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Maclaurin series of sin(x)

AP.CALC:
LIM‑8 (EU)
,
LIM‑8.E (LO)
,
LIM‑8.E.1 (EK)
,
LIM‑8.F (LO)
,
LIM‑8.F.2 (EK)
Approximating sin(x) with a Maclaurin series (which is like a Taylor polynomial centered at x=0 with infinitely many terms). It turns out that this series is exactly the same as the function itself! Created by Sal Khan.

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  • blobby green style avatar for user Greg Savage
    How come he plugged in zero into all of the derivatives but not into the x's with exponents?
    (3 votes)
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    • blobby green style avatar for user Ekulwahs
      The polynomial p(X) is a representation of a funtion f(x). SO if you wanted to find the value of cos(0.1) it would be almost impossible without a calculator to use f(0,1).

      So instead they found a way to manipulate f(x) into a p(x) polynomial form. that is to say that the f(x) and p(x) graphs ( where p(x) is to the nth approximation) would look the same.

      And if the graphs look the same, they will give you the same corresponding y values for whatever value of x you you need to find the answer for.

      so you would be able to plug that (0.1) value into the p(x) polynomial ie: p(0.1) and find a value with relatively simple arithmetic.

      eg: for f(x)= cos(x), what would the 4th approximation of cos(0.1) be?

      p(x)= 1-x^2/2!+(0.1)^4/4! ( I just plugged 0.1 wherever there is an x )

      p(0.1)=1-(0.1)^2/2+x^4/24 = 0.995....

      the actual value of cos(0.1)=.999998476

      which is pretty close, the acuracy will increase the higher the degree of your polynomial.

      I hope that helped you Greg.
      (18 votes)
  • old spice man green style avatar for user Pedro
    How is Maclaurin and Taylor Series related to Complex Number's Polar Form, given that they share the same playlist?

    I'm honestly not getting what the relation between these two might be.
    (9 votes)
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    • duskpin seedling style avatar for user john
      Ooooold question, but my guess is that :
      1. complex numbers are deeply connected to trigonometry (as you can see when studying the polar form)
      2. these series include the approximation for cos(x) and sin(x).

      I'm doing a class on Complex Calculus and series are a part of the program.
      (0 votes)
  • blobby green style avatar for user imoraesflivon
    Is the maclaurin series related someway to the parity of mathematical functions?
    (5 votes)
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  • female robot grace style avatar for user M
    What are the applications of studying this in real life?
    (1 vote)
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  • blobby green style avatar for user mclusky313
    What is the application of series? Can you use it to approximate the equation of an unknown function? Sorry if my question is stupid, I'm a bit lost; I'll go back and review the previous videos. :/
    (3 votes)
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    • blobby green style avatar for user Emery Poole
      This process can only be done on a known function, and it is very difficult unless the derivatives can be calculated easily or repeat. It is very useful for things like sine and cosine functions, where it is impossible to calculate things like sin(0.237) without using the taylor or maclaurin series or actually drawing a circle and taking measurements. This also has many applications in sound processing.
      (5 votes)
  • leaf yellow style avatar for user Narek Kazarian
    Maclaurin Taylor Series at 0 for sinx - odd powers over odd factorials
    Maclaurin Taylor Series at 0 for cosx - even powers over even factorials

    Is this is any way related to the fact that sinx is an odd function while cosx is an even function?

    Or are the two ideas completely irrelevant?
    (5 votes)
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    • piceratops ultimate style avatar for user Zachary Conrad
      Yeah. Odd power functions have odd symmetry and even power functions have even symmetry. So a polynomial comprised of only odd power functions still has odd symmetry and likewise for even. Since the cosine function has even symmetry its polynomial representation cannot have any odd powers and likewise for sine.
      (2 votes)
  • piceratops ultimate style avatar for user Brandon.Hoeksema
    So I'm wondering what presuppositions must be made in order to come up with these trigonometric equations in terms of only x. All I can see is that we have to know what the derivatives of these trig functions are. So could somebody tell me what presuppositions must be made and perhaps how to prove that they are true? I just never understood how we can know what a sine or cosine of an angle is without like measuring lengths of a triangle or something (which probably wouldn't be too accurate).
    (4 votes)
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  • leaf green style avatar for user Kristupas Stumbrys
    Can you see the property of sin(x)^2 + cos(x)^2 = 1 using Taylor series?
    (3 votes)
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    • leaf orange style avatar for user Rohen Shah
      Great question!
      In theory, you could if you find the taylor polynomial if "Sin(x)^2" and the Taylor polynomial of "Cos(x)^2", and if you add them all the terms except for the number "1" would cancel. So, this would verify that sin(x)^2+cos(x)^2 = 1
      (2 votes)
  • male robot hal style avatar for user Quazars
    I understand that you can essentially rewrite the functions by using this method but I don't understand why it has to be zero. From what I know a MacLaurin series is when x=0 but a Taylor series can be any other value or am I incorrect? What does that value represent? Also is there a proof somewhere of this method?
    (2 votes)
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    • leafers seed style avatar for user Travis Bartholome
      You're right; the center doesn't have to be 0, it's just often very convenient to use 0 because it reduces the number of terms we have to handle. And yes, a Maclaurin series is just a particular kind of Taylor series that is centered at 0 (it's the same theorem).

      That number, 0 or whatever you choose, represents the "center" of the series; it's the point around which we're building our successive approximations of the function. Taylor's formula is about creating better and better polynomial approximations of a function based on its derivative behavior at a given point; so we choose the point in which we're interested and construct a polynomial approximation around that point. When we make an approximation, we also have to consider what values of x allow that approximation to actually equal the desired value whenever we sum the infinite series. Sometimes the approximation will converge for all values of x, and sometimes it will only converge in a finite interval around the center that we choose; it depends on the function. But for the approximations that don't converge for all x, it's useful to be able to move our "window" of convergence so that it encompasses the value that we want to approximate.

      Say we're approximating ln(e + 0.1). For one thing, we can't use a Maclaurin series because the function isn't even defined at 0. We might choose a Taylor series centered at x = e rather than at x = 1 because at x = 1, the approximation will only converge on the interval (0, 2), which doesn't include our value (about 2.8).

      As far as proofs go, I'd check the Wikipedia page for Taylor's Theorem or http://math.stackexchange.com/questions/481661/simplest-proof-of-taylors-theorem .

      I hope that helps.
      (3 votes)
  • leaf green style avatar for user shivani mishra
    how can i graph all the derivatives of sinx?
    (1 vote)
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Video transcript

In the last video, we took the Maclaurin series of cosine of x. We approximated it using this polynomial. And we saw this pretty interesting pattern. Let's see if we can find a similar pattern if we try to approximate sine of x using a Maclaurin series. And once again, a Maclaurin series is really the same thing as a Taylor series, where we are centering our approximation around x is equal to 0. So it's just a special case of a Taylor series. So let's take f of x in this situation to be equal to sine of x. And let's do the same thing that we did with cosine of x. Let's just take the different derivatives of sine of x really fast. So if you have the first derivative of sine of x, is just cosine of x. The second derivative of the sine of x is the derivative of cosine of x, which is negative sine of x. The third derivative is going to be the derivative of this. So I'll just write a 3 in parentheses there, instead of doing prime, prime, prime. So the third derivative is the derivative of this, which is negative cosine of x. The fourth derivative is the derivative of this, which is positive sine of x again. So you see, just like cosine of x, it kind of cycles after you take the derivative enough times. And we care-- in order to do the Maclaurin, series-- we care about evaluating the function, and each of these derivatives at x is equal to 0. So let's do that. So for this, let me do this in a different color, not that same blue. I'll do it in this purple color. So f-- that's hard to see, I think So let's do this other blue color. So f of 0, in this situation, is 0. And f, the first derivative evaluated at 0, is 1. Cosine of 0 is 1. Negative sine of 0 is going to be 0. So f prime prime, the second derivative evaluated at 0 is 0. The third derivative evaluated at 0 is negative 1. Cosine of 0 is 1. You have a negative out there. It is negative 1. And then the fourth derivative evaluated at 0 is going to be 0 again. And we could keep going, but once again, it seems like there's a pattern. 0, 1, 0, negative 1, 0, then you're going to go back to positive 1. So on and so forth. So let's find its polynomial representation using the Maclaurin series. And just a reminder, this one up here, this was approximately cosine of x. And you'll get closer and closer to cosine of x. I'm not rigorously showing you how close, in that it's definitely the exact same thing as cosine of x, but you get closer and closer and closer to cosine of x as you keep adding terms here. And if you go to infinity, you're going to be pretty much at cosine of x. Now let's do the same thing for sine of x. So I'll pick a new color. This green should be nice. So this is our new p of x. So this is approximately going to be sine of x, as we add more and more terms. And so the first term here, f of 0, that's just going to be 0. So we're not even going to need to include that. The next term is going to be f prime of 0, which is 1, times x. So it's going to be x. Then the next term is f prime, the second derivative at 0, which we see here is 0. Let me scroll down a little bit. It is 0. So we won't have the second term. This third term right here, the third derivative of sine of x evaluated at 0, is negative 1. So we're now going to have a negative 1. Let me scroll down so you can see this. Negative 1-- this is negative 1 in this case-- times x to the third over 3 factorial. And then the next term is going to be 0, because that's the fourth derivative. The fourth derivative evaluated at 0 is the next coefficient. We see that that is going to be 0, so it's going to drop off. And what you're going to see here-- and actually maybe I haven't done enough terms for you, for you to feel good about this. Let me do one more term right over here. Just so it becomes clear. f of the fifth derivative of x is going to be cosine of x again. The fifth derivative-- we'll do it in that same color, just so it's consistent-- the fifth derivative evaluated at 0 is going to be 1. So the fourth derivative evaluated at 0 is 0, then you go to the fifth derivative evaluated at 0, it's going to be positive 1. And if I kept doing this, it would be positive 1-- I have to write the 1 as the coefficient-- times x to the fifth over 5 factorial. So there's something interesting going on here. For cosine of x, I had 1, essentially 1 times x to the 0. Then I don't have x to the first power. I don't have x to the odd powers, actually. And then I just essentially have x to all of the even powers. And whatever power it is, I'm dividing it by that factorial. And then the signs keep switching. I shouldn't say this is an even power, because 0 really isn't. Well, I guess you can view it as an even number, because-- well I won't go into all of that. But it's essentially 0, 2, 4, 6, so on and so forth. So this is interesting, especially when you compare to this. This is all of the odd powers. This is x to the first over 1 factorial. I didn't write it here. This is x to the third over 3 factorial plus x to the fifth over 5 factorial. Yeah, 0 would be an even number. Anyway, my brain is in a different place right now. And you could keep going. If we kept this process up, you would then keep switching sines. X to the seventh over 7 factorial plus x to the ninth over 9 factorial. So there's something interesting here. You once again see this kind of complimentary nature between sine and cosine here. You see almost this-- they're kind of filling each other's gaps over here. Cosine of x is all of the even powers of x divided by that power's factorial. Sine of x, when you take its polynomial representation, is all of the odd powers of x divided by its factorial, and you switch sines. In the next video, I'll do e to the x. And what's really fascinating is that e to the x starts to look like a little bit of a combination here, but not quite. And you really do get the combination when you involve imaginary numbers. And that's when it starts to get really, really mind blowing.