Learn about Polynomial approximations of functions - part 5 Video

Get it off your chest – be the first to comment on this video!

Learn about Polynomial approximations of functions - part 5

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 am doing this and you’ll see in a few videos from now when we come with the—when we come up with the grand conclusion. So anyway, let’s just set f(x) and you might just want to do this yourself instead of watching me do it because I should be pretty self explanatory now, now that you saw cosine of x but—and then you could check your work. You can pause right now and then you can check to see that we have the same answer and you’ll probably be right and I probably made a careless mistake. So f(x) = sin x. So, first of all let’s just figure out all of the derivatives of sine of x and we can already guess it. It probably cycles similar to cosine of x, slight variation. So what’s the derivative of sine of x? Well, that is just cosine of x. What is the second derivative? I am just going to stop bringing parentheses around these numbers. The second derivative; well that’s just derivative of that; it’s –sin x. The third derivative of x—well I guess I’ll put the parentheses so you don’t think it’s f3 × x. What is the third derivative? Well, that’s just going to be the minus cosine, right? Because the derivative of sine is cosine, then we have that minus sine there. Then 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 sine of x. And then the cycle continues, right? All right, so we know the derivatives and we could just keep going. So let’s evaluate the derivatives of our function of sine of x at x = 0. So f(0)—well that’s sine of zero; what sine of zero? We’ll find that zero is zero. f prime of zero is equal to cosine of zero; well that’s equal to one. The second derivative at zero, that’s minus sine of zero. Well sine of zero is still zero, so that’s zero. 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 zero; that’s cosine of zero is one, we have a minus sine, so it’s minus one. This you already know the fourth derivative is zero; sine of 0. And then it starts to cycle again. In the fifth derivative; zero, you go to one. So we start with zero, then positive one, then zero, then minus one, then zero, then positive one. So every other number’s a zero. Every other—I guess you could say coefficient in the Maclaurin series is a zero; the coefficient where you don’t include the factorial term. And then the ones in between oscillate between positive one and negative one. 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 either the x really is equal to those functions over the entire domain. I might to do it later. Frankly, I’ve been thinking about the proof myself. I haven’t—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 am 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 leap of faith that the Maclaurin series representation just don’t approximate those functions around zero, that when you take the infinite series, it actually equals the function. So the sine of x—the Maclaurin series representation is going to be equal to—well, f(0); well that’s zero—+ f of—so the first derivative. It’s going to be 1 × x1 over 1 factorial which is just one. This is just one. And then we have—this is just plus zero. And then we have minus. And now where this is the third derivative, right? So -1 × x3 over 3 factorial. Now we have a zero. And then we have a plus 1. Now this is now the fifth derivative. So x5 over 5 factorial. Now, we’ll just keep going but I think you see the pattern. And so we write it, rewrite it. Sin x = 0. So, we get x1. So that is just—well, x – x3 over three factorial plus x5 over five factorial. And then you can imagine the pattern we’d go minus. We’re just taking the odd numbers; x7 over 7 factorial + x9 over 9 factorial – x11 over 11 factorial. And we’ll just keep going, right? And so we’ll just keep oscillating in sine and that’s a bit of upon and we use all the odd exponents. So, if I were to write that in sigma notation, I would say, well that equals the sigma notation often is the hard part. Well once again, let’s see. The first term when the term is zero—well, when the—this is the first term, right? And we get a positive sign. So we want—because we’re not oscillating sine, we’re probably going to take negative one to some powers. So it will be -1n+1 because we want every—so let’s see if that works. If this is the first term, this will be a—no, no, no, no, that won’t work. It will be to the 2n+1. And actually I think I should have done that in the previous video too. I think it should have been -12n, not -1n. I am sorry for that mistake, right. So it’s -12n+1 × x2n—oh no, no, sorry. I was right in the previous videos. See I’m confusing myself because I don’t count the zero terms. So let me—probably help me to write this 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 zero. And so the first term is positive. So, it will be -1n, right? Because -10 is 1, right? So that is positive and then the second term is negative, then positive, negative, right?—times—and so what is—the zeros term is x, so it has to be x—let me see, 2n+1. Does that work? Right. Because the first term would then be three, right. x2n+1 over 2n+1 factorial. It’s almost easier when you just write it out like that. Well that is 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 cos x = 1 - x2 over two factorial + x4 over four factorial – x6 over 6 factorial + x8 over 8 factorial. So, they are almost the opposite, right? They almost compliment each other. The cosine; these are all of the even exponents, right? And even factorials and sine is all of the odd exponents and—this is x0, right? It is because—that’s why you get one here. And in sine, it’s all of the odd exponents and all of the odd factorials and the denominator. So that by itself I think is pretty neat and what is—especially—just another you know, father 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 π over 2, which is all they are, right? If you were to graph it, they’re just shifted 90° to the left, to the right of each other. You can actually represent them differently by essentially picking the odd or even—I guess terms of kind of this factorial polynomial series, whatever you want to call it but anyway, it does not 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.

Khan Academy

The Khan Academy is a not-for-profit organization with the mission of providing a high quality education to anyone, anywhere.

http://www.khanacademy.org/

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 am doing this and you’ll see in a few videos from now when we come with the—when we come up with the grand conclusion.

So anyway, let’s just set f(x) and you might just want to do this yourself instead of watching me do it because I should be pretty self explanatory now, now that you saw cosine of x but—and then you could check your work. You can pause right now and then you can check to see that we have the same answer and you’ll probably be right and I probably made a careless mistake.

So f(x) = sin x. So, first of all let’s just figure out all of the derivatives of sine of x and we can already guess it. It probably cycles similar to cosine of x, slight variation. So what’s the derivative of sine of x? Well, that is just cosine of x. What is the second derivative? I am just going to stop bringing parentheses around these numbers. The second derivative; well that’s just derivative of that; it’s –sin x. The third derivative of x—well I guess I’ll put the parentheses so you don’t think it’s f3 × x. What is the third derivative?

Well, that’s just going to be the minus cosine, right? Because the derivative of sine is cosine, then we have that minus sine there. Then 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 sine of x. And then the cycle continues, right?

All right, so we know the derivatives and we could just keep going. So let’s evaluate the derivatives of our function of sine of x at x = 0. So f(0)—well that’s sine of zero; what sine of zero? We’ll find that zero is zero. f prime of zero is equal to cosine of zero; well that’s equal to one.

The second derivative at zero, that’s minus sine of zero. Well sine of zero is still zero, so that’s zero. 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 zero; that’s cosine of zero is one, we have a minus sine, so it’s minus one. This you already know the fourth derivative is zero; sine of 0. And then it starts to cycle again.

In the fifth derivative; zero, you go to one. So we start with zero, then positive one, then zero, then minus one, then zero, then positive one. So every other number’s a zero. Every other—I guess you could say coefficient in the Maclaurin series is a zero; the coefficient where you don’t include the factorial term. And then the ones in between oscillate between positive one and negative one.

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 either the x really is equal to those functions over the entire domain. I might to do it later. Frankly, I’ve been thinking about the proof myself. I haven’t—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 am 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 leap of faith that the Maclaurin series representation just don’t approximate those functions around zero, that when you take the infinite series, it actually equals the function.

So the sine of x—the Maclaurin series representation is going to be equal to—well, f(0); well that’s zero—+ f of—so the first derivative. It’s going to be 1 × x1 over 1 factorial which is just one. This is just one. And then we have—this is just plus zero. And then we have minus. And now where this is the third derivative, right? So -1 × x3 over 3 factorial. Now we have a zero. And then we have a plus 1. Now this is now the fifth derivative. So x5 over 5 factorial. Now, we’ll just keep going but I think you see the pattern. And so we write it, rewrite it.

Sin x = 0. So, we get x1. So that is just—well, x – x3 over three factorial plus x5 over five factorial. And then you can imagine the pattern we’d go minus. We’re just taking the odd numbers; x7 over 7 factorial + x9 over 9 factorial – x11 over 11 factorial. And we’ll just keep going, right? And so we’ll just keep oscillating in sine and that’s a bit of upon and we use all the odd exponents.

So, if I were to write that in sigma notation, I would say, well that equals the sigma notation often is the hard part. Well once again, let’s see. The first term when the term is zero—well, when the—this is the first term, right? And we get a positive sign. So we want—because we’re not oscillating sine, we’re probably going to take negative one to some powers. So it will be -1n+1 because we want every—so let’s see if that works. If this is the first term, this will be a—no, no, no, no, that won’t work. It will be to the 2n+1.

And actually I think I should have done that in the previous video too. I think it should have been -12n, not -1n. I am sorry for that mistake, right. So it’s -12n+1 × x2n—oh no, no, sorry. I was right in the previous videos. See I’m confusing myself because I don’t count the zero terms.

So let me—probably help me to write this 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 zero. And so the first term is positive. So, it will be -1n, right? Because -10 is 1, right? So that is positive and then the second term is negative, then positive, negative, right?—times—and so what is—the zeros term is x, so it has to be x—let me see, 2n+1. Does that work? Right. Because the first term would then be three, right. x2n+1 over 2n+1 factorial. It’s almost easier when you just write it out like that.

Well that is 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 cos x = 1 - x2 over two factorial + x4 over four factorial – x6 over 6 factorial + x8 over 8 factorial. So, they are almost the opposite, right? They almost compliment each other. The cosine; these are all of the even exponents, right? And even factorials and sine is all of the odd exponents and—this is x0, right? It is because—that’s why you get one here. And in sine, it’s all of the odd exponents and all of the odd factorials and the denominator.

So that by itself I think is pretty neat and what is—especially—just another you know, father 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 π over 2, which is all they are, right? If you were to graph it, they’re just shifted 90° to the left, to the right of each other.

You can actually represent them differently by essentially picking the odd or even—I guess terms of kind of this factorial polynomial series, whatever you want to call it but anyway, it does not 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.

Transcription by: Scribe4you Transcription Services