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In the last video, we[br]took the Maclaurin series

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of cosine of x.

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We approximated it[br]using this polynomial.

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And we saw this pretty[br]interesting pattern.

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Let's see if we can[br]find a similar pattern

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if we try to approximate sine[br]of x using a Maclaurin series.

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And once again, a[br]Maclaurin series

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is really the same thing[br]as a Taylor series,

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where we are centering[br]our approximation

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around x is equal to 0.

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So it's just a special[br]case of a Taylor series.

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So let's take f of[br]x in this situation

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to be equal to sine of x.

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And let's do the same thing[br]that we did with cosine of x.

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Let's just take the[br]different derivatives

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of sine of x really fast.

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So if you have the first[br]derivative of sine of x,

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is just cosine of x.

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The second derivative[br]of the sine of x

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is the derivative of cosine of[br]x, which is negative sine of x.

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The third derivative is going[br]to be the derivative of this.

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So I'll just write[br]a 3 in parentheses

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there, instead of doing[br]prime, prime, prime.

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So the third derivative[br]is the derivative

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of this, which is[br]negative cosine of x.

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The fourth derivative[br]is the derivative

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of this, which is[br]positive sine of x again.

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So you see, just like cosine[br]of x, it kind of cycles

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after you take the[br]derivative enough times.

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And we care-- in order to do[br]the Maclaurin, series-- we care

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about evaluating the function,[br]and each of these derivatives

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at x is equal to 0.

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So let's do that.

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So for this, let me do this[br]in a different color, not

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that same blue.

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I'll do it in this purple color.

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So f-- that's hard[br]to see, I think

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So let's do this[br]other blue color.

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So f of 0, in this[br]situation, is 0.

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And f, the first derivative[br]evaluated at 0, is 1.

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Cosine of 0 is 1.

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Negative sine of 0[br]is going to be 0.

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So f prime prime, the second[br]derivative evaluated at 0 is 0.

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The third derivative[br]evaluated at 0 is negative 1.

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Cosine of 0 is 1.

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You have a negative out there.

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It is negative 1.

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And then the fourth[br]derivative evaluated at 0

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is going to be 0 again.

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And we could keep going,[br]but once again, it

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seems like there's a pattern.

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0, 1, 0, negative[br]1, 0, then you're

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going to go back to positive 1.

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So on and so forth.

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So let's find its[br]polynomial representation

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using the Maclaurin series.

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And just a reminder,[br]this one up here,

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this was approximately[br]cosine of x.

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And you'll get closer and[br]closer to cosine of x.

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I'm not rigorously[br]showing you how

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close, in that it's definitely[br]the exact same thing as cosine

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of x, but you get closer[br]and closer and closer

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to cosine of x as you[br]keep adding terms here.

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And if you go to[br]infinity, you're

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going to be pretty[br]much at cosine of x.

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Now let's do the same[br]thing for sine of x.

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So I'll pick a new color.

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This green should be nice.

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So this is our new p of x.

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So this is approximately[br]going to be

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sine of x, as we add[br]more and more terms.

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And so the first term here, f[br]of 0, that's just going to be 0.

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So we're not even going[br]to need to include that.

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The next term is[br]going to be f prime

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of 0, which is 1, times x.

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So it's going to be x.

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Then the next term is f[br]prime, the second derivative

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at 0, which we see here is 0.

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Let me scroll down a little bit.

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It is 0.

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So we won't have[br]the second term.

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This third term right[br]here, the third derivative

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of sine of x evaluated[br]at 0, is negative 1.

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So we're now going[br]to have a negative 1.

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Let me scroll down[br]so you can see this.

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Negative 1-- this is[br]negative 1 in this case--

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times x to the third[br]over 3 factorial.

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And then the next[br]term is going to be 0,

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because that's the[br]fourth derivative.

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The fourth derivative evaluated[br]at 0 is the next coefficient.

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We see that that is going to be[br]0, so it's going to drop off.

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And what you're[br]going to see here--

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and actually maybe I haven't[br]done enough terms for you,

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for you to feel good about this.

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Let me do one more[br]term right over here.

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Just so it becomes clear.

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f of the fifth[br]derivative of x is

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going to be cosine of x again.

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The fifth derivative-- we'll[br]do it in that same color,

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just so it's consistent-- the[br]fifth derivative evaluated at 0

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is going to be 1.

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So the fourth derivative[br]evaluated at 0 is 0,

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then you go to the fifth[br]derivative evaluated at 0,

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it's going to be positive 1.

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And if I kept doing this,[br]it would be positive

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1-- I have to write the 1[br]as the coefficient-- times x

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to the fifth over 5 factorial.

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So there's something[br]interesting going on here.

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For cosine of x, I had 1,[br]essentially 1 times x to the 0.

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Then I don't have x[br]to the first power.

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I don't have x to the[br]odd powers, actually.

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And then I just essentially have[br]x to all of the even powers.

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And whatever power it is, I'm[br]dividing it by that factorial.

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And then the sines[br]keep switching.

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I shouldn't say this is an even[br]power, because 0 really isn't.

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Well, I guess you can[br]view it as an even number,

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because-- well I won't[br]go into all of that.

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But it's essentially 0, 2,[br]4, 6, so on and so forth.

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So this is[br]interesting, especially

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when you compare to this.

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This is all of the odd powers.

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This is x to the first[br]over 1 factorial.

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I didn't write it here.

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This is x to the[br]third over 3 factorial

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plus x to the fifth[br]over 5 factorial.

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Yeah, 0 would be an even number.

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Anyway, my brain is in a[br]different place right now.

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And you could keep going.

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If we kept this process[br]up, you would then

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keep switching sines.

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X to the seventh over[br]7 factorial plus x

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to the ninth over 9 factorial.

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So there's something[br]interesting here.

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You once again see this[br]kind of complimentary nature

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between sine and cosine here.

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You see almost[br]this-- they're kind

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of filling each[br]other's gaps over here.

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Cosine of x is all[br]of the even powers

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of x divided by that[br]power's factorial.

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Sine of x, when you take its[br]polynomial representation,

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is all of the odd powers of[br]x divided by its factorial,

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and you switch sines.

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In the next video,[br]I'll do e to the x.

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And what's really[br]fascinating is that e

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to the x starts to look like[br]a little bit of a combination

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here, but not quite.

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And you really do[br]get the combination

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when you involve[br]imaginary numbers.

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And that's when it starts to[br]get really, really mind blowing.