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Write the first four nonzero terms of the McLaurin series for e to the x.

Use the McLaurin series for e to the x to write the third degree Taylor polynomial for G of x, which is equal to e to the x * F of x about x equal to 0.

So McLaurin series, if that looks familiar to you, watch the videos on Khan Academy on McLaurin series. I'll give a little bit of a primer here.

If we're taking the McLaurin series of f of x, that's going to be equal to f of 0. We could view it as f of 0 * x to the 0 over 0 factorial (if you assume 0 factorial is equal to 1), and then plus f prime of 0 * x to the 1 over 1 factorial, plus f prime prime (the second derivative) evaluated at zero * x squared over 2 factorial.

I think you see the pattern here: plus f prime prime prime (the third derivative) evaluated at zero of x cubed over 3 factorial. I think you see where this is going.

And of course, this first one, x to the 0 over 0 factorial, that's just one, so oftentimes it will just be written as f of zero. This term, 1 factorial, is just one, so oftentimes you'll just see it written as f prime of 0 * x, so on and so forth.

So, and actually, let me just write approximately right over there. And so let's do it for e to the x.

So e to the x is approximately equal to, well, it's going to be e to the 0, which is 1, plus you might already know that if f of x is equal to e to the x, then f prime of x is also equal to e to the x. That's one of the magical things about e to the x: the slope of the tangent line at any point is equal to the value, is equal to the x value there.

And if you take the third derivative or second derivative, it's also, you can take as many derivatives as you want, you still get e to the x. That's one of the special things about e. So the first derivative evaluated at zero, well, that's still e to the 0 power times x to the 1, so plus x.

And then we have the second derivative evaluated at zero, well, that's still one, so it's going to be times, or so it's 1 * x squared over 2 factorial. So we could say plus x squared over 2 + x squared / 2 (2 factorial is 2 * 1), and then plus, once again, the third derivative evaluated at zero, that's just e to the x evaluated at zero, which is e to the 0, which is 1, so x cubed over 3 factorial.

So plus x cubed over 3 factorial. We could write it as 3 factorial, or that's 3 * 2 * 1, that's equal to 6.

And then we keep going. So we just wrote the first four nonzero terms of the McLaurin series for e to the x. That's 1, 2, 3, four nonzero terms.

Now we want to use, so let me just underline that, that's part of what they're asking us to do. Then they say use the McLaurin series for e to the x to write the third degree Taylor polynomial for G of x, which is equal to the product of e to the x and F of x.

So what I'm going to do is I'm going to write our original f of x. Let's write down the first few terms of it, and then what we could do is think about, well, how can we multiply those two polynomials?

And we just have to know enough about the multiplication of those two polynomials to just get us our terms that are no higher than third degree.

So f of x is approximately equal to, let's see, it's x, I have a bad memory, it is x minus 3x squared, x minus 3x squared, and then plus 3x cubed + 3x cubed + 3x cubed plus, and actually, you could say minus if you like because then it's going to be, you're going to have, it's plus minus plus minus, however you want to do it, and that's enough.

And why do I feel confident that it's enough? Well, we only want to write the third degree Taylor polynomial.

So if we multiply these and we involve terms that are higher than third degree, well, that's going to give us the higher terms in our polynomial that are higher than third degree.

So let's just think about it. What's going to be the product? So e to the x * F of x, that's going to be approximately equal to, well, let's see, we are going to multiply this infinite polynomial times this infinite polynomial, and that might seem intimidating to you at first.

But what you could do is you could go for each of these terms, start multiplying it times each of these terms out here. Essentially, you know, when you're multiplying polynomials, you're just repeatedly doing the distributive property, so you take this and distribute it onto that. But we should only worry about the terms up to third degree because anything beyond that, well, that's going to add up to a higher than third degree term.

So x * 1 is x, x * x squared or, sorry, x * x is x squared, so plus x squared. x * (x squared over 2) is x to the 3 over 2, so I'll just write plus 12 x cubed.

And I'm going to stop there because if I do x times that, that's going to give me a fourth degree term, and I don't want to worry about that. We are writing the third degree Taylor polynomial.

So that's x, x squared, x cubed / 2. So now let's distribute the -3x squared, and I'll use another color here just to help explain it a little bit.

So if you multiply that times 1, that's going to be -3x squared. I'm just going to a second line here so I can line things up and add them nicely.

And then this time this is -3x cubed. Once again, I'm going to stop there; if I multiply these two, I'm going to get a fourth degree term, and I don't care about the fourth degree terms.

And then, and then let's do, let's worry about this guy. And so let's start distributing. So if I multiply it times this one, I'm going to get 3x cubed.

And I'm going to stop there because then if I start multiplying it times that guy, that's going to give me a fourth degree term and then a fifth degree term and then a sixth degree term, which I don't need to worry about.

So these are all of the pieces that are going to make up that third degree polynomial.

And so what is that going to be equal to? What is that going to be equal to? And this is a little bit, this was a little bit of, you know, getting your math intuition for multiplying infinite polynomials.

Let's see, you're going to have x, and then you're going to have this is 1x squared minus 1 and a half x squared, so that's going to be -12 x squared.

And let's see here: you have +2 minus 1 and a half, which would give you 2, and then +3 plus, oh, no, sorry, 12 minus 1 and a half, which would be -1 + 3 is +2.

So we could say e to the x * F of x, the third degree Taylor polynomial for this, is x - 12x squared + 2x cubed, and we are done.

So this was a little bit tricky; you had to appreciate how to, but it's not calculus; it's a little bit of just algebra, of just appreciating, okay, I need a third degree. I don't have to distribute this times an infinite number of terms, and because at first you might say, well, that's super hard; how do I multiply two infinite polynomials? The key is we only worry about the third degree, up to the third degree.