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Welcome back.

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Well, in the last video we took the Maclaurin series for

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cosine of x, the Maclaurin series representation.

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So I guess we might as well do the same for sine of x.

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And I'm sounding very nonchalant, but I have a reason

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behind why I'm doing this, and you'll see in a few videos from

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now, when we come up with the grand conclusion.

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So anyway, let's just set up f of x.

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And you might just want to do this yourself instead of

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watching me do it, because it should be pretty

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self-explanatory now that you saw cosine of x.

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And then you could check your work.

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You can pause right now, and then you can check to see

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that we got the same answer.

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You'll probably be right and I probably made

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a careless mistake.

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So f of x is equal to sine of x.

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First of all, let's just figure out all of the

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derivatives of sine of x.

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We can already guess that it probably cycles similar to

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cosine of x, with slight variation.

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So what's the derivative of sine of x?

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Well, that's just cosine of x.

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What's the second derivative?

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I'm just going to stop putting parentheses

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around these numbers.

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The second derivative, well, that's just derivative of that.

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It's minus sine of x.

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The third derivative of x.

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Well, I guess I'll put the parentheses so you don't think

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it's f to the third times x.

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The third derivative, well, that's just going to be the

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minus cosine, rightu0008?

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Because the derivative of sine is cosine, but then we have

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that minus sine there.

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And the fourth derivative.

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The derivative of cosine is minus sine, but we have a minus

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there, so we get back to sine.

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And then the cycle continues.

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So the fifth derivative is just going to be cosine of x, or

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just the derivative of the sine of x.

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And then the cycle continues, right?

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All right.

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So we know the derivatives, and we can just keep going.

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So let's evaluate the derivatives of our function

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of sine of x at x equals 0.

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f of 0.

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Well, that's sine of zero.

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What's sine of 0?

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Well, sine of 0 is 0.

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f prime of 0 is equal to cosine of 0.

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That's equal to 1.

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The second derivative at 0, that's minus sine of 0.

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Well, sine of 0 is still 0.

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So that's 0.

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You can see it's a very similar pattern to what we saw in the

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Maclaurin series for cosine of x.

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And then the third derivative evaluated at 0.

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That's cosine of 0 is 1.

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But we have a minus sine, so that's minus 1.

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This you already know.

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The fourth derivative is 0.

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Sine of 0 is 0.

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And then it starts to cycle again.

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The fifth derivative, 0 equal to 1.

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So we start with 0, then positive 1, then 0, then minus

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1, then 0, then positive 1.

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Every other number is a 0.

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Every other, I guess you could say, coefficient in the

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Maclaurin series is a 0.

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The coefficient when you don't include the factorial term.

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And then the ones in between oscillate between positive

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1 and negative 1.

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So what would be the Maclaurin series for sine of x?

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The Maclaurin series representation?

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So we could say that sine of x -- And remember, I haven't

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proven to you that the Maclaurin series representation

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of sine of x or cosine of x or e to the x really is equal to

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those functions over the entire domain.

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I might do that later.

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Frankly, I've been thinking about the proof myself.

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It hasn't been completely intuitive how to do that proof.

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Although if you test them out, it does seem to

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make a lot of sense.

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But you shouldn't just take my word for it.

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I'm going to look up the proof and I will prove

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it to you, eventually.

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But for now, you just have to take it as a bit of a leap of

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faith that the Maclaurin series representations just don't

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approximate those functions around 0, that when you take

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the infinite series, it actually equals the function.

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So sine of x.

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The Maclaurin series representation is going to be

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equal to -- well, f of 0, that's 0 plus f of -- so the

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first derivative, it's going to be 1 times x to the 1 over 1

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factorial, which is just 1.

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This is just 1.

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And then we have -- this is just plus 0 -- and then we have

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minus -- and now this is the third derivative -- so, minus 1

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times x to the third over 3 factorial.

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And then you have a 0.

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Then we have a plus 1.

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And this is now the fifth derivative, so x to the

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fifth over five factorial.

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And we'll just keep going, but I think you see the pattern

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as we write and rewrite it.

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Sine of x is equal to 0, so we get x to the first, so that's

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just x minus x to the third over 3 factorial plus x to

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the fifth over 5 factorial.

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And then you can imagine the pattern.

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We'd go minus -- we're just taking the odd numbers -- x to

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the seventh over 7 factorial plus x to the ninth over 9

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factorial minus x to the eleventh over 11 factorial,

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and we'll just keep going.

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And so we'll just keep oscillating in sine -- that's

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a bit of a pun -- and we use all of the odd exponents.

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So if I were to write that in sigma notation, and sigma

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notation often is the hard part.

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Well once again, the first term, when the term is 0 --

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this is the first term, right -- we get a positive sine.

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Because we're going to oscillate in sine, we're

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probably going to take negative 1 to some power.

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It'll be negative 1 to the n plus 1.

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So let's see if that works.

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If this is the first term, this will be a -- no,

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no, no, that won't work.

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It'll be to the 2n plus 1.

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Actually, I think I should have done that in the

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previous video, too.

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I think it should have been negative 1 to the 2n,

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not negative 1 to the n.

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I'm sorry for that mistake.

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So it's negative 1 to the 2n plus 1 times x to the 2n.

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Oh, no, no, sorry, I was right in the previous video.

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See, I'm confusing myself, because I don't

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count the 0 terms.

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It would probably help me to write the sigma down first.

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So this is equal to -- as you can see I do all of this in

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real time -- infinity from n is equal to 0.

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And so, the first term is positive, so it'll be

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negative 1 to the n, right?

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Because negative 1 to the 0 power is 1, right?

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So that's positive and then the second term is negative, then

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positive, negative, right?

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And so, the zeroth term is x, so it has to be x to the --

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let me see -- 2n plus 1.

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Does that work?

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Right, because the first term would then be 3.

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Right. x to the 2n plus 1 over 2n plus 1 factorial.

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It's almost easier when you just write it out like that.

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Well, that's pretty interesting.

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But what's even more interesting is if you see the

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similarity between the Maclaurin series representation

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for sine of x, and then the representation we figured

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out for cosine of x in the previous video.

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We figured out that cosine of x is equal to 1 minus x squared

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over 2 factorial plus x to the fourth over 4 factorial minus x

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to the sixth over six factorial plus x to the eighth

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over 8 factorial.

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So they're almost the opposite, right?

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They almost complement each other.

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The cosine, these are all of the even exponents, right?

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And even factorials.

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And sine is all of the odd exponents, because this

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is x to the 0, right?

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So that's why you get 1 here.

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And in sine, it's all of the odd exponents and all of the

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odd factorials in the denominator.

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So that by itself, I think, is pretty neat.

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What is especially neat, just another fodder for thought, is

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that we know from trigonometry that sine is just a shifted

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cosine function or that cosine is just a shifted

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sine function.

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But what's neat is by shifting it by pi over 2 -- which is all

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they are, right, if you were to graph it, they're just shifted

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90 degrees to the left or the right of each other -- you can

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actually represent them differently by essentially

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picking the odd or even terms of this factorial polynomial

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series, whatever you want to call it.

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But anyway.

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Doesn't matter if you didn't understand what I said at

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the end, as long as you appreciate how cool this is.

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I will see you in the next video.

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