x, so it has to be x to the --
- let me see -- 2n plus 1.
- Does that work?
- Right, because the first term would then be 3.
- Right. x to the 2n plus 1 over 2n plus 1 factorial.
- It's almost easier when you just write it out like that.
- Well, that's pretty interesting.
- But what's even more interesting is if you see the
- similarity between the Maclaurin series representation
- for sine of x, and then the representation we figured
- out for cosine of x in the previous video.
- We figured out that cosine of x is equal to 1 minus x squared
- over 2 factorial plus x to the fourth over 4 factorial minus x
- to the sixth over six factorial plus x to the eighth
- over 8 factorial.
- So they're almost the opposite, right?
- They almost complement each other.
- The cosine, these are all of the even exponents, right?
- And even factorials.
- And sine is all of the odd exponents, because this
- is x to the 0, right?
- So that's why you get 1 here.
- And in sine, it's all of the odd exponents and all of the
- odd factorials in the denominator.
- So that by itself, I think, is pretty neat.
- What is especially neat, just another fodder for thought, is
- that we know from trigonometry that sine is just a shifted
- cosine function or that cosine is just a shifted
- sine function.
- But what's neat is by shifting it by pi over 2 -- which is all
- they are, right, if you were to graph it, they're just shifted
- 90 degrees to the left or the right of each other -- you can
- actually represent them differently by essentially
- picking the odd or even terms of this factorial polynomial
- series, whatever you want to call it.
- But anyway.
- Doesn't matter if you didn't understand what I said at
- the end, as long as you appreciate how cool this is.
- I will see you in the next video.
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