We in the last video we took the Macloren series of representation of e to the x. Now let’s do it with a couple of other functions and we’ll see in a few videos it all fits together like a giant puzzle.

The following video will tell you about Polynomial approximation of functions (part 4). You can use youtube to mp3 tool to convert the video to mp3.

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## Polynomial approximation of functions (part 4)

Let’s do cosine of x. Let’s set f of x, f of x is equal to cosine of x. What’s f prime of x? What’s the first derivative of cosine of x? Well that just equals minus sine of x.

Minus sine of x. What’s the second derivative? Well that’s just minus times derivative of sine of x. So the derivative of sine of x is cosine x, it’s minus cosine of x. And what’s the third derivative? f 3 of x.

The derivative of cosine x is minus sine of x, we already have a minus here so it becomes positive sine of x. What’s f 4 of x? The fourth derivative of x? It equals cosine of x again. As we keep taking derivatives we’ll keep repeating and the pattern will go on, right? The fifth derivative of x, the fifth derivative of this function, the fourth is the same as a function, so the fifth is going to be the same as the first derivative.

cosine of x is sine of x. Hopefully you see the pattern. We’re going to do the Macloren representation, which is a specific example of the Taylor series where we figure out the values of the derivatives at x is equal to zero. So let’s do that right now.

So f of zero, let me do it in to another color to fend off monotony. f of zero. What’s cosine of zero? Cosine of zero is 1. f prime of zero is equal to sine of– well not minus sine of zero, but what sine of zero? Sine of zero is zero, so minus zero is still zero, so this is zero.

f prime prime of zero. Cosine of zero we already know is one. We have a negative sine here, so it becomes a minus one.

The third derivative at x is equal to zero. Sine of zero is zero. So this is zero.

I think you might start to see a pattern emerging. The fourth derivative at zero. Cosine of zero is equal to 1.

And then the fifth derivative, this is hard to read but you get the point is just zero again. So what’s the pattern as we take the derivatives? 1, zero, minus 1, zero, 1, zero. So it alternates between zero and 1. So 1, zero, minus 1, zero, positive, zero, negative, zero, positive.

So every other number is zero and in between them we alternate between a positive 1 and a negative 1. So now let’s use that information to figure out them the Macloren series representation. So we proved, hopefully, we didn’t prove it definitely converges over the entire domain of the function. You have to take my word for it.

We’ll experiment a little bit with a graphing calculator in a few videos. We said that this representation– and it should make intuitive sense, because when you take the infinite Macloren series, when you take that infinite sum, you’re essentially creating a function where that function is equal to your original function at the point you chose. In the case of a Macloren we’re picking x equals zero, and it equals every derivative of this function. Just intuitively it seems, well if a function equals something at a point and every one of its derivatives is also equal to the function at that point, maybe those functions are equal to each other.

I haven’t proven that to you yet. We know that the representation is a sum from n is equal to zero to infinity of the nth derivative evaluated at zero. A Macloren series is a specific case of a Taylor series.

We actually haven’t done anything with Taylor series, I was hoping to get there later. But the Macloren series is a really cool one because it’s going to show us all these relationships between e and cosine and sine and eventually i and pi and you will find it exciting. The Macloren is that times x to the n over n factorial. That’s what we said it was.

So if this is our f of x, f of x is cosine of x, what does this turn into? Well, f of x is equal to, it equals f of zero times x to the zero over zero factorial, that’s just one, right? Plus, now we’re at n equals 1. It’s the first derivative at zero. f prime of zero, well that’s just equal to zero. And who cares what that– that would be x to the first over 1, right? Now we’re at the second derivative.

The second derivative at zero is minus 1. Minus 1 times x squared over 2 factorial plus the third derivative at zero. The third derivative at zero we figured out was zero.

Zero who cares what that is. It would have been x to the third over 3 factorial. And then what’s the fourth derivative? The fourth derivative at zero is just equal to 1. So we have times 1 and then we’re at x to the fourth over 4 factorial.

Let me see if I can write this a little bit neater. The next one, the fifth derivative at zero times x to the fifth over 5 factorial. We’ll keep going.

Let me write this, clean this up and hopefully the pattern merges if it hasn’t emerged already. f of x is equal to cosine of x is equal to– let met get rid of the zeros– 1 and then we have minus x squared over 2 factorial. This term, this goes away. This is a zero term.

And the next one is a positive. Plus x to the fourth over 4 factorial. And the fifth term goes away.

But then the cycle continues. The next one is going to be minus. Because we had minus 1 plus 1.

It’s going to be minus x to the sixth over 6 factorial. You could take the sixth derivative. You’ll see that the derivative of minus sine of x is minus cosine of x, that’s where we get the minus 1 from.

And they we’re going to go plus. So we’re just taking all the even terms. x to the eighth over 8 factorial minus x to the 10th over 10 factorial. We could just keep going on and on and on.

And so we have a situation where we can rewrite cosine of x is equal to the sum, if you believe that this Macloren series actually does converge to cosine of x over the entire domain of x, that’s kind of an assumption we’re making. Hopefully one day we will have the tools set to actually prove that as well. From n is equal to zero. So what’s happening? We’re taking all of the even powers.

So we could say x to the 2n, that ensures that no matter what value of n I put in here I get an even numbers. So we’ll go to the zeroth power then the second power, over 2n factorial. So that takes care of going from 1 to x squared over 2, to x to the fourth over 4 factorial, 6 over 6 factorial, et cetera. But now we have to make it switched signs like that.

Well let’s just multiply it negative 1. Let’s see what we can do. Negative 1 to the– so we want the first term to be positive, the second term to be negative.

So we could say times minus 1 to the n plus 1. Let’s see if that works. When m is zero what’s negative 1 to the n plus 1? zero, it would be minus 1.

And then when it’s 1– When it’s zero– no, no it’s just going to be negative 1 to the n. Because when it’s zero, negative 1 to zero is 1. When it’s 1, negative 1. So this will work out.

Negative 1 to the n is cosine. You could try it out. This is the n is equal to zero. We need to switch colors.

That’s n is equal to zero and here we get x to the zero over zero factorial, which is 1. We have negative 1 to the zero is 1, so that becomes 1. When n is equal to 1, this becomes x squared over 2 factorial, we have negative 1 to the 1 power, so that’s where you get the negative 1.

And then when n is equal to 2, the negative 1 squared becomes positive again. So the negative 1 is what provides the alternating numbers. So pretty neat. We just figured out another way to represent cosine of x.

And it might be looking a little bit interesting to you that this representation kind of resembles part of the representation of e to the x. What’s the difference between this and the e to the x? e to the x had the odd exponent terms and it didn’t switch signs. But other than that, they’re pretty much the same. So in the next video we’ll do sine of x and then we’ll try to put it all together.

I’ll see you soon.

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